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Qm Math Composites

1 · Prerequisites

2 · Summary

The composite mathematical layer of the non-relativistic quantum mechanics collection, all of it mathematics: completed Hilbert tensor products and their bounded maps, density operators with their scalar spectral probabilities, partial traces and Schmidt decompositions, trace-class cyclicity and the sigma-spectral layer, effects, instruments, Kraus families and their dilations, the compatible-observable joint projection-valued measure, the uncertainty and finite-sampling inequalities, permutation sectors, the Euclidean L² product unitary, and the exact free-Gaussian propagation with its complex-Gaussian input. Every item on this page carries domain: mathematics and depends only on mathematical suppliers; nothing here asserts a physical interpretation, an instrument realization or an empirical claim, and those remain with the consuming quantum-mechanics pages.

The tensor layer begins with Hilbert tensor construction: inner product, completion and product bases, which builds the algebraic tensor product of two complex Hilbert spaces, proves that the induced sesquilinear form is an inner product with orthonormal coordinate expansions, and completes it to a Hilbert space in which product bases of orthonormal families have dense span. The rank-one operators $T_{u, v} x = ⟨ x, v ⟩ u$ are the basic trace-class model: Rank-one operator calculus computes $∥ T_{u, v} ∥ = ∥ T_{u, v} ∥_{1} = ∥ u ∥ ∥ v ∥$ , $tr (T_{u, v}) = ⟨ u, v ⟩$ , the adjoint $T_{u, v}^{*} = T_{v, u}$ , positivity of $T_{x, x}$ , the product identities with bounded and trace-class factors and trace-norm continuity of $(u, v) \mapsto T_{u, v}$ .

On it, Bounded tensor maps: extension, norm estimate, adjoint and commutation establishes the amplifications $A \otimes I$ and $I \otimes B$ , their adjoints, norms and extensions, and Permutation sector projections on the n-fold tensor power turns the finite symmetric group action on an iterated product into orthogonal projections onto the symmetric and antisymmetric sectors with the transposition character identities. Euclidean L² product unitary: L²(R^a)⊗̂L²(R^b) ≅ L²(R^{a+b}) identifies the natural isometry $L^{2} (X) \hat{\otimes} L^{2} (Y) \to L^{2} (X \times Y)$ for the Euclidean product, with Fubini and Tonelli as its analytic inputs, and Density operators and scalar spectral probability defines density operators, proves that $μ_{ρ} (B) = tr (ρ E (B))$ is a probability measure in every expansion $ρ = \sum_{n} w_{n} T_{u_{n}, u_{n}}$ , and records the scalar spectral moment identities for a self-adjoint operator with its projection-valued measure.

The finite-dimensional and trace-class layer comprises Finite partial trace, Schmidt decomposition and finite no-signalling, which constructs the partial trace through its defining trace identity, proves positivity and trace preservation and the Schmidt decomposition with the purity criterion $tr ρ^{2} = \sum_{r} s_{r}^{4}$ ; Nuclear bounded cyclicity for trace-class operators, which computes $B T_{u, v} = T_{B u, v}$ , $T_{u, v} B = T_{u, B^{*} v}$ and cyclicity of the trace against bounded operators through nuclear representations; Separable partial trace and its basic properties, which extends the partial trace to the completed tensor product of arbitrary Hilbert spaces and proves its uniqueness, positivity, trace preservation and contractivity; and Infinite Schmidt decomposition and the pure product criterion, which proves the Schmidt expansion for arbitrary vectors of $H \hat{\otimes} K$ by constructing countable factor supports, and the reduction and purity formulae for unit vectors. On this basis Finite effects and instruments: normalization and non-determination of the update introduces finite effects and instruments with their Lüders examples and probability normalization, Abstract trace-class instruments: effects and Heisenberg duality proves the duality between abstract trace-class instruments and their effects together with the adjoint instrument $J_{B}^{*}$ , Finite Naimark compression: a finite POVM as a compression of a PVM constructs a finite PVM on a larger tensor space whose compression by an isometry gives the specified finite POVM, and Countable Kraus normalization and trace-norm convergence proves trace-norm convergence, positivity and trace preservation for countable Kraus families with strong completeness.

The infinite and continuous instrument layer is Finite-ancilla complete positivity of countable Kraus maps, which defines finite-ancilla complete positivity and proves that finite-dimensional ancilla amplifications of normalized countable Kraus maps preserve self-adjoint positive trace-class inputs; Countable Kraus isometric dilation and partial-trace recovery, which builds the isometric dilation $W ξ = \sum_{a} M_{a} ξ \otimes e_{a}$ with the repair of the normalization needed for it; Local trace-preserving Kraus maps do not change the opposite reduced state, which proves that a local instrument with $\sum_{a} M_{a}^{*} M_{a} = I$ preserves the reduced state of the other factor; Marker overlap contraction of the two-path cross term, which bounds the overlap of marker states by the received trace-class data; and Continuous measurable Kraus instruments: isometry, POVM and trace-norm integration, which integrates a strongly measurable Kraus family over a sigma-finite measure space to obtain an instrument with the conditional update on events of positive probability.

The analytic layer begins with Complex Gaussian continuation, which evaluates $\int e^{- z ∣ x ∣^{2}} d x = (π / z)^{n / 2}$ on $Re z > 0$ by holomorphic continuation, and Momentum Fourier Jacobian: the unitary momentum representation, which fixes the pointwise momentum formula $ψ_{p} (p) = (2 π ℏ)^{- d / 2} \int e^{- i p \cdot x / ℏ} ψ (x) d x$ for Schwartz inputs and its unitary extension $U = D_{ℏ} F_{2} : L^{2} (d x) \to L^{2} (d p)$ with $∥ U ψ ∥_{2} = ∥ ψ ∥_{2}$ . Real multiplication operators and free plus bounded-potential Schrödinger operators proves self-adjointness of real multiplication operators, of the free operator $H_{0} = F_{2}^{*} M_{c ∣ ξ ∣^{2} + b} F_{2}$ on its $W^{2, 2}$ domain, and of $H_{0} + V$ for bounded measurable real $V$ via Kato–Rellich, with the strongly continuous unitary group generated by the self-adjoint operator; on those results Exact free Gaussian propagation computes the exact free evolution of the Gaussian packet $ψ_{t} = (2 π σ^{2})^{- 1 / 4} (1 + i τ)^{- 1 / 2} \exp (- (x - x_{0} - p_{0} t / m)^{2} / (4 σ^{2} (1 + i τ)) + i p_{0} (x - x_{0}) / ℏ - i p_{0}^{2} t / (2 m ℏ))$ with mean $x_{0} + p_{0} t / m$ , variances $σ^{2} (1 + τ^{2})$ and $ℏ^{2} / (4 σ^{2})$ and $Δ Q Δ P = \frac{ℏ}{2} \sqrt{1 + τ^{2}}$ . Weak-form Robertson uncertainty on a common domain supplies the weak-form Robertson bound on a common domain, with the Schwartz position-momentum case $∣ Im ⟨ x_{Q}, y_{P} ⟩ ∣ = c / 2$ , and The CHSH bound and the finite-sampling law for independent trials records the CHSH inequality for the four-binary-outcome family together with the finite-sampling deviation bound behind its statistical reading.

The compatible-measurement layer is Joint PVM construction for strongly commuting regular projection-valued measures, which constructs the unique regular joint projection-valued measure $G$ of strongly commuting regular projection-valued measures $E, F$ with $G (B \times C) = E (B) F (C)$ , and Compatible Lüders repeatability, which computes the sequential Lüders probabilities $tr (ρ P_{i} Q_{j})$ , proves that the conditional state $P_{i} Q_{j} ρ P_{i} Q_{j} / tr (ρ P_{i} Q_{j})$ does not depend on the order of the two compatible sharp measurements, and derives repeatability for the stipulated ideal instrument. All items are draft mathematics under construction; their physical consumers add the measurement, evolution and apparatus assumptions, and no laboratory realization of any of these instruments is claimed here.

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

LemmaStatement: AI-adaptedProof: AI-adaptedOpen item page →

Hilbert tensor construction: inner product, completion and product bases

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $H, K$ be complex Hilbert spaces with the pairing linear in the first argument (Real and complex inner-product spaces and their induced length, Hilbert space). Regard $H$ as a right $C$ -module by $u \cdot c : = c u$ and $K$ as a left $C$ -module by $c \cdot v : = c v$ . Write $H ⊙ K$ for the complex vector space of The tensor product $M \otimes_{R} N$ from the additive group underlying the free $Z$ -module on $M \times N$ , elementary tensors, and finite tensor sums applied to these modules: its underlying abelian group is the quotient $F / H_{0}$ of that definition, the class of $e_{(u, v)}$ is written $u \otimes v$ , every element of $H ⊙ K$ is a finite sum of elementary tensors $u \otimes v$ , and the defining relations make $(u, v) \mapsto u \otimes v$ $C$ -bilinear. Define, for finite sums,

$⟨ \sum_{i} u_{i} \otimes v_{i}, \sum_{j} u_{j}^{'} \otimes v_{j}^{'} ⟩ : = \sum_{i, j} ⟨ u_{i}, u_{j}^{'} ⟩_{H} ⟨ v_{i}, v_{j}^{'} ⟩_{K} .$

Then:

(a) well-definedness and sesquilinearity: the displayed value is independent of the representatives of its two arguments and defines a Hermitian sesquilinear form on $H ⊙ K$ , linear in the first argument and conjugate-linear in the second;

(b) positive definiteness and orthonormal coordinates: $⟨ z, z ⟩ \geq 0$ for every $z \in H ⊙ K$ , with $⟨ z, z ⟩ = 0$ exactly for $z = 0$ ; more precisely, if $U_{0} \subseteq H$ and $V_{0} \subseteq K$ are finite-dimensional subspaces with orthonormal bases $e_{1}, \dots, e_{p}$ and $f_{1}, \dots, f_{q}$ , then every $z$ in the span of the elementary tensors from $U_{0} \times V_{0}$ has a unique expansion $z = \sum_{a, b} c_{a b} e_{a} \otimes f_{b}$ and $⟨ z, z ⟩ = \sum_{a, b} ∣ c_{a b} ∣^{2}$ ; in particular the assignment is an inner product on $H ⊙ K$ ;

(c) completion: the norm completion $(\hat{H ⊙ K}, κ)$ of $H ⊙ K$ (Completion of a normed space) exists and, with the unique inner product extending the given one, is a complex Hilbert space; $κ$ is a linear isometry with dense image, $∥ κ (u \otimes v) ∥ = ∥ u ∥ ∥ v ∥$ , and any two completions of $H ⊙ K$ are identified by a unique surjective linear isometry fixing $H ⊙ K$ pointwise;

(d) product bases: if $(e_{a})_{a \in A}$ and $(f_{b})_{b \in B}$ are orthonormal families in $H$ and $K$ whose closed linear spans are $H$ and $K$ , then $(κ (e_{a} \otimes f_{b}))_{(a, b) \in A \times B}$ is an orthonormal family in $\hat{H ⊙ K}$ whose closed linear span is $\hat{H ⊙ K}$ ; if $A$ and $B$ are countable then $\hat{H ⊙ K}$ is separable.

The notation $H \hat{\otimes} K : = \hat{H ⊙ K}$ is used in the rest of this pair. More precisely, if $(Y, ι)$ and $(Z, j)$ are any two norm completions of $H ⊙ K$ , there is a unique surjective linear isometry $U : Y \to Z$ with $U \circ ι = j$ . No assertion is made about unbounded operators or algebraic sums of non-completed tensor products.

Facts & Assumptions

Given: Complex Hilbert spaces $H, K$ with pairing linear in the first argument; the algebraic tensor product $H ⊙ K$ of The tensor product $M \otimes_{R} N$ from the additive group underlying the free $Z$ -module on $M \times N$ , elementary tensors, and finite tensor sums over $R = C$ ; the displayed sesquilinear form; finite sums of elementary tensors.

[F1]

The tensor product $M \otimes_{R} N$ is the quotient of the free $Z$ -module $F = Z^{(M \times N)}$ by the subgroup generated by $e_{(m + m^{'}, n)} - e_{(m, n)} - e_{(m^{'}, n)}$ , $e_{(m, n + n^{'})} - e_{(m, n)} - e_{(m, n^{'})}$ and $e_{(m r, n)} - e_{(m, r n)}$ ; its elements are the cosets $m \otimes n$ , every tensor is a finite sum of elementary tensors, the relations $(m + m^{'}) \otimes n = m \otimes n + m^{'} \otimes n$ , $m \otimes (n + n^{'}) = m \otimes n + m \otimes n^{'}$ and $(m r) \otimes n = m \otimes (r n)$ hold, and $0 \otimes n = 0 = m \otimes 0$ (The tensor product $M \otimes_{R} N$ from the additive group underlying the free $Z$ -module on $M \times N$ , elementary tensors, and finite tensor sums).

[F2]

A completion of a normed space $X$ is a pair $(\hat{X}, ι)$ with $\hat{X}$ Banach, $ι : X \to \hat{X}$ a linear isometry and $ι [X]$ dense; and if $X$ is an inner product space, its norm completion carries a unique inner product extending the given one whose induced length is the completion norm, making it a Hilbert space (Completion of a normed space, The norm completion of an inner-product space is a Hilbert space).

[F3]

The inner product conventions of the library are: linear in the first argument, conjugate-linear in the second, positive definite, and the induced length is $∥ v ∥ = ⟨ v, v ⟩^{1 / 2}$ (Real and complex inner-product spaces and their induced length, The induced length is a norm).

[F4]

Every finite-dimensional complex inner product space has an orthonormal basis (Every finite-dimensional real or complex inner product space has an orthonormal basis).

[F5]

For an orthonormal family $(g_{i})_{i \in I}$ in a Hilbert space the finite partial sums $P_{F} x = \sum_{i \in F} ⟨ x, g_{i} ⟩ g_{i}$ converge to $x$ for every $x$ exactly when the closed linear span of the family is the whole space (Orthonormal families, complete orthonormal systems and Hilbert bases, Parseval equivalences for an orthonormal family).

[F6]

In a complex inner product space, $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ , and $∥ x ∥ = 0$ implies $x = 0$ (Cauchy–Schwarz: $∣ ⟨ u, v ⟩ ∣ \leq ∥ u ∥ ∥ v ∥$ , with equality exactly for linearly dependent vectors, Orthogonality and the orthogonal complement).

Proof

technique · direct: descend the coordinate functionals through the defining relations, verify the sesquilinear form and its positive definiteness in orthonormal coordinates of the finite-dimensional factor spans, complete, and extend the product-basis density by the norm estimate for elementary products

1.1givenF1algebra

[The quotient is a complex vector space] Since $C$ is commutative, for each $c \in C$ the $Z$ -linear self-map $L_{c}$ of $F$ induced by $e_{(u, v)} \mapsto e_{(c u, v)}$ maps every defining generator of $H_{0}$ to a defining generator: $L_{c} (e_{(u + u^{'}, v)} - e_{(u, v)} - e_{(u^{'}, v)}) = e_{(c u + c u^{'}, v)} - e_{(c u, v)} - e_{(c u^{'}, v)}$ is the first relation at $(c u, c u^{'}, v)$ , the second family is preserved in the same way, and $L_{c} (e_{(u r, v)} - e_{(u, r v)}) = e_{(c (u r), v)} - e_{(c u, r v)}$ is the third relation at $(c u, r, v)$ because $c (u r) = (c u) r$ . Hence $L_{c} [H_{0}] \subseteq H_{0}$ and $L_{c}$ descends to a $Z$ -linear map of $F / H_{0}$ ; because $L_{c} L_{c^{'}} = L_{c c^{'}}$ and the actions distribute over the abelian group, $F / H_{0}$ becomes a $C$ -vector space with $c \cdot (u \otimes v) = (c u) \otimes v$ ; the third relation gives $(c u) \otimes v = u \otimes (c v)$ , so the quotient is the $C$ -bilinear tensor product $H ⊙ K$ .

1.2givenF1F3algebra

[Well-definedness and sesquilinearity] For fixed $(u_{0}, v_{0})$ the map $(u, v) \mapsto ⟨ u, u_{0} ⟩_{H} ⟨ v, v_{0} ⟩_{K}$ is $C$ -linear on $H \times K$ (both factors are linear in their first argument by [F3]) and vanishes on each generating relation of $H_{0}$ : for the additive relations by bilinearity of the pairings, and for $(u r, v) - (u, r v)$ because $⟨ u r, u_{0} ⟩ ⟨ v, v_{0} ⟩ = ⟨ u, u_{0} ⟩ ⟨ r v, v_{0} ⟩$ holds in the commutative field $C$ with the $C$ -module structures. Hence it descends to a $C$ -linear functional of the first tensor argument, and the two-variable form descends in its first argument; interchanging the roles and using conjugate symmetry of both pairings gives that it also descends in the second argument and is conjugate-linear there. Finally $⟨ z, w ⟩ = \overline{⟨ w, z ⟩}$ because both factor pairings have this property and complex conjugation is multiplicative.

2.1step 1.1step 1.2F4F6algebra

[Positive definiteness and unique orthonormal coordinates] Let $z = \sum_{i = 1}^{n} u_{i} \otimes v_{i}$ , let $U_{0} = span {u_{i}}$ and $V_{0} = span {v_{i}}$ , and choose orthonormal bases $e_{1}, \dots, e_{p}$ of $U_{0}$ and $f_{1}, \dots, f_{q}$ of $V_{0}$ by [F4]. Write $u_{i} = \sum_{a} α_{i a} e_{a}$ and $v_{i} = \sum_{b} β_{i b} f_{b}$ ; bilinearity of $\otimes$ (step 1.1) gives $z = \sum_{a, b} c_{a b} e_{a} \otimes f_{b}$ with $c_{a b} = \sum_{i} α_{i a} β_{i b}$ . By the defining formula of the Statement and orthonormality, $⟨ z, z ⟩ = \sum_{a, b, c, d} c_{a b} \overline{c_{c d}} ⟨ e_{a}, e_{c} ⟩ ⟨ f_{b}, f_{d} ⟩ = \sum_{a, b} ∣ c_{a b} ∣^{2} \geq 0$ . If $z = 0$ , then this same displayed formula gives $\sum_{a, b} ∣ c_{a b} ∣^{2} = ⟨ z, z ⟩ = 0$ , so every $c_{a b} = 0$ , because a sum of nonnegative real numbers vanishes only if every term does; conversely, if $c_{a b} = 0$ for all $a, b$ then bilinearity gives $z = \sum_{a, b} 0 \cdot e_{a} \otimes f_{b} = 0$ . Hence $⟨ z, z ⟩ = 0$ exactly when $z = 0$ , and applying the same argument to the difference of two expansions shows that the coefficients relative to orthonormal families are unique.

3.1step 2.1F2F3constructalgebra

[Completion and the norm of elementary tensors] By steps 1.2 and 2.1, $(H ⊙ K, ⟨ \cdot, \cdot ⟩)$ is a complex inner product space with induced norm $∥ z ∥ = ⟨ z, z ⟩^{1 / 2}$ by [F3]. Its norm completion $(\hat{H ⊙ K}, κ)$ exists by [F2], and by [F2] it carries a unique inner product extending the given one whose induced length is the completion norm; with it $\hat{H ⊙ K}$ is a Hilbert space and $κ$ is a linear isometry with dense image. The norm of an elementary tensor is $∥ κ (u \otimes v) ∥^{2} = ⟨ u \otimes v, u \otimes v ⟩ = ⟨ u, u ⟩_{H} ⟨ v, v ⟩_{K} = ∥ u ∥^{2} ∥ v ∥^{2}$ , hence $∥ κ (u \otimes v) ∥ = ∥ u ∥ ∥ v ∥$ . To prove the completion uniqueness asserted above, let $(Y, ι)$ and $(Z, j)$ be completions. The rule $ι (z) \mapsto j (z)$ is a well-defined linear isometry on the dense subspace $ι [H ⊙ K]$ . For $y \in Y$ , choose $z_{n} \in H ⊙ K$ with $ι (z_{n}) \to y$ and define $U (y) : = \lim_{n} j (z_{n})$ ; the isometry makes $(j (z_{n}))$ Cauchy and makes this limit independent of the approximating sequence. Thus $U$ is a linear isometry with $U \circ ι = j$ . Its image is closed because $Y$ is complete and contains the dense set $j [H ⊙ K]$ , so it is onto; any two such continuous extensions agree on a dense subspace and hence everywhere.

4.1step 3.1F1algebra

[Orthonormal family of products] Let $(e_{a})_{a \in A}$ , $(f_{b})_{b \in B}$ be orthonormal with closed linear spans $H$ and $K$ . For pairs $(a, b) \neq (a^{'}, b^{'})$ at least one index differs, so $⟨ κ (e_{a} \otimes f_{b}), κ (e_{a^{'}} \otimes f_{b^{'}}) ⟩ = ⟨ e_{a}, e_{a^{'}} ⟩_{H} ⟨ f_{b}, f_{b^{'}} ⟩_{K} = 0$ , and for equal pairs the value is $1$ ; hence the products are orthonormal in $\hat{H ⊙ K}$ .

5.1step 3.1step 4.1F5F6algebra∎

[Density of the product span] Fix $ε > 0$ and $z = \sum_{i = 1}^{n} u_{i} \otimes v_{i} \in H ⊙ K$ . By [F5] applied to $(e_{a})$ and to $(f_{b})$ there are finite sets $A_{i} \subseteq A$ , $B_{i} \subseteq B$ with $∥ u_{i} - \sum_{a \in A_{i}} ⟨ u_{i}, e_{a} ⟩ e_{a} ∥ < ε / (n (∥ v_{i} ∥ + 1))$ and $∥ v_{i} - \sum_{b \in B_{i}} ⟨ v_{i}, f_{b} ⟩ f_{b} ∥ < ε / (n (∥ u_{i} ∥ + 1))$ ; write $u_{i}^{'}, v_{i}^{'}$ for the finite sums and $z^{'} = \sum_{i} u_{i}^{'} \otimes v_{i}^{'}$ . Then $z - z^{'} = \sum_{i} ((u_{i} - u_{i}^{'}) \otimes v_{i} + u_{i}^{'} \otimes (v_{i} - v_{i}^{'}))$ , and by the triangle inequality, $∥ u_{i}^{'} ∥ \leq ∥ u_{i} ∥$ (Bessel, [F5]) and the norm formula of step 3.1, $∥ z - z^{'} ∥ \leq \sum_{i} (∥ u_{i} - u_{i}^{'} ∥ ∥ v_{i} ∥ + ∥ u_{i} ∥ ∥ v_{i} - v_{i}^{'} ∥) < 2 ε$ . Hence every element of the dense subspace $κ [H ⊙ K]$ is approximated by finite sums $\sum c_{a b} κ (e_{a} \otimes f_{b})$ , so the closed linear span of the product family contains $κ [H ⊙ K]$ and therefore coincides with $\hat{H ⊙ K}$ ; by [F5] the family is a complete orthonormal system. If $A$ and $B$ are countable, then $A \times B$ is countable and the closed span of a countable family is separable, so $\hat{H ⊙ K}$ is separable.

LemmaStatement: Literature-sourcedProof: AI-adaptedOpen item page →

Rank-one operator calculus

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $H$ be a complex Hilbert space with the pairing linear in the first argument (Hilbert space, Real and complex inner-product spaces and their induced length), and for $u, v \in H$ let

$T_{u, v} x : = ⟨ x, v ⟩ u (x \in H) .$

Then:

(a) $T_{u, v} \in B (H)$ with $∥ T_{u, v} ∥ = ∥ u ∥ ∥ v ∥$ (A bounded linear operator between normed spaces, The operator norm as the least bound and as the unit-sphere or unit-ball supremum), and the assignment $(u, v) \mapsto T_{u, v}$ is linear in $u$ and conjugate-linear in $v$ ;

(b) $T_{u, v}$ is trace class (Trace class operator) with $∥ T_{u, v} ∥_{1} = ∥ u ∥ ∥ v ∥$ and $tr (T_{u, v}) = ⟨ u, v ⟩$ ;

(c) $T_{u, v}^{*} = T_{v, u}$ (The Hilbert-space adjoint of a bounded operator); $T_{x, x}$ is a positive bounded operator in the sense of Self-adjoint, positive, unitary and normal operators, with $⟨ T_{x, x} y, y ⟩ = ∣ ⟨ y, x ⟩ ∣^{2} \geq 0$ for every $y \in H$ , and $tr (T_{x, x}) = ∥ x ∥^{2}$ , $∥ T_{x, x} ∥_{1} = ∥ x ∥^{2}$ ;

(d) for bounded $B \in B (H)$ , $B T_{u, v} = T_{B u, v}$ and $T_{u, v} B = T_{u, B^{*} v}$ ;

(e) for trace-class $S$ , $S T_{u, v} = T_{S u, v}$ , $T_{u, v} S = T_{u, S^{*} v}$ and $tr (S T_{x, y}) = ⟨ S x, y ⟩ = tr (T_{x, y} S)$ ;

(f) the map $(u, v) \mapsto T_{u, v}$ is continuous $H \times H \to S_{1} (H)$ for the trace norm, so in particular $v \mapsto T_{v, v}$ and $v \mapsto T_{u, v}$ are trace-norm continuous for fixed $u$ .

Scope. Only the rank-one calculus is asserted; no spectral, compactness or ideal statement beyond (b) is claimed, and no physical interpretation is attached.

Facts & Assumptions

Given: A complex Hilbert space $H$ with pairing linear in the first argument; vectors $u, v, x, y, w, z \in H$ ; a bounded operator $B$ ; a trace-class operator $S$ .

[F1]

A compact operator $T$ is trace class when $\sum_{n} s_{n} (T) < \infty$ , with trace norm $∥ T ∥_{1} = \sum_{n} s_{n} (T)$ ; then $∥ T ∥_{1} \geq ∥ T ∥$ , the zero operator is trace class, and every bounded finite-rank operator is compact (Trace class operator, Bounded finite rank operators are compact).

[F2]

$T$ is trace class if and only if it has a nuclear representation $T x = \sum_{j} ⟨ x, u_{j} ⟩ v_{j}$ converging in operator norm with $\sum_{j} ∥ u_{j} ∥ ∥ v_{j} ∥ < \infty$ ; then $∥ T ∥_{1} = \inf \sum_{j} ∥ u_{j} ∥ ∥ v_{j} ∥$ over all such representations, the infimum being attained (Nuclear series characterizes trace norm).

[F3]

For every nuclear representation $T = \sum_{j} T_{v_{j}, u_{j}}$ of a trace-class operator, the scalar series $\sum_{j} ⟨ v_{j}, u_{j} ⟩$ converges absolutely to a number $tr (T)$ independent of the representation, the trace is linear in the trace-class variable, and $∣ tr (T) ∣ \leq ∥ T ∥_{1}$ (Trace is absolutely convergent and basis independent).

[F4]

The Hilbert adjoint is characterised by $⟨ B x, y ⟩ = ⟨ x, B^{*} y ⟩$ for all $x, y$ and is unique; the pairing is linear in the first argument and conjugate-linear in the second with $⟨ x, y ⟩ = \overline{⟨ y, x ⟩}$ , and $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ (The Hilbert-space adjoint of a bounded operator, Real and complex inner-product spaces and their induced length, Cauchy–Schwarz: $∣ ⟨ u, v ⟩ ∣ \leq ∥ u ∥ ∥ v ∥$ , with equality exactly for linearly dependent vectors).

[F5]

On the trace-class operators $∥ \cdot ∥_{1}$ is a norm, they form a Banach space, and $∥ T ∥ \leq ∥ T ∥_{1}$ and $∥ A T B ∥_{1} \leq ∥ A ∥ ∥ T ∥_{1} ∥ B ∥$ (Trace class is a two sided Banach operator ideal).

[F6]

The operator norm is $∥ B ∥ = \sup_{∥ x ∥ \leq 1} ∥ B x ∥$ for bounded $B$ ; a bounded operator is positive when its quadratic-form values are real and nonnegative (The operator norm as the least bound and as the unit-sphere or unit-ball supremum, A bounded linear operator between normed spaces, Self-adjoint, positive, unitary and normal operators).

Proof

1.1givenF4F6algebra

[Boundedness, norm and linearity] $T_{u, v}$ is linear and $∥ T_{u, v} x ∥ = ∣ ⟨ x, v ⟩ ∣ ∥ u ∥ \leq ∥ x ∥ ∥ v ∥ ∥ u ∥$ by [F4], so $∥ T_{u, v} ∥ \leq ∥ u ∥ ∥ v ∥$ [F6]; if $u \neq 0$ and $v \neq 0$ , taking $x = v / ∥ v ∥$ gives $T_{u, v} x = ∥ v ∥ u$ and hence $∥ T_{u, v} ∥ \geq ∥ u ∥ ∥ v ∥$ , while for $u = 0$ or $v = 0$ the operator vanishes and both sides are $0$ . Linearity in $u$ and conjugate-linearity in $v$ are immediate from the definition and the pairing conventions [F4].

2.1step 1.1F1F2F5algebra

[Trace class and trace norm] The range of $T_{u, v}$ is contained in $C u$ , so $T_{u, v}$ has rank at most one, is finite rank, compact [F1]; its one-term nuclear representation is $T_{u, v} = \sum_{j} T_{v_{j}, u_{j}}$ with $v_{1} = u$ , $u_{1} = v$ and the remaining terms zero, so $\sum_{j} ∥ u_{j} ∥ ∥ v_{j} ∥ = ∥ v ∥ ∥ u ∥$ and [F2] first establishes trace class and gives $∥ T_{u, v} ∥_{1} \leq ∥ u ∥ ∥ v ∥$ , while $∥ T_{u, v} ∥_{1} \geq ∥ T_{u, v} ∥ = ∥ u ∥ ∥ v ∥$ by [F5] and step 1.1. Hence $∥ T_{u, v} ∥_{1} = ∥ u ∥ ∥ v ∥$ .

3.1step 2.1F3algebra

[The trace] Applying [F3] to the one-term nuclear representation of step 2.1 gives $tr (T_{u, v}) = \sum_{j} ⟨ v_{j}, u_{j} ⟩ = ⟨ u, v ⟩$ . In particular $tr (T_{x, x}) = ⟨ x, x ⟩ = ∥ x ∥^{2}$ and, with step 2.1, $∥ T_{x, x} ∥_{1} = ∥ x ∥^{2}$ .

3.2step 1.1step 2.1F4F6algebra

[Adjoint, positivity and one-sided products] For all $x, y$ , $⟨ T_{u, v} x, y ⟩ = ⟨ x, v ⟩ ⟨ u, y ⟩ = ⟨ x, ⟨ y, u ⟩ v ⟩$ by [F4], so $T_{u, v}^{*} = T_{v, u}$ ; in particular $T_{x, x}$ is self-adjoint, and $⟨ T_{x, x} y, y ⟩ = ∣ ⟨ y, x ⟩ ∣^{2} \geq 0$ for every $y$ , so it is positive by [F6]. For bounded $B$ : $(B T_{u, v}) x = B (⟨ x, v ⟩ u) = ⟨ x, v ⟩ B u = T_{B u, v} x$ and $(T_{u, v} B) x = ⟨ B x, v ⟩ u = ⟨ x, B^{*} v ⟩ u = T_{u, B^{*} v} x$ , using the defining property of the adjoint [F4].

3.3step 1.1step 2.1F5algebra

[Trace-norm continuity] By the bilinearity of step 1.1, $T_{u, v} - T_{u^{'}, v^{'}} = T_{u - u^{'}, v} + T_{u^{'}, v - v^{'}}$ ; by [F5] and the estimates of steps 1.1 and 2.1, $∥ T_{u, v} - T_{u^{'}, v^{'}} ∥_{1} \leq ∥ u - u^{'} ∥ ∥ v ∥ + ∥ u^{'} ∥ ∥ v - v^{'} ∥$ , which tends to $0$ whenever $(u^{'}, v^{'}) \to (u, v)$ in $H \times H$ ; hence $(u, v) \mapsto T_{u, v}$ is continuous into $S_{1} (H)$ , and fixing one slot gives the stated continuity. This is (f).

4.1step 3.1step 3.2F3F4algebra∎

[Products with trace-class operators] If $S$ is trace class then $(S T_{u, v}) x = ⟨ x, v ⟩ S u = T_{S u, v} x$ and $(T_{u, v} S) x = ⟨ S x, v ⟩ u = ⟨ x, S^{*} v ⟩ u = T_{u, S^{*} v} x$ , so $S T_{u, v} = T_{S u, v}$ and $T_{u, v} S = T_{u, S^{*} v}$ [F4]; applying step 3.1 to the rank-one operators $T_{S u, v}$ and $T_{u, S^{*} v}$ gives $tr (S T_{x, y}) = ⟨ S x, y ⟩$ and $tr (T_{x, y} S) = ⟨ x, S^{*} y ⟩ = ⟨ S x, y ⟩$ . This is (e); (a)-(d) are steps 1.1, 2.1, 3.1 and 3.2, and the scope remark records that nothing beyond the rank-one calculus is asserted.

LemmaStatement: AI-adaptedProof: AI-adaptedOpen item page →

Bounded tensor maps: extension, norm estimate, adjoint and commutation

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $H, H^{'}, K, K^{'}$ be complex Hilbert spaces and let $S \in B (H, H^{'})$ , $T \in B (K, K^{'})$ be bounded linear operators (A bounded linear operator between normed spaces). Write $H \hat{\otimes} K$ for the completion of the algebraic tensor product constructed in Hilbert tensor construction: inner product, completion and product bases, with inclusion $κ$ .

(a) There is a unique bounded linear operator $S \otimes T : H \hat{\otimes} K \to H^{'} \hat{\otimes} K^{'}$ with $(S \otimes T) κ (u \otimes v) = κ^{'} (S u \otimes T v)$ for all $u \in H$ , $v \in K$ , and it satisfies $∥ S \otimes T ∥ \leq ∥ S ∥ ∥ T ∥$ (The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

(b) For Hilbert spaces $H^{''}, K^{''}$ and bounded $S^{'} : H^{''} \to H$ , $T^{'} : K^{''} \to K$ , one has $(S \otimes T) (S^{'} \otimes T^{'}) = (S S^{'}) \otimes (T T^{'})$ . In particular, for endomorphisms $S : H \to H$ , $T : K \to K$ , the endomorphisms $S \otimes I_{K}$ and $I_{H} \otimes T$ commute and their composite is $S \otimes T$ .

(d) If $S, T$ are unitary, then so is $S \otimes T$ . No extension of unbounded maps, and no algebraic manipulation of domains of unbounded operators, is asserted.

Facts & Assumptions

Given: Complex Hilbert spaces $H, H^{'}, K, K^{'}$ ; bounded operators $S \in B (H, H^{'})$ , $T \in B (K, K^{'})$ ; in (b), composable $S^{'} : H^{''} \to H$ , $T^{'} : K^{''} \to K$ ; the completions $H \hat{\otimes} K$ , $H^{'} \hat{\otimes} K^{'}$ of Hilbert tensor construction: inner product, completion and product bases with inclusions $κ, κ^{'}$ ; the elementary tensors $u \otimes v$ .

[F1]

The construction of Hilbert tensor construction: inner product, completion and product bases gives: the algebraic tensor product $H ⊙ K$ with $u \otimes v$ bilinear; a positive definite inner product with $⟨ u \otimes v, u^{'} \otimes v^{'} ⟩ = ⟨ u, u^{'} ⟩_{H} ⟨ v, v^{'} ⟩_{K}$ ; the completion $H \hat{\otimes} K$ with dense inclusion $κ$ ; unique orthonormal-coordinate expansions with $∥ \sum_{a, b} c_{a b} e_{a} \otimes f_{b} ∥^{2} = \sum_{a, b} ∣ c_{a b} ∣^{2}$ for orthonormal families $(e_{a}) \subseteq H$ , $(f_{b}) \subseteq K$ spanning finite-dimensional subspaces; and $∥ κ (u \otimes v) ∥ = ∥ u ∥ ∥ v ∥$ .

[F2]

If $X$ is a normed space, $D \subseteq X$ a dense normed subspace, $Y$ a Banach space and $L : D \to Y$ bounded linear, then $L$ has a unique bounded linear extension to $X$ of the same norm (A bounded linear map from a dense normed subspace into a Banach space extends uniquely with the same norm).

[F3]

A linear map is bounded exactly when it admits a finite bound $∥ L x ∥ \leq C ∥ x ∥$ , and the operator norm is the least such bound (A bounded linear operator between normed spaces, The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

[F4]

The Hilbert adjoint $L^{*}$ of a bounded operator $L \in B (X, Y)$ is the unique bounded operator with $⟨ L x, y ⟩_{Y} = ⟨ x, L^{*} y ⟩_{X}$ for all $x \in X$ , $y \in Y$ ; it satisfies $(L_{1} L_{2})^{*} = L_{2}^{*} L_{1}^{*}$ , $L^{* *} = L$ and $∥ L^{*} ∥ = ∥ L ∥$ (The Hilbert-space adjoint of a bounded operator, Hilbert-adjoint identities).

Proof

1.1givenF1algebra

[The bilinear extension on the algebraic tensor product] Put $R_{0} (u \otimes v) : = κ^{'} (S u \otimes T v)$ and extend $C$ -linearly to finite sums. This is well defined and $C$ -linear on $H ⊙ K$ because $(u, v) \mapsto S u \otimes T v$ is $C$ -bilinear, composition of the linear maps $S, T$ with the bilinear tensor map; thus $R_{0}$ is the unique linear map on $H ⊙ K$ with $R_{0} (u \otimes v) = κ^{'} (S u \otimes T v)$ .

2.1step 1.1F1F3algebra

[One-sided norm estimates] Let $z = \sum_{i = 1}^{n} u_{i} \otimes v_{i} \in H ⊙ K$ . Let $V_{0} = span {v_{i}}$ and let $f_{1}, \dots, f_{q}$ be an orthonormal basis of $V_{0}$ ; grouping the expansion of each $v_{i}$ in this basis gives $z = \sum_{b = 1}^{q} u_{b} \otimes f_{b}$ with $u_{b} \in span {u_{i}}$ . By the norm formula of [F1] and $∥ S u_{b} ∥ \leq ∥ S ∥ ∥ u_{b} ∥$ , $∥ (S \otimes I) z ∥^{2} = \sum_{b} ∥ S u_{b} ∥^{2} \leq ∥ S ∥^{2} \sum_{b} ∥ u_{b} ∥^{2} = ∥ S ∥^{2} ∥ z ∥^{2}$ , so the map $S \otimes I_{K}$ on elementary tensors is bounded with norm at most $∥ S ∥$ ; symmetrically $∥ (I_{H} \otimes T) z ∥ \leq ∥ T ∥ ∥ z ∥$ . Hence $∥ R_{0} z ∥ \leq ∥ S ∥ ∥ T ∥ ∥ z ∥$ for all $z \in H ⊙ K$ , because $R_{0}$ is the composite of the two one-sided maps on elementary tensors and the estimates compose.

3.1step 1.1step 2.1F2

[Extension to the completions] By [F2] applied to the dense subspace $H ⊙ K$ of $H \hat{\otimes} K$ and the Banach space $H^{'} \hat{\otimes} K^{'}$ , the map $R_{0}$ has a unique bounded linear extension $S \otimes T : H \hat{\otimes} K \to H^{'} \hat{\otimes} K^{'}$ of the same norm; by step 2.1 its norm is at most $∥ S ∥ ∥ T ∥$ , and uniqueness also shows that $(S \otimes T) κ = R_{0}$ . This is clause (a).

4.1step 3.1F2algebra

[Products and commutation] For $u \otimes v$ one has $(S \otimes T) (S^{'} \otimes T^{'}) (u \otimes v) = κ^{'} (S S^{'} u \otimes T T^{'} v) = (S S^{'} \otimes T T^{'}) (u \otimes v)$ ; the two maps are bounded and agree on the dense subspace $H'' ⊙ K''$ , hence everywhere by uniqueness of bounded extensions, giving the first identity of (b) for elementary tensors in $H'' ⊙ K''$ . In the endomorphism case, taking $T^{'} = I_{K}$ , $S^{'} = I_{H}$ and using $I \otimes I = I$ shows $(S \otimes I_{K}) (I_{H} \otimes T) = S \otimes T$ ; applying the same identity with the roles exchanged gives $(I_{H} \otimes T) (S \otimes I_{K}) = S \otimes T$ , since $S I_{H} = S$ and $I_{K} T = T$ .

4.2step 3.1F1F4algebra

[Adjoint] For $u \in H$ , $u^{'} \in H^{'}$ and $v \in K$ , $v^{'} \in K^{'}$ , $⟨ (S \otimes T) κ (u \otimes v), κ^{'} (u^{'} \otimes v^{'}) ⟩ = ⟨ S u, u^{'} ⟩_{H^{'}} ⟨ T v, v^{'} ⟩_{K^{'}} = ⟨ u, S^{*} u^{'} ⟩_{H} ⟨ v, T^{*} v^{'} ⟩_{K} = ⟨ κ (u \otimes v), (S^{*} \otimes T^{*}) κ^{'} (u^{'} \otimes v^{'}) ⟩$ , using [F4] and the pairing formula of [F1]. Both sides are sesquilinear in the elementary tensors and continuous in each argument, and the elementary tensors span a dense subspace, so the identity holds on all of $H \hat{\otimes} K \times H^{'} \hat{\otimes} K^{'}$ ; by the uniqueness in [F4], $(S \otimes T)^{*} = S^{*} \otimes T^{*}$ .

5.1step 4.1step 4.2algebra∎

[Unitary case] If $S, T$ are unitary then $S^{*} S = I$ , $S S^{*} = I$ , $T^{*} T = I$ , $T T^{*} = I$ , so by steps 4.1 and 4.2 $(S \otimes T)^{*} (S \otimes T) = S^{*} S \otimes T^{*} T = I \otimes I = I$ and likewise $(S \otimes T) (S \otimes T)^{*} = I$ , whence $S \otimes T$ is unitary.

LemmaStatement: AI-adaptedProof: AI-adaptedOpen item page →

Density operators and scalar spectral probability

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )) and let $H$ be a nonzero complex Hilbert space (Hilbert space). Recall that $T_{u, v} \in B (H)$ denotes the rank-one operator $T_{u, v} x = ⟨ x, v ⟩ u$ (Rank-one operator calculus).

(a) A density operator on $H$ is a self-adjoint positive trace-class operator $ρ$ with $tr ρ = 1$ (Trace class operator, Trace of a trace class operator, Self-adjoint, positive, unitary and normal operators). If $(u_{n})_{n \in J}$ is a finite or countable orthonormal family, with $J$ a finite initial segment of the positive integers or all positive integers, and $w_{n} \geq 0$ with $\sum_{n} w_{n} = 1$ , then the sum $ρ : = \sum_{n} w_{n} T_{u_{n}, u_{n}}$ converges in trace norm to a density operator with $tr ρ = \sum_{n} w_{n} = 1$ . Conversely every density operator $ρ$ admits such an expansion whose weights are its positive eigenvalues counted with multiplicity and whose vectors are orthonormal eigenvectors; distinct expansions of the same $ρ$ exist when a positive eigenspace has dimension at least two.

(b) Let $E$ be a projection-valued measure on a measurable space $(X, Σ)$ acting on $H$ (Projection valued measure). Then $μ_{ρ} (B) : = tr (ρ E (B))$ is a well-defined probability measure on $(X, Σ)$ (Measures on sigma-algebras, Probability measures and probability spaces); it is given by the absolutely convergent series $μ_{ρ} (B) = \sum_{n} w_{n} ∥ E (B) u_{n} ∥^{2}$ in every expansion of (a), and reduces to $μ_{ρ} (B) = ∥ E (B) u ∥^{2}$ for a pure density operator $ρ = T_{u, u}$ with $∥ u ∥ = 1$ .

(c) Assume in addition the Axiom of Choice (The Axiom of Choice). Let $A$ be a self-adjoint operator with projection-valued measure $E_{A}$ (Spectral theorem for unbounded self-adjoint operators (PVM form)) and let $ψ \in H$ have $∥ ψ ∥ = 1$ and $ψ \in D (A)$ . Here $∣ A ∣^{1 / 2}$ denotes the unbounded PVM integral $f (E_{A})$ for $f (λ) = ∣ λ ∣^{1 / 2}$ , with domain $D (f (E_{A})) = {ψ : \int ∣ λ ∣ d E_{A, ψ} < \infty}$ as in Integral of a measurable function against a projection-valued measure. Then

$\int_{R} λ d E_{A, ψ} (λ) = ⟨ A ψ, ψ ⟩, \int_{R} λ^{2} d E_{A, ψ} (λ) = ∥ A ψ ∥^{2},$

where $E_{A, ψ} (B) = ⟨ E_{A} (B) ψ, ψ ⟩$ (Scalar and complex measures from a pvm); moreover $\int_{R} ∣ λ ∣ d E_{A, ψ} (λ) < \infty$ exactly when $ψ \in D (∣ A ∣^{1 / 2})$ , and then the mean $\int λ d E_{A, ψ}$ is finite. No product $A ρ$ of an unbounded operator with a density operator is used, and no physical interpretation of $μ_{ρ}$ or of the moments is asserted.

Facts & Assumptions

Given: A nonzero complex Hilbert space $H$ ; orthonormal vectors $u_{n}$ and weights $w_{n} \geq 0$ ; a density operator $ρ$ ; a PVM $E$ on $(X, Σ)$ ; in clause (c) a self-adjoint $A$ with PVM $E_{A}$ and a unit vector $ψ \in D (A)$ .

[F1]

$T_{u, v} x = ⟨ x, v ⟩ u$ is trace class with $∥ T_{u, v} ∥_{1} = ∥ u ∥ ∥ v ∥$ and $tr (T_{u, v}) = ⟨ u, v ⟩$ ; $T$ is trace class iff it is a norm limit of a nuclear series $T = \sum_{j} T_{v_{j}, u_{j}}$ with $\sum_{j} ∥ v_{j} ∥ ∥ u_{j} ∥ < \infty$ , and then $∥ T ∥_{1} = \inf \sum_{j} ∥ v_{j} ∥ ∥ u_{j} ∥$ over such representations (Rank-one operator calculus, Nuclear series characterizes trace norm).

[F2]

Trace class is a two-sided Banach operator ideal: $∥ A T B ∥_{1} \leq ∥ A ∥ ∥ T ∥_{1} ∥ B ∥$ , and the trace is linear on the trace-class operators with $∣ tr T ∣ \leq ∥ T ∥_{1}$ (Trace class is a two sided Banach operator ideal, Trace is absolutely convergent and basis independent).

[F3]

If $T$ is compact and self-adjoint, its nonzero eigenvalues are real with finite multiplicity and accumulate only at $0$ , the closed span of their eigenspaces is $(\ker T)^{⊥} = \overline{ran T}$ , and $T x = \sum_{λ} λ P_{λ} x$ converges in norm, where $P_{λ}$ is the orthogonal projection onto the eigenspace of $λ$ (Spectral theorem for compact self adjoint operators).

[F4]

If $T \in B (H)$ is self-adjoint and positive and trace class, then $s_{n} (T) = λ_{n}$ and $tr T = \sum_{n \geq 1} λ_{n} = ∥ T ∥_{1}$ , the sum running over the positive eigenvalues with multiplicity; in particular a positive trace-class operator is compact (Trace of a positive operator is the sum of its eigenvalues, Trace class operator).

[F5]

A PVM $E$ is a map into orthogonal projections with $E (\emptyset) = 0$ , $E (X) = I$ , $E (B \cap C) = E (B) E (C)$ and strong countable additivity; for every $x$ the set function $E_{x} (B) = ⟨ E (B) x, x ⟩$ is a positive measure with $E_{x} (X) = ∥ x ∥^{2} \leq ∥ x ∥^{2}$ and $E (B)$ is an orthogonal projection, so $⟨ E (B) x, x ⟩ = ∥ E (B) x ∥^{2} \geq 0$ (Projection valued measure, Scalar and complex measures from a pvm).

[F6]

Assume the Axiom of Choice. For a self-adjoint $A$ with PVM $E_{A}$ , $D (A) = {ψ : \int λ^{2} d E_{A, ψ} < \infty}$ and $A ψ = \int λ d E_{A} (λ) ψ$ ; the integral against a PVM defines $f (A)$ on $D (f (A)) = {ψ : \int ∣ f ∣^{2} d E_{A, ψ} < \infty}$ (Spectral theorem for unbounded self-adjoint operators (PVM form), Integral of a measurable function against a projection-valued measure). For bounded $f$ and $x \in H$ , $⟨ Φ_{E} (f) x, x ⟩ = \int f d E_{x}$ and $∥ Φ_{E} (f) x ∥^{2} = \int ∣ f ∣^{2} d E_{x}$ (Bounded borel pvm integral).

[F7]

If nonnegative measurable $f_{n}$ increase pointwise to $f$ , then $\int f_{n} d μ ↑ \int f d μ$ (Monotone convergence for the integral); if measurable complex-valued $f_{n}, f$ satisfy $f_{n} \to f$ almost everywhere and $∣ f_{n} ∣ \leq g$ almost everywhere for a nonnegative measurable integrable $g$ , then $\int f_{n} d μ \to \int f d μ$ (Dominated convergence).

Proof

1.1givenF1F2F3F4choosealgebra

[Expansion of a density operator] Let $ρ$ be a density operator. By [F4] and [F1] it is compact, self-adjoint and positive, so [F3] applies: its positive eigenvalues $μ_{1} > μ_{2} > \dots$ have finite multiplicities, accumulate only at $0$ , and $ρ x = \sum_{k} μ_{k} P_{k} x$ with $P_{k}$ the orthogonal projection onto $E_{μ_{k}} (ρ)$ . Each $E_{μ_{k}} (ρ)$ is finite-dimensional, so it has an orthonormal basis by Every finite-dimensional real or complex inner product space has an orthonormal basis; using Countable Choice to select the finite bases and listing their vectors in a finite or countable sequence $(u_{n})$ with the corresponding eigenvalues $w_{n} > 0$ indexed by a finite initial segment or the positive integers as appropriate and enumerated by [F3] (the multiset is finite or countably infinite) gives an orthonormal family with $ρ x = \sum_{n} w_{n} ⟨ x, u_{n} ⟩ u_{n}$ , that is, $ρ = \sum_{n} w_{n} T_{u_{n}, u_{n}}$ in operator norm, and [F4] gives $tr ρ = \sum_{n} w_{n} = ∥ ρ ∥_{1} = 1$ . Their rank-one trace norms sum to one by [F1], so the partial sums also converge in trace norm by completeness [F2]; that limit equals $ρ$ because both limits agree in operator norm.

1.2givenF1F2algebra

[Weighted sums are density operators] Conversely let $(u_{n})$ be orthonormal and $w_{n} \geq 0$ with $\sum_{n} w_{n} = 1$ . The series $\sum_{n} w_{n} T_{u_{n}, u_{n}}$ is nuclear with $\sum_{n} w_{n} ∥ u_{n} ∥ ∥ u_{n} ∥ = \sum_{n} w_{n} < \infty$ , and its trace norms sum to one by [F1]. Completeness of trace class [F2] therefore makes its partial sums converge in trace norm to a trace-class operator $ρ$ with $∥ ρ ∥_{1} \leq 1$ ; this also gives operator-norm convergence; it is self-adjoint, and $⟨ ρ x, x ⟩ = \sum_{n} w_{n} ∣ ⟨ x, u_{n} ⟩ ∣^{2} \geq 0$ by continuity of the pairing and nonnegativity of each term, so $ρ$ is positive; [F1] and [F2] give $tr ρ = \sum_{n} w_{n} ⟨ u_{n}, u_{n} ⟩ = \sum_{n} w_{n} = 1$ by linearity and absolute convergence. Hence $ρ$ is a density operator. Distinct orthonormal bases of a single positive eigenspace produce different families $(u_{n})$ with the same sum, so expansions are not unique.

2.1step 1.1step 1.2F1F2F5F7algebra

[The scalar law of a PVM] Let $E$ be a PVM on $(X, Σ)$ and let $ρ = \sum_{n} w_{n} T_{u_{n}, u_{n}}$ be an expansion from step 1.1 (any expansion gives the same value below by step 1.2 and [F2]). For $B \in Σ$ , $T_{u_{n}, u_{n}} E (B) = T_{u_{n}, E (B) u_{n}}$ , so by linearity of the trace and trace-norm convergence of the series (traced against the bounded $E (B)$ ), $tr (ρ E (B)) = \sum_{n} w_{n} tr (T_{u_{n}, E (B) u_{n}}) = \sum_{n} w_{n} ⟨ u_{n}, E (B) u_{n} ⟩ = \sum_{n} w_{n} ∥ E (B) u_{n} ∥^{2}$ , the series converging absolutely with value in $[0, 1]$ because the finite subsums are bounded by $\sum w_{n} ∥ u_{n} ∥^{2} = 1$ by [F5]. Setting $B = X$ gives $μ_{ρ} (X) = \sum_{n} w_{n} = 1$ . If $B_{k}$ are pairwise disjoint with union $B$ , then for each $n$ the identity $∥ E (B) u_{n} ∥^{2} = \sum_{k} ∥ E (B_{k}) u_{n} ∥^{2}$ holds by strong additivity of $E$ and continuity of the norm (Projection valued measure); summing the resulting absolutely convergent nonnegative double series by [F7] gives $μ_{ρ} (B) = \sum_{k} μ_{ρ} (B_{k})$ . Thus $μ_{ρ}$ is a probability measure; for $ρ = T_{u, u}$ with $∥ u ∥ = 1$ step 1.2 applies with $w_{1} = 1$ , giving $μ_{ρ} (B) = ∥ E (B) u ∥^{2}$ .

3.1givenF6F7algebra∎

[Moments for a self-adjoint operator, under AC] Assume the Axiom of Choice and let $ψ \in D (A)$ with $∥ ψ ∥ = 1$ and $E_{A}$ the PVM of $A$ . By [F6], $\int λ^{2} d E_{A, ψ} < \infty$ and, for the bounded measurable truncations $λ_{n} : = λ 1_{{∣ λ ∣ \leq n}}$ , $Φ_{E_{A}} (λ_{n}) ψ \to A ψ$ in norm. The bounded quadratic identity gives $∥ Φ_{E_{A}} (λ_{n}) ψ ∥^{2} = \int λ^{2} 1_{{∣ λ ∣ \leq n}} d E_{A, ψ}$ . These nonnegative measurable integrands increase to $λ^{2}$ , so monotone convergence [F7] and norm continuity give $\int λ^{2} d E_{A, ψ} = \lim_{n} ∥ Φ_{E_{A}} (λ_{n}) ψ ∥^{2} = ∥ A ψ ∥^{2}$ . Since $ψ \in D (A)$ the integral $\int ∣ λ ∣ d E_{A, ψ}$ is finite, because $∣ λ ∣ \leq (1 + λ^{2}) / 2$ and $\int d E_{A, ψ} = 1$ ; the same truncations $λ_{n}$ satisfy $∣ λ_{n} ∣ \leq ∣ λ ∣$ and $λ_{n} \to λ$ pointwise, so Dominated convergence applied to the finite measures $E_{A, ψ}$ gives $\int λ d E_{A, ψ} = \lim_{n} \int λ_{n} d E_{A, ψ}$ . Pairing the identity $A ψ = \lim_{n} Φ_{E_{A}} (λ_{n}) ψ$ (the defining truncation limit of Integral of a measurable function against a projection-valued measure) with $ψ$ and using $⟨ Φ_{E_{A}} (λ_{n}) ψ, ψ ⟩ = \int λ_{n} d E_{A, ψ}$ from [F6] yields $⟨ A ψ, ψ ⟩ = \int λ d E_{A, ψ} (λ)$ . Finally [F6] applied to $f = ∣ λ ∣^{1 / 2}$ gives $\int ∣ λ ∣ d E_{A, ψ} < \infty$ exactly when $ψ \in D (∣ A ∣^{1 / 2})$ , and in that case the mean integral is finite.

LemmaStatement: AI-adaptedProof: AI-adaptedOpen item page →

Weak-form Robertson uncertainty on a common domain

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $H$ be a complex Hilbert space (Hilbert space) and let $A, B$ be symmetric densely defined operators on $H$ (Symmetric, self-adjoint and essentially self-adjoint operators, Adjoint of a densely defined operator). Let $u \in D (A) \cap D (B)$ have $∥ u ∥ = 1$ , and put

$a : = ⟨ A u, u ⟩, b : = ⟨ B u, u ⟩, x : = (A - a) u, y : = (B - b) u, Δ A : = ∥ x ∥, Δ B : = ∥ y ∥ .$

Then:

(a) $a$ and $b$ are real, $x, y \in H$ , and $Δ A Δ B \geq ∣ ⟨ x, y ⟩ ∣ \geq ∣ Im ⟨ x, y ⟩ ∣$ ;

(b) (refined form) $(Δ A)^{2} (Δ B)^{2} \geq [Re ⟨ x, y ⟩]^{2} + [Im ⟨ x, y ⟩]^{2}$ ;

(c) if in addition $u \in D (A B) \cap D (B A)$ , then $([A, B] u, u) = ⟨ A B u, u ⟩ - ⟨ B A u, u ⟩ = - 2 i Im ⟨ A u, B u ⟩ = - 2 i Im ⟨ x, y ⟩$ , and consequently $Δ A Δ B \geq \frac{1}{2} ∣ ([A, B] u, u) ∣$ ;

(d) for $c > 0$ , $H = L^{2} (R)$ and the operators $Q f (x) = x f (x)$ , $P f (x) = - i c f^{'} (x)$ defined on the Schwartz space $S (R)$ : every unit $f \in S (R)$ satisfies $(Q P - P Q) f = i c f$ , the numbers $a_{Q} = \int_{R} x ∣ f (x) ∣^{2} d x$ and $b_{P} = ⟨ P f, f ⟩$ are real, and with $x_{Q} : = (Q - a_{Q}) f$ , $y_{P} : = (P - b_{P}) f$ one has $∣ Im ⟨ x_{Q}, y_{P} ⟩ ∣ = c / 2$ ; hence $Δ Q Δ P \geq c / 2$ in the sense of the centered norms.

No assertion is made about products $A B$ off $D (A B)$ and the statement produces no energy-time inequality.

Facts & Assumptions

Given: A complex Hilbert space $H$ with pairing linear in the first argument; symmetric densely defined $A, B$ ; a unit vector $u \in D (A) \cap D (B)$ ; in clause (d) the operators $Q, P$ on $S (R)$ .

[F1]

For a symmetric operator $T$ (so $T \subseteq T^{*}$ ) and $v \in D (T)$ , $⟨ T v, v ⟩ = ⟨ v, T v ⟩ = \overline{⟨ T v, v ⟩}$ , so the number is real (Symmetric, self-adjoint and essentially self-adjoint operators, Adjoint of a densely defined operator).

[F2]

If $T$ is symmetric and $v \in D (T)$ , then $(T - λ) v \in H$ for every real $λ$ ; and $∥ (T - λ) v ∥^{2} = ∥ T v ∥^{2} - 2 λ Re ⟨ T v, v ⟩ + λ^{2} ∥ v ∥^{2}$ , so centering by the real mean $λ = ⟨ T v, v ⟩$ is legitimate (Symmetric, self-adjoint and essentially self-adjoint operators).

[F3]

In a complex inner product space, $∣ ⟨ v, w ⟩ ∣ \leq ∥ v ∥ ∥ w ∥$ and $∣ ⟨ v, w ⟩ ∣^{2} = (Re ⟨ v, w ⟩)^{2} + (Im ⟨ v, w ⟩)^{2}$ (Cauchy–Schwarz: $∣ ⟨ u, v ⟩ ∣ \leq ∥ u ∥ ∥ v ∥$ , with equality exactly for linearly dependent vectors).

[F4]

$S (R)$ consists of $C^{\infty}$ complex functions all of whose polynomial-weighted derivatives are bounded; in particular $f, f^{'}, x f, x f^{'} \in L^{2} (R)$ and $x f (x) \overline{f (x)} \to 0$ as $x \to \pm \infty$ for every $f \in S (R)$ (Schwartz space and its seminorms).

[F5]

Sums, scalar multiples and products of differentiable real functions satisfy $(f g)^{'} = f^{'} g + f g^{'}$ , and the derivative of a complex-valued differentiable function is computed componentwise (Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ ).

[F6]

For differentiable real $u, v$ on $[a, b]$ with integrable derivatives, $\int_{a}^{b} u v^{'} = u (b) v (b) - u (a) v (a) - \int_{a}^{b} u^{'} v$ (If $u, v$ are differentiable on $[a, b]$ with $u^{'}, v^{'}$ integrable, then $\int_{a}^{b} u v^{'} = u (b) v (b) - u (a) v (a) - \int_{a}^{b} u^{'} v$ ); a continuous real function on an interval has a primitive and $\int_{a}^{b} G^{'} = G (b) - G (a)$ for it (Every continuous function on an interval has a primitive; two primitives differ by a constant; and $\int_{a}^{b} f = G (b) - G (a)$ for any primitive $G$ ); the improper integral over an unbounded interval is the limit of the compact-interval integrals (Improper integrals over unbounded intervals).

Proof

1.1givenF1F2algebra

[Reality of the means] By [F1] applied to $T = A$ and $T = B$ , $a = ⟨ A u, u ⟩$ and $b = ⟨ B u, u ⟩$ are real. Hence $x = A u - a u \in H$ and $y = B u - b u \in H$ , and $Δ A = ∥ x ∥$ , $Δ B = ∥ y ∥$ are the norms of the centered vectors; the centering by the real means is the construction recorded in [F2].

1.2givenF4F5F6algebra

[Boundary-vanishing pairing identity on the Schwartz space] Let $f, g \in S (R)$ . Applying [F6] on $[- R, R]$ to the real and imaginary parts of $f$ and $\overline{g}$ and summing the four identities gives $\int_{- R}^{R} f^{'} \overline{g} = [f \overline{g}]_{- R}^{R} - \int_{- R}^{R} f \overline{g^{'}}$ ; by [F4] the boundary term $f (R) \overline{g} (R) - f (- R) \overline{g} (- R)$ tends to $0$ as $R \to \infty$ , while both improper integrals converge absolutely because $f^{'} \overline{g}$ and $f \overline{g^{'}}$ are products of Schwartz functions; taking the limit in [F6]'s finite identity therefore gives $\int_{R} f^{'} \overline{g} = - \int_{R} f \overline{g^{'}}$ .

2.1step 1.1F3algebra

[Cauchy-Schwarz bounds] By [F3], $Δ A Δ B = ∥ x ∥ ∥ y ∥ \geq ∣ ⟨ x, y ⟩ ∣$ , and $∣ ⟨ x, y ⟩ ∣ \geq ∣ Im ⟨ x, y ⟩ ∣$ ; squaring the first identity of [F3] gives $(Δ A)^{2} (Δ B)^{2} \geq (Re ⟨ x, y ⟩)^{2} + (Im ⟨ x, y ⟩)^{2}$ . This is (a) and (b).

2.2step 1.2F4F5F6algebra

[The position-momentum commutator] For $f \in S (R)$ the product rule [F5] applied componentwise gives $(x f)^{'} (x) = f (x) + x f^{'} (x)$ and hence $(Q P - P Q) f (x) = x (- i c f^{'} (x)) - (- i c) (f (x) + x f^{'} (x)) = i c f (x)$ , so $Q P - P Q = i c I$ on $S (R)$ . Moreover $\int_{R} (x ∣ f ∣^{2})^{'} (x) d x = 0$ : on $[- R, R]$ the evaluation clause of [F6] gives $\int_{- R}^{R} (x ∣ f ∣^{2})^{'} d x = x ∣ f ∣^{2} ∣_{- R}^{R}$ , which tends to $0$ by [F4], and $(x ∣ f ∣^{2})^{'} = ∣ f ∣^{2} + 2 Re (x f^{'} \overline{f})$ is continuous and dominated by an integrable Schwartz bound, so its improper integral over $R$ exists by [F6] and equals this limit. Expanding and using $\int_{R} ∣ f ∣^{2} = 1$ yields $\int_{R} x f^{'} \overline{f} d x + \int_{R} x f \overline{f^{'}} d x = - 1$ ; the second integral equals $\overline{\int_{R} x f^{'} \overline{f} d x}$ by complex conjugation, so $2 Re \int_{R} x f^{'} \overline{f} d x = - 1$ .

3.1step 1.1step 2.1algebra

[Commutator pairing] Assume $u \in D (A B) \cap D (B A)$ . Using symmetry of $A$ with $B u \in D (A)$ and symmetry of $B$ with $A u \in D (B)$ , $⟨ A B u, u ⟩ = ⟨ B u, A u ⟩$ and $⟨ B A u, u ⟩ = ⟨ A u, B u ⟩$ , so $([A, B] u, u) = ⟨ B u, A u ⟩ - ⟨ A u, B u ⟩ = \overline{⟨ A u, B u ⟩} - ⟨ A u, B u ⟩ = - 2 i Im ⟨ A u, B u ⟩$ . Since $a, b \in R$ , expanding $⟨ x, y ⟩ = ⟨ (A - a) u, (B - b) u ⟩$ gives $⟨ x, y ⟩ = ⟨ A u, B u ⟩ - a b$ and therefore $Im ⟨ x, y ⟩ = Im ⟨ A u, B u ⟩$ ; combining with step 2.1 gives $Δ A Δ B \geq \frac{1}{2} ∣ ([A, B] u, u) ∣$ .

4.1step 2.1step 2.2F4algebra∎

[The uncertainty product for the Schwartz pair] Let $f \in S (R)$ have $∥ f ∥ = 1$ and put $a_{Q} = \int_{R} x ∣ f ∣^{2} d x \in R$ , $b_{P} = ⟨ P f, f ⟩ = - i c \int_{R} f^{'} \overline{f} d x$ ; step 1.2 with $g = f$ gives $\int_{R} f^{'} \overline{f} = - \overline{\int_{R} f^{'} \overline{f}}$ , so this number is purely imaginary and $b_{P}$ is real. With $x_{Q} = (Q - a_{Q}) f$ and $y_{P} = (P - b_{P}) f$ , both in $L^{2} (R)$ by [F4], one computes $⟨ x_{Q}, y_{P} ⟩ = ⟨ Q f, P f ⟩ - a_{Q} b_{P}$ , using $⟨ Q f, f ⟩ = a_{Q}$ , $⟨ f, P f ⟩ = \overline{⟨ P f, f ⟩} = b_{P}$ and $∥ f ∥ = 1$ . Now $Im ⟨ Q f, P f ⟩$ : writing $w = \int_{R} x f \overline{f^{'}} d x$ and using $⟨ Q f, P f ⟩ = \int_{R} x f \overline{(- i c f^{'})} = i c w$ , and $w + \overline{w} = \int_{R} x (∣ f ∣^{2})^{'} = - 1$ from step 2.2, so $Re w = - \frac{1}{2}$ and $Im ⟨ Q f, P f ⟩ = c Re w = - \frac{c}{2}$ . Hence $∣ Im ⟨ x_{Q}, y_{P} ⟩ ∣ = c / 2$ , and the Cauchy-Schwarz computation of step 2.1 applied to the explicitly centered vectors $x_{Q}, y_{P}$ gives $Δ Q Δ P \geq c / 2$ .

TheoremStatement: AI-adaptedProof: AI-adaptedOpen item page →

Real multiplication operators and free plus bounded-potential Schrödinger operators

Statement

Assume the Axiom of Countable Choice and the Axiom of Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ ), The Axiom of Choice), let $d \geq 1$ and work on the complex Hilbert space $H = L^{2} (R^{d}; C)$ , whose completeness follows from Riesz-Fischer completeness of $L^{p}$ for $1 \leq p \leq \infty$ and whose Hilbert-space structure is defined in Hilbert space.

(a) For a real measurable finite-valued function $a : R^{d} \to R$ define $M_{a} f : = a f$ on $D (M_{a}) : = {f \in H : a f \in H}$ . Then $M_{a}$ is a densely defined self-adjoint operator on $H$ (Adjoint of a densely defined operator, Symmetric, self-adjoint and essentially self-adjoint operators).

(b) Let $c > 0$ , $b \in R$ , and let $F_{2}$ be the unitary Plancherel transform with the convention $\hat{f} (ξ) = \int f (x) e^{- 2 π i x \cdot ξ} d x$ on $L^{1}$ (Fourier transform on complex L1 classes, Plancherel theorem). Define

$H_{0} : = F_{2}^{*} M_{c ∣ ξ ∣^{2} + b} F_{2}, D (H_{0}) : = {f \in H : ∣ ξ ∣^{2} F_{2} f \in H} .$

Then $H_{0}$ is self-adjoint, $D (H_{0})$ coincides with the Sobolev space $W^{2, 2} (R^{d}; C)$ as a set (with equivalent norms, Integer-order Sobolev spaces and their norms, Integer-order W^{k,2} and H^k agree with equivalent norms), $H_{0} \geq b$ in the quadratic-form sense, and for every Schwartz function $f \in S (R^{d})$ , $H_{0} f = - (c / 4 π^{2}) Δ f + b f$ .

(c) If $V : R^{d} \to R$ is measurable with $∥ V ∥_{\infty} < \infty$ and $H : = H_{0} + M_{V}$ with domain $D (H_{0})$ , then $H$ is self-adjoint on $D (H_{0})$ , is bounded below, and the assignment $t \mapsto e^{i t H}$ is a strongly continuous one-parameter unitary group whose generator is $i H$ : the limit $\lim_{t \to 0} \frac{1}{t} (e^{i t H} ψ - ψ)$ exists exactly for $ψ \in D (H)$ and equals $i H ψ$ (A self-adjoint operator generates a strongly continuous unitary group).

(d) Scope: the argument uses only boundedness and measurability of $V$ ; it proves nothing for $V \propto - 1 / ∣ x ∣$ , $V \propto ∣ x ∣^{2}$ or the delta interaction, does not identify $H_{0}$ with any measured free-particle Hamiltonian, and asserts no physical interpretation.

Facts & Assumptions

Given: $d \geq 1$ ; $H = L^{2} (R^{d}; C)$ ; a real measurable finite-valued $a$ ; constants $c > 0$ , $b \in R$ ; a real bounded measurable $V$ .

[F1]

If measurable $f_{n} \to f$ pointwise and $∣ f_{n} ∣ \leq g$ with $\int g < \infty$ , then $\int ∣ f_{n} - f ∣ \to 0$ (Dominated convergence).

[F2]

For a densely defined operator $T$ , $D (T^{*})$ consists of those $y$ for which $x \mapsto ⟨ T x, y ⟩$ is bounded on $D (T)$ ; $T$ is symmetric when $⟨ T x, y ⟩ = ⟨ x, T y ⟩$ for $x, y \in D (T)$ , and self-adjoint when $T = T^{*}$ including domains (Adjoint of a densely defined operator, Symmetric, self-adjoint and essentially self-adjoint operators).

[F3]

$F_{2} : L^{2} (R^{d}) \to L^{2} (R^{d})$ is a surjective linear isometry extending the $L^{1}$ transform $\hat{f} (ξ) = \int f (x) e^{- 2 π i x \cdot ξ} d x$ , and it preserves the inner product (Plancherel theorem, Fourier transform on complex L1 classes).

[F4]

For $f \in S (R^{d})$ the Plancherel transforms of $f$ and of its distributional Laplacian satisfy $F_{2} (Δ f) (ξ) = - 4 π^{2} ∣ ξ ∣^{2} F_{2} f (ξ)$ for a.e. $ξ$ (Distributional derivatives are polynomial Fourier multipliers, Schwartz space and its seminorms).

[F5]

$W^{2, 2} (R^{d}; C)$ equals the set ${f \in L^{2} : ⟨ ξ ⟩^{2} F_{2} f \in L^{2}}$ and the two norms are equivalent; consequently ${f : ∣ ξ ∣^{2} F_{2} f \in L^{2}} = W^{2, 2}$ as sets (Integer-order W^{k,2} and H^k agree with equivalent norms, Integer-order Sobolev spaces and their norms).

[F6]

Let $A$ be self-adjoint and $B$ symmetric with $D (A) \subseteq D (B)$ and $A$ -bound less than one; then $A + B$ on $D (A)$ is self-adjoint and, when $A \geq γ$ with admissible pair $(a, b)$ for $B$ , $A + B \geq γ - \max {a ∣ γ ∣ + b, b / (1 - a)}$ (Kato-Rellich theorem, Relative boundedness with respect to an operator).

[F7]

If $T$ is self-adjoint with spectral PVM $E$ and $U (t) = e^{i t T}$ , then $U$ is a strongly continuous one-parameter unitary group, $U (t) D (T) = D (T)$ , and $\lim_{t \to 0} \frac{1}{t} (U (t) ψ - ψ)$ exists exactly for $ψ \in D (T)$ , where it equals $i T ψ$ (A self-adjoint operator generates a strongly continuous unitary group).

[F8]

For Lebesgue measure and $p = 2$ , the quotient space $L^{2} (R^{d})$ is complete, so with its usual first-variable-linear inner product it is a complex Hilbert space (Riesz-Fischer completeness of $L^{p}$ for $1 \leq p \leq \infty$ , Hilbert space).

Proof

1.1givenF1F2F8algebra

[Density and symmetry of the multiplier] By [F8], the defined $L^{2}$ space $H$ is a complex Hilbert space. For $f \in H$ and $n \geq 1$ put $f_{n} : = f 1_{{∣ a ∣ \leq n}}$ . Then $a f_{n}$ is bounded by $n ∣ f_{n} ∣$ , so $f_{n} \in D (M_{a})$ , and $f_{n} \to f$ in $H$ by [F1] applied to $∣ f_{n} - f ∣^{2} \leq ∣ f ∣^{2}$ ; hence $D (M_{a})$ is dense. For $f, g \in D (M_{a})$ the integral $\int a f \overline{g}$ is defined and equals $\int f \overline{a g}$ because $a$ is real-valued, so $⟨ M_{a} f, g ⟩ = ⟨ f, M_{a} g ⟩$ : $M_{a}$ is symmetric.

2.1step 1.1F2algebra

[Adjoint domain of the multiplier] Let $g \in D (M_{a}^{*})$ with representing vector $h = M_{a}^{*} g$ , so $\int a f \overline{g} = \int f \overline{h}$ for all $f \in D (M_{a})$ . Fix $n$ and restrict to $f \in H$ vanishing off $E_{n} : = {∣ a ∣ \leq n}$ ; every such $f$ lies in $D (M_{a})$ , and the identity reads $\int_{E_{n}} (a \overline{g} - \overline{h}) f = 0$ for all $f \in L^{2} (E_{n})$ , so $h = a g$ a.e. on $E_{n}$ ; as $E_{n} ↑ R^{d}$ , $h = a g$ a.e., and since $h \in H$ , $a g \in H$ and $g \in D (M_{a})$ . Conversely every $g \in D (M_{a})$ satisfies the adjoint identity with $h = M_{a} g$ . Hence $D (M_{a}^{*}) = D (M_{a})$ and $M_{a}^{*} = M_{a}$ : $M_{a}$ is self-adjoint, giving (a).

3.1step 2.1F2F3algebra

[Unitary conjugation] Let $T : = M_{c ∣ ξ ∣^{2} + b}$ , self-adjoint on $D (T) = {g : c ∣ ξ ∣^{2} g \in H}$ by step 2.1 applied to the real finite-valued function $c ∣ ξ ∣^{2} + b$ . Define $H_{0} : = F_{2}^{*} T F_{2}$ on $D (H_{0}) = F_{2}^{- 1} D (T) = {f : ∣ ξ ∣^{2} F_{2} f \in H}$ . For $x, y \in D (H_{0})$ , $⟨ H_{0} x, y ⟩ = ⟨ T F_{2} x, F_{2} y ⟩$ by unitarity of $F_{2}$ [F3]. If $y \in D (H_{0}^{*})$ , the functional $x \mapsto ⟨ H_{0} x, y ⟩$ is bounded on $D (H_{0})$ ; conjugating by the unitary $F_{2}$ shows that $g \mapsto ⟨ T g, F_{2} y ⟩$ is bounded on $D (T)$ , so $F_{2} y \in D (T^{*}) = D (T)$ and hence $y \in D (H_{0})$ with $H_{0}^{*} y = F_{2}^{*} T^{*} F_{2} y = H_{0} y$ ; the reverse inclusion is immediate from the symmetry identity. Thus $H_{0}$ is self-adjoint.

4.1step 3.1F3F4F5algebra

[Domain, lower bound and free action] By [F5], $D (H_{0}) = W^{2, 2} (R^{d})$ as a set with equivalent norms. For $f \in D (H_{0})$ the pairing with $f$ is computed in the $ξ$ -variable, where multiplication by $c ∣ ξ ∣^{2} + b \geq b$ gives $⟨ H_{0} f, f ⟩ = \int (c ∣ ξ ∣^{2} + b) ∣ F_{2} f (ξ) ∣^{2} d ξ \geq b ∥ F_{2} f ∥^{2} = b ∥ f ∥^{2}$ , so $H_{0} \geq b$ in the quadratic-form sense. For a Schwartz function $f$ , [F4] gives $F_{2} (- (c / 4 π^{2}) Δ f + b f) (ξ) = (c ∣ ξ ∣^{2} + b) F_{2} f (ξ)$ a.e.; both sides are continuous representatives, and Plancherel inversion identifies them as elements of $H$ , so $H_{0} f = - (c / 4 π^{2}) Δ f + b f$ . This is (b).

5.1step 4.1F6algebra

[Bounded perturbation by Kato-Rellich] Let $V$ be real with $∥ V ∥_{\infty} < \infty$ and let $M_{V}$ be the bounded multiplication operator on $H$ . Then $D (H_{0}) \subseteq D (M_{V}) = H$ , $M_{V}$ is symmetric because $V$ is real, and $∥ M_{V} f ∥ \leq ∥ V ∥_{\infty} ∥ f ∥$ for every $f \in D (H_{0})$ , so the $H_{0}$ -bound of $M_{V}$ is $0 < 1$ with admissible pair $(0, ∥ V ∥_{\infty})$ (Relative boundedness with respect to an operator). By [F6] the operator $H = H_{0} + M_{V}$ on $D (H_{0})$ is self-adjoint and, since $H_{0} \geq b$ , satisfies $H \geq b - \max {0 \cdot ∣ b ∣ + ∥ V ∥_{\infty}, ∥ V ∥_{\infty}} = b - ∥ V ∥_{\infty}$ , hence is bounded below.

6.1step 5.1F7∎

[Unitary dynamics] Applying [F7] to the self-adjoint $H$ of step 5.1 yields that $t \mapsto e^{i t H}$ is a strongly continuous one-parameter unitary group, $e^{i t H} D (H) = D (H)$ , and the difference quotients converge exactly on $D (H)$ ; the physical convention $t \mapsto e^{- i t H}$ is the same group at negative time. No claim in this item extends to unbounded or singular $V$ , which are excluded by the boundedness and measurability hypotheses recorded in [F6].

LemmaStatement: AI-adaptedProof: AI-adaptedOpen item page →

The CHSH bound and the finite-sampling law for independent trials

Statement

No choice principle is used. Let $(Ω, F, P)$ be a probability space (Probability measures and probability spaces).

(a) CHSH bound. Let $A_{0}, A_{1}, B_{0}, B_{1} : Ω \to {- 1, 1}$ be random variables with values in ${- 1, 1}$ (Random elements and real random variables). Then the four expectations exist and

$∣ E [A_{0} B_{0}] + E [A_{0} B_{1}] + E [A_{1} B_{0}] - E [A_{1} B_{1}] ∣ \leq 2.$

(b) Positive-probability sequences. Let $N \geq 1$ , $m \geq 1$ , $p_{1}, \dots, p_{m} \geq 0$ with $\sum_{j = 1}^{m} p_{j} = 1$ , and let $X_{1}, \dots, X_{N} : Ω \to {1, \dots, m}$ be mutually independent with $P (X_{k} = j) = p_{j}$ for all $k, j$ (Pairwise and mutual independence of finite-valued random variables). Then for every sequence $(j_{1}, \dots, j_{N}) \in {1, \dots, m}^{N}$ with $p_{j_{k}} > 0$ for every $k$ ,

$P (X_{k} = j_{k} for every k) = \prod_{k = 1}^{N} p_{j_{k}} > 0.$

Consequently every event disjoint from this outcome has probability at most $1 - \prod_{k} p_{j_{k}} < 1$ : any criterion that rejects at least one such sequence has positive failure probability under this sampling model.

(c) Binomial frequency. Let $S$ be binomial $(N, p)$ with $N \geq 1$ and $p \in [0, 1]$ (Bernoulli random variables and binomial random variables as sums of independent Bernoulli trials). Then the frequency $S / N$ satisfies $E [S / N] = p$ , $Var (S / N) = p (1 - p) / N$ , and for every $ε > 0$

$P (∣ S / N - p ∣ \geq ε) \leq \frac{p (1 - p)}{N ε^{2}} .$

(d) Multinomial count law. The number of sequences $(j_{1}, \dots, j_{N})$ with prescribed fibre sizes $n_{j} : = # {k : j_{k} = j}$ is the multinomial coefficient $(\binom{N}{n_{1}, \dots, n_{m}})$ (The multinomial coefficient $(\binom{n}{k_{0}, \dots, k_{m - 1}})$ as the number of ordered partitions of an $n$ -set into blocks of prescribed sizes), and each such sequence has probability $\prod_{j = 1}^{m} p_{j}^{n_{j}}$ ; in this displayed product use the zero-count convention $p^{0} : = 1$ for every $p \in [0, 1]$ , including $p = 0$ . Hence the probability that the count vector equals $(n_{1}, \dots, n_{m})$ is $(\binom{N}{n_{1}, \dots, n_{m}}) \prod_{j} p_{j}^{n_{j}}$ .

No statistical inference, empirical claim or measured frequency is asserted; the lemma only computes probabilities of an explicitly specified independent sampling model.

Facts & Assumptions

Given: A probability space $(Ω, F, P)$ ; variables $A_{i}, B_{j}$ with values in ${- 1, 1}$ ; independent ${1, \dots, m}$ -valued $X_{1}, \dots, X_{N}$ ; a binomial $(N, p)$ variable $S$ .

[F1]

A probability measure $P$ is countably additive on disjoint families, hence finitely additive (Measures on sigma-algebras, Probability measures and probability spaces); a real random variable is a measurable $Ω \to R$ (Random elements and real random variables).

[F2]

$E [a X + b Y] = a E [X] + b E [Y]$ for integrable $X, Y$ and scalars $a, b$ ; $E [X] \leq E [Y]$ almost surely for integrable real $X \leq Y$ ; $∣ E [X] ∣ \leq E [∣ X ∣]$ (Linearity, monotonicity, and the modulus bound for expectation, Expectation of a nonnegative or integrable random variable).

[F3]

A finite family $(X_{i})$ of finite-valued random variables is mutually independent when $P (X_{j} = x_{j} for all j \in J) = \prod_{j \in J} P (X_{j} = x_{j})$ for every nonempty $J$ and every selection $x_{j} \in X_{j} (Ω)$ (Pairwise and mutual independence of finite-valued random variables).

[F4]

A binomial $(N, p)$ variable is a sum of $N$ mutually independent Bernoulli $(p)$ variables; $E [S] = N p$ and $Var (S) = N p (1 - p)$ (Bernoulli random variables and binomial random variables as sums of independent Bernoulli trials, A Bernoulli $(p)$ variable has mean $p$ and variance $p (1 - p)$ ; a binomial $(n, p)$ variable has mean $n p$ and variance $n p (1 - p)$ ).

[F5]

For a real random variable, $Var (X) = E [(X - E [X])^{2}]$ when $X$ is square-integrable, and $P (∣ X - E [X] ∣ \geq a) \leq Var (X) / a^{2}$ for $a > 0$ (Moments, variance, and covariance on a probability space, Chebyshev's inequality for random variables).

[F6]

The multinomial coefficient $(\binom{N}{n_{1}, \dots, n_{m}})$ is the number of colourings $c : {1, \dots, N} \to {1, \dots, m}$ with $∣ c^{- 1} [{j}] ∣ = n_{j}$ for every $j$ , defined only when $\sum_{j} n_{j} = N$ (The multinomial coefficient $(\binom{n}{k_{0}, \dots, k_{m - 1}})$ as the number of ordered partitions of an $n$ -set into blocks of prescribed sizes).

Proof

1.1givenF2algebra

[Pointwise CHSH identity] Fix $ω \in Ω$ . Since $B_{0} (ω), B_{1} (ω) \in {- 1, 1}$ , the numbers $B_{0} (ω) + B_{1} (ω)$ and $B_{0} (ω) - B_{1} (ω)$ are $\pm 2$ or $0$ , and exactly one of them is nonzero: the first is $0$ precisely when $B_{1} (ω) = - B_{0} (ω)$ , in which case the second is $\pm 2$ , and otherwise the first is $\pm 2$ and the second $0$ . Hence $A_{0} (B_{0} + B_{1}) + A_{1} (B_{0} - B_{1})$ equals $\pm 2$ at every $ω$ , so its absolute value is bounded by $2$ pointwise and $E [A_{0} B_{0}] + E [A_{0} B_{1}] + E [A_{1} B_{0}] - E [A_{1} B_{1}] = E [A_{0} (B_{0} + B_{1}) + A_{1} (B_{0} - B_{1})]$ by linearity [F2], the variables being bounded hence integrable. The monotonicity and modulus bound of [F2] applied to the pointwise bound $- 2 \leq A_{0} (B_{0} + B_{1}) + A_{1} (B_{0} - B_{1}) \leq 2$ give the stated $\leq 2$ bound.

1.2givenF1F3algebra

[Positive-probability sequences] Since $p_{j_{k}} > 0$ , each event ${X_{k} = j_{k}}$ is nonempty and $j_{k} \in X_{k} (Ω)$ . By mutual independence [F3] applied to the full index set $J = {1, \dots, N}$ and the selection $x_{k} = j_{k}$ , $P (X_{k} = j_{k} for all k) = \prod_{k} P (X_{k} = j_{k}) = \prod_{k} p_{j_{k}}$ , which is a product of positive reals and hence positive. If an event $R$ is disjoint from ${X_{k} = j_{k} for all k}$ , then $P (R) \leq 1 - \prod_{k} p_{j_{k}} < 1$ by additivity and complementation [F1].

1.3givenF2F4F5algebra

[Binomial frequency moments] By [F4], $S$ is a sum of $N$ mutually independent Bernoulli $(p)$ variables, with $E [S] = N p$ and $Var (S) = N p (1 - p)$ . Linearity [F2] gives $E [S / N] = \frac{1}{N} E [S] = p$ ; and by the definition of variance [F5], $Var (S / N) = E [(S / N - p)^{2}] = \frac{1}{N^{2}} E [(S - N p)^{2}] = \frac{1}{N^{2}} Var (S) = \frac{p (1 - p)}{N}$ . Applying the Chebyshev inequality [F5] to the square-integrable variable $S / N$ with parameter $ε > 0$ gives $P (∣ S / N - p ∣ \geq ε) \leq p (1 - p) / (N ε^{2})$ .

2.1givenF1F3F6algebra∎

[Multinomial count law] A sequence $(j_{1}, \dots, j_{N})$ is exactly a function $c : {1, \dots, N} \to {1, \dots, m}$ , with fibre sizes $n_{j}$ ; by [F6] the number of such functions with prescribed fibre sizes is $(\binom{N}{n_{1}, \dots, n_{m}})$ . If some $j_{k} \notin X_{k} (Ω)$ , the sequence event is empty, and $p_{j_{k}} = P (X_{k} = j_{k}) = 0$ , so its probability and $\prod_{k} p_{j_{k}}$ are both zero. Otherwise every selected label lies in the corresponding image, and independence [F3] gives probability $\prod_{k} p_{j_{k}}$ . Thus in either case each sequence has probability $\prod_{k} p_{j_{k}} = \prod_{j} p_{j}^{n_{j}}$ , because the exponent of $p_{j}$ is the number of $k$ with $j_{k} = j$ , namely $n_{j}$ ; when $n_{j} = 0$ the sequence contains no factor $p_{j}$ , and the displayed product convention in (d) assigns that factor the value $1$ , also when $p_{j} = 0$ . The count event is the disjoint union of the corresponding sequence events, so finite additivity [F1] gives probability $(\binom{N}{n_{1}, \dots, n_{m}}) \prod_{j} p_{j}^{n_{j}}$ .

TheoremStatement: AI-adaptedProof: AI-adaptedOpen item page →

Finite partial trace, Schmidt decomposition and finite no-signalling

Statement

Let $V, W$ be nonzero finite-dimensional complex inner product spaces with orthonormal bases $(e_{1}, \dots, e_{m})$ of $V$ and $(f_{1}, \dots, f_{n})$ of $W$ , let $V \otimes W$ be their tensor product with the inner product of Hilbert tensor construction: inner product, completion and product bases, and let $tr$ denote the basis-independent trace of an endomorphism (The basis-independent trace of an endomorphism of a finite-dimensional vector space).

(a) For every endomorphism $R$ of $V \otimes W$ there is a unique endomorphism ${Tr}_{W} R$ of $V$ with

$tr (({Tr}_{W} R) A) = tr (R (A \otimes I_{W})) for every A \in End (V),$

namely ${Tr}_{W} R = \sum_{j = 1}^{n} φ_{j} \circ R \circ ι_{j}$ , where $ι_{j} (v) = v \otimes f_{j}$ and $φ_{j} (x \otimes w) = ⟨ w, f_{j} ⟩ x$ ; it does not depend on the orthonormal basis $(f_{j})$ of $W$ , and the analogous definition gives ${Tr}_{V} R$ with $tr (({Tr}_{V} R) B) = tr (R (I_{V} \otimes B))$ .

(b) ${Tr}_{W}$ is linear, positive (it maps positive endomorphisms to positive endomorphisms) and trace preserving on density matrices: if $R \geq 0$ and $tr R = 1$ then ${Tr}_{W} R \geq 0$ and $tr {Tr}_{W} R = 1$ ; and ${Tr}_{W} T_{Ψ, Ψ} = C C^{*}$ for every $Ψ \in V \otimes W$ , where $C : W \to V$ is the coefficient map determined by $Ψ = \sum_{j} C f_{j} \otimes f_{j}$ .

$Ψ = \sum_{r = 1}^{k} s_{r} v_{r} \otimes u_{r},$

with $k \leq \min (m, n)$ , orthonormal families $(v_{r}) \subseteq V$ , $(u_{r}) \subseteq W$ and $s_{1} \geq \dots \geq s_{k} > 0$ uniquely determined by $Ψ$ as the positive singular values of the coefficient map; moreover $∥ Ψ ∥^{2} = \sum_{r} s_{r}^{2}$ . For $∥ Ψ ∥ = 1$ the reduced density $ρ = {Tr}_{W} T_{Ψ, Ψ}$ satisfies $tr ρ^{2} = \sum_{r} s_{r}^{4}$ , and $tr ρ^{2} = 1$ holds exactly when $k = 1$ , i.e. exactly when $Ψ$ is a product vector; mixed-state separability is not decided by this criterion.

(d) Finite nonselective no-signalling. If $(M_{a})_{a = 1}^{p}$ are endomorphisms of $W$ with $\sum_{a} M_{a}^{*} M_{a} = I_{W}$ then for every endomorphism $R$ of $V \otimes W$

${Tr}_{W} (\sum_{a} (I_{V} \otimes M_{a}) R (I_{V} \otimes M_{a}^{*})) = {Tr}_{W} R .$

This is an identity for the specified finite local instrument; selective conditioning (dividing one summand by its nonzero trace) is not covered.

Facts & Assumptions

Given: Finite-dimensional complex inner product spaces $V, W$ with orthonormal bases $(e_{i}), (f_{j})$ ; an endomorphism $R$ of $V \otimes W$ ; in (c) a vector $Ψ$ ; in (d) endomorphisms $M_{a}$ of $W$ .

[F1]

The tensor product $V \otimes W$ with the inner product $⟨ u \otimes v, u^{'} \otimes v^{'} ⟩ = ⟨ u, u^{'} ⟩_{V} ⟨ v, v^{'} ⟩_{W}$ is a finite-dimensional inner product space; $(e_{i} \otimes f_{j})_{i, j}$ is an orthonormal basis, and every tensor has a unique expansion $\sum_{i, j} c_{i j} e_{i} \otimes f_{j}$ with $∥ Ψ ∥^{2} = \sum_{i, j} ∣ c_{i j} ∣^{2}$ ; the inclusion and the bilinear relations are those of Hilbert tensor construction: inner product, completion and product bases.

[F2]

For endomorphisms $S : V_{1} \to V_{2}$ and $T : V_{2} \to V_{1}$ of finite-dimensional spaces, $tr (S T) = tr (T S)$ ; the trace of an endomorphism is computed as the trace of its matrix in any ordered basis (For $A \in M_{m \times n} (F)$ and $B \in M_{n \times m} (F)$ , $tr (A B) = tr (B A)$ , The basis-independent trace of an endomorphism of a finite-dimensional vector space, Coordinate columns $[v]_{B}$ and matrices $[T]_{B}^{C}$ of linear maps relative to ordered bases).

[F3]

Every linear map $C : W \to V$ of finite-dimensional inner product spaces has a singular value decomposition: orthonormal bases $(u_{1}, \dots, u_{n})$ of $W$ and $(v_{1}, \dots, v_{m})$ of $V$ and $s_{1} \geq \dots \geq s_{k} > 0$ with $C w = \sum_{r = 1}^{k} s_{r} ⟨ w, u_{r} ⟩ v_{r}$ (Every linear map between finite-dimensional real or complex inner product spaces admits a singular value decomposition); the positive numbers $s_{r}^{2}$ are the nonzero eigenvalues of $C C^{*}$ , counted with multiplicity (Complex spectral theorem: a normal endomorphism of a finite-dimensional complex inner product space has an orthonormal eigenbasis, and conversely, A normal endomorphism is a sum of its eigenvalues times pairwise orthogonal projections, and each spectral projection is a polynomial in the endomorphism, Self-adjoint and normal endomorphisms of a finite-dimensional real or complex inner product space).

[F4]

Non-negative means $⟨ T v, v ⟩ \geq 0$ for all $v$ (Non-negative and positive operators); every finite-dimensional inner product space has an orthonormal basis (Every finite-dimensional real or complex inner product space has an orthonormal basis).

Proof

1.1givenF1F2algebra

[Existence and uniqueness of the partial trace] Define ${Tr}_{W} R : = \sum_{j = 1}^{n} φ_{j} R ι_{j}$ , a linear endomorphism of $V$ . Computed in the basis $(e_{i} \otimes f_{j})$ and using [F2], the matrix of $φ_{j}$ contracts the $W$ -index into $f_{j}$ , and the total matrix of ${Tr}_{W} R$ has entries $({Tr}_{W} R)_{i i^{'}} = \sum_{j} R_{i j, i^{'} j}$ ; hence, with $(A)_{i^{'} i^{''}}$ the matrix of $A$ , $tr (({Tr}_{W} R) A) = \sum_{i, i^{'}, j} R_{i j, i^{'} j} A_{i^{'} i} = tr (R (A \otimes I_{W}))$ . Uniqueness: if $S$ and $S^{'}$ both satisfy the pairing identity, then $tr ((S - S^{'}) A) = 0$ for all $A$ ; taking $A = T_{α, β}$ (the rank-one endomorphism $v \mapsto ⟨ v, β ⟩ α$ ) for $α, β$ ranging over a basis of $V$ and using [F1] and [F2] extracts every matrix entry of $S - S^{'}$ , so $S = S^{'}$ . Basis independence: the defining identity involves only the basis-independent trace [F2]; equivalently the formula with another orthonormal basis $(g_{j})$ yields the same operator by the uniqueness just proved.

2.1step 1.1givenF4algebra

[Positivity and trace preservation] If $R \geq 0$ then for every $x \in V$ , $⟨ ({Tr}_{W} R) x, x ⟩ = \sum_{j} ⟨ R (x \otimes f_{j}), x \otimes f_{j} ⟩ \geq 0$ because $R \geq 0$ [F4] and each summand is a nonnegative real; hence ${Tr}_{W} R \geq 0$ . Taking $A = I_{V}$ in the defining identity gives $tr {Tr}_{W} R = tr R$ ; consequently a density matrix $R$ (positive, $tr R = 1$ , in the sense of Density operators and scalar spectral probability) has ${Tr}_{W} R \geq 0$ and trace $1$ , so $({Tr}_{W})$ maps density matrices to density matrices.

2.2step 1.1F1F2algebra

[Reduced state of a pure vector] Let $Ψ = \sum_{i, j} c_{i j} e_{i} \otimes f_{j}$ and define $C : W \to V$ by $C f_{j} = \sum_{i} c_{i j} e_{i}$ , so that $Ψ = \sum_{j} C f_{j} \otimes f_{j}$ by [F1] (insert the expansions of the $C f_{j}$ ). For the rank-one endomorphism $T_{Ψ, Ψ}$ and $A \in End (V)$ , the defining identity of step 1.1 gives $tr (({Tr}_{W} T_{Ψ, Ψ}) A) = tr (T_{Ψ, Ψ} (A \otimes I)) = ⟨ (A \otimes I) Ψ, Ψ ⟩$ , and substituting the expansion yields $\sum_{i, i^{'}} ⟨ A e_{i}, e_{i^{'}} ⟩ \sum_{j} c_{i j} \overline{c_{i^{'} j}} = tr (A C C^{*})$ , using the matrix of $C C^{*}$ in the basis $(e_{i})$ . Since this holds for all $A$ and the operator is determined by its pairings with all $A$ (step 1.1), ${Tr}_{W} T_{Ψ, Ψ} = C C^{*}$ .

3.1step 2.2F1F3algebra

[Schmidt expansion with the complex convention] Apply [F3] to the coefficient map $C$ : there are orthonormal right singular vectors $q_{r} \in W$ and left singular vectors $v_{r} \in V$ such that $C f_{j} = \sum_{r} s_{r} ⟨ f_{j}, q_{r} ⟩ v_{r}$ . Define $u_{r} : = \sum_{j} ⟨ f_{j}, q_{r} ⟩ f_{j}$ , the coordinate conjugate of $q_{r}$ in the fixed basis $(f_{j})$ . The coordinate conjugation reverses inner products, so $⟨ u_{r}, u_{t} ⟩ = \overline{⟨ q_{r}, q_{t} ⟩} = δ_{r t}$ . Substituting the formula for $C f_{j}$ into $Ψ = \sum_{j} C f_{j} \otimes f_{j}$ gives $Ψ = \sum_{r} s_{r} v_{r} \otimes u_{r}$ . Thus the claimed Schmidt expansion holds, with singular values $s_{r}$ of $C$ . Changing the factor bases changes the coefficient matrix by unitary matrices (with transpose rather than adjoint on the second factor); hence the singular values remain unchanged. Orthogonality of the elementary Schmidt products gives $∥ Ψ ∥^{2} = \sum_{r} s_{r}^{2}$ .

4.1step 2.2step 3.1algebra

[Purity and the product criterion] By steps 2.2 and 3.1, $ρ = {Tr}_{W} T_{Ψ, Ψ} = C C^{*}$ has eigenvalues $s_{r}^{2}$ with multiplicity by [F3], so for $∥ Ψ ∥ = 1$ we have $tr ρ^{2} = \sum_{r} s_{r}^{4}$ and $tr ρ = 1 = \sum_{r} s_{r}^{2}$ . Since all $s_{r}$ are positive, $\sum_{r} s_{r}^{4} \leq (\sum_{r} s_{r}^{2})^{2} = 1$ with equality exactly when at most one $s_{r}$ is nonzero, i.e. $k = 1$ ; and $k = 1$ means $Ψ = s_{1} v_{1} \otimes u_{1}$ is a product vector, conversely a product vector has $C$ of rank one by step 2.2. This is the stated criterion; it concerns pure vectors only.

5.1step 1.1F2algebra∎

[Finite nonselective no-signalling] Let $(M_{a})$ satisfy $\sum_{a} M_{a}^{*} M_{a} = I_{W}$ . For $A \in End (V)$ , using the defining identity of step 1.1 and the identity $(I \otimes M_{a}^{*}) (A \otimes I) = (A \otimes M_{a}^{*})$ together with cyclicity [F2] inside the finite trace, $tr (({Tr}_{W} \sum_{a} (I \otimes M_{a}) R (I \otimes M_{a}^{*})) A) = \sum_{a} tr (R (A \otimes M_{a}^{*} M_{a})) = tr (R (A \otimes \sum_{a} M_{a}^{*} M_{a})) = tr (R (A \otimes I)) = tr (({Tr}_{W} R) A)$ ; since this holds for every $A$ , the uniqueness in step 1.1 gives ${Tr}_{W} (\sum_{a} (I \otimes M_{a}) R (I \otimes M_{a}^{*})) = {Tr}_{W} R$ .

LemmaStatement: AI-adaptedProof: AI-adaptedOpen item page →

Finite effects and instruments: normalization and non-determination of the update

Statement

Let $H$ be a finite-dimensional complex Hilbert space of dimension $d \geq 1$ (Hilbert space), with adjoints as in The adjoint $T^{*} : W \to V$ is characterised by $⟨ T v, w ⟩_{W} = ⟨ v, T^{*} w ⟩_{V}$ and the finite trace as in The basis-independent trace of an endomorphism of a finite-dimensional vector space. Recall that an endomorphism $F$ is an effect when $F \geq 0$ and $I - F \geq 0$ , i.e. $0 \leq F \leq I$ in the quadratic-form order (Non-negative and positive operators).

(a) Definitions. A finite POVM is a finite family $(F_{a})_{a = 1}^{p}$ , $p \geq 1$ , of effects with $\sum_{a} F_{a} = I$ . An instrument here is a finite family $(M_{a, r})_{1 \leq a \leq p, 1 \leq r \leq q_{a}}$ of endomorphisms of $H$ , where each $q_{a}$ is a non-negative integer; its outcome operators and update maps are $F_{a} : = \sum_{r} M_{a, r}^{*} M_{a, r}$ and $I_{a} : End (H) \to End (H)$ , $I_{a} (R) : = \sum_{r} M_{a, r} R M_{a, r}^{*}$ , with empty sums equal to zero. It is normalized when $\sum_{a, r} M_{a, r}^{*} M_{a, r} = I$ . For a normalized instrument the $F_{a}$ are its outcome effects, as proved in (b); without normalization they need not be effects.

(b) If $(F_{a})$ is a finite POVM and $ρ$ a density operator (Density operators and scalar spectral probability) then $p_{a} : = tr (ρ F_{a}) \geq 0$ and $\sum_{a} p_{a} = 1$ . If $(M_{a, r})$ is a normalized instrument then the $F_{a}$ form a finite POVM, $tr I_{a} (ρ) = p_{a}$ , each $I_{a} (ρ)$ is non-negative, and whenever $p_{a} > 0$ the normalized update $I_{a} (ρ) / p_{a}$ is a density operator.

(c) Ideal Lüders case. For an orthogonal projection $P$ (The orthogonal projection $P_{W} v$ is the $W$ -component in $V = W \oplus W^{⊥}$ ) put $M_{0} = P$ , $M_{1} = I - P$ . Then $\sum_{j} M_{j}^{*} M_{j} = P + (I - P) = I$ , the effects are $F_{0} = P$ , $F_{1} = I - P$ , and $I_{j} (ρ) = P ρ P$ or $(I - P) ρ (I - P)$ .

(d) Non-determination. Effects alone do not determine the update: there are two normalized instruments on $C^{2}$ with the same outcome effects $F_{0} = diag (1, 0)$ , $F_{1} = diag (0, 1)$ whose updates differ on the same input.

Facts & Assumptions

Given: A finite-dimensional complex Hilbert space $H$ ; a density operator $ρ$ ; a finite POVM $(F_{a})$ ; a normalized instrument $(M_{a, r})$ ; an orthogonal projection $P$ .

[F2]

A self-adjoint endomorphism $T$ is non-negative exactly when $⟨ T v, v ⟩ \geq 0$ for all $v$ (Non-negative and positive operators); its eigenvalues are then non-negative by evaluation on eigenvectors. For non-negative $ρ$ , [F3] gives $(M ρ M^{*})^{*} = M ρ M^{*}$ , and the adjoint identity gives $⟨ M ρ M^{*} x, x ⟩ = ⟨ ρ M^{*} x, M^{*} x ⟩ \geq 0$ , so $M ρ M^{*}$ is non-negative.

[F3]

For endomorphisms $S, T$ of a finite-dimensional space, $tr (S T) = tr (T S)$ (For $A \in M_{m \times n} (F)$ and $B \in M_{n \times m} (F)$ , $tr (A B) = tr (B A)$ , The basis-independent trace of an endomorphism of a finite-dimensional vector space). The finite adjoint identities are justified directly: choose an orthonormal basis, available by applying Complex spectral theorem: a normal endomorphism of a finite-dimensional complex inner product space has an orthonormal eigenbasis, and conversely to $I$ . If $A = (a_{i j})$ is an operator matrix, then $A^{*} = {\overline{A}}^{T}$ satisfies the defining identity of The adjoint $T^{*} : W \to V$ is characterised by $⟨ T v, w ⟩_{W} = ⟨ v, T^{*} w ⟩_{V}$ by expanding the two inner products in coordinates. Uniqueness follows because two adjoints $B, C$ give $⟨ x, (B - C) y ⟩ = 0$ for all $x, y$ , hence $(B - C) y = 0$ . Matrix conjugation and transposition now give $M^{* *} = M$ , $(A B)^{*} = B^{*} A^{*}$ and $(A + B)^{*} = A^{*} + B^{*}$ . Thus these finite identities require no Countable Choice; the general Hilbert-space theorem Hilbert-adjoint identities is not invoked.

[F4]

A density operator means a self-adjoint positive trace-class $ρ$ with trace $1$ (Density operators and scalar spectral probability), where positive means non-negative as in that supplier. Write $T_{u, v} x : = ⟨ x, v ⟩ u$ , with the inner product linear in its first argument, as in the same supplier. In the finite-dimensional $H$ used here, the finite-dimensional spectral theorem (Complex spectral theorem: a normal endomorphism of a finite-dimensional complex inner product space has an orthonormal eigenbasis, and conversely) gives an orthonormal eigenbasis $u_{1}, \dots, u_{d}$ for the self-adjoint, hence normal, $ρ$ . Put $w_{n} : = ⟨ ρ u_{n}, u_{n} ⟩ \geq 0$ by [F2]. The finite trace (The basis-independent trace of an endomorphism of a finite-dimensional vector space) gives $\sum_{n = 1}^{d} w_{n} = tr ρ = 1$ , and expansion in this basis gives $ρ = \sum_{n = 1}^{d} w_{n} T_{u_{n}, u_{n}}$ . This finite argument uses no Countable Choice and does not invoke the general density expansion.

[F5]

An endomorphism $U$ with $U^{*} U = I$ is a linear isometry; every orthogonal projection $P$ satisfies $P^{2} = P = P^{*}$ (Linear isometries, and orthogonal or unitary operators on finite-dimensional inner product spaces, The orthogonal projection $P_{W} v$ is the $W$ -component in $V = W \oplus W^{⊥}$ ).

Proof

1.1givenF2F3F4algebra

[POVM probabilities] By the finite-dimensional argument in [F4], $ρ = \sum_{n = 1}^{d} w_{n} T_{u_{n}, u_{n}}$ with $(u_{n})$ orthonormal and $\sum_{n} w_{n} = 1$ . Then $p_{a} = tr (ρ F_{a}) = \sum_{n} w_{n} ⟨ F_{a} u_{n}, u_{n} ⟩$ by linearity of the finite trace [F3], and each term is non-negative because $F_{a} \geq 0$ [F2]; hence $p_{a} \geq 0$ . Summing over $a$ and using linearity, $\sum_{a} p_{a} = \sum_{n} w_{n} ⟨ (\sum_{a} F_{a}) u_{n}, u_{n} ⟩ = \sum_{n} w_{n} = 1$ .

1.2givenF3F5algebra

[Lüders instance] For $M_{0} = P$ and $M_{1} = I - P$ with $P^{2} = P = P^{*}$ [F5], $M_{0}^{*} M_{0} = P^{2} = P$ and $M_{1}^{*} M_{1} = (I - P)^{2} = I - P$ , so $\sum_{j} M_{j}^{*} M_{j} = I$ and the effects are $F_{0} = P$ , $F_{1} = I - P$ ; the updates are $I_{0} (ρ) = P ρ P$ , $I_{1} (ρ) = (I - P) ρ (I - P)$ .

2.1step 1.1F2F3F4algebra

[Update traces and positivity] Let $F_{a} = \sum_{r} M_{a, r}^{*} M_{a, r}$ . Then $tr I_{a} (ρ) = \sum_{r} tr (M_{a, r} ρ M_{a, r}^{*}) = \sum_{r} tr (ρ M_{a, r}^{*} M_{a, r}) = tr (ρ F_{a}) = p_{a}$ , by cyclicity [F3]. For each $r$ , $(M_{a, r} ρ M_{a, r}^{*})^{*} = M_{a, r} ρ M_{a, r}^{*}$ because $ρ = ρ^{*}$ ; hence $I_{a} (ρ)$ is self-adjoint. For $x \in H$ , $⟨ I_{a} (ρ) x, x ⟩ = \sum_{r} ⟨ M_{a, r} ρ M_{a, r}^{*} x, x ⟩ = \sum_{r} ⟨ ρ (M_{a, r}^{*} x), M_{a, r}^{*} x ⟩ \geq 0$ because $ρ \geq 0$ ; thus $I_{a} (ρ) \geq 0$ . If $p_{a} > 0$ , then $I_{a} (ρ) / p_{a}$ has trace $1$ and is positive, hence is a density operator in the sense of [F4]. The effects of a normalized instrument form a POVM: each $M_{a, r}^{*} M_{a, r}$ is non-negative, and both $F_{a}$ and $I - F_{a} = \sum_{(b, r) : b \neq a} M_{b, r}^{*} M_{b, r}$ are non-negative, so $0 \leq F_{a} \leq I$ . This is (b).

3.1step 2.1F3F5algebra∎

[Two instruments with equal effects] On $H = C^{2}$ with orthonormal basis $(e_{1}, e_{2})$ let $F_{0} = T_{e_{1}, e_{1}}$ , $F_{1} = T_{e_{2}, e_{2}}$ and let $σ$ be the swap $σ e_{1} = e_{2}$ , $σ e_{2} = e_{1}$ , so that $σ^{*} σ = I$ by [F5]. Instrument A: $M_{0, 1} : = F_{0}$ , $M_{1, 1} : = F_{1}$ . Instrument B: $M_{0, 1} : = σ F_{0}$ , $M_{1, 1} : = F_{1}$ . Both are normalized: $F_{0}^{*} F_{0} + F_{1}^{*} F_{1} = F_{0} + F_{1} = I$ , and $(σ F_{0})^{*} (σ F_{0}) = F_{0} σ^{*} σ F_{0} = F_{0}^{2} = F_{0}$ , so $F_{1} + F_{0} = I$ . Their effect families coincide, but their updates differ on $ρ = \frac{1}{2} I$ : $I_{0}^{A} (ρ) = \frac{1}{2} F_{0} = \frac{1}{2} diag (1, 0)$ while $I_{0}^{B} (ρ) = \frac{1}{2} σ F_{0} σ = \frac{1}{2} diag (0, 1)$ , by $σ T_{e_{1}, e_{1}} σ = T_{σ e_{1}, σ e_{1}} = T_{e_{2}, e_{2}}$ . This is (d).

LemmaStatement: AI-adaptedProof: AI-adaptedOpen item page →

Permutation sector projections on the n-fold tensor power

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $H$ be a complex Hilbert space, $n \geq 1$ , and $H^{\otimes n}$ the $n$ -fold completed tensor power ( $H^{\otimes 1} : = H$ , $H^{\otimes (k + 1)} : = H^{\otimes k} \hat{\otimes} H$ as in Hilbert tensor construction: inner product, completion and product bases). For $v_{0}, \dots, v_{n - 1} \in H$ write $v_{0} \otimes \dots \otimes v_{n - 1} \in H^{\otimes n}$ for the iterated elementary tensor.

(a) For every $π$ in the symmetric group $S_{n}$ (The finite symmetric group $S_{n}$ , one-line notation, and cycle notation) there is a unique unitary $U_{π} \in B (H^{\otimes n})$ with

$U_{π} (v_{0} \otimes \dots \otimes v_{n - 1}) = v_{π^{- 1} (0)} \otimes \dots \otimes v_{π^{- 1} (n - 1)},$

and $U_{π} U_{σ} = U_{π σ}$ , $U_{π}^{*} = U_{π^{- 1}}$ .

(b) The operators

$P_{+} : = \frac{1}{n!} \sum_{π \in S_{n}} U_{π}, P_{-} : = \frac{1}{n!} \sum_{π \in S_{n}} sgn (π) U_{π}$

(with $sgn$ the sign homomorphism, The sign is a homomorphism $S_{n} \to {+ 1, - 1}$ , surjective exactly when $n \geq 2$ ) are orthogonal projections.

(c) $U_{π} P_{+} = P_{+} U_{π} = P_{+}$ and $U_{π} P_{-} = P_{-} U_{π} = sgn (π) P_{-}$ for every $π$ ; consequently the range of $P_{+}$ is the subspace of vectors fixed by every $U_{π}$ , and the range of $P_{-}$ is the subspace of vectors transforming by $sgn$ .

(d) If $v_{0}, \dots, v_{n - 1}$ have a repeated factor, that is, $v_{i} = v_{j}$ for some distinct $i, j \in {0, \dots, n - 1}$ , then $P_{-} (v_{0} \otimes \dots \otimes v_{n - 1}) = 0$ .

Selecting the symmetric or antisymmetric sector for a particle species is a physical postulate, not a theorem of this construction; no spin-statistics statement is proved here.

Facts & Assumptions

Given: A complex Hilbert space $H$ ; $n \geq 1$ ; the tensor powers $H^{\otimes k}$ ; the symmetric group $S_{n}$ and its sign homomorphism.

[F1]

The completed tensor product construction provides: the algebraic tensor product with bilinear elementary tensors, the inner product with $⟨ u \otimes v, u^{'} \otimes v^{'} ⟩ = ⟨ u, u^{'} ⟩ ⟨ v, v^{'} ⟩$ , the dense inclusion, unique orthonormal-coordinate expansions and $∥ u \otimes v ∥ = ∥ u ∥ ∥ v ∥$ (Hilbert tensor construction: inner product, completion and product bases).

[F2]

Bounded operators $S, T$ extend uniquely to $S \otimes T$ with $∥ S \otimes T ∥ \leq ∥ S ∥ ∥ T ∥$ , the adjoint is $S^{*} \otimes T^{*}$ , and compositions multiply; unitary factors have unitary tensor product (Bounded tensor maps: extension, norm estimate, adjoint and commutation).

[F3]

Any parenthesized iterated tensor product represents multilinear maps: for a multilinear map $Φ : H^{n} \to Z$ there is a unique linear $L$ with $L (v_{1} \otimes \dots \otimes v_{n}) = Φ (v_{1}, \dots, v_{n})$ and different parenthesizations agree (Finite iterated tensor products represent multilinear maps independently of parenthesization).

[F4]

$sgn : S_{n} \to {+ 1, - 1}$ is a homomorphism with $sgn (π σ) = sgn (π) sgn (σ)$ and $sgn (π^{- 1}) = sgn (π)$ ; $A_{n} = \ker sgn$ consists of the even permutations (The sign is a homomorphism $S_{n} \to {+ 1, - 1}$ , surjective exactly when $n \geq 2$ , The alternating group $A_{n} = \ker (sgn)$ of even permutations, For $n \geq 2$ , sign is the unique nontrivial homomorphism $S_{n} \to {+ 1, - 1}$ ).

Proof

1.1givenF1F3constructalgebra

[Permutation isometries on the dense algebraic core] The span of iterated elementary tensors is dense in $H^{\otimes n}$ , by induction on the completion construction [F1]. On the algebraic iterated tensor product, the multilinear assignment $(v_{0}, \dots, v_{n - 1}) \mapsto v_{π^{- 1} (0)} \otimes \dots \otimes v_{π^{- 1} (n - 1)}$ induces a linear map by [F3]. It descends isometrically to its image in the completed tensor power: for elementary tensors $v = v_{0} \otimes \dots \otimes v_{n - 1}$ and $w = w_{0} \otimes \dots \otimes w_{n - 1}$ , the inner product of their permuted images is $\prod_{k = 0}^{n - 1} ⟨ v_{π^{- 1} (k)}, w_{π^{- 1} (k)} ⟩ = \prod_{k = 0}^{n - 1} ⟨ v_{k}, w_{k} ⟩ = ⟨ v, w ⟩$ by [F1]. Sesquilinearity gives the same identity for finite linear combinations, so any algebraic relation with zero norm has zero-norm image. The isometry therefore extends uniquely by continuity to $U_{π}$ on the completion. For $n = 1$ it is the identity, and the zero Hilbert space is included.

2.1step 1.1F1F2algebra

[Group law and unitarity] Direct substitution of indices in the elementary formula gives $U_{π} U_{σ} = U_{π σ}$ and $U_{e} = I$ on the dense algebraic core; continuity gives these equalities on the whole completion. Thus $U_{π^{- 1}} U_{π} = I = U_{π} U_{π^{- 1}}$ , so the isometry of step 1.1 is surjective and unitary, with $U_{π}^{*} = U_{π^{- 1}}$ . Any bounded operator with the same elementary action agrees with this one on a dense span and hence everywhere, proving uniqueness. No choice of permutation word or zeroth tensor power is needed.

3.1step 2.1F4algebra

[The projections are orthogonal projections] With coefficients $c_{+} (π) = 1$ and $c_{-} (π) = sgn (π)$ , write $P_{\pm} : = \frac{1}{n!} \sum_{π} c_{\pm} (π) U_{π}$ as in the Statement, use $U_{π} U_{σ} = U_{π σ}$ and reindex $τ = π σ$ : for $P_{+}$ this gives $P_{+}^{2} = \frac{1}{(n!)^{2}} \sum_{π, σ} U_{π σ} = \frac{1}{(n!)^{2}} \sum_{τ} n! U_{τ} = P_{+}$ , and for $P_{-}$ , since $σ = π^{- 1} τ$ and by [F4] $sgn (π) sgn (σ) = sgn (τ)$ , it gives $P_{-}^{2} = \frac{1}{(n!)^{2}} \sum_{π, σ} sgn (π) sgn (σ) U_{π σ} = \frac{1}{(n!)^{2}} \sum_{τ} n! sgn (τ) U_{τ} = P_{-}$ ; moreover $P_{\pm}^{*} = \frac{1}{n!} \sum_{π} c_{\pm} (π) U_{π^{- 1}} = \frac{1}{n!} \sum_{τ} c_{\pm} (τ) U_{τ} = P_{\pm}$ because $U_{π}^{*} = U_{π^{- 1}}$ and $sgn (π^{- 1}) = sgn (π)$ [F4]. Hence $P_{\pm}$ are orthogonal projections.

4.1step 2.1step 3.1F4algebra

[Ranges and transformation laws] For every $π$ , $U_{π} P_{+} = \frac{1}{n!} \sum_{σ} U_{π σ} = \frac{1}{n!} \sum_{τ} U_{τ} = P_{+}$ and $U_{π} P_{-} = \frac{1}{n!} \sum_{σ} sgn (σ) U_{π σ} = sgn (π) \frac{1}{n!} \sum_{τ} sgn (τ) U_{τ} = sgn (π) P_{-}$ , using $sgn (σ) sgn (π σ) = sgn (π)$ [F4]; by self-adjointness of the projections these identities also hold with the order reversed. Consequently, if $u \in ran P_{+}$ then $U_{π} u = P_{+} u = u$ for all $π$ , and if $u \in ran P_{-}$ then $U_{π} u = sgn (π) u$ ; conversely $U_{π} u = u$ for all $π$ gives $P_{+} u = u$ , and $U_{π} u = sgn (π) u$ for all $π$ gives $P_{-} u = \frac{1}{n!} \sum_{π} sgn (π)^{2} u = u$ .

5.1step 2.1step 3.1F4algebra∎

[Vanishing on repeated factors] Suppose $v_{i} = v_{j}$ for distinct $i, j \in {0, \dots, n - 1}$ and let $τ$ be the transposition swapping those positions. Then $U_{τ} (v_{0} \otimes \dots \otimes v_{n - 1}) = v_{0} \otimes \dots \otimes v_{n - 1}$ because the swap exchanges equal vectors. Pairing the terms of the sum defining $n! P_{-}$ as ${π, π τ}$ over a set of coset representatives of the subgroup ${e, τ}$ gives, for each pair, $sgn (π) U_{π} v + sgn (π τ) U_{π τ} v = sgn (π) U_{π} v - sgn (π) U_{π} U_{τ} v = 0$ by $U_{π τ} = U_{π} U_{τ}$ and $U_{τ} v = v$ . Hence $P_{-} v = 0$ for the tensor with a repeated factor.

LemmaStatement: AI-adaptedProof: AI-adaptedOpen item page →

Euclidean L² product unitary: L²(R^a)⊗̂L²(R^b) ≅ L²(R^{a+b})

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )), let $a, b \geq 1$ , and let $λ_{n}$ denote Lebesgue measure on $L (R^{n})$ as defined in Lebesgue measurable sets, the family $L (R^{n})$ , and the restricted set function $λ_{n}$ and completed from Borel measure as in $L (R^{n})$ is exactly the completion of the restriction of $λ_{n}$ to the Borel sets. The product measure $λ_{a} \times λ_{b}$ exists by For sigma-finite factors, the product measure exists, has the rectangle formula, is sigma-finite, and is unique, and under $R^{a + b} = R^{a} \times R^{b}$ , $λ_{a + b}$ is its completion (The Euclidean Lebesgue measure is the completion of the product of the factor Lebesgue measures). Write $H_{a} = L^{2} (R^{a}, λ_{a})$ and $H_{b} = L^{2} (R^{b}, λ_{b})$ , with $L^{2}$ understood as the almost-everywhere quotient of square-integrable measurable functions (The space $L^{p} (μ)$ as the quotient by null functions). Then there is a unique linear isometry

$Θ : H_{a} \hat{\otimes} H_{b} ⟶ L^{2} (R^{a + b})$

with $Θ (f \otimes g) (x, y) = f (x) g (y)$ for all $f \in H_{a}$ , $g \in H_{b}$ and a.e. $(x, y)$ , and $Θ$ is onto: it is a surjective linear isometry, hence unitary. Vector-valued or spin factors follow by finite component sums. No identification of non-Euclidean product measures is asserted.

Facts & Assumptions

Given: $a, b \geq 1$ ; $f, g, f^{'}, g^{'}$ in the respective $L^{2}$ spaces; the completed tensor product $H_{a} \hat{\otimes} H_{b}$ of Hilbert tensor construction: inner product, completion and product bases.

[F1]

$∥ f ∥_{2} = (\int ∣ f ∣^{2} d λ_{n})^{1 / 2}$ , the complex $L^{2}$ pairing is $⟨ f, f^{'} ⟩ = \int f \overline{f^{'}} d λ_{n}$ , and $\int ∣ f f^{'} ∣ d λ_{n} \leq ∥ f ∥_{2} ∥ f^{'} ∥_{2}$ (The space $L^{p} (μ)$ as the quotient by null functions, The complex $L^{2}$ pairing on equivalence classes, Cauchy-Schwarz inequality for $L^{2}$ ). The completed tensor product carries the inner product with $⟨ f \otimes g, f^{'} \otimes g^{'} ⟩ = ⟨ f, f^{'} ⟩ ⟨ g, g^{'} ⟩$ and the elementary tensors span a dense subspace (Hilbert tensor construction: inner product, completion and product bases).

[F2]

The completed product measure $λ_{a + b}$ is the completion of $λ_{a} \times λ_{b}$ ; Fubini and Tonelli hold for functions measurable in this completed product. Every Lebesgue-measurable factor has a Borel representative equal almost everywhere, so products of factor representatives define completed-product-measurable functions, independently of the chosen representatives (The Euclidean Lebesgue measure is the completion of the product of the factor Lebesgue measures, Tonelli and Fubini for the completed product, with only almost-everywhere section measurability, A function measurable for a completion is almost everywhere equal to one measurable for the original sigma-algebra, The Borel product of R^m and R^n is the Borel sigma-algebra of R^{m+n}).

[F3]

Finite positive Borel measures agreeing on rational half-open boxes and on their total mass agree on the Borel $σ$ -algebra; rational half-open boxes form a $π$ -system generating it (Finite measures agreeing on a generating pi-system and on the whole space are equal, Pi-systems, Measurable rectangles in a product of measurable spaces, For n at least one, open sets, closed sets, compact sets, open balls, boxes, rational open boxes, and rational half-open boxes generate the Borel sigma-algebra on R^n, The measure with density $f$ relative to $μ$ ).

[F4]

A bounded linear map on a dense subspace extends uniquely with the same norm, provided its target is Banach; each $L^{2} (λ_{n})$ is complete by Riesz–Fischer (A bounded linear map from a dense normed subspace into a Banach space extends uniquely with the same norm, Riesz-Fischer completeness of $L^{p}$ for $1 \leq p \leq \infty$ ).

[F5]

Lebesgue measure on $R^{n}$ is the completion of its restriction to the Borel sets, and every real Lebesgue-measurable function equals a Borel-measurable function almost everywhere under Countable Choice ( $L (R^{n})$ is exactly the completion of the restriction of $λ_{n}$ to the Borel sets, A function measurable for a completion is almost everywhere equal to one measurable for the original sigma-algebra).

[F6]

If a linear subspace $M$ of a Hilbert space has $M^{⊥} = {0}$ , then $\overline{M} = M^{⊥ ⊥}$ is the whole space (The double orthogonal complement of a subspace is its closure).

Proof

1.1givenF1F2algebra

[Isometry on finite sums] For $f \in L^{2} (λ_{a})$ and $g \in L^{2} (λ_{b})$ choose measurable representatives. By [F2], replacing them by Borel representatives changes the product only on null strips in the completed product, and the product class is well defined. For finite sums $z = \sum_{i = 1}^{n} α_{i} f_{i} \otimes g_{i}$ and $z^{'} = \sum_{j = 1}^{m} β_{j} f_{j}^{'} \otimes g_{j}^{'}$ , each $f_{i} \overline{f_{j}^{'}} \in L^{1} (λ_{a})$ and $g_{i} \overline{g_{j}^{'}} \in L^{1} (λ_{b})$ by [F1], so the product is integrable in the completed product by Fubini [F2], and $\int_{R^{a + b}} f_{i} (x) \overline{f_{j}^{'} (x)} g_{i} (y) \overline{g_{j}^{'} (y)} d λ_{a + b} (x, y) = ⟨ f_{i}, f_{j}^{'} ⟩ ⟨ g_{i}, g_{j}^{'} ⟩ .$ The map $(f, g) \mapsto [f (x) g (y)]$ is bilinear, hence descends through the defining tensor relations to a linear map $Θ_{0}$ on the algebraic tensor product. Summing with coefficients $α_{i} \overline{β_{j}}$ gives $⟨ Θ_{0} z, Θ_{0} z^{'} ⟩ = ⟨ z, z^{'} ⟩$ by [F1]. Taking $z^{'} = z$ gives $Θ_{0} z = 0 \Rightarrow z = 0$ , so $Θ_{0}$ is well defined and isometric on the dense algebraic tensor product.

1.2givenF2F3F5algebra

[Finite-positive-measure uniqueness in the first variable] Let $h \in L^{2} (R^{a + b})$ be orthogonal to every product of bounded-box indicators, so $\int_{B \times C} h = 0$ for all boxes $B, C$ . For each bounded rational half-open box $B \subseteq R^{b}$ , define $h_{B} (x) : = \int_{B} h (x, y) d y$ for a.e. $x$ . Cauchy-Schwarz gives $∣ h_{B} (x) ∣^{2} \leq ∣ B ∣ \int_{B} ∣ h (x, y) ∣^{2} d y$ , and Tonelli [F2] shows $h_{B} \in L^{2} (R^{a})$ . Fubini [F2] gives $\int_{C} h_{B} = \int_{C \times B} h = 0$ for every rational half-open box $C \subseteq R^{a}$ . Fix a bounded rational half-open box $K \subseteq R^{a}$ and write $u = Re h_{B}$ , $v = Im h_{B}$ . By [F5], choose Borel representatives $\tilde{u}, \tilde{v}$ equal a.e. to $u, v$ . The family $P_{K} : = {K \cap C : C is a rational half-open box in R^{a}}$ is a $π$ -system containing $K$ and generating the relative Borel $σ$ -algebra of $K$ by [F3]. The positive and negative parts ${\tilde{u}}_{+}, {\tilde{u}}_{-}, {\tilde{v}}_{+}, {\tilde{v}}_{-}$ are integrable on $K$ because $K$ has finite measure and $h_{B} \in L^{2}$ . The finite positive Borel measures $A \mapsto \int_{A} {\tilde{u}}_{+}$ and $A \mapsto \int_{A} {\tilde{u}}_{-}$ agree on $P_{K}$ , including the total mass at $K$ , since $\int_{A} u = 0$ there; likewise the measures from ${\tilde{v}}_{+}$ and ${\tilde{v}}_{-}$ agree. By [F3] these equalities hold for every Borel $A \subseteq K$ . Taking the Borel sets where $\tilde{u} > 0$ , $\tilde{u} < 0$ , $\tilde{v} > 0$ and $\tilde{v} < 0$ shows $u = v = 0$ a.e. on $K$ . The countable rational half-open boxes $K$ cover $R^{a}$ , so $h_{B} = 0$ a.e. on $R^{a}$ for each rational half-open $B$ .

2.1step 1.2F2F3F5F6algebra

[Zero orthogonal complement] There are countably many rational half-open boxes $B \subseteq R^{b}$ , and step 1.2 gives $h_{B} = 0$ a.e. for each; completed-product Fubini [F2] also gives $h (x, \cdot) \in L^{2} (R^{b})$ for a.e. $x$ . Intersecting these countably many full-measure sets, for a.e. $x$ the section is in $L^{2}$ and $\int_{B} h (x, y) d y = 0$ for every rational half-open $B$ . Fix such an $x$ . On each bounded rational half-open box $K \subseteq R^{b}$ , choose Borel representatives of the real and imaginary parts of $h (x, \cdot)$ by [F5]. The positive and negative parts of these representatives define finite positive Borel measures on $K$ ; as in step 1.2 they agree on the $π$ -system of rational half-open subboxes, including $K$ , so [F3] makes the positive and negative measures equal on every Borel subset of $K$ . Their positive and negative level sets then show the real and imaginary parts vanish a.e. on $K$ . The countably many $K$ cover $R^{b}$ , hence $h (x, \cdot) = 0$ a.e. for a.e. $x$ , and Tonelli [F2] gives $h = 0$ a.e. Thus $Ran Θ_{0}^{⊥} = {0}$ ; since $Ran Θ_{0}$ is a linear subspace, [F6] gives $\overline{Ran Θ_{0}} = Ran Θ_{0}^{⊥ ⊥} = L^{2} (λ_{a + b})$ .

3.1step 1.1step 2.1F1F4∎

[Extension and surjectivity] By [F4], whose Banach-space hypothesis holds by Riesz–Fischer, the isometry $Θ_{0}$ on the dense algebraic tensor product extends uniquely to a linear isometry $Θ$ of the completion. Its range is closed (isometries have closed range: if $Θ z_{n} \to w$ then $(z_{n})$ is Cauchy, so $z_{n} \to z$ and $w = Θ z$ ) and dense by step 2.1, hence is all of $L^{2} (λ_{a + b})$ . Thus $Θ$ is a surjective linear isometry, i.e. unitary, with the stated action on elementary tensors.

LemmaStatement: AI-adaptedProof: AI-adaptedOpen item page →

Nuclear bounded cyclicity for trace-class operators

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $H$ be a complex Hilbert space (Hilbert space) and for $u, v \in H$ let $T_{u, v} x = ⟨ x, v ⟩ u$ (Rank-one operator calculus).

(a) For every bounded $B \in B (H)$ (A bounded linear operator between normed spaces),

$B T_{u, v} = T_{B u, v}, T_{u, v} B = T_{u, B^{*} v},$

with the Hilbert adjoint $B^{*}$ (The Hilbert-space adjoint of a bounded operator).

(b) Let $T \in B (H)$ be trace class and let $T = \sum_{j} T_{u_{j}, v_{j}}$ be a nuclear representation, i.e. the partial sums converge to $T$ in operator norm and $\sum_{j} ∥ u_{j} ∥ ∥ v_{j} ∥ < \infty$ (Nuclear series characterizes trace norm, Trace class operator). Then for every bounded $B$

$tr (B T) = \sum_{j} ⟨ B u_{j}, v_{j} ⟩, tr (T B) = \sum_{j} ⟨ u_{j}, B^{*} v_{j} ⟩,$

both series converging absolutely with $∣ tr (B T) ∣ \leq ∥ B ∥ \sum_{j} ∥ u_{j} ∥ ∥ v_{j} ∥$ , and $tr (B T) = tr (T B)$ (Trace is absolutely convergent and basis independent). In particular $tr (B T) = tr (T B)$ for every trace-class $T$ and bounded $B$ , and $tr (T_{u, v}) = ⟨ u, v ⟩$ . No cyclic permutation of unbounded operators is asserted.

Facts & Assumptions

Given: A complex Hilbert space $H$ ; vectors $u, v, u_{j}, v_{j} \in H$ ; bounded operators $B$ ; a trace-class $T$ with nuclear representation $T = \sum_{j} T_{u_{j}, v_{j}}$ .

[F1]

$T_{u, v} \in B (H)$ is trace class, $∥ T_{u, v} ∥_{1} = ∥ u ∥ ∥ v ∥$ , $tr (T_{u, v}) = ⟨ u, v ⟩$ , and $T_{u, v}^{*} = T_{v, u}$ (Rank-one operator calculus).

[F2]

$T$ is trace class exactly when it has a nuclear representation $T = \sum_{j} T_{u_{j}, v_{j}}$ with $\sum_{j} ∥ u_{j} ∥ ∥ v_{j} ∥ < \infty$ converging in operator norm; and then $∥ T ∥_{1} = \inf \sum_{j} ∥ u_{j} ∥ ∥ v_{j} ∥$ over all such representations (Nuclear series characterizes trace norm).

[F3]

The trace-class operators form a two-sided Banach operator ideal: $B T \in S_{1} (H)$ for bounded $B$ and trace-class $T$ , with $∥ B T ∥_{1} \leq ∥ B ∥ ∥ T ∥_{1}$ for the operator norm (Trace class is a two sided Banach operator ideal, The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

[F4]

The trace is linear on trace-class operators and $∣ tr S ∣ \leq ∥ S ∥_{1}$ (Trace is absolutely convergent and basis independent).

[F5]

Bounded adjoints obey $∥ B^{*} ∥ = ∥ B ∥$ (Hilbert-adjoint identities). The Hilbert adjoint satisfies $⟨ B x, y ⟩ = ⟨ x, B^{*} y ⟩$ and $T_{u, v}^{*} = T_{v, u}$ (The Hilbert-space adjoint of a bounded operator, Rank-one operator calculus).

Proof

1.1givenF5algebra

[Rank-one multiplication] For $x \in H$ , $(B T_{u, v}) x = B (⟨ x, v ⟩ u) = ⟨ x, v ⟩ B u = T_{B u, v} x$ , so $B T_{u, v} = T_{B u, v}$ ; and $(T_{u, v} B) x = ⟨ B x, v ⟩ u = ⟨ x, B^{*} v ⟩ u = T_{u, B^{*} v} x$ by [F5], so $T_{u, v} B = T_{u, B^{*} v}$ . Both are bounded because $B$ is and the displayed maps have the stated form.

2.1step 1.1F1F2F3algebra

[Trace-norm convergence of the multiplier series] Write $T_{n} : = \sum_{j \leq n} T_{u_{j}, v_{j}}$ , so $∥ T - T_{n} ∥ \to 0$ . Then $B T_{n} = \sum_{j \leq n} T_{B u_{j}, v_{j}}$ by step 1.1, and $\sum_{j} ∥ T_{B u_{j}, v_{j}} ∥_{1} = \sum_{j} ∥ B u_{j} ∥ ∥ v_{j} ∥ \leq ∥ B ∥ \sum_{j} ∥ u_{j} ∥ ∥ v_{j} ∥ < \infty$ by [F1]; since the trace-class operators are complete in trace norm [F3], the series $\sum_{j} T_{B u_{j}, v_{j}}$ converges in trace norm, and its limit agrees with $B T$ because trace-norm convergence implies operator-norm convergence and $B T_{n} \to B T$ in operator norm. Hence $B T = \sum_{j} T_{B u_{j}, v_{j}}$ in trace norm, with $∥ B T ∥_{1} \leq ∥ B ∥ \sum_{j} ∥ u_{j} ∥ ∥ v_{j} ∥$ .

3.1step 2.1F1F4algebra

[Trace of the left product] By [F4], $tr (B T) = \lim_{n} tr (B T_{n}) = \lim_{n} \sum_{j \leq n} tr (T_{B u_{j}, v_{j}}) = \lim_{n} \sum_{j \leq n} ⟨ B u_{j}, v_{j} ⟩$ , because the trace is linear on the finite sums $B T_{n}$ and continuous for the trace norm, and $B T_{n} \to B T$ in trace norm by step 2.1. Thus $tr (B T) = \sum_{j} ⟨ B u_{j}, v_{j} ⟩$ , and the series converges absolutely since $∣ ⟨ B u_{j}, v_{j} ⟩ ∣ = ∣ tr (T_{B u_{j}, v_{j}}) ∣ \leq ∥ B u_{j} ∥ ∥ v_{j} ∥$ by [F1]; the bound $∣ tr (B T) ∣ \leq ∥ B ∥ \sum_{j} ∥ u_{j} ∥ ∥ v_{j} ∥$ follows from [F4] and step 2.1.

4.1step 1.1step 2.1step 3.1F1F2F3F4F5algebra∎

[Right multiplication and cyclicity] By step 1.1, $T_{n} B = \sum_{j \leq n} T_{u_{j}, B^{*} v_{j}}$ , and $\sum_{j} ∥ u_{j} ∥ ∥ B^{*} v_{j} ∥ \leq ∥ B ∥ \sum_{j} ∥ u_{j} ∥ ∥ v_{j} ∥ < \infty$ by [F5]. Exactly as in step 2.1, trace-norm completeness gives convergence of this series to $T B$ , since its operator-norm limit is $T B$ . Trace continuity and the rank-one trace formula therefore give $tr (T B) = \sum_{j} ⟨ u_{j}, B^{*} v_{j} ⟩$ with absolute convergence. Each summand equals $⟨ B u_{j}, v_{j} ⟩$ by the adjoint identity, so step 3.1 gives $tr (T B) = tr (B T)$ for the original arbitrary nuclear representation. Every trace-class operator has such a representation by [F2]; the stated rank-one trace identity is [F1]. Only bounded multipliers were used.

LemmaStatement: AI-adaptedProof: AI-adaptedOpen item page →

Complex Gaussian continuation

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). For $z \in C$ with $Re z > 0$ and $n \geq 1$ let

$G_{n} (z) : = \int_{R^{n}} e^{- z ∣ x ∣^{2}} d x,$

where $d x$ is product Lebesgue measure and the complex integral is computed from its real and imaginary parts. Equivalently, the absolutely convergent improper multiple integral used here is the cube-truncation limit $\lim_{R \to \infty} \int_{[- R, R]^{n}} e^{- z ∣ x ∣^{2}} d x$ of these finite Lebesgue integrals. For $n = 1$ it also equals the sum of the two separately convergent one-sided improper Riemann integrals at $0$ , computed componentwise (Improper integrals over unbounded intervals); this agreement is proved below. Then:

(a) $G_{n} (z)$ converges absolutely for every $z$ with $Re z > 0$ , and $G_{1}$ is holomorphic on the right half-plane $Ω : = {z \in C : Re z > 0}$ with $G_{1}^{'} (z) = - \int_{R} x^{2} e^{- z x^{2}} d x$ (Complex differentiability at a point, the complex derivative, holomorphic functions, and entire functions, A complex domain is a nonempty connected open subset of $C$ );

(b) with $Log$ the principal logarithm and the branch $z^{- 1 / 2} : = \exp (- \frac{1}{2} Log z)$ on the slit plane (Complex logarithms, the principal logarithm, and principal and multivalued complex powers, The principal logarithm is the normalised holomorphic branch on the slit plane),

$G_{1} (z) = \sqrt{π} z^{- 1 / 2}, G_{1} (z)^{2} = \frac{π}{z} (z \in Ω),$

the value on the positive real axis being $G_{1} (t) = \sqrt{π / t}$ for $t > 0$ ;

(c) for every $n \geq 1$ and $z \in Ω$ , $G_{n} (z) = G_{1} (z)^{n} = (π / z)^{n / 2}$ , the power being the branch $(π / z)^{n / 2} = \exp (\frac{n}{2} (\log π - Log z))$ .

In particular the factor $(1 + i τ)^{- 1 / 2}$ is defined by this branch for every real $τ$ , since $Re (1 + i τ) = 1 > 0$ . No asymptotic or distributional interpretation is asserted.

Facts & Assumptions

Given: $n \geq 1$ ; a complex parameter $z$ with $Re z > 0$ ; the functions $G_{n}$ of the Statement; the principal logarithm $Log$ .

[F1]

For $n \geq 1$ , $t > 0$ and $ξ \in R^{n}$ , $F (e^{- π t ∣ x ∣^{2}}) (ξ) = t^{- n / 2} e^{- π ∣ ξ ∣^{2} / t}$ ; at $ξ = 0$ this reads $\int_{R^{n}} e^{- π t ∣ x ∣^{2}} d x = t^{- n / 2}$ , and substituting $s = π t$ gives $\int_{R^{n}} e^{- s ∣ x ∣^{2}} d x = (π / s)^{n / 2}$ . Every polynomial times a positive real Gaussian is absolutely integrable (Euclidean Gaussian transform with the 2π normalization).

[F2]

On a fixed measure space, if measurable complex-valued $f_{n}, f$ satisfy $f_{n} \to f$ almost everywhere and $∣ f_{n} ∣ \leq g$ almost everywhere for one nonnegative measurable $g$ with $\int g < \infty$ , then $\int ∣ f_{n} - f ∣ \to 0$ (Dominated convergence). For a parameter $h \to 0$ , apply this to each sequence of nonzero parameters tending to zero.

[F3]

A function is holomorphic when it is complex differentiable at every point of its domain; the derivative is the limit of difference quotients (Complex differentiability at a point, the complex derivative, holomorphic functions, and entire functions). A complex domain is a nonempty connected open set (A complex domain is a nonempty connected open subset of $C$ ).

[F4]

The principal logarithm $Log$ is holomorphic on the slit plane $S = C ∖ (- \infty, 0]$ , satisfies $\exp (Log z) = z$ and $Log z = \log ∣ z ∣ + i Arg z$ with $Arg z \in (- π, π)$ , and $Log t = \ln t$ for $t > 0$ ; complex powers are defined by $z^{a} = \exp (a Log z)$ (Complex logarithms, the principal logarithm, and principal and multivalued complex powers, The principal logarithm is the normalised holomorphic branch on the slit plane).

[F5]

If two holomorphic functions on a complex domain agree on a set with an accumulation point in the domain, they agree everywhere (Identity theorem for holomorphic functions).

[F6]

Tonelli's theorem: for a product-measurable nonnegative function on a $σ$ -finite product, the iterated and double integrals agree (Tonelli's theorem for nonnegative measurable functions on a sigma-finite product).

[F7]

For continuous $f : R^{n} \to C$ with $\int ∣ f ∣ < \infty$ , the improper multiple integral used here is $\lim_{R \to \infty} \int_{[- R, R]^{n}} f$ , with finite Lebesgue integrals computed componentwise. Dominated convergence (Dominated convergence) applied to $1_{[- R, R]^{n}} f$ identifies this limit with the product Lebesgue integral: for any sequence $R_{j} \to \infty$ the indicators tend pointwise to $1$ and are dominated by $∣ f ∣$ , hence every such sequence has the same limit. For $n = 1$ , the real and imaginary parts are bounded and Riemann integrable on every compact interval by A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion, and their finite Riemann and Lebesgue integrals agree by A bounded Riemann integrable function on a closed bounded interval is Lebesgue measurable and has the same integral. Applying dominated convergence separately to $1_{[0, R]} f$ as $R \to \infty$ and $1_{[R, 0]} f$ as $R \to - \infty$ therefore proves convergence of both one-sided improper integrals of Improper integrals over unbounded intervals; the same argument for $∣ f ∣$ proves absolute convergence. Their sum is the full integral because the overlap ${0}$ has Lebesgue measure zero. Existence of a symmetric limit alone does not imply this agreement.

Proof

1.1givenF1F7algebra

[Absolute convergence] Let $z \in Ω$ and $δ : = Re z > 0$ . For every $x$ , $∣ e^{- z ∣ x ∣^{2}} ∣ = e^{- δ ∣ x ∣^{2}}$ , whose integral over $R^{n}$ is finite by [F1] with $s = δ$ ; hence the real and imaginary parts of $x \mapsto e^{- z ∣ x ∣^{2}}$ are absolutely improperly integrable and $G_{n} (z)$ is well defined by [F7]. The same bound holds uniformly on compact subsets of $Ω$ , where $δ$ can be taken positive uniformly.

2.1step 1.1F2F3F7algebra

[Holomorphy of $G_{1}$ ] Fix $z \in Ω$ and $0 < δ < Re z$ ; let $∣ h ∣ < δ / 2$ . For the function $w \mapsto e^{w}$ the second-order Taylor remainder satisfies $∣ e^{w} - 1 - w ∣ \leq ∣ w ∣^{2} e^{∣ w ∣}$ (the power series of $e^{w} - 1 - w$ has its first term $\frac{w^{2}}{2}$ and the remaining terms are bounded by $\frac{∣ w ∣^{2}}{2} (e^{∣ w ∣} - 1) \leq ∣ w ∣^{2} e^{∣ w ∣}$ ). Applying this with $w = - h x^{2}$ gives $∣ e^{- z x^{2}} \frac{e^{- h x^{2}} - 1 + h x^{2}}{h} ∣ \leq ∣ h ∣ x^{4} e^{(∣ h ∣ - Re z) x^{2}} \leq ∣ h ∣ x^{4} e^{- δ x^{2} / 2}$ for all real $x$ and all $∣ h ∣ < δ / 2$ , and $x \mapsto x^{4} e^{- δ x^{2} / 2}$ has finite integral; the remainder functions are continuous, hence measurable, converge pointwise to zero, and for $∣ h ∣ < δ / 2$ are dominated by the nonnegative integrable function $(δ / 2) x^{4} e^{- δ x^{2} / 2}$ . By [F2] along every sequence $h \to 0$ , their integrals over $R$ tend to zero. Thus the difference quotient $(G_{1} (z + h) - G_{1} (z)) / h$ tends, as $h \to 0$ , to $- \int_{R} x^{2} e^{- z x^{2}} d x$ . Hence $G_{1}$ is complex differentiable at every $z \in Ω$ , with the displayed derivative; by [F3] it is holomorphic on $Ω$ .

2.2step 1.1F1F3F4algebra

[The candidate branch] Put $H (z) : = \sqrt{π} \exp (- \frac{1}{2} Log z)$ for $z$ in the slit plane $S \supseteq Ω$ . By [F4], $Log$ is holomorphic on $S$ with ${Log}^{'} z = 1 / z$ , and the complex exponential is entire with derivative itself, so by the chain rule for complex derivatives $H$ is holomorphic on the complex domain $S$ (The chain rule for complex derivatives, The complex exponential is entire and its complex derivative is itself); also $H (z)^{2} = π \exp (- Log z) = π / z$ because $\exp (Log z) = z$ . On the positive axis, $Log t = \ln t$ [F4], so $H (t) = \sqrt{π} t^{- 1 / 2} = \sqrt{π / t}$ , which equals $G_{1} (t) = \int_{R} e^{- t x^{2}} d x$ by [F1] with $n = 1$ , $s = t$ .

3.1step 2.1step 2.2F3F5algebra

[Agreement by the identity theorem] The right half-plane $Ω$ is nonempty, open and convex (if $Re z_{1}, Re z_{2} > 0$ then $Re ((1 - λ) z_{1} + λ z_{2}) > 0$ for $0 \leq λ \leq 1$ ), hence a complex domain, and it is contained in $S$ , so both $G_{1}$ (step 2.1) and $H$ (step 2.2) are holomorphic on $Ω$ . They agree at every $t > 0$ (step 2.2), and the set $(0, \infty)$ has accumulation points in $Ω$ ; by [F5] applied to the complex domain $Ω$ , $G_{1} = H$ on $Ω$ , giving clause (b).

4.1step 3.1F4F6algebra∎

[Factorization] For $z \in Ω$ write $e^{- z ∣ x ∣^{2}} = \prod_{j = 1}^{n} e^{- z x_{j}^{2}}$ ; the absolute value factors as $\prod_{j} e^{- Re (z) x_{j}^{2}}$ , a product of nonnegative integrable functions of the single coordinates. Tonelli's theorem [F6] applied to the nonnegative function $x \mapsto \prod_{j} e^{- Re (z) x_{j}^{2}}$ shows that the multiple integral of $∣ e^{- z ∣ x ∣^{2}} ∣$ factorizes as the product of the one-dimensional integrals and is finite; hence $x \mapsto e^{- z ∣ x ∣^{2}}$ is integrable on the product, and the Fubini theorem for $L^{1}$ functions (Fubini's theorem for L^1 functions on a sigma-finite product) applied to its real and imaginary parts gives $G_{n} (z) = \prod_{j = 1}^{n} G_{1} (z) = G_{1} (z)^{n}$ . By step 3.1, $G_{1} (z)^{n} = (π / z)^{n / 2}$ in the branch notation of [F4]. This is clause (c), and $Re (1 + i τ) = 1 > 0$ shows the branch is defined at $1 + i τ$ .

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Momentum Fourier Jacobian: the unitary momentum representation

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )), let $d \geq 1$ and $ℏ > 0$ . Write $F_{2}$ for the unitary Plancherel transform extending $\hat{f} (ξ) = \int_{R^{d}} f (x) e^{- 2 π i x \cdot ξ} d x$ (Fourier transform on complex L1 classes, Plancherel theorem); the $L^{2}$ spaces are Hilbert spaces by Riesz-Fischer completeness of $L^{p}$ for $1 \leq p \leq \infty$ and Hilbert space. For $ψ \in S (R^{d})$ , denote by $\hat{ψ}$ the continuous representative of $F_{2} ψ$ given by that absolutely convergent integral.

(a) For $ψ \in S (R^{d})$ the momentum wavefunction

$ψ_{p} (p) : = (2 π ℏ)^{- d / 2} \int_{R^{d}} e^{- i p \cdot x / ℏ} ψ (x) d x$

is well defined, bounded and continuous, and $ψ_{p} (p) = (2 π ℏ)^{- d / 2} \hat{ψ} (p / (2 π ℏ))$ for every $p$ ; here $\hat{ψ}$ is the stated continuous representative, not a point evaluation of an $L^{2}$ equivalence class.

(b) The dilation operator $D_{ℏ} g (p) : = (2 π ℏ)^{- d / 2} g (p / (2 π ℏ))$ is a surjective linear isometry of $L^{2} (R^{d}; C)$ onto itself, with inverse $D_{ℏ}^{- 1} g (ξ) = (2 π ℏ)^{d / 2} g (2 π ℏ ξ)$ .

(c) The formula $U_{0} ψ : = ψ_{p}$ defines a linear isometry $S (R^{d}) \to L^{2} (R^{d})$ (for the Lebesgue spaces $d x$ and $d p$ ) that extends uniquely to a surjective linear isometry $U : L^{2} (R^{d}, d x) \to L^{2} (R^{d}, d p)$ ; on Schwartz functions $U = D_{ℏ} F_{2}$ , and for every $ψ \in L^{2} (R^{d})$ ,

$∥ U ψ ∥_{L^{2} (d p)} = ∥ ψ ∥_{L^{2} (d x)} .$

The normalization statement is: the substitution $ξ = p / (2 π ℏ)$ carries $d ξ = (2 π ℏ)^{- d} d p$ , and the factor $(2 π ℏ)^{- d / 2}$ in $U_{0}$ is exactly what makes $U$ an isometry with respect to $d p$ . No physical identification of $U ψ$ with a measured momentum distribution is made.

Facts & Assumptions

Given: $d \geq 1$ , $ℏ > 0$ ; a Schwartz function $ψ \in S (R^{d})$ ; the Plancherel transform $F_{2}$ and the $L^{1}$ transform $\hat{f} (ξ) = \int f (x) e^{- 2 π i x \cdot ξ} d x$ .

[F1]

$F_{2}$ is a surjective complex-linear isometry $L^{2} (R^{d}) \to L^{2} (R^{d})$ that extends the $L^{1}$ Fourier transform $\hat{f} (ξ) = \int f (x) e^{- 2 π i x \cdot ξ} d x$ (Plancherel theorem, Fourier transform on complex L1 classes).

[F2]

If $f_{n} \to f$ pointwise and $∣ f_{n} ∣ \leq g$ for a fixed integrable $g$ , then $\int f_{n} \to \int f$ (Dominated convergence).

[F3]

An invertible linear map $T$ of $R^{d}$ scales Lebesgue measure by $∣ \det T ∣$ on measurable sets (A linear map $T$ of $R^{n}$ sends Lebesgue measurable sets to Lebesgue measurable sets, with $λ_{n} (T [E]) = ∣ \det T ∣ λ_{n} (E)$ when $T$ is invertible and $T [E]$ Lebesgue null when it is not); for nonnegative measurable $h$ the integral identity $\int h (T x) d x = ∣ \det T ∣^{- 1} \int h (y) d y$ follows by approximation with simple functions and monotone convergence (Monotone convergence for the integral).

[F4]

$S (R^{d})$ contains $C_{c}^{\infty} (R^{d})$ , which is dense in $L^{2} (R^{d})$ (Smooth compactly supported functions of an open set are dense in $L^{2}$ , Schwartz space and its seminorms); in particular every Schwartz function lies in $L^{1} \cap L^{2}$ . The $L^{2}$ space is complete by Riesz-Fischer completeness of $L^{p}$ for $1 \leq p \leq \infty$ .

[F5]

A bounded linear map from a dense subspace of a normed space into a Banach space extends uniquely with the same norm (A bounded linear map from a dense normed subspace into a Banach space extends uniquely with the same norm).

Proof

1.1givenF2F4algebra

[Absolute convergence and continuity] For $ψ \in S (R^{d})$ , $∣ e^{- i p \cdot x / ℏ} ψ (x) ∣ = ∣ ψ (x) ∣$ is integrable by [F4], so the defining integral converges absolutely for every $p$ and $∣ ψ_{p} (p) ∣ \leq (2 π ℏ)^{- d / 2} ∥ ψ ∥_{1}$ . Continuity in $p$ follows from [F2] applied to $e^{- i p_{n} \cdot x / ℏ} ψ \to e^{- i p \cdot x / ℏ} ψ$ with the integrable dominating function $∣ ψ ∣$ .

2.1step 1.1F1algebra

[Identification with the Plancherel transform] For $ψ \in S$ , put $ξ = p / (2 π ℏ)$ , so $2 π x \cdot ξ = p \cdot x / ℏ$ ; comparing the defining integral of $ψ_{p}$ with the $L^{1}$ Fourier transform [F1] gives $ψ_{p} (p) = (2 π ℏ)^{- d / 2} \hat{ψ} (ξ) = (2 π ℏ)^{- d / 2} F_{2} ψ (p / (2 π ℏ))$ , the last equality because $ψ \in L^{1}$ and $F_{2}$ extends $\hat{\cdot}$ . Hence $U_{0} = D_{ℏ} \circ F_{2}$ on $S (R^{d})$ .

2.2step 1.1F3algebra

[The dilation is a unitary] For $g \in L^{2}$ the substitution $p = (2 π ℏ) ξ$ is an invertible linear map with determinant $(2 π ℏ)^{d}$ , so by [F3] applied to $h = ∣ g ∣^{2}$ , $\int ∣ D_{ℏ} g (p) ∣^{2} d p = (2 π ℏ)^{- d} \int ∣ g (p / (2 π ℏ)) ∣^{2} d p = (2 π ℏ)^{- d} (2 π ℏ)^{d} \int ∣ g (ξ) ∣^{2} d ξ = ∥ g ∥^{2}$ . Thus $D_{ℏ}$ is a linear isometry; its inverse $D_{ℏ}^{- 1} g (ξ) = (2 π ℏ)^{d / 2} g (2 π ℏ ξ)$ is likewise an isometry and the two compositions are the identity, so $D_{ℏ}$ is a surjective isometry.

3.1step 2.1step 2.2F1algebra

[Norm preservation on Schwartz functions] By steps 2.1 and 2.2 and unitarity of $F_{2}$ [F1], $∥ ψ_{p} ∥_{2} = ∥ D_{ℏ} F_{2} ψ ∥_{2} = ∥ F_{2} ψ ∥_{2} = ∥ ψ ∥_{2}$ for $ψ \in S (R^{d})$ ; moreover $U_{0}$ is linear on $S (R^{d})$ because the integral and the dilation are linear.

4.1step 2.2step 3.1F1F4F5algebra∎

[Unique unitary extension] The subspace $S (R^{d})$ is dense in $L^{2}$ because it contains the dense subspace $C_{c}^{\infty} (R^{d})$ [F4]; by [F5] the isometry $U_{0}$ on $S (R^{d})$ has a unique bounded linear extension $U : L^{2} (R^{d}, d x) \to L^{2} (R^{d}, d p)$ with the same norm, so $∥ U ψ ∥ = ∥ ψ ∥$ for every $ψ \in L^{2}$ . Surjectivity: for arbitrary $h \in L^{2} (R^{d}, d p)$ , the vector $ψ : = F_{2}^{*} D_{ℏ}^{- 1} h$ lies in $L^{2}$ , and since $U$ agrees with $D_{ℏ} F_{2}$ on the dense set $S$ while both sides are bounded, $U ψ = h$ ; hence $U$ is onto. The norm identity of clause (c) is exactly $∥ U ψ ∥_{2} = ∥ ψ ∥_{2}$ , and the normalization discussion is the determinant computation of step 2.2 read at $g = F_{2} ψ$ .

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Separable partial trace and its basic properties

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $H, K$ be nonzero complex Hilbert spaces (Hilbert space) and let $H \hat{\otimes} K$ be their completed tensor product (Hilbert tensor construction: inner product, completion and product bases); write $T_{x, z} x^{'} = ⟨ x^{'}, z ⟩ x$ for rank-one operators (Rank-one operator calculus) and $tr$ for the trace of a trace-class operator (Trace is absolutely convergent and basis independent).

(a) There is a unique linear map ${Tr}_{K} : S_{1} (H \hat{\otimes} K) \to S_{1} (H)$ with

$tr (({Tr}_{K} T) A) = tr (T (A \otimes I_{K}))$

for every trace-class $T$ and every $A \in B (H)$ ; it satisfies $∥ {Tr}_{K} T ∥_{1} \leq ∥ T ∥_{1}$ and is independent of any chosen orthonormal basis of $K$ .

(b) ${Tr}_{K}$ maps self-adjoint positive trace-class operators to self-adjoint positive trace-class operators (positivity in the bounded-operator sense of Self-adjoint, positive, unitary and normal operators), is trace preserving ( $tr {Tr}_{K} T = tr T$ ), and maps density operators to density operators (Density operators and scalar spectral probability).

The analogous map ${Tr}_{H}$ satisfies $tr (({Tr}_{H} T) A) = tr (T (I_{H} \otimes A))$ for $A \in B (K)$ . No claim is made about instruments or conditioning.

Facts & Assumptions

Given: Nonzero complex Hilbert spaces $H, K$ ; vectors $u, u^{'} \in H$ , $v, v^{'} \in K$ ; a trace-class $T$ on $H \hat{\otimes} K$ with a nuclear representation $T = \sum_{j} T_{w_{j}, z_{j}}$ , $w_{j}, z_{j} \in H \hat{\otimes} K$ ; a bounded $A \in B (H)$ .

[F1]

$T_{x, z}$ is trace class (Trace class operator) with $∥ T_{x, z} ∥_{1} = ∥ x ∥ ∥ z ∥$ , $tr (T_{x, z}) = ⟨ x, z ⟩$ , and $T_{x, z} B = T_{x, B^{*} z}$ , $B T_{x, z} = T_{B x, z}$ for bounded $B$ ; for trace-class $T$ , $tr (T B) = tr (B T)$ and $tr (T) = \sum_{j} ⟨ w_{j}, z_{j} ⟩$ for every nuclear representation (Rank-one operator calculus, Nuclear bounded cyclicity for trace-class operators, Trace is absolutely convergent and basis independent).

[F2]

$T$ is trace class exactly when $T = \sum_{j} T_{w_{j}, z_{j}}$ with $\sum_{j} ∥ w_{j} ∥ ∥ z_{j} ∥ < \infty$ converging in operator norm (The operator norm as the least bound and as the unit-sphere or unit-ball supremum), and then $∥ T ∥_{1} = \inf \sum_{j} ∥ w_{j} ∥ ∥ z_{j} ∥$ ; $S_{1}$ is complete in trace norm and a two-sided ideal with $∥ B T B^{'} ∥_{1} \leq ∥ B ∥ ∥ T ∥_{1} ∥ B^{'} ∥$ (Nuclear series characterizes trace norm, Trace class is a two sided Banach operator ideal).

[F3]

If $S \in S_{1} (H)$ and $x, y \in H$ then $tr (S T_{x, y}) = ⟨ S x, y ⟩$ ; if $tr (S A) = 0$ for all bounded $A$ then $S = 0$ (Nuclear bounded cyclicity for trace-class operators, Rank-one operator calculus).

[F4]

Algebraic tensors are dense in the completed product, with the orthogonal coordinate norm formula; a closed span of a given countable sequence has a finite/countable orthonormal basis by Gram–Schmidt. Bounded tensor maps extend and satisfy the adjoint and norm identities; Cauchy–Schwarz bounds scalar coefficient sums (Hilbert tensor construction: inner product, completion and product bases, A Hilbert space with a dense sequence has a finite or countable orthonormal basis, Bounded tensor maps: extension, norm estimate, adjoint and commutation, Cauchy–Schwarz: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ , with equality exactly for dependent pairs).

Proof

1.1givenF1algebra

[The rank-one values] For $T_{u \otimes v, u^{'} \otimes v^{'}}$ and bounded $A$ , $T (A \otimes I) = T_{u \otimes v, (A \otimes I)^{*} (u^{'} \otimes v^{'})} = T_{u \otimes v, A^{*} u^{'} \otimes v^{'}}$ by [F1], whose trace is $⟨ u \otimes v, A^{*} u^{'} \otimes v^{'} ⟩ = ⟨ u, A^{*} u^{'} ⟩ ⟨ v, v^{'} ⟩ = ⟨ A u, u^{'} ⟩ ⟨ v, v^{'} ⟩$ . On the other hand $tr ((⟨ v, v^{'} ⟩ T_{u, u^{'}}) A) = ⟨ v, v^{'} ⟩ tr (T_{u, u^{'}} A) = ⟨ v, v^{'} ⟩ ⟨ A u, u^{'} ⟩$ by [F1]. Hence the assignment ${Tr}_{K} T_{u \otimes v, u^{'} \otimes v^{'}} : = ⟨ v, v^{'} ⟩ T_{u, u^{'}}$ satisfies the defining identity on rank-one tensors, and $∥ {Tr}_{K} T_{u \otimes v, u^{'} \otimes v^{'}} ∥_{1} \leq ∥ v ∥ ∥ v^{'} ∥ ∥ u ∥ ∥ u^{'} ∥ = ∥ u \otimes v ∥ ∥ u^{'} \otimes v^{'} ∥$ by [F1].

2.1step 1.1F1F2F3F4chooseconstructalgebra

[Arbitrary rank-one vectors] Fix $w, z \in H \hat{\otimes} K$ . By density and Countable Choice choose sequences of finite algebraic tensors converging to these two vectors; collect their countably many second-factor vectors and let $K_{0}$ be their closed linear span. Finite rational complex combinations form an explicit dense sequence in $K_{0}$ , so [F4] provides a finite/countable orthonormal basis $(f_{b})$ of $K_{0}$ . The maps taking $\sum_{i} u_{i} \otimes v_{i}$ to $\sum_{i} ⟨ v_{i}, f_{b} ⟩ u_{i}$ have norm at most one by orthogonal coordinates and extend to the completion. Their values $w_{b}, z_{b} \in H$ satisfy $w = \sum_{b} w_{b} \otimes f_{b}$ , $z = \sum_{b} z_{b} \otimes f_{b}$ and $\sum_{b} ∥ w_{b} ∥^{2} = ∥ w ∥^{2}$ , $\sum_{b} ∥ z_{b} ∥^{2} = ∥ z ∥^{2}$ : the finite-coordinate projections are contractions, converge on algebraic tensors in $H ⊙ K_{0}$ , and hence converge on its closure, which contains $w, z$ . Define $Q (w, z) : = \sum_{b} T_{w_{b}, z_{b}}$ . By [F1], [F2] and Cauchy–Schwarz, this converges in trace norm and $∥ Q (w, z) ∥_{1} \leq \sum_{b} ∥ w_{b} ∥ ∥ z_{b} ∥ \leq ∥ w ∥ ∥ z ∥$ . Finite-coordinate truncations and step 1.1 give $tr (Q (w, z) A) = ⟨ (A \otimes I) w, z ⟩ = tr (T_{w, z} (A \otimes I))$ ; rank-one trace-norm continuity and bounded tensor maps [F4] pass this identity to limits. The identity uniquely determines $Q (w, z)$ by [F3], so it is independent of the supports, basis and approximants, is sesquilinear in $w, z$ , and agrees with the elementary assignment.

3.1step 2.1F1F2F3F4algebra

[Extension to trace class] For a nuclear representation $T = \sum_{j} T_{w_{j}, z_{j}}$ , define ${Tr}_{K} T : = \sum_{j} Q (w_{j}, z_{j})$ . Step 2.1 and [F2] give trace-norm convergence and the bound $∥ {Tr}_{K} T ∥_{1} \leq \inf \sum_{j} ∥ w_{j} ∥ ∥ z_{j} ∥ = ∥ T ∥_{1}$ . The defining trace pairing follows by summing that of step 2.1, using trace-norm continuity and $∥ A \otimes I ∥ \leq ∥ A ∥$ . If two nuclear representations are used, both images have the same pairings against every bounded $A$ , and their difference vanishes by [F3]. Thus the definition is independent of representation. The same uniqueness proves linearity, since the sum and scalar multiple of images satisfy the pairing for the sum and scalar multiple of inputs. This proves (a) and the elementary formula (c).

4.1step 2.1step 3.1F1F2algebra

[Positivity and trace preservation] Step 2.1 gives $Q (w, w) = \sum_{b} T_{w_{b}, w_{b}}$ , which is self-adjoint positive since $⟨ T_{w_{b}, w_{b}} x, x ⟩ = ∣ ⟨ x, w_{b} ⟩ ∣^{2} \geq 0$ and trace-norm convergence implies operator-norm convergence. Every self-adjoint positive trace-class input of positive trace becomes a density operator after division by its trace, so it has the positive spectral rank-one expansion of Density operators and scalar spectral probability; a zero-trace self-adjoint positive trace-class input is zero by Trace of a positive operator is the sum of its eigenvalues. The bounded map of step 3.1 carries the partial sums of this expansion to positive sums of the corresponding $Q (w, w)$ and hence carries its limit to a self-adjoint positive operator. Taking $A = I_{H}$ in the defining pairing gives $tr {Tr}_{K} T = tr T$ . Hence density operators map to density operators.

5.1step 3.1F3∎

[Uniqueness] If two linear maps ${Tr}_{K}, {Tr}_{K}^{'}$ satisfy the defining identity for all $T$ and $A$ , then for fixed $T$ the difference $S = {Tr}_{K} T - {Tr}_{K}^{'} T$ satisfies $tr (S A) = 0$ for all bounded $A$ , so $S = 0$ by [F3]. Hence the map is unique and, in particular, independent of any basis; the symmetrically defined ${Tr}_{H}$ has the stated identity.

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Countable Kraus normalization and trace-norm convergence

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $H$ be a complex Hilbert space (Hilbert space) and let $(M_{a})_{a \geq 1}$ be bounded operators on $H$ with

$\sum_{a = 1}^{N} M_{a}^{*} M_{a} ⟶ I strongly as N \to \infty .$

Then:

(a) $\sum_{a \geq 1} ∥ M_{a} x ∥^{2} = ∥ x ∥^{2}$ for every $x \in H$ ;

(b) for every self-adjoint positive trace-class $ρ$ (positivity as in Self-adjoint, positive, unitary and normal operators) the series $\sum_{a \geq 1} M_{a} ρ M_{a}^{*}$ converges in trace norm to a self-adjoint positive trace-class operator $Φ (ρ)$ with $tr Φ (ρ) = tr ρ$ ;

(c) $Φ$ extends uniquely to a linear positive map $Φ : S_{1} (H) \to S_{1} (H)$ with $tr Φ (ρ) = tr ρ$ for all trace-class $ρ$ and $∥ Φ (ρ) ∥_{1} \leq ∥ ρ ∥_{1}$ for self-adjoint $ρ$ ; it maps density operators to density operators. No claim is made that every positive trace-preserving map has such a family.

Facts & Assumptions

Given: A complex Hilbert space $H$ ; bounded operators $M_{a}$ with strong completeness $\sum_{a \leq N} M_{a}^{*} M_{a} \to I$ ; a positive trace-class $ρ$ .

[F1]

For every $x$ , $⟨ (\sum_{a \leq N} M_{a}^{*} M_{a}) x, x ⟩ \to ⟨ x, x ⟩$ and the partial sums are increasing (each term $M_{a}^{*} M_{a} \geq 0$ ), so the nonnegative series $\sum_{a} ∥ M_{a} x ∥^{2} = \sum_{a} ⟨ M_{a}^{*} M_{a} x, x ⟩$ has value $∥ x ∥^{2}$ (Hilbert space).

[F2]

$tr (M_{a} ρ M_{a}^{*}) = tr (M_{a}^{*} M_{a} ρ)$ ; $T_{x, z} B = T_{x, B^{*} z}$ ; $T$ is trace class exactly when it is a nuclear series; trace class is a complete two-sided ideal; the trace is linear, continuous for $∥ \cdot ∥_{1}$ and $tr T = \sum_{j} tr T_{w_{j}, z_{j}}$ for nuclear representations (Nuclear bounded cyclicity for trace-class operators, Rank-one operator calculus, Trace class is a two sided Banach operator ideal, Trace is absolutely convergent and basis independent).

[F3]

A self-adjoint positive trace-class operator with positive trace becomes a density operator after dividing by its trace and has the spectral expansion $ρ = \sum_{n} w_{n} T_{u_{n}, u_{n}}$ with orthonormal $u_{n}$ , $w_{n} \geq 0$ and $tr ρ = \sum_{n} w_{n}$ ; a zero-trace self-adjoint positive trace-class operator is zero. Conversely such sums (with $\sum w_{n} < \infty$ ) are self-adjoint positive trace class (Density operators and scalar spectral probability, Trace of a positive operator is the sum of its eigenvalues, Self-adjoint, positive, unitary and normal operators).

[F4]

A self-adjoint compact operator has an expansion $\sum_{λ} λ P_{λ}$ and $tr (T^{\pm}) = \sum_{\pm λ > 0} ∣ λ ∣$ ; for a positive trace-class operator $S$ , $∥ S ∥_{1} = tr S$ (Spectral theorem for compact self adjoint operators, Trace of a positive operator is the sum of its eigenvalues, Trace class is a two sided Banach operator ideal, Trace class iff product of two Hilbert Schmidt operators).

Proof

1.1givenF1algebra

[Norm identity, including the zero space] If $H = {0}$ , then every $M_{a}$ and every trace-class input is zero, so the unique map on $S_{1} (H) = {0}$ satisfies (a)-(c). Suppose henceforth that $H \neq {0}$ . For every $x$ , $\sum_{a \leq N} ∥ M_{a} x ∥^{2}$ is the increasing sequence $⟨ (\sum_{a \leq N} M_{a}^{*} M_{a}) x, x ⟩$ , so its supremum is $⟨ x, x ⟩ = ∥ x ∥^{2}$ by strong convergence and continuity of the inner product in $N$ ; by [F1] this supremum is $\sum_{a} ∥ M_{a} x ∥^{2}$ .

2.1step 1.1F2F3F4algebra

[Positive input] Let $ρ = \sum_{n} w_{n} T_{u_{n}, u_{n}}$ be the spectral expansion [F3]. The partial sums $Φ_{N} (ρ) : = \sum_{a \leq N} M_{a} ρ M_{a}^{*}$ are positive trace-class with $tr Φ_{N} (ρ) = \sum_{a \leq N} \sum_{n} w_{n} ∥ M_{a} u_{n} ∥^{2}$ by [F2]; the double sum increases with $N$ , and its limit is $\sum_{n} w_{n} \sum_{a} ∥ M_{a} u_{n} ∥^{2} = \sum_{n} w_{n} ∥ u_{n} ∥^{2} = tr ρ$ by step 1.1 and Tonelli for nonnegative series. Hence the increasing positive sequence $Φ_{N} (ρ)$ has traces increasing to $tr ρ$ , so for $N \geq M$ we have $∥ Φ_{N} (ρ) - Φ_{M} (ρ) ∥_{1} = tr Φ_{N} (ρ) - tr Φ_{M} (ρ) \to 0$ ([F4] with positivity of the difference); by completeness of $S_{1} (H)$ [F2] the series converges in trace norm to a trace-class $Φ (ρ) \geq 0$ (the trace-norm limit of positive operators is positive) with $tr Φ (ρ) = tr ρ$ .

3.1step 1.1step 2.1F2F3F4algebra

[Linear extension by nuclear convergence] For rank-one input, $M_{a} T_{u, v} M_{a}^{*} = T_{M_{a} u, M_{a} v}$ by [F2]. Step 1.1 and Cauchy–Schwarz give $\sum_{a} ∥ M_{a} u ∥ ∥ M_{a} v ∥ \leq ∥ u ∥ ∥ v ∥$ . Thus for a nuclear input $T = \sum_{j} T_{u_{j}, v_{j}}$ the double series of Kraus rank-one images is absolutely summable in trace norm, with total norm at most $\sum_{j} ∥ u_{j} ∥ ∥ v_{j} ∥$ . Its sum is the trace-norm limit of $Φ_{N} (T) = \sum_{a \leq N} M_{a} T M_{a}^{*}$ , by interchanging absolutely convergent sums and the bounded two-sided ideal action. This limit depends only on $T$ , so defines a linear map $Φ$ extending the positive-input map of step 2.1. Taking the nuclear infimum (Nuclear series characterizes trace norm) gives the bound $∥ Φ (T) ∥_{1} \leq ∥ T ∥_{1}$ for every input. For rank-one inputs, the trace sum is $\sum_{a} ⟨ M_{a} u, M_{a} v ⟩ = ⟨ u, v ⟩$ , by strong completeness and the adjoint identity; extending through the absolutely convergent nuclear sum proves $tr Φ (T) = tr T$ . The positive trace-class operators linearly span trace class by the self-adjoint real/imaginary and positive/negative spectral decompositions in [F4], so no other linear extension agreeing with step 2.1 is possible.

4.1step 2.1step 3.1F4∎

[Norm bound and conclusion] For self-adjoint $ρ$ with spectral decomposition $ρ = ρ^{+} - ρ^{-}$ write $T = ρ$ . Then $Φ (ρ) = Φ (ρ^{+}) - Φ (ρ^{-})$ by construction, so $∥ Φ (ρ) ∥_{1} \leq tr Φ (ρ^{+}) + tr Φ (ρ^{-}) = tr ρ^{+} + tr ρ^{-} = ∥ ρ ∥_{1}$ using step 2.1 and [F4]. Positivity for arbitrary $ρ \geq 0$ is step 2.1, and trace preservation was shown in step 3.1; hence density operators map to density operators.

LemmaStatement: AI-adaptedProof: AI-adaptedOpen item page →

Finite-ancilla complete positivity of countable Kraus maps

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $H$ be a complex Hilbert space (Hilbert space), let $(M_{a})_{a \geq 1}$ be bounded operators on $H$ with $\sum_{a \leq N} M_{a}^{*} M_{a} \to I$ strongly, and let $Φ (ρ) = \sum_{a} M_{a} ρ M_{a}^{*}$ be the map of Countable Kraus normalization and trace-norm convergence. Here a map on trace-class operators is completely positive on finite-dimensional ancillas when, for every finite-dimensional complex inner product space $C$ , its matrixwise amplification $I_{C} \otimes Φ$ preserves self-adjoint positive trace-class operators on $C \hat{\otimes} H$ . Then for every finite-dimensional complex inner product space $C$ the amplification

$I_{C} \otimes Φ : S_{1} (C \hat{\otimes} H) \to S_{1} (C \hat{\otimes} H)$

is well defined by the trace-norm convergent series $\sum_{a} (I_{C} \otimes M_{a}) R (I_{C} \otimes M_{a}^{*})$ and preserves positivity: every self-adjoint positive trace-class $R$ (positivity in the sense of Self-adjoint, positive, unitary and normal operators) has $I_{C} \otimes Φ (R)$ self-adjoint positive. Hence $Φ$ is completely positive on every finite-dimensional ancilla in the stated sense.

Facts & Assumptions

Given: $H$ , $(M_{a})$ and $Φ$ as in the Statement; a finite-dimensional complex inner product space $C$ ; a positive trace-class $R$ on $C \hat{\otimes} H$ .

[F1]

The completed tensor product $C \hat{\otimes} H$ with the inner product $⟨ c \otimes h, c^{'} \otimes h^{'} ⟩ = ⟨ c, c^{'} ⟩ ⟨ h, h^{'} ⟩$ is a complex Hilbert space with dense span of elementary tensors; for bounded $S$ the amplification $I_{C} \otimes S$ is bounded with $∥ I_{C} \otimes S ∥ \leq ∥ S ∥$ (Hilbert tensor construction: inner product, completion and product bases, Bounded tensor maps: extension, norm estimate, adjoint and commutation).

[F2]

For every $R \geq 0$ in $S_{1}$ and every bounded $V$ , $V R V^{*} \geq 0$ and $tr (V R V^{*}) = tr (V^{*} V R)$ ; the trace-class operators are complete in trace norm (Nuclear bounded cyclicity for trace-class operators, Trace class is a two sided Banach operator ideal, Trace class operator).

[F3]

A positive trace-class operator $R$ on a Hilbert space has a spectral expansion $R = \sum_{n} w_{n} T_{ξ_{n}, ξ_{n}}$ with orthonormal $ξ_{n}$ , $w_{n} \geq 0$ and $\sum_{n} w_{n} = tr R$ (Density operators and scalar spectral probability).

[F4]

The partial sums $S_{N} = \sum_{a \leq N} M_{a}^{*} M_{a}$ satisfy $0 \leq S_{N} \leq I$ and converge strongly to $I$ . For finite-dimensional $C$ with orthonormal basis $(e_{k})_{k = 1}^{d}$ , every vector in $C \hat{\otimes} H$ has the form $ξ = \sum_{k} e_{k} \otimes h_{k}$ , so $I_{C} \otimes S_{N}$ acts coordinatewise, is a positive contraction, and converges strongly to $I$ (Bounded tensor maps: extension, norm estimate, adjoint and commutation, Hilbert tensor construction: inner product, completion and product bases).

Proof

1.1givenF1F4algebra

[The amplified family is strongly complete] For each $a$ , $(I_{C} \otimes M_{a})^{*} (I_{C} \otimes M_{a}) = I_{C} \otimes M_{a}^{*} M_{a}$ by [F1]. If $(e_{k})_{k = 1}^{d}$ is an orthonormal basis of $C$ and $ξ = \sum_{k} e_{k} \otimes h_{k}$ , then ${∥ (I_{C} \otimes \sum_{a \leq N} M_{a}^{*} M_{a} - I) ξ ∥}^{2} = \sum_{k = 1}^{d} {∥ (S_{N} - I) h_{k} ∥}^{2} ⟶ 0$ by the assumed strong convergence on $H$ . Thus the amplified family satisfies the hypothesis of Countable Kraus normalization and trace-norm convergence on $C \hat{\otimes} H$ , and that lemma applies to it.

2.1step 1.1F1F2F3algebra∎

[Positivity] If $C = {0}$ or $H = {0}$ , then $C \hat{\otimes} H = {0}$ , the only trace-class input is $R = 0$ , and the asserted series and its limit are zero, so the conclusion holds. Suppose henceforth that $C, H \neq {0}$ . For $R \geq 0$ and each $a$ , $(I_{C} \otimes M_{a}) R (I_{C} \otimes M_{a})^{*} \geq 0$ by [F2]; the partial sums are positive, and the trace-norm limit is positive because $⟨ S x, x ⟩ = \lim ⟨ S_{N} x, x ⟩ \geq 0$ for trace-norm convergent $S_{N} \to S$ . Alternatively, applying the spectral argument of Countable Kraus normalization and trace-norm convergence to the amplified family: by [F3], $R = \sum_{n} w_{n} T_{ξ_{n}, ξ_{n}}$ with $ξ_{n} \in C \hat{\otimes} H$ , the partial sums $\sum_{a \leq N} (I_{C} \otimes M_{a}) R (I_{C} \otimes M_{a})^{*} = \sum_{n} w_{n} \sum_{a \leq N} T_{(I_{C} \otimes M_{a}) ξ_{n}, (I_{C} \otimes M_{a}) ξ_{n}}$ are positive with traces $\sum_{n} w_{n} \sum_{a \leq N} ∥ (I_{C} \otimes M_{a}) ξ_{n} ∥^{2} ↑ \sum_{n} w_{n} ∥ ξ_{n} ∥^{2} = tr R$ by step 1.1, so they converge in trace norm to a positive limit with trace $tr R$ , and the series converges to $I_{C} \otimes Φ (R)$ .

LemmaStatement: AI-adaptedProof: AI-adaptedOpen item page →

Local trace-preserving Kraus maps do not change the opposite reduced state

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $H, K$ be complex Hilbert spaces, let $(M_{a})_{a \geq 1}$ be bounded operators on $K$ with $\sum_{a \leq N} M_{a}^{*} M_{a} \to I_{K}$ strongly, and let $R$ be a trace-class operator on $H \hat{\otimes} K$ . In this statement, $R \geq 0$ means that $R$ is self-adjoint and $⟨ R x, x ⟩ \geq 0$ for every $x$ , as in Self-adjoint, positive, unitary and normal operators. Then the series (when either factor is ${0}$ , use the zero partial-trace maps ${Tr}_{K} : S_{1} ({0}) \to S_{1} (H)$ and ${Tr}_{H} : S_{1} ({0}) \to S_{1} (K)$ ; when both factors are nonzero, use the partial traces of Separable partial trace and its basic properties)

$\sum_{a \geq 1} (I_{H} \otimes M_{a}) R (I_{H} \otimes M_{a}^{*})$

converges in trace norm, its sum $Φ (R)$ is trace class with $tr Φ (R) = tr R$ and $Φ (R) \geq 0$ if $R \geq 0$ , and the reduced state on $H$ is unchanged:

${Tr}_{K} Φ (R) = {Tr}_{K} R .$

By symmetry, a local map $R \mapsto \sum_{a} (N_{a} \otimes I_{K}) R (N_{a}^{*} \otimes I_{K})$ on the first factor also leaves ${Tr}_{H} R$ fixed. Selective conditioning (one summand normalized by its nonzero trace) is not covered.

Facts & Assumptions

Given: $H, K$ ; the family $(M_{a})$ ; a trace-class $R$ on $H \hat{\otimes} K$ with nuclear representation $R = \sum_{j} T_{w_{j}, z_{j}}$ ; a bounded $A \in B (H)$ .

[F1]

The amplified family $(I_{H} \otimes M_{a})$ consists of bounded operators on $H \hat{\otimes} K$ with $(I_{H} \otimes M_{a})^{*} (I_{H} \otimes M_{a}) = I_{H} \otimes M_{a}^{*} M_{a}$ , and $\sum_{a \leq N} (I_{H} \otimes M_{a})^{*} (I_{H} \otimes M_{a}) \to I$ strongly: on finite sums of elementary tensors this is the strong convergence of $\sum_{a \leq N} M_{a}^{*} M_{a}$ , and the general case follows by approximation and $∥ I_{H} \otimes S_{N} ∥ \leq 1$ (Bounded tensor maps: extension, norm estimate, adjoint and commutation, Hilbert tensor construction: inner product, completion and product bases).

[F2]

Under the strong completeness hypothesis, $\sum_{a} V_{a} R V_{a}^{*}$ converges in trace norm for every trace-class $R$ , is positive for positive $R$ , and preserves the trace, where $V_{a} : = I_{H} \otimes M_{a}$ ; moreover the identity $tr (\sum_{a} V_{a} R V_{a}^{*}) = tr R$ is the trace-preservation clause (Countable Kraus normalization and trace-norm convergence, Finite-ancilla complete positivity of countable Kraus maps).

[F3]

When $H, K \neq {0}$ , ${Tr}_{K}$ is the unique linear map on $S_{1} (H \hat{\otimes} K)$ with $tr (({Tr}_{K} T) A) = tr (T (A \otimes I_{K}))$ for all bounded $A$ , and it satisfies $∥ {Tr}_{K} T ∥_{1} \leq ∥ T ∥_{1}$ (Separable partial trace and its basic properties); the Statement supplies the zero-map convention when a factor is zero.

[F4]

For a bounded operator $B$ and trace-class $T$ , $B T$ is trace class with $∥ B T ∥_{1} \leq ∥ B ∥ ∥ T ∥_{1}$ , and $tr (B T)$ is linear in $T$ and continuous in trace norm (Trace class is a two sided Banach operator ideal, Trace is absolutely convergent and basis independent). In addition, every trace-class $T$ has a nuclear representation $T = \sum_{j} T_{w_{j}, z_{j}}$ , and for bounded $B$ the trace formulas $tr (T B) = \sum_{j} ⟨ w_{j}, B^{*} z_{j} ⟩ = tr (B T)$ hold with absolute convergence (Nuclear bounded cyclicity for trace-class operators, Facts [F2] and Statement (b)).

Proof

1.1givenF1F2

[Convergence of the local map] If $H = {0}$ or $K = {0}$ , then $R = 0$ and every summand $(I_{H} \otimes M_{a}) R (I_{H} \otimes M_{a})^{*}$ is zero; hence $Φ (R) = 0$ and both sides of ${Tr}_{K} Φ (R) = {Tr}_{K} R$ are zero by the convention in the Statement. Assume henceforth $H, K \neq {0}$ . By [F1] the amplified family satisfies the strong completeness hypothesis of [F2] on the Hilbert space $H \hat{\otimes} K$ , so the series $\sum_{a} V_{a} R V_{a}^{*}$ with $V_{a} = I_{H} \otimes M_{a}$ converges in trace norm, its sum $Φ (R)$ is trace class with $tr Φ (R) = tr R$ , and $Φ (R) \geq 0$ for $R \geq 0$ .

2.1step 1.1F4algebra

[Pairing computation] For bounded $A$ and $V_{a} = I_{H} \otimes M_{a}$ , apply the bounded cyclicity of [F4] to the trace-class operator $T = R V_{a}^{*} (A \otimes I_{K})$ and bounded $V_{a}$ . It gives $tr (V_{a} R V_{a}^{*} (A \otimes I_{K})) = tr (R V_{a}^{*} (A \otimes I_{K}) V_{a})$ with $V_{a}^{*} (A \otimes I_{K}) V_{a} = (I_{H} \otimes M_{a}^{*}) (A \otimes I_{K}) (I_{H} \otimes M_{a}) = A \otimes M_{a}^{*} M_{a}$ . For each $N$ , sum this identity over $a \leq N$ ; trace-norm convergence of the series in step 1.1 and trace-norm continuity of the bounded trace pairing [F4] give $tr (Φ (R) (A \otimes I_{K})) = \lim_{N \to \infty} tr (R (A \otimes S_{N}))$ , where $S_{N} = \sum_{a \leq N} M_{a}^{*} M_{a}$ .

3.1step 2.1F1F4algebra

[Passing to the limit in the completeness sum] Let $R = \sum_{j} T_{w_{j}, z_{j}}$ be nuclear. For each $N$ , $tr (R (A \otimes \sum_{a \leq N} M_{a}^{*} M_{a})) = \sum_{j} ⟨ w_{j}, (A \otimes S_{N})^{*} z_{j} ⟩$ with $S_{N} : = \sum_{a \leq N} M_{a}^{*} M_{a}$ , an absolutely convergent series with $∣ ⟨ w_{j}, (A \otimes S_{N})^{*} z_{j} ⟩ ∣ \leq ∥ A ∥ ∥ w_{j} ∥ ∥ z_{j} ∥$ ; since $(A \otimes S_{N})^{*} \to A^{*} \otimes I$ strongly, each term converges to $⟨ w_{j}, (A^{*} \otimes I) z_{j} ⟩$ , and the summable majorant gives $\sum_{j} ⟨ w_{j}, (A \otimes S_{N})^{*} z_{j} ⟩ \to \sum_{j} ⟨ w_{j}, (A^{*} \otimes I) z_{j} ⟩ = tr (R (A \otimes I_{K}))$ . Hence $tr (Φ (R) (A \otimes I_{K})) = tr (R (A \otimes I_{K}))$ .

4.1step 3.1F3∎

[Conclusion] By [F3], $tr (({Tr}_{K} Φ (R)) A) = tr (Φ (R) (A \otimes I_{K})) = tr (R (A \otimes I_{K})) = tr (({Tr}_{K} R) A)$ for every bounded $A$ ; since the pairings with all $A$ determine a trace-class operator by the uniqueness in [F3], ${Tr}_{K} Φ (R) = {Tr}_{K} R$ . The statement for the first factor is obtained by exchanging the roles of $H$ and $K$ .

TheoremStatement: AI-adaptedProof: AI-adaptedOpen item page →

Infinite Schmidt decomposition and the pure product criterion

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )) and let $H, K$ be nonzero complex Hilbert spaces (Hilbert space) with $H \hat{\otimes} K$ their completed tensor product (Hilbert tensor construction: inner product, completion and product bases). A product vector means an elementary tensor $v \otimes u$ with $v \in H$ and $u \in K$ . Then for every $Ψ \in H \hat{\otimes} K$ there are an at most countable index set $J$ , orthonormal families $(v_{r})_{r \in J} \subseteq H$ and $(u_{r})_{r \in J} \subseteq K$ , and numbers $s_{r} > 0$ with

$Ψ = \sum_{r \in J} s_{r} v_{r} \otimes u_{r} (norm convergence in H \hat{\otimes} K),$

and $∥ Ψ ∥^{2} = \sum_{r} s_{r}^{2}$ . The $s_{r}$ are the positive square roots of the nonzero eigenvalues of $C C^{*}$ , where $C : \overline{K_{0}} \to H_{0}$ is the linear coefficient operator on countable closed factor supports of $Ψ$ , defined below; if $J$ is infinite then $s_{r} \to 0$ . For $∥ Ψ ∥ = 1$ the reduced density $ρ = {Tr}_{K} T_{Ψ, Ψ}$ has $tr ρ^{2} = \sum_{r} s_{r}^{4}$ , and $tr ρ^{2} = 1$ holds exactly when exactly one $s_{r}$ is nonzero, i.e. exactly when $Ψ$ is a product vector. No assertion about separability of mixed density operators is made here.

For clarity, choose finite algebraic tensors converging to $Ψ$ and let $H_{0} \subseteq H$ , $K_{0} \subseteq K$ be the closed spans of their factor vectors. These supports have finite/countable orthonormal bases, and $Ψ \in H_{0} \hat{\otimes} K_{0}$ . The conjugate Hilbert space $\overline{K_{0}}$ consists of symbols $\overset{ˉ}{z}$ , with $a \overset{ˉ}{z} : = \overline{\overset{ˉ}{a} z}$ and $⟨ \overset{ˉ}{z}, \overset{ˉ}{w} ⟩_{\overline{K_{0}}} : = ⟨ w, z ⟩_{K_{0}}$ . For an orthonormal basis $(f_{j})$ of $K_{0}$ , write $Ψ = \sum_{j} h_{j} \otimes f_{j}$ and set $C {\overset{ˉ}{f}}_{j} = h_{j}$ , extending linearly and continuously. The positive singular values and the nonzero spectrum of $C C^{*}$ are independent of these supports and bases, as the reduced-state identity proved below shows.

Facts & Assumptions

Given: Countable Choice, arbitrary nonzero complex Hilbert spaces $H, K$ , and $Ψ \in H \hat{\otimes} K$ . The supports, bases and coefficient map are constructed in the proof; no countable basis of the ambient spaces is assumed.

[F1]

Algebraic tensors are dense in their Hilbert completion, and orthogonal coordinates have the squared-norm sum formula. A given dense sequence in a Hilbert space yields a finite/countable orthonormal basis by Gram–Schmidt (Hilbert tensor construction: inner product, completion and product bases, A Hilbert space with a dense sequence has a finite or countable orthonormal basis, Parseval equivalences for an orthonormal family).

[F2]

Bounded finite-rank operators are compact, and an operator-norm limit of compact operators into a Hilbert space is compact under Countable Choice. Hilbert adjoints satisfy the adjoint identity and $(C^{*})^{*} = C$ ; Cauchy–Schwarz bounds coordinate sums (Bounded finite rank operators are compact, Norm limit of compact operators is compact, Hilbert-adjoint identities, Cauchy–Schwarz: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ , with equality exactly for dependent pairs).

[F3]

A compact positive self-adjoint operator has positive eigenvalues of finite multiplicity, accumulating only at zero, and its nonzero eigenspaces span the orthogonal complement of its kernel. A positive trace-class operator has trace and trace norm equal to the sum of its positive eigenvalues (Spectral theorem for compact self adjoint operators, Trace of a positive operator is the sum of its eigenvalues).

[F4]

Rank-one operators have trace norm $∥ T_{u, v} ∥_{1} = ∥ u ∥ ∥ v ∥$ and trace $⟨ u, v ⟩$ . The partial trace is bounded in trace norm with the defining bounded-operator pairing and elementary rank-one formula. Bounded tensor maps extend with their norm and adjoint identities (Rank-one operator calculus, Separable partial trace and its basic properties, Bounded tensor maps: extension, norm estimate, adjoint and commutation, Trace class is a two sided Banach operator ideal).

Proof

1.1givenF1F4chooseconstructalgebra

[Countable supports and coefficients] By density and Countable Choice choose algebraic tensors $Ψ_{n} \to Ψ$ with finite factor decompositions. Their countably many factor vectors span closed subspaces $H_{0}, K_{0}$ with explicit dense sequences of finite rational complex combinations; [F1] supplies finite/countable orthonormal bases. The inclusion of $H_{0} ⊙ K_{0}$ in $H \hat{\otimes} K$ is isometric and extends to its closed completion, which contains $Ψ$ . Choose the basis $(f_{j})$ of $K_{0}$ . As in the orthogonal contraction construction of Separable partial trace and its basic properties, coefficient contractions define $h_{j} \in H_{0}$ with $Ψ = \sum_{j} h_{j} \otimes f_{j}$ and $\sum_{j} ∥ h_{j} ∥^{2} = ∥ Ψ ∥^{2}$ . All sums here may be finite or empty, including $Ψ = 0$ .

2.1step 1.1F1F2F4constructalgebra

[The typed compact coefficient operator] Define $C (\sum_{j} z_{j} {\overset{ˉ}{f}}_{j}) : = \sum_{j} z_{j} h_{j}$ first on finite sums in $\overline{K_{0}}$ . Cauchy–Schwarz gives $∥ C z ∥ \leq (\sum_{j} ∥ h_{j} ∥^{2})^{1 / 2} ∥ z ∥$ , so it extends to all of $\overline{K_{0}}$ by continuity and completeness. The finite truncations $C_{N}$ have finite rank and $∥ C - C_{N} ∥ \leq (\sum_{j > N} ∥ h_{j} ∥^{2})^{1 / 2} \to 0$ , so [F2] makes $C$ compact. The adjoint has the correctly typed formula $C^{*} x = \sum_{j} ⟨ x, h_{j} ⟩ {\overset{ˉ}{f}}_{j}$ for $x \in H_{0}$ , hence $C C^{*} x = \sum_{j} ⟨ x, h_{j} ⟩ h_{j}$ . The finite sums $\sum_{j \leq N} T_{h_{j}, h_{j}}$ converge in trace norm by [F4], with trace sum $\sum_{j} ∥ h_{j} ∥^{2}$ ; their operator-norm limit is $C C^{*}$ , so $C C^{*}$ is positive trace class with $tr C C^{*} = ∥ Ψ ∥^{2}$ .

3.1step 1.1step 2.1F4algebra

[Reduced-state identity] Let $Ψ_{N} = \sum_{j \leq N} h_{j} \otimes f_{j}$ . The elementary partial-trace formula gives ${Tr}_{K} T_{Ψ_{N}, Ψ_{N}} = \sum_{j \leq N} T_{h_{j}, h_{j}}$ , regarding the latter operators on $H$ as zero on $H_{0}^{⊥}$ . Rank-one continuity and boundedness of the partial trace [F4] pass to the limit and give ${Tr}_{K} T_{Ψ, Ψ} = C C^{*}$ extended by zero. Thus the nonzero eigenvalues of $C C^{*}$ are intrinsic to $Ψ$ , independent of all auxiliary supports and bases.

4.1step 2.1step 3.1F1F2F3chooseconstructalgebra

[Singular vectors in the conjugate factor] By [F3] choose orthonormal eigenvectors $v_{r} \in H_{0}$ for the positive eigenvalues $s_{r}^{2}$ of $C C^{*}$ , counted with multiplicity; finite-dimensional eigenspace bases are supplied by [F1] and their countable selection uses the stated Countable Choice. Define ${\overset{ˉ}{u}}_{r} : = s_{r}^{- 1} C^{*} v_{r} \in \overline{K_{0}}$ . Then $⟨ {\overset{ˉ}{u}}_{r}, {\overset{ˉ}{u}}_{t} ⟩ = s_{r}^{- 1} s_{t}^{- 1} ⟨ v_{r}, C C^{*} v_{t} ⟩ = δ_{r t}$ , so the actual vectors $u_{r} \in K_{0}$ are orthonormal too. By step 2.1 their coefficients obey $u_{r} = \sum_{j} s_{r}^{- 1} ⟨ h_{j}, v_{r} ⟩ f_{j}$ . The trace formula [F3] gives $\sum_{r} s_{r}^{2} = tr C C^{*} = ∥ Ψ ∥^{2}$ , and infinitely many positive eigenvalues tend to zero.

5.1step 1.1step 4.1F1F3algebra

[Reconstruction] If $g \in \ker C C^{*}$ then $0 = ⟨ C C^{*} g, g ⟩ = ∥ C^{*} g ∥^{2}$ , so all $⟨ g, h_{j} ⟩$ vanish. Thus every $h_{j}$ is in the closed span of the $v_{r}$ by [F3], and $h_{j} = \sum_{r} ⟨ h_{j}, v_{r} ⟩ v_{r}$ . The orthonormal series $\sum_{r} s_{r} v_{r} \otimes u_{r}$ converges in tensor norm because $\sum_{r} s_{r}^{2} < \infty$ . Its $f_{j}$ coefficient is $\sum_{r} s_{r} ⟨ u_{r}, f_{j} ⟩ v_{r} = \sum_{r} ⟨ h_{j}, v_{r} ⟩ v_{r} = h_{j}$ by step 4.1. The orthogonal coefficient norm formula [F1] therefore identifies its sum with $Ψ$ . If $Ψ = 0$ , the positive spectrum and the Schmidt family are empty, giving the zero expansion.

6.1step 2.1step 3.1step 4.1step 5.1F3F4algebra∎

[Purity and product criterion] For $∥ Ψ ∥ = 1$ , steps 3.1–5.1 show that the reduced density has eigenvalues $s_{r}^{2}$ and $\sum_{r} s_{r}^{2} = 1$ , hence $tr ρ^{2} = \sum_{r} s_{r}^{4}$ . The difference $(\sum_{r} s_{r}^{2})^{2} - \sum_{r} s_{r}^{4} = 2 \sum_{r < t} s_{r}^{2} s_{t}^{2}$ vanishes exactly when there is one positive coefficient. That case makes the expansion a product vector. Conversely for a nonzero product $a \otimes b$ , its coefficients are $h_{j} = ⟨ b, f_{j} ⟩ a$ , so $C$ has rank one and $C C^{*}$ has one positive eigenvalue; for a unit product its reduced purity is one. All ambient Hilbert spaces remain arbitrary; only the supports of this particular vector were made countable. This establishes every claimed clause.

LemmaStatement: AI-adaptedProof: AI-adaptedOpen item page →

Abstract trace-class instruments: effects and Heisenberg duality

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $H$ be a nonzero complex Hilbert space (Hilbert space) and let $(X, Σ)$ be a measurable space. In this item, a positive trace-class operator means a self-adjoint trace-class operator $T$ with $⟨ T x, x ⟩ \geq 0$ for every $x \in H$ ; a bounded effect means a self-adjoint $F \in B (H)$ with $0 \leq ⟨ F x, x ⟩ \leq ∥ x ∥^{2}$ for every $x \in H$ .

By an abstract instrument on $(X, Σ)$ we mean a map $J : Σ \to B (S_{1} (H))$ into bounded linear maps of the trace-class operators such that (i) each $J_{B}$ is completely positive (every finite amplification $I_{C^{n}} \otimes J_{B}$ preserves positivity on positive trace-class operators); (ii) for pairwise disjoint $B_{n}$ with union $B$ , $J_{B} (ρ) = \sum_{n} J_{B_{n}} (ρ)$ in trace norm for every trace-class $ρ$ ; (iii) $J_{X}$ is trace preserving.

Then:

(a) for every $B \in Σ$ there is a unique effect $F_{B} \in B (H)$ , $0 \leq F_{B} \leq I$ (Self-adjoint, positive, unitary and normal operators), with $tr (J_{B} (T)) = tr (T F_{B})$ for every trace-class $T$ ;

(b) $F_{X} = I$ , and $B \mapsto F_{B}$ is weakly countably additive: for pairwise disjoint $B_{n}$ with union $B$ and all $x, y \in H$ , $⟨ F_{B} x, y ⟩ = \sum_{n} ⟨ F_{B_{n}} x, y ⟩$ ;

(c) for every $B$ and every bounded $A \in B (H)$ there is a unique bounded operator $J_{B}^{*} (A)$ with $tr (J_{B} (T) A) = tr (T J_{B}^{*} (A))$ for every trace-class $T$ ; the map $A \mapsto J_{B}^{*} (A)$ is linear and bounded with $J_{B}^{*} (A) \geq 0$ for $A \geq 0$ and $J_{X}^{*} (I) = I$ ;

(d) for a density operator $ρ$ and $B \in Σ$ the number $p (B) = tr (J_{B} (ρ)) = tr (ρ F_{B})$ lies in $[0, 1]$ , and $p$ is countably additive. Whether an experimental procedure realizes such an instrument is a separate model assumption; the conditional update $J_{B} (ρ) / p (B)$ is defined only when $p (B) > 0$ .

No universal representation theorem for arbitrary instruments is asserted.

Facts & Assumptions

Given: A nonzero complex Hilbert space $H$ ; an abstract instrument $J$ as defined; bounded $A$ ; vectors $x, y, u, v$ .

[F1]

$T_{x, y}$ is trace class with $tr (T_{x, y}) = ⟨ x, y ⟩$ , $∥ T_{x, y} ∥_{1} = ∥ x ∥ ∥ y ∥$ ; $T_{x, y}^{*} = T_{y, x}$ ; $tr (S T_{x, y}) = ⟨ S x, y ⟩$ for trace-class $S$ ; and $∣ tr T ∣ \leq ∥ T ∥_{1}$ (Rank-one operator calculus, Nuclear bounded cyclicity for trace-class operators, Trace is absolutely convergent and basis independent).

[F2]

Trace-class operators form a Banach space in trace norm, and the trace is linear and trace-norm continuous; for bounded $B$ , $∥ B T ∥_{1} \leq ∥ B ∥ ∥ T ∥_{1}$ (Trace class is a two sided Banach operator ideal).

[F3]

A bounded linear functional $f$ on $H$ has a unique Riesz representative $y$ with $f (x) = ⟨ x, y ⟩$ (Riesz representation for Hilbert spaces). Applying this to the conjugate of each bounded conjugate-linear section of a sesquilinear form represents that form by a unique bounded operator. If the diagonal values of the form are real, polarization makes it Hermitian, and the representing operator is self-adjoint.

[F4]

Complete positivity implies positivity of $J_{B}$ by the definition in this Statement. A self-adjoint positive trace-class operator has an orthonormal rank-one expansion with nonnegative weights summing to its trace, and its trace is the sum of its positive eigenvalues (Density operators and scalar spectral probability, Trace of a positive operator is the sum of its eigenvalues).

Proof

1.1givenF1F2algebra

[The functional and its sesquilinear form] Fix $B \in Σ$ . The map $f_{B} (T) : = tr (J_{B} (T))$ is a bounded linear functional on $S_{1} (H)$ with $∣ f_{B} (T) ∣ \leq ∥ J_{B} ∥ ∥ T ∥_{1}$ by [F2]. Define $b_{B} (u, v) : = f_{B} (T_{u, v})$ on rank-one operators. Because $T_{u, v}$ is linear in $u$ and conjugate-linear in $v$ and $f_{B}$ is linear, $b_{B}$ is sesquilinear; by [F1], $∣ b_{B} (u, v) ∣ \leq ∥ J_{B} ∥ ∥ u ∥ ∥ v ∥$ .

2.1step 1.1F1F2F3F4algebra

[The effect] For fixed $u$ , Riesz representation of $v \mapsto \overline{b_{B} (u, v)}$ gives a unique vector $F_{B} u$ with $b_{B} (u, v) = ⟨ F_{B} u, v ⟩$ ; the sesquilinear bound in step 1.1 makes $u \mapsto F_{B} u$ a bounded linear operator. By [F4], $J_{B} (T_{u, u})$ is self-adjoint positive trace class and has nonnegative trace, so $b_{B} (u, u) \geq 0$ . The complement relation $J_{X} = J_{B} + J_{X ∖ B}$ and positivity of $J_{X ∖ B}$ give $b_{B} (u, u) \leq b_{X} (u, u) = tr (T_{u, u}) = ∥ u ∥^{2}$ . Thus the diagonal form values are real, so polarization makes $b_{B}$ Hermitian and $F_{B}$ self-adjoint; the two inequalities show it is an effect in the stated sense. Linearity and trace-norm continuity [F1, F2] extend $tr (J_{B} (T_{u, v})) = ⟨ F_{B} u, v ⟩ = tr (T_{u, v} F_{B})$ to every trace-class $T$ . Uniqueness follows because all pairings $⟨ F_{B} u, v ⟩$ are fixed. This is (a).

3.1step 2.1F1F2algebra

[Normalization and countable additivity] For $T = T_{u, v}$ and $B = X$ the trace-preservation clause of the instrument gives $⟨ F_{X} u, v ⟩ = tr (J_{X} (T_{u, v})) = tr (T_{u, v}) = ⟨ u, v ⟩$ , so $F_{X} = I$ . For pairwise disjoint $B_{n}$ with union $B$ , the instrument's countable additivity says $\sum_{n \leq N} J_{B_{n}} (T) \to J_{B} (T)$ in trace norm; by linearity and trace-norm continuity of the functional $tr (\cdot F_{B})$ -pairing [F1, F2], $tr (T_{u, v} F_{B}) = tr (J_{B} (T_{u, v})) = \lim_{N} \sum_{n \leq N} tr (J_{B_{n}} (T_{u, v})) = \sum_{n} ⟨ F_{B_{n}} u, v ⟩$ by step 2.1. As $u, v$ are arbitrary, this is $⟨ F_{B} u, v ⟩ = \sum_{n} ⟨ F_{B_{n}} u, v ⟩$ , which is (b).

4.1step 2.1step 3.1F1F2F3F4algebra

[Heisenberg dual] Fix $B$ and bounded $A$ . Define $c_{A} (u, v) : = tr (J_{B} (T_{u, v}) A)$ . This is sesquilinear in $(u, v)$ with $∣ c_{A} (u, v) ∣ \leq ∥ J_{B} ∥ ∥ A ∥ ∥ u ∥ ∥ v ∥$ by [F1, F2]; by the Riesz construction in [F3] there is a unique bounded $J_{B}^{*} (A)$ with $c_{A} (u, v) = ⟨ J_{B}^{*} (A) u, v ⟩$ and $∥ J_{B}^{*} (A) ∥ \leq ∥ J_{B} ∥ ∥ A ∥$ . Then for rank-one $T_{u, v}$ , $tr (J_{B} (T_{u, v}) A) = ⟨ J_{B}^{*} (A) u, v ⟩ = tr (T_{u, v} J_{B}^{*} (A))$ by [F1]; both sides are linear and trace-norm continuous in $T$ [F2], so the identity holds for all trace-class $T$ . Linearity of $A \mapsto J_{B}^{*} (A)$ follows by uniqueness of the representing operator. If $A \geq 0$ and $x \in H$ , then $⟨ J_{B}^{*} (A) x, x ⟩ = tr (J_{B} (T_{x, x}) A) \geq 0$ : by [F4], $J_{B} (T_{x, x}) = \sum_{n} w_{n} T_{u_{n}, u_{n}}$ with $w_{n} \geq 0$ , and the trace is $\sum_{n} w_{n} ⟨ A u_{n}, u_{n} ⟩$ , a convergent nonnegative sum. Its diagonal values are real, so polarization also gives $J_{B}^{*} (A)^{*} = J_{B}^{*} (A)$ . Finally $J_{X}^{*} (I) = I$ by $tr (J_{X} (T) I) = tr T = tr (T I)$ and uniqueness. This is (c).

5.1step 2.1step 3.1F1F2F4algebra∎

[Probabilities] For a density operator $ρ = \sum_{n} w_{n} T_{u_{n}, u_{n}}$ , step 2.1 gives $p (B) = tr (J_{B} (ρ)) = tr (ρ F_{B}) = \sum_{n} w_{n} ⟨ F_{B} u_{n}, u_{n} ⟩$ . Since $0 \leq F_{B} \leq I$ , each diagonal term lies in $[0, 1]$ and $\sum_{n} w_{n} = 1$ , so $p (B) \in [0, 1]$ . Trace-norm additivity of $J$ and trace continuity imply countable additivity of $p$ , and $p (X) = 1$ by trace preservation. Positivity and normalization of $J_{B} (ρ) / p (B)$ hold precisely when $p (B) > 0$ . Whether the prescribed instrument is experimentally realized remains a separate model assumption.

LemmaStatement: AI-adaptedProof: AI-adaptedOpen item page →

Finite Naimark compression: a finite POVM as a compression of a PVM

Statement

Assume the Axiom of Countable Choice and the Axiom of Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ ), The Axiom of Choice). Let $H$ be a nonzero complex Hilbert space (Hilbert space) and $(F_{1}, \dots, F_{n})$ a finite POVM on $H$ : a finite family of bounded self-adjoint operators $0 \leq F_{a} \leq I$ with $\sum_{a} F_{a} = I$ (Finite effects and instruments: normalization and non-determination of the update). Let $C = C^{n}$ with standard orthonormal basis $(e_{a})$ and form $H \hat{\otimes} C$ (Hilbert tensor construction: inner product, completion and product bases). Then

$W : H \to H \hat{\otimes} C, W ψ : = \sum_{a = 1}^{n} F_{a}^{1 / 2} ψ \otimes e_{a},$

is a linear isometry, the operators $P_{a} : = I_{H} \otimes T_{e_{a}, e_{a}}$ are orthogonal projections with $\sum_{a} P_{a} = I$ (a finite PVM on ${1, \dots, n}$ , The Hilbert orthogonal projection onto a closed subspace), and

$W^{*} P_{a} W = F_{a} (a = 1, \dots, n) .$

The square roots $F_{a}^{1 / 2}$ are the positive square roots supplied by the bounded functional calculus. The construction asserts no physical realization of the ancilla.

Facts & Assumptions

Given: $H$ ; a finite POVM $(F_{a})$ ; the ancilla $C = C^{n}$ with orthonormal basis $(e_{a})$ .

[F1]

$0 \leq F_{a} \leq I$ ; for the bounded self-adjoint $F_{a}$ the continuous functional calculus provides a unique isometric unital star-homomorphism $f \mapsto f (F_{a})$ ; in particular $\sqrt{F_{a}}$ is bounded, self-adjoint, positive, satisfies ${\sqrt{F_{a}}}^{2} = F_{a}$ and commutes with $F_{a}$ (Continuous functional calculus for bounded self adjoint operators, Borel functional calculus for a bounded normal operator, Finite effects and instruments: normalization and non-determination of the update).

[F2]

The completed tensor product has the inner product $⟨ u \otimes c, u^{'} \otimes c^{'} ⟩ = ⟨ u, u^{'} ⟩ ⟨ c, c^{'} ⟩$ , the elementary tensors span a dense subspace, and bounded operators extend as $S \otimes T$ with adjoint $S^{*} \otimes T^{*}$ (Hilbert tensor construction: inner product, completion and product bases, Bounded tensor maps: extension, norm estimate, adjoint and commutation).

[F3]

$T_{e_{a}, e_{a}}$ is the orthogonal projection of $C$ onto $C e_{a}$ ; $T_{e_{a}, e_{a}} T_{e_{b}, e_{b}} = δ_{a b} T_{e_{a}, e_{a}}$ and $\sum_{a} T_{e_{a}, e_{a}} = I_{C}$ (Finite effects and instruments: normalization and non-determination of the update, The Hilbert orthogonal projection onto a closed subspace, Hilbert projections are linear, self-adjoint and contractive).

Proof

1.1givenF1F2F3algebra

[The isometry] The map $ψ \mapsto \sum_{a} F_{a}^{1 / 2} ψ \otimes e_{a}$ is well defined and linear; for finite sums of elementary tensors the inner product is computed by [F2], and for $ψ \in H$ , $∥ W ψ ∥^{2} = \sum_{a, b} ⟨ F_{a}^{1 / 2} ψ, F_{b}^{1 / 2} ψ ⟩ ⟨ e_{a}, e_{b} ⟩ = \sum_{a} ∥ F_{a}^{1 / 2} ψ ∥^{2} = \sum_{a} ⟨ F_{a} ψ, ψ ⟩ = ⟨ ψ, ψ ⟩$ by [F1] and [F3]. Hence $W$ is an isometry.

2.1step 1.1F2F3algebra

[The projections] The operators $P_{a} = I_{H} \otimes T_{e_{a}, e_{a}}$ are orthogonal projections by [F2] and [F3]; distinct ones are orthogonal and $\sum_{a} P_{a} = I_{H} \otimes I_{C} = I$ by [F3].

3.1step 1.1F1F2algebra∎

[Compression] Compute the adjoint directly: for $ξ \in H$ , vectors $ξ_{a} \in H$ and scalars $z_{a}$ , $⟨ W ξ, \sum_{a} z_{a} ξ_{a} \otimes e_{a} ⟩ = \sum_{a} \overline{z_{a}} ⟨ F_{a}^{1 / 2} ξ, ξ_{a} ⟩ = \sum_{a} \overline{z_{a}} ⟨ ξ, F_{a}^{1 / 2} ξ_{a} ⟩$ by [F2] and [F1], so $W^{*} (\sum_{a} z_{a} ξ_{a} \otimes e_{a}) = \sum_{a} z_{a} F_{a}^{1 / 2} ξ_{a}$ ; in particular $W^{*} (ξ \otimes e_{a}) = F_{a}^{1 / 2} ξ$ . Then $P_{a} W ψ = F_{a}^{1 / 2} ψ \otimes e_{a}$ and $W^{*} P_{a} W ψ = F_{a}^{1 / 2} F_{a}^{1 / 2} ψ = F_{a} ψ$ by [F1].

LemmaStatement: AI-adaptedProof: AI-adaptedOpen item page →

Countable Kraus isometric dilation and partial-trace recovery

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $H$ be a nonzero complex Hilbert space (Hilbert space) and let $(M_{a, r})_{a, r \geq 1}$ be bounded operators on $H$ with

$\sum_{a \leq A, r \leq R} M_{a, r}^{*} M_{a, r} ⟶ I strongly as A, R \to \infty .$

Let $(e_{a, r})_{a, r \geq 1}$ be the standard orthonormal basis of $ℓ^{2}$ and form $H \hat{\otimes} ℓ^{2}$ (Hilbert tensor construction: inner product, completion and product bases). Then

$W : H \to H \hat{\otimes} ℓ^{2}, W ψ : = \sum_{a, r \geq 1} M_{a, r} ψ \otimes e_{a, r}$

is a linear isometry; for each outcome label $a$ the strong limit $P_{a} : = \sum_{r \geq 1} I_{H} \otimes T_{e_{a, r}, e_{a, r}}$ is an orthogonal projection (The Hilbert orthogonal projection onto a closed subspace), the family $(P_{a})_{a \geq 1}$ consists of pairwise orthogonal projections with $\sum_{a} P_{a} = I$ strongly, and

$W^{*} P_{a} W = \sum_{r \geq 1} M_{a, r}^{*} M_{a, r} = : F_{a};$

moreover the partial trace of the dilated state over $ℓ^{2}$ recovers the Kraus instrument: for every trace-class $ρ$ on $H$ ,

${Tr}_{ℓ^{2}} (W ρ W^{*}) = \sum_{a, r \geq 1} M_{a, r} ρ M_{a, r}^{*},$

the series converging in trace norm and the right side being the map of Countable Kraus normalization and trace-norm convergence. This realizes the specified family; it is not a classification of all channels.

Facts & Assumptions

Given: $H$ ; the family $(M_{a, r})$ with the stated strong completeness; the orthonormal basis $(e_{a, r})$ of $ℓ^{2}$ .

[F1]

For $ψ \in H$ , the nonnegative double series $\sum_{a, r} ∥ M_{a, r} ψ ∥^{2}$ has value $∥ ψ ∥^{2}$ , and $∥ W ψ ∥^{2} = \sum_{a, r} ∥ M_{a, r} ψ ∥^{2}$ whenever the partial sums converge (Countable Kraus normalization and trace-norm convergence, Hilbert tensor construction: inner product, completion and product bases, Parseval equivalences for an orthonormal family).

[F2]

The operators $T_{e_{a, r}, e_{a, r}}$ are orthogonal projections onto $C e_{a, r}$ , pairwise orthogonal, with $\sum_{a, r} T_{e_{a, r}, e_{a, r}} = I$ strongly; a strong sum of pairwise orthogonal projections is the orthogonal projection onto the closed span of their ranges, and telescoping strong sums behave additively (Weak and strong additivity of orthogonal projections, The Hilbert orthogonal projection onto a closed subspace, Finite effects and instruments: normalization and non-determination of the update).

[F3]

${Tr}_{ℓ^{2}}$ is the trace-norm continuous linear map with defining identity $tr (({Tr}_{ℓ^{2}} T) A) = tr (T (A \otimes I))$ for bounded $A$ , and on rank-one operators ${Tr}_{ℓ^{2}} T_{u \otimes e, u^{'} \otimes e^{'}} = ⟨ e, e^{'} ⟩ T_{u, u^{'}}$ (Separable partial trace and its basic properties).

[F4]

$tr (T_{x, z}) = ⟨ x, z ⟩$ , and for bounded $B$ , $T_{x, z} B = T_{x, B^{*} z}$ (Rank-one operator calculus, Nuclear bounded cyclicity for trace-class operators).

Proof

1.1givenF1algebra

[The dilation is an isometry] The series defining $W ψ$ converges: the partial sums $\sum_{a \leq A, r \leq R} M_{a, r} ψ \otimes e_{a, r}$ are Cauchy because for $A^{'} \geq A$ , $R^{'} \geq R$ the squared norm of the difference is the corresponding nonnegative subsum, which is bounded by the tail of the summable series $\sum_{a, r} ∥ M_{a, r} ψ ∥^{2}$ ; by [F1] the limit $W ψ$ has $∥ W ψ ∥^{2} = \sum_{a, r} ∥ M_{a, r} ψ ∥^{2} = ∥ ψ ∥^{2}$ , so $W$ is linear and isometric.

2.1step 1.1F2algebra

[Outcome projections] By [F2] each $P_{a} = \sum_{r} I \otimes T_{e_{a, r}, e_{a, r}}$ is the orthogonal projection onto $H \hat{\otimes} \overset{}{span} {e_{a, r} : r \geq 1}$ , the strong sums over $a$ telescope to $I$ , and $P_{a} P_{b} = δ_{a b} P_{a}$ .

3.1step 2.1F1algebra

[Compression] As in the finite case, $W^{*} (ξ \otimes e_{a, r}) = M_{a, r}^{*} ξ$ because $W$ is an isometry with the displayed action, so $W^{*} P_{a} W ψ = \sum_{r} M_{a, r}^{*} M_{a, r} ψ$ for every $ψ$ , the strong limit defining $F_{a}$ .

4.1step 1.1step 3.1F1F3F4algebra∎

[Partial-trace recovery by truncations] For $x, y \in H$ and a finite rectangle $D$ of index pairs put $w_{D} : = \sum_{(a, r) \in D} M_{a, r} x \otimes e_{a, r}$ and $z_{D} : = \sum_{(a, r) \in D} M_{a, r} y \otimes e_{a, r}$ . Then $w_{D} \to W x$ , $z_{D} \to W y$ by step 1.1, and rank-one trace-norm continuity gives $T_{w_{D}, z_{D}} \to T_{W x, W y} = W T_{x, y} W^{*}$ . For each finite rectangle the elementary formula [F3] gives ${Tr}_{ℓ^{2}} T_{w_{D}, z_{D}} = \sum_{(a, r) \in D} T_{M_{a, r} x, M_{a, r} y}$ ; Cauchy–Schwarz and [F1] give $\sum_{a, r} ∥ M_{a, r} x ∥ ∥ M_{a, r} y ∥ \leq ∥ x ∥ ∥ y ∥$ , so these diagonal terms converge absolutely in trace norm. Taking limits using [F3] proves the Kraus formula on rank-one inputs. Both sides are bounded linear maps on trace class: the left side by the two-sided ideal bound for $W, W^{*}$ and [F3], and the right side by nuclear expansion and the preceding rank-one bound. Finite rank-one sums are trace-norm dense by the nuclear characterization in Nuclear bounded cyclicity for trace-class operators, so the identity extends to every trace-class input and equals the map of Countable Kraus normalization and trace-norm convergence.

LemmaStatement: AI-adaptedProof: AI-adaptedOpen item page →

Continuous measurable Kraus instruments: isometry, POVM and trace-norm integration

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $(X, Σ, μ)$ be a $σ$ -finite measure space, $H$ a nonzero separable complex Hilbert space, and $(M_{r})_{r \geq 1}$ bounded operator-valued maps on $X$ such that:

(i) for every $ψ \in H$ the map $x \mapsto M_{r} (x) ψ$ is strongly measurable for each $r$ (Strongly measurable Banach-valued function) and

$\int_{X} \sum_{r \geq 1} ∥ M_{r} (x) ψ ∥^{2} d μ (x) = ∥ ψ ∥^{2} (ψ \in H) .$

Here $E : = ℓ^{2} (H)$ is the space of sequences $z = (z_{r})_{r \geq 1}$ with $\sum_{r} ∥ z_{r} ∥^{2} < \infty$ , with inner product $⟨ z, w ⟩_{E} : = \sum_{r} ⟨ z_{r}, w_{r} ⟩_{H}$ . Define $L^{2} (X, μ; E)$ as the vector space of strongly measurable fields $f : X \to E$ with $\int_{X} ∥ f (x) ∥_{E}^{2} d μ < \infty$ , modulo equality almost everywhere; its inner product is $⟨ [f], [g] ⟩ : = \int_{X} ⟨ f (x), g (x) ⟩_{E} d μ$ . The proof establishes that both spaces are Hilbert spaces before using adjoints or PVMs on this target.

Then:

(a) the formula $W ψ (x) : = (M_{r} (x) ψ)_{r}$ defines a linear isometry $W : H \to L^{2} (X, μ; ℓ^{2} (H))$ ;

(b) for $B \in Σ$ the multiplication operator $P (B)$ by $1_{B}$ is a projection-valued measure on $(X, Σ)$ , and $F (B) : = W^{*} P (B) W$ is a normalized POVM. Here a POVM is a weakly countably additive map $F : Σ \to B (H)$ whose values are self-adjoint positive contractions, with $F (\emptyset) = 0$ and $F (X) = I$ ; positivity is in the sense of Self-adjoint, positive, unitary and normal operators;

(c) for every self-adjoint positive trace-class $ρ = \sum_{n} w_{n} T_{u_{n}, u_{n}}$ (Self-adjoint, positive, unitary and normal operators, Trace class operator, Density operators and scalar spectral probability) the Bochner integral

$I_{B} (ρ) : = \sum_{n \geq 1} w_{n} \int_{B} \sum_{r \geq 1} T_{M_{r} (x) u_{n}, M_{r} (x) u_{n}} d μ (x)$

converges in trace norm and defines a positive trace-class operator with $tr I_{B} (ρ) = tr (ρ F (B))$ ; the map $B \mapsto I_{B} (ρ)$ is countably additive in trace norm and $I_{X}$ is trace preserving, so $I$ is an abstract instrument in the sense of Abstract trace-class instruments: effects and Heisenberg duality; moreover $tr (I_{B} (ρ) A) = \sum_{r} \int_{B} tr (M_{r} (x) ρ M_{r} (x)^{*} A) d μ (x)$ for bounded $A$ , and finite-dimensional ancilla amplifications preserve positivity;

(d) the conditional update at an event $B$ with $p (B) = tr (ρ F (B)) > 0$ is $I_{B} (ρ) / p (B)$ . The normalization is integrated over events; no pointwise identity $\sum_{r} M_{r} (x)^{*} M_{r} (x) = I$ is assumed. Universal normal-instrument representation is not claimed.

Facts & Assumptions

Given: $(X, Σ, μ)$ , $H$ , $(M_{r})$ with the normalization identity; a self-adjoint positive trace-class $ρ = \sum_{n} w_{n} T_{u_{n}, u_{n}}$ .

[F1]

$∥ (M_{r} (x) ψ)_{r} ∥_{ℓ^{2} (H)}^{2} = \sum_{r} ∥ M_{r} (x) ψ ∥^{2}$ ; the normalization hypothesis says the Bochner square norm is $∥ ψ ∥^{2}$ , so the resulting equivalence class belongs to $L^{2}$ and has norm $∥ ψ ∥$ (Bochner-integrable function, The space $L^{p} (μ)$ as the quotient by null functions, Hilbert tensor construction: inner product, completion and product bases).

[F2]

For $T \in S_{1} (H)$ the rank-one map $v \mapsto T_{v, v}$ is trace-norm continuous with $∥ T_{v, v} ∥_{1} = ∥ v ∥^{2}$ ; $T_{x, z} B = T_{x, B^{*} z}$ and $∥ A T B ∥_{1} \leq ∥ A ∥ ∥ T ∥_{1} ∥ B ∥$ (Rank-one operator calculus, Trace class is a two sided Banach operator ideal, Nuclear bounded cyclicity for trace-class operators).

[F3]

A strongly measurable Banach-valued function is Bochner integrable exactly when the integral of its norm is finite; the Bochner integral is norm-continuous and dominated convergence holds under an integrable majorant (Bochner integrability criterion, Bochner integral norm inequality, Bochner dominated convergence theorem).

[F4]

Tonelli's theorem permits interchanging the nonnegative sum over $r$ and the integral over $X$ ; the trace pairing satisfies $tr (T_{x, z}) = ⟨ x, z ⟩$ and the polarisation/positivity properties of the previous items (Tonelli's theorem for nonnegative measurable functions on a sigma-finite product, Rank-one operator calculus, Separable partial trace and its basic properties).

[F5]

Nuclear representations converge in trace norm as well as operator norm; finite rank-one sums are trace-norm dense, and $tr (T_{u, v} A) = ⟨ A u, v ⟩$ for bounded $A$ . Cauchy–Schwarz applies both to the countable coordinate sum and to the scalar integral (Nuclear series characterizes trace norm, Trace is absolutely convergent and basis independent, Rank-one operator calculus, Cauchy–Schwarz: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ , with equality exactly for dependent pairs).

[F6]

Monotone convergence applies to nonnegative scalar functions; scalar $L^{2}$ Cauchy–Schwarz and Minkowski give the integral product and sum bounds. Equality almost everywhere means equality outside a measurable null set (Monotone convergence for the integral, Cauchy-Schwarz inequality for $L^{2}$ , Minkowski's inequality for integrals, including $p = \infty$ , Measure-null sets and almost-everywhere statements relative to a measure).

Proof

1.1givenF1F3F6chooseconstructalgebra

[Construction and completeness of the Hilbert target] Coordinate Cauchy–Schwarz makes the inner product of $E = ℓ^{2} (H)$ absolutely convergent and gives its squared norm $\sum_{r} ∥ z_{r} ∥^{2}$ . If $(z^{(n)})$ is Cauchy in this norm, each coordinate converges in the complete space $H$ to $z_{r}$ ; taking limits of finite squared sums shows $\sum_{r} ∥ z_{r} ∥^{2} < \infty$ and $∥ z^{(n)} - z ∥_{E} \to 0$ . Thus $E$ is Hilbert. Strongly measurable $E$ -valued fields have measurable norm and inner-product functions by their simple approximants; scalar Cauchy–Schwarz [F6] gives absolute integrability of the target inner product. This integral descends to the almost-everywhere quotient, is positive definite there, and has the stated norm; scalar Minkowski gives its triangle inequality. For completeness, take a Cauchy sequence of classes and a subsequence with representatives $f_{k}$ satisfying $∥ f_{k + 1} - f_{k} ∥_{2} \leq 2^{- k}$ ; Countable Choice permits the countable representative selection. Set $d_{k} = f_{k + 1} - f_{k}$ . For $g_{N} : = \sum_{k \leq N} ∥ d_{k} (\cdot) ∥_{E}$ , scalar Minkowski gives $∥ g_{N} ∥_{2} \leq \sum_{k \leq N} 2^{- k} \leq 1$ . Monotone convergence [F6] applied to $g_{N}^{2}$ gives $g : = \sum_{k} ∥ d_{k} ∥_{E} \in L^{2}$ and $g < \infty$ almost everywhere. Hence $f_{1} + \sum_{k} d_{k}$ converges pointwise off a measurable null set in complete $E$ ; define its limit as zero on that set. The limit is strongly measurable: the countably many simple approximant ranges for all $f_{k}$ have a separable closed span, norm distances to a countable dense subset are measurable, and choosing the nearest of the first $N$ dense points gives finite-valued simple approximants converging pointwise. Scalar Minkowski and monotone convergence also give $∥ f - f_{k} ∥_{2} \leq \sum_{j \geq k} 2^{- j} \to 0$ , and $f \in L^{2}$ since $∥ f ∥_{E} \leq ∥ f_{1} ∥_{E} + g$ . The original Cauchy sequence converges to the same class by the triangle inequality. Therefore the defined target is a Hilbert space; the construction uses no assumption that $μ (X)$ is finite.

2.1step 1.1givenF1F3F6constructalgebra

[The $L^{2}$ isometry] Fix $ψ \in H$ . The measurable nonnegative function $g_{ψ} (x) : = \sum_{r} ∥ M_{r} (x) ψ ∥^{2}$ has finite integral $∥ ψ ∥^{2}$ , so it is finite outside a null set. On that set the finite coordinate truncations converge in $ℓ^{2} (H)$ ; define the vector field to be zero on the exceptional null set. The truncations are strongly measurable and their pointwise limit is strongly measurable, as follows directly by approximation with simple functions (or by separability of $ℓ^{2} (H)$ and measurable coordinates). The field belongs to $L^{2} (X; ℓ^{2} (H))$ because its squared norm is $g_{ψ}$ almost everywhere. Its equivalence class defines $W ψ$ , independently of the null-set representative. For any finite collection of vectors, the coordinate linear identities hold outside the union of their exceptional sets, so $W$ is linear as a map of equivalence classes; normalization gives $∥ W ψ ∥ = ∥ ψ ∥$ , hence $W$ is a linear isometry. No $L^{1}$ integrability of this raw vector field is needed.

3.1step 2.1F2F3F4F5algebra

[The PVM and the effects] Multiplication by $1_{B}$ on $L^{2}$ is a bounded self-adjoint projection, with $P (X) = I$ and $P (B \cap C) = P (B) P (C)$ . For disjoint $B_{j}$ with union $B$ , put $U_{N} = ⋃_{j \leq N} B_{j}$ . The squared $L^{2}$ norm of $(P (B) - \sum_{j \leq N} P (B_{j})) f$ is $\int_{B ∖ U_{N}} ∥ f (x) ∥^{2} d μ$ ; this tends to zero because $1_{U_{N}} ∥ f ∥^{2} ↑ 1_{B} ∥ f ∥^{2}$ and monotone convergence applies. Thus $P$ is a PVM. Compression gives self-adjoint positive contractions $F (B) = W^{*} P (B) W$ , normalization $F (X) = I$ , and weak countable additivity. Equivalently, for every $u, v$ , $⟨ F (B) u, v ⟩ = \int_{B} \sum_{r} ⟨ M_{r} (x) u, M_{r} (x) v ⟩ d μ$ , with absolute convergence by Cauchy–Schwarz and normalization. This is the normalized POVM asserted in (b).

4.1step 2.1step 3.1F2F3F4F5constructalgebra

[Integrable cross fields] For $u, v \in H$ , each $x \mapsto T_{M_{r} (x) u, M_{r} (x) v}$ is strongly measurable in trace norm by [F2]. By Cauchy–Schwarz twice and normalization, $\int_{X} \sum_{r} ∥ M_{r} (x) u ∥ ∥ M_{r} (x) v ∥ d μ \leq ∥ u ∥ ∥ v ∥ .$ Thus, outside a null set, the rank-one series converges absolutely in trace norm. Its sum $K_{u, v} (x)$ is strongly measurable and Bochner integrable by [F3], with integrated norm at most $∥ u ∥ ∥ v ∥$ . Define $E_{B} (u, v) : = \int_{B} K_{u, v} (x) d μ$ . This is sesquilinear, satisfies $∥ E_{B} (u, v) ∥_{1} \leq ∥ u ∥ ∥ v ∥$ , and has pairing $tr (E_{B} (u, v) A) = \sum_{r} \int_{B} ⟨ A M_{r} (x) u, M_{r} (x) v ⟩ d μ$ for bounded $A$ , by the boundedness of trace pairings and absolute integrability. In particular $E_{B} (u, u)$ is positive and $tr E_{B} (u, u) = ⟨ F (B) u, u ⟩$ .

5.1step 3.1step 4.1F2F3F4F5constructalgebra

[The bounded linear instrument] For a nuclear representation $T = \sum_{j} T_{u_{j}, v_{j}}$ , define $I_{B} (T) : = \sum_{j} E_{B} (u_{j}, v_{j})$ , converging in trace norm. The bound of step 4.1 gives $∥ I_{B} (T) ∥_{1} \leq \sum_{j} ∥ u_{j} ∥ ∥ v_{j} ∥$ . This is independent of representation: for a nuclear representation of zero, Tonelli and step 4.1 allow exchanging its $j$ sum, the $r$ sum and each bounded- $A$ trace pairing with the integral. For each fixed $r, x$ , the two-sided ideal bound makes $\sum_{j} M_{r} (x) T_{u_{j}, v_{j}} M_{r} (x)^{*} = M_{r} (x) 0 M_{r} (x)^{*} = 0$ in trace norm. Therefore all the resulting pairings vanish; rank-one test operators and [F5] show the image is zero. The difference of two representations is such a zero representation, proving independence. Hence $I_{B}$ is linear, bounded with norm at most one by taking the nuclear infimum, and positive by positive spectral rank-one expansion. For positive $ρ = \sum_{n} w_{n} T_{u_{n}, u_{n}}$ its definition is exactly the Bochner formula in (c), and absolute integrability gives the stated bounded- $A$ pairing. Taking $A = I$ gives $tr I_{B} (T) = tr (T F (B))$ first for rank-one inputs by step 3.1 and then for all inputs by nuclear continuity; in particular $I_{X}$ is trace preserving.

6.1step 4.1step 5.1F2F3F4F5algebra

[Countable additivity and complete positivity] For disjoint $B_{j}$ with union $B$ , the fields $1_{U_{N}} K_{u, v}$ , where $U_{N} = ⋃_{j \leq N} B_{j}$ , converge pointwise almost everywhere to $1_{B} K_{u, v}$ and are dominated in norm by the integrable function $∥ K_{u, v} ∥_{1}$ ; Bochner dominated convergence [F3] gives $E_{B} (u, v) = \sum_{j} E_{B_{j}} (u, v)$ in trace norm. For a nuclear input $T = \sum_{k} T_{u_{k}, v_{k}}$ , [F2] and step 4.1 give In fact, $\sum_{j} \sum_{k} ∥ E_{B_{j}} (u_{k}, v_{k}) ∥_{1} \leq \sum_{k} \int_{B} \sum_{r} ∥ M_{r} (x) u_{k} ∥ ∥ M_{r} (x) v_{k} ∥ d μ \leq \sum_{k} ∥ u_{k} ∥ ∥ v_{k} ∥ < \infty$ . Thus the two sums may be interchanged in trace norm, and the omitted event tails tend to zero; this proves countable additivity of $I_{B} (T)$ in trace norm. For each finite ancilla $C^{n}$ , act on $C^{n} \hat{\otimes} H$ with the pointwise Kraus maps $I \otimes M_{r} (x)$ . For a vector $ξ = \sum_{k = 1}^{n} e_{k} \otimes ξ_{k}$ , the integrated squared output norm is $\sum_{k} \int_{X} \sum_{r} ∥ M_{r} (x) ξ_{k} ∥^{2} d μ = ∥ ξ ∥^{2}$ . The same construction as steps 4.1–5.1 therefore maps each self-adjoint positive rank-one $T_{ξ, ξ}$ to an integral of positive rank-one operators and extends positively to every self-adjoint positive trace-class amplified input. Expanding its finite ancilla matrix entries shows this extension equals the finite amplification of $I_{B}$ . Thus every $I_{B}$ is completely positive, and together with additivity and trace preservation this establishes the full abstract instrument, not merely its restriction to positive inputs.

7.1step 5.1step 6.1F2algebra∎

[Conditional updates] For a density operator $ρ$ , $I_{B} (ρ) \geq 0$ and $p (B) = tr I_{B} (ρ) = tr (ρ F (B))$ by step 5.1. If $p (B) > 0$ , division gives a positive operator of trace one and hence the claimed density operator; if $p (B) = 0$ no normalized update is defined. Nothing here claims a normalized singleton density or a universal representation theorem.

LemmaStatement: AI-adaptedProof: AI-adaptedOpen item page →

Marker overlap contraction of the two-path cross term

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $(D, A, μ)$ be a $σ$ -finite measure space, let $L^{2} (D, μ)$ denote the quotient of square-integrable measurable functions modulo almost-everywhere equality (The space $L^{p} (μ)$ as the quotient by null functions), let $K$ be a complex Hilbert space (Hilbert space) with unit vectors $e_{1}, e_{2}$ , let $a, b \in L^{2} (D, μ)$ be unit vectors, and let $α, β \in C$ with $∣ α ∣^{2} + ∣ β ∣^{2} = 1$ . Choose measurable representatives of $a, b$ ; all displayed pointwise expressions are understood almost everywhere, and the integrals below are independent of the representatives. Put

$Ψ : = α a \otimes e_{1} + β b \otimes e_{2} \in L^{2} (D) \hat{\otimes} K .$

Then:

(a) the reduced positive trace-class weight $ρ : = {Tr}_{K} T_{Ψ, Ψ}$ on $L^{2} (D)$ is

$ρ = ∣ α ∣^{2} T_{a, a} + ∣ β ∣^{2} T_{b, b} + α \overline{β} ⟨ e_{1}, e_{2} ⟩ T_{a, b} + \overline{α} β ⟨ e_{2}, e_{1} ⟩ T_{b, a};$

(b) for every measurable $B \subseteq D$ with finite measure, writing $P_{B}$ for the multiplication operator by $1_{B}$ ,

$tr (ρ P_{B}) = \int_{B} (∣ α ∣^{2} ∣ a ∣^{2} + ∣ β ∣^{2} ∣ b ∣^{2} + 2 Re (α \overline{β} ⟨ e_{1}, e_{2} ⟩ a (x) \overline{b (x)})) d μ;$

(c) the marker overlap contracts the cross term: $∣ ⟨ e_{1}, e_{2} ⟩ ∣ \leq 1$ , and if $e_{1} ⊥ e_{2}$ the cross term in (b) vanishes, leaving the incoherent weight $∣ α ∣^{2} ∣ a ∣^{2} + ∣ β ∣^{2} ∣ b ∣^{2}$ .

The weight satisfies $tr ρ = ∥ Ψ ∥^{2} = 1 + 2 Re (α \overline{β} ⟨ a, b ⟩ ⟨ e_{1}, e_{2} ⟩)$ ; it is a density operator precisely when $∥ Ψ ∥ = 1$ . The stated assumptions do not force this normalization. These are identities for the specified vector and its unnormalized weight; they assert no measured detector law and cover no selective conditioning.

Facts & Assumptions

Given: $D, K, e_{1}, e_{2}, a, b, α, β, Ψ$ as in the Statement; a measurable $B \subseteq D$ with finite measure.

[F0]

Under the assumed Countable Choice, complex $L^{2} (D, μ)$ is complete for its $L^{2}$ norm and has the first-variable-linear inner product $⟨ f, g ⟩ = \int_{D} f \overline{g} d μ$ inducing that norm (Complex completeness, density, and inner product: the consumer interface). Thus it is a complex Hilbert space. The given unit vectors $a, b$ make it nonzero; the unit $e_{1}$ makes $K$ nonzero, so both tensor and partial-trace suppliers apply.

[F1]

The completed tensor product $L^{2} (D) \hat{\otimes} K$ carries the inner product $⟨ f \otimes u, f^{'} \otimes u^{'} ⟩ = ⟨ f, f^{'} ⟩ ⟨ u, u^{'} ⟩$ and $∥ f \otimes u ∥ = ∥ f ∥ ∥ u ∥$ (Hilbert tensor construction: inner product, completion and product bases).

[F2]

${Tr}_{K}$ is linear with ${Tr}_{K} T_{f \otimes u, f^{'} \otimes u^{'}} = ⟨ u, u^{'} ⟩ T_{f, f^{'}}$ ; $T_{f, f^{'}} g = ⟨ g, f^{'} ⟩ f$ ; and $tr (T_{f, f^{'}} P_{B}) = ⟨ f, P_{B} f^{'} ⟩ = \int_{B} f \overline{f^{'}} d μ$ for the self-adjoint multiplication operator $P_{B}$ (Separable partial trace and its basic properties, Nuclear bounded cyclicity for trace-class operators).

[F3]

$∣ ⟨ e_{1}, e_{2} ⟩ ∣ \leq 1$ (Cauchy–Schwarz: $∣ ⟨ u, v ⟩ ∣ \leq ∥ u ∥ ∥ v ∥$ , with equality exactly for linearly dependent vectors); the functions $a \overline{b}$ , $∣ a ∣^{2}$ , $∣ b ∣^{2}$ are integrable on $B$ of finite measure by Cauchy-Schwarz (Cauchy-Schwarz inequality for $L^{2}$ ).

Proof

1.1givenF0F1F2algebra

[The reduced state] Expanding $T_{Ψ, Ψ}$ into its four rank-one summands and applying the rank-one rule [F2] termwise gives ${Tr}_{K} T_{Ψ, Ψ} = ∣ α ∣^{2} T_{a, a} + ∣ β ∣^{2} T_{b, b} + α \overline{β} ⟨ e_{1}, e_{2} ⟩ T_{a, b} + \overline{α} β ⟨ e_{2}, e_{1} ⟩ T_{b, a}$ , which is (a). The trace-preserving partial trace [F2] gives $tr ρ = ∥ Ψ ∥^{2}$ ; expanding the tensor norm by [F1] gives the displayed normalization formula. Positivity follows from positivity of $T_{Ψ, Ψ}$ and the positive partial trace [F2], so trace one is exactly the density condition.

2.1step 1.1F2F3algebra

[The bin weight] By [F2], $tr (ρ P_{B}) = ∣ α ∣^{2} \int_{B} ∣ a ∣^{2} + ∣ β ∣^{2} \int_{B} ∣ b ∣^{2} + α \overline{β} ⟨ e_{1}, e_{2} ⟩ \int_{B} a \overline{b} + \overline{α} β ⟨ e_{2}, e_{1} ⟩ \int_{B} b \overline{a}$ ; since the last two integrals are conjugates of one another (because $⟨ e_{2}, e_{1} ⟩ = \overline{⟨ e_{1}, e_{2} ⟩}$ and $\int_{B} b \overline{a} = \overline{\int_{B} a \overline{b}}$ ), their sum is $2 Re (α \overline{β} ⟨ e_{1}, e_{2} ⟩ \int_{B} a \overline{b})$ , giving (b).

3.1step 2.1F3algebra∎

[The overlap bound] By [F3], $∣ ⟨ e_{1}, e_{2} ⟩ ∣ \leq 1$ . Thus the modulus of the complex cross-term coefficient $α \overline{β} ⟨ e_{1}, e_{2} ⟩$ is at most $∣ α β ∣$ , and the real cross-term contribution at $x$ is bounded in magnitude by $2 ∣ α β ∣ ∣ a (x) b (x) ∣$ . A complex phase may rotate that contribution, so no comparison of its real part with the real part for overlap one is asserted. When $e_{1} ⊥ e_{2}$ , the overlap is zero and the cross term vanishes, leaving exactly the diagonal sum, as claimed in (c).

LemmaStatement: AI-adaptedProof: AI-adaptedOpen item page →

Exact free Gaussian propagation

Statement

Assume the Axiom of Countable Choice and the Axiom of Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ ), The Axiom of Choice). Fix $m > 0$ , $ℏ > 0$ , $σ > 0$ and real $x_{0}, p_{0}$ , and let

$ψ_{0} (x) : = (2 π σ^{2})^{- 1 / 4} \exp (- \frac{(x - x_{0})^{2}}{4 σ^{2}}) \exp (i \frac{p_{0} (x - x_{0})}{ℏ}), x \in R .$

Let $H_{0}$ be the free operator of clause (b) of Real multiplication operators and free plus bounded-potential Schrödinger operators with $c : = 2 π^{2} ℏ^{2} / m$ and $b : = 0$ , i.e.

$H_{0} : = F_{2}^{*} M_{2 π^{2} ℏ^{2} ξ^{2} / m} F_{2}, D (H_{0}) = {f \in L^{2} (R) : ξ^{2} F_{2} f \in L^{2} (R)},$

which is self-adjoint with $D (H_{0}) = W^{2, 2} (R; C)$ and $H_{0} f = - \frac{ℏ^{2}}{2 m} Δ f$ for $f \in S (R)$ . Let $U$ be the momentum transform of Momentum Fourier Jacobian: the unitary momentum representation, the surjective isometry $L^{2} (R, d x) \to L^{2} (R, d p)$ with $U = D_{ℏ} \circ F_{2}$ on $S (R)$ . For $t \in R$ put $τ : = \frac{ℏ t}{2 m σ^{2}}$ and

$ψ_{t} (x) : = (2 π σ^{2})^{- 1 / 4} (1 + i τ)^{- 1 / 2} \exp (- \frac{(x - x_{0} - p_{0} t / m)^{2}}{4 σ^{2} (1 + i τ)} + i \frac{p_{0} (x - x_{0})}{ℏ} - i \frac{p_{0}^{2} t}{2 m ℏ}),$

where $z^{- 1 / 2} : = \exp (- \frac{1}{2} Log z)$ is the principal branch of Complex logarithms, the principal logarithm, and principal and multivalued complex powers, legitimate at $z = 1 + i τ$ because $Re (1 + i τ) = 1 > 0$ . Let $Q f (x) = x f (x)$ and $P f (x) = - i ℏ f^{'} (x)$ act on $S (R)$ . Then:

(a) $ψ_{0} \in S (R)$ , $∥ ψ_{0} ∥_{2} = 1$ , and for every $t$ one has $ψ_{t} \in S (R) \subseteq D (H_{0})$ ; the map $t \mapsto ψ_{t}$ is continuously differentiable $R \to L^{2} (R)$ with $ψ_{t} \in D (H_{0})$ and $\frac{d}{d t} ψ_{t} = - \frac{i}{ℏ} H_{0} ψ_{t}$ , and consequently $ψ_{t} = e^{- i t H_{0} / ℏ} ψ_{0}$ , where $e^{- i t H_{0} / ℏ} : = e^{i t T}$ with $T : = - H_{0} / ℏ$ denotes the unitary group generated by the self-adjoint operator $T$ (A self-adjoint operator generates a strongly continuous unitary group).

(b) Position density. $∣ ψ_{t} (x) ∣^{2} = (2 π σ^{2} (1 + τ^{2}))^{- 1 / 2} \exp (- (x - x_{0} - p_{0} t / m)^{2} / (2 σ^{2} (1 + τ^{2})))$ ; it is a probability density of mean $x_{0} + p_{0} t / m$ and variance $σ^{2} (1 + τ^{2})$ .

(c) Momentum density. $∣ U ψ_{t} (p) ∣^{2} = \sqrt{2 / π} (σ / ℏ) \exp (- 2 σ^{2} (p - p_{0})^{2} / ℏ^{2}) = ∣ U ψ_{0} (p) ∣^{2}$ for almost every $p$ ; it is a probability density of mean $p_{0}$ and variance $ℏ^{2} / (4 σ^{2})$ , independent of $t$ .

(d) Standard deviations. With $Δ Q : = ∥ (Q - ⟨ Q ψ_{t}, ψ_{t} ⟩) ψ_{t} ∥_{2}$ and $Δ P : = ∥ (P - ⟨ P ψ_{t}, ψ_{t} ⟩) ψ_{t} ∥_{2}$ one has $Δ Q = σ \sqrt{1 + τ^{2}}$ , $Δ P = ℏ / (2 σ)$ and

$Δ Q Δ P = \frac{ℏ}{2} \sqrt{1 + τ^{2}} \geq \frac{ℏ}{2},$

with equality exactly at $t = 0$ ; the lower bound $ℏ / 2$ is the one supplied by clause (d) of Weak-form Robertson uncertainty on a common domain with $c = ℏ$ .

(e) Scope. These are exact conditional calculations for the free model defined by $H_{0}$ ; no measurement, instrument or laboratory preparation is described, and nothing here asserts that a physical wave packet is Gaussian.

Facts & Assumptions

Given: $m, ℏ, σ > 0$ , $x_{0}, p_{0} \in R$ , the vector $ψ_{0}$ above, the operator $H_{0}$ and the transform $U$ .

[F1]

$S (R)$ consists of the $C^{\infty}$ functions all of whose polynomial-weighted derivatives are bounded; every Schwartz function lies in $L^{1} \cap L^{2}$ (Schwartz space and its seminorms).

[F2]

$U : L^{2} (R, d x) \to L^{2} (R, d p)$ is a surjective complex-linear isometry with $U = D_{ℏ} F_{2}$ on $S (R)$ , where $D_{ℏ} g (p) = (2 π ℏ)^{- 1 / 2} g (p / (2 π ℏ))$ ; being a surjective isometry, $U$ preserves inner products and $U^{*} = U^{- 1}$ (Momentum Fourier Jacobian: the unitary momentum representation).

[F3]

If $Re z > 0$ then $\int_{R} e^{- z x^{2}} d x = \sqrt{π} z^{- 1 / 2}$ for the branch $z^{- 1 / 2} = \exp (- \frac{1}{2} Log z)$ , and for $G_{1} (z) : = \int_{R} e^{- z x^{2}} d x$ one has $G_{1}^{'} (z) = - \int_{R} x^{2} e^{- z x^{2}} d x$ (Complex Gaussian continuation(a),(b)).

[F4]

$H_{0}$ is self-adjoint, $D (H_{0}) = W^{2, 2} (R; C) = {f \in L^{2} : ξ^{2} F_{2} f \in L^{2}}$ with equivalent norms on this set, and $H_{0} f = - \frac{ℏ^{2}}{2 m} Δ f$ for $f \in S (R)$ (Real multiplication operators and free plus bounded-potential Schrödinger operators(b)). Here $F_{2}$ is the Plancherel transform extending the $L^{1}$ transform $\hat{f} (ξ) = \int f (x) e^{- 2 π i x ξ} d x$ ; it is a surjective isometry preserving inner products (Plancherel theorem, Fourier transform on complex L1 classes).

[F5]

For a self-adjoint operator $T$ the operators $V (t) : = e^{i t T}$ form a strongly continuous one-parameter unitary group with $V (t) D (T) = D (T)$ , $T V (t) = V (t) T$ on $D (T)$ , and $\lim_{s \to 0} \frac{1}{s} (V (s) x - x) = i T x$ exactly for $x \in D (T)$ (A self-adjoint operator generates a strongly continuous unitary group).

[F6]

Dominated convergence: if $f_{n} \to f$ pointwise and $∣ f_{n} ∣ \leq g$ for a fixed integrable $g$ , then $\int f_{n} \to \int f$ (Dominated convergence). A continuous real function on an interval has a primitive and $\int_{a}^{b} G^{'} = G (b) - G (a)$ for it (Every continuous function on an interval has a primitive; two primitives differ by a constant; and $\int_{a}^{b} f = G (b) - G (a)$ for any primitive $G$ ); the integral over $R$ is the limit of the compact-interval integrals when it exists (Improper integrals over unbounded intervals); $(f g)^{'} = f^{'} g + f g^{'}$ (Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ ); for differentiable $u, v$ with integrable derivatives on $[a, b]$ , $\int_{a}^{b} u v^{'} = u (b) v (b) - u (a) v (a) - \int_{a}^{b} u^{'} v$ (If $u, v$ are differentiable on $[a, b]$ with $u^{'}, v^{'}$ integrable, then $\int_{a}^{b} u v^{'} = u (b) v (b) - u (a) v (a) - \int_{a}^{b} u^{'} v$ ). Substitutions: an invertible linear $T : R \to R$ scales Lebesgue measure by $∣ \det T ∣$ (A linear map $T$ of $R^{n}$ sends Lebesgue measurable sets to Lebesgue measurable sets, with $λ_{n} (T [E]) = ∣ \det T ∣ λ_{n} (E)$ when $T$ is invertible and $T [E]$ Lebesgue null when it is not), and $λ_{1} (E + h) = λ_{1} (E)$ for every Lebesgue measurable $E$ and $h \in R$ (Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation); hence for every nonnegative measurable $H$ and every real $x_{0}$ , $\int_{R} H (x - x_{0}) d x = \int_{R} H (y) d y$ , by approximation with simple functions and monotone convergence (Monotone convergence for the integral).

[F7]

For every unit $f \in S (R)$ and every $c > 0$ , with $Q f (x) = x f (x)$ and $P_{c} f (x) = - i c f^{'} (x)$ , one has $Δ Q Δ P_{c} \geq c / 2$ , the differences being centered norms (Weak-form Robertson uncertainty on a common domain(d)).

Proof

1.1F1F3F6algebra

[The initial packet is a unit Schwartz vector] The derivative $\partial_{x}^{k} ψ_{0}$ equals $P_{k} (x) ψ_{0} (x)$ for a polynomial $P_{k}$ : this follows by induction with the product rule [F6] applied to the three factors $(2 π σ^{2})^{- 1 / 4}$ , $e^{- (x - x_{0})^{2} / (4 σ^{2})}$ and the bounded factor $e^{i p_{0} (x - x_{0}) / ℏ}$ , each differentiation of the last factor multiplying it by the constant $i p_{0} / ℏ$ . Since $x \mapsto x^{j} P_{k} (x) e^{- (x - x_{0})^{2} / (4 σ^{2})}$ is bounded for all $j, k$ (the exponential decays faster than any polynomial), $ψ_{0} \in S (R)$ [F1]. By [F6] $\int_{R} ∣ ψ_{0} (x) ∣^{2} d x = (2 π σ^{2})^{- 1 / 2} \int_{R} e^{- y^{2} / (2 σ^{2})} d y = (2 π σ^{2})^{- 1 / 2} \sqrt{π} (2 σ^{2})^{1 / 2} = 1$ , using [F3] with $z = 1 / (2 σ^{2})$ and the substitution $x = x_{0} + y$ .

1.2F3F6algebra

[The complex shift identity] For $Re a > 0$ and $b \in R$ one has $\int_{R} e^{- a y^{2}} e^{i b y} d y = \sqrt{π / a} e^{- b^{2} / (4 a)}$ with $\sqrt{π / a} : = \sqrt{π} a^{- 1 / 2}$ : setting $F (b) : = \int_{R} e^{- a y^{2} + i b y} d y$ (absolutely convergent since $∣ e^{- a y^{2} + i b y} ∣ = e^{- Re (a) y^{2}}$ is integrable) and $G_{R} (b) : = \int_{- R}^{R} e^{- a y^{2} + i b y} d y$ , the fundamental theorem and [F6] give $\frac{d}{d b} G_{R} (b) = i \int_{- R}^{R} y e^{- a y^{2} + i b y} d y$ , and since $∣ y ∣ e^{- Re (a) y^{2}}$ is integrable, $G_{R} \to F$ pointwise with the difference quotients converging, so $F$ is differentiable with $F^{'} (b) = i \int_{R} y e^{- a y^{2} + i b y} d y$ ; integrating $\partial_{y} (e^{- a y^{2} + i b y}) = (- 2 a y + i b) e^{- a y^{2} + i b y}$ over $[- R, R]$ gives $\int_{- R}^{R} (- 2 a y + i b) e^{- a y^{2} + i b y} d y = e^{- a R^{2} + i b R} - e^{- a R^{2} - i b R}$ , of modulus at most $2 e^{- Re (a) R^{2}} \to 0$ , so [F6] as $R \to \infty$ gives $- 2 a F_{y} (b) + i b F (b) = 0$ with $F_{y} (b) : = \int_{R} y e^{- a y^{2} + i b y} d y$ , hence $F_{y} = \frac{i b}{2 a} F$ and $F^{'} = - \frac{b}{2 a} F$ ; the product rule [F6] gives $\frac{d}{d b} [F (b) e^{b^{2} / (4 a)}] = 0$ , so the product is constant (fundamental theorem applied to real and imaginary parts [F6]) and its value at $b = 0$ is $F (0) = \int_{R} e^{- a y^{2}} d y = \sqrt{π} a^{- 1 / 2}$ by [F3].

1.3F2F4algebra

[The free operator in momentum coordinates] Let $B$ be the multiplication operator $(B φ) (p) = \frac{p^{2}}{2 m} φ (p)$ on $D (B) : = {φ \in L^{2} (d p) : p^{2} φ \in L^{2}}$ ; then $U^{*} B U = H_{0}$ with $D (H_{0}) = U^{*} D (B)$ , and $H_{0} U^{*} = U^{*} B$ , because $D_{ℏ}^{*} = D_{ℏ}^{- 1}$ and $(D_{ℏ}^{- 1} M_{g} D_{ℏ} h) (ξ) = g (2 π ℏ ξ) h (ξ)$ for every Borel $g$ and every $h$ in the domain of $D_{ℏ}^{- 1} M_{g} D_{ℏ}$ , so with $g (p) = p^{2} / (2 m)$ one has $U^{*} B U = F_{2}^{*} D_{ℏ}^{*} M_{p^{2} / (2 m)} D_{ℏ} F_{2} = F_{2}^{*} M_{2 π^{2} ℏ^{2} ξ^{2} / m} F_{2} = H_{0}$ as operators [F2, F4]; for the domains, $f \in D (H_{0})$ means $ξ^{2} F_{2} f \in L^{2}$ , which by $U f (p) = (2 π ℏ)^{- 1 / 2} F_{2} f (p / (2 π ℏ))$ is equivalent to $p^{2} U f \in L^{2}$ (the substitution $ξ = p / (2 π ℏ)$ and the constants $(2 π ℏ)^{\pm 2}$ preserve $L^{2}$ -membership), i.e. to $U f \in D (B)$ , so $D (H_{0}) = U^{*} D (B)$ , the identity $U^{*} B U = H_{0}$ holds on it, and multiplying on the right by $U^{*}$ gives $H_{0} U^{*} = U^{*} B U U^{*} = U^{*} B$ .

2.1step 1.1step 1.2F2F4F6algebra

[Fourier transform of the initial packet] Since $ψ_{0} \in L^{1} \cap L^{2}$ [step 1.1], $F_{2} ψ_{0}$ is computed by the $L^{1}$ integral [F4]; substituting $x = x_{0} + y$ [F6] and using $ψ_{0} (x_{0} + y) = (2 π σ^{2})^{- 1 / 4} e^{- y^{2} / (4 σ^{2})} e^{i p_{0} y / ℏ}$ gives $F_{2} ψ_{0} (ξ) = (2 π σ^{2})^{- 1 / 4} e^{- 2 π i x_{0} ξ} \int_{R} e^{- y^{2} / (4 σ^{2})} e^{i (p_{0} / ℏ - 2 π ξ) y} d y = (2 π σ^{2})^{- 1 / 4} e^{- 2 π i x_{0} ξ} 2 σ \sqrt{π} e^{- σ^{2} (p_{0} / ℏ - 2 π ξ)^{2}}$ by step 1.2 with $a = 1 / (4 σ^{2})$ and $b = p_{0} / ℏ - 2 π ξ$ ; since $ψ_{0} \in S (R)$ , $U ψ_{0} = D_{ℏ} F_{2} ψ_{0}$ [F2], so $φ_{0} : = U ψ_{0}$ satisfies $φ_{0} (p) = (2 π ℏ)^{- 1 / 2} F_{2} ψ_{0} (p / (2 π ℏ)) = K e^{- i x_{0} p / ℏ} e^{- σ^{2} (p - p_{0})^{2} / ℏ^{2}}$ with $K : = (2 π ℏ)^{- 1 / 2} (2 π σ^{2})^{- 1 / 4} 2 σ \sqrt{π}$ and $K^{2} = \sqrt{2 / π} σ / ℏ$ .

2.2step 1.1F1F4algebra

[The packet is Schwartz and its modulus is the spreading Gaussian] Putting $z_{t} : = 1 / (1 + i τ)$ (so $Re z_{t} = 1 / (1 + τ^{2}) > 0$ ) and $c_{t} : = x_{0} + p_{0} t / m$ , one has $∣ ψ_{t} (x) ∣ = (2 π σ^{2})^{- 1 / 4} ∣ 1 + i τ ∣^{- 1 / 2} e^{- Re (z_{t}) (x - c_{t})^{2} / (4 σ^{2})} = (2 π σ^{2})^{- 1 / 4} (1 + τ^{2})^{- 1 / 4} e^{- (x - c_{t})^{2} / (4 σ^{2} (1 + τ^{2}))}$ ; the remaining $x$ -dependent phase is $\exp (- i Im (z_{t}) (x - c_{t})^{2} / (4 σ^{2}) + i p_{0} (x - x_{0}) / ℏ)$ , which has modulus one. By the product rule, each $x$ -derivative of $ψ_{t}$ is a polynomial in $x$ times $ψ_{t}$ , so the displayed decaying modulus makes every polynomial-weighted derivative bounded and $ψ_{t} \in S (R)$ ; in particular $ψ_{t} \in L^{1} \cap L^{2}$ and $ψ_{t} \in D (H_{0})$ by [F4].

3.1step 1.2step 2.1step 2.2F2F4F6algebra

[Momentum coordinates of the packet] With $A_{t} : = (2 π σ^{2})^{- 1 / 4} (1 + i τ)^{- 1 / 2}$ , $a_{t} : = 1 / (4 σ^{2} (1 + i τ))$ and $b_{ξ} : = p_{0} / ℏ - 2 π ξ$ , substituting $x = c_{t} + y$ [F6] in the $L^{1}$ integral [F4] of the packet $ψ_{t} = A_{t} e^{- i p_{0}^{2} t / (2 m ℏ)} e^{- a_{t} (x - c_{t})^{2}} e^{i (p_{0} / ℏ) (x - x_{0})}$ [step 2.2] gives $F_{2} ψ_{t} (ξ) = A_{t} \sqrt{π / a_{t}} e^{i p_{0}^{2} t / (2 m ℏ)} e^{- 2 π i ξ c_{t}} e^{- b_{ξ}^{2} / (4 a_{t})}$ by step 1.2 with $a = a_{t}$ (legal since $Re a_{t} > 0$ ); since $ψ_{t} \in S (R)$ , $U ψ_{t} = D_{ℏ} F_{2} ψ_{t}$ [F2], and with $p = 2 π ℏ ξ$ , $\sqrt{π / a_{t}} = 2 σ \sqrt{π} (1 + i τ)^{1 / 2}$ , $A_{t} (1 + i τ)^{1 / 2} = (2 π σ^{2})^{- 1 / 4}$ and $1 / (4 a_{t}) = σ^{2} (1 + i τ)$ one gets $U ψ_{t} (p) = K e^{i p_{0}^{2} t / (2 m ℏ)} e^{- i p c_{t} / ℏ} e^{- σ^{2} (p - p_{0})^{2} / ℏ^{2}} e^{- i t (p - p_{0})^{2} / (2 m ℏ)}$ with $K$ as in step 2.1.

3.2step 2.2F3F6algebra

[Position density, mean and variance] By step 2.2, $∣ ψ_{t} (x) ∣^{2} = (2 π σ^{2} (1 + τ^{2}))^{- 1 / 2} e^{- (x - c_{t})^{2} / (2 d_{t}^{2})}$ with $c_{t} = x_{0} + p_{0} t / m$ and $d_{t} : = σ \sqrt{1 + τ^{2}}$ ; substituting $x = c_{t} + d_{t} u$ and using $\int_{R} e^{- u^{2} / 2} d u = \sqrt{2 π}$ (that is [F3] with $z = 1 / 2$ ) gives $\int_{R} ∣ ψ_{t} ∣^{2} = 1$ , while the centering $\int_{R} (x - c_{t}) ∣ ψ_{t} (x) ∣^{2} d x = d_{t} \int_{R} u \frac{e^{- u^{2} / 2}}{\sqrt{2 π}} d u = 0$ and $\int_{R} (x - c_{t})^{2} ∣ ψ_{t} (x) ∣^{2} d x = d_{t}^{2} \int_{R} u^{2} \frac{e^{- u^{2} / 2}}{\sqrt{2 π}} d u = d_{t}^{2}$ , since $\int_{R} u^{2} e^{- u^{2} / 2} d u = - G_{1}^{'} (1 / 2) = \sqrt{2 π}$ by [F3]; hence the density has mean $c_{t} = x_{0} + p_{0} t / m$ and variance $σ^{2} (1 + τ^{2})$ , which is clause (b).

4.1step 2.1step 3.1F2algebra

[The packet is the free evolution of the momentum Gaussian] With $θ_{t} (p) : = t p^{2} / (2 m ℏ)$ and $φ_{t} : = e^{- i θ_{t}} φ_{0}$ , steps 2.1 and 3.1 give $U ψ_{t} (p) = e^{- i t p^{2} / (2 m ℏ)} φ_{0} (p) = φ_{t} (p)$ for every $p$ , because the two exponents differ by $i (p_{0}^{2} t / (2 m ℏ) - p (x_{0} + p_{0} t / m) / ℏ - t (p - p_{0})^{2} / (2 m ℏ) + p x_{0} / ℏ + t p^{2} / (2 m ℏ)) = i t (p_{0}^{2} - 2 p p_{0} - (p - p_{0})^{2} + p^{2}) / (2 m ℏ) = 0$ ; hence $U ψ_{t} = φ_{t}$ and, $U$ being a surjective isometry [F2], $ψ_{t} = U^{*} φ_{t}$ .

4.2step 2.1step 3.1step 3.2F3F6algebra

[Momentum density, mean and variance] By steps 2.1 and 3.1, $∣ U ψ_{t} (p) ∣^{2} = ∣ φ_{0} (p) ∣^{2} = K^{2} e^{- 2 σ^{2} (p - p_{0})^{2} / ℏ^{2}} = \sqrt{2 / π} \frac{σ}{ℏ} e^{- (p - p_{0})^{2} / (2 w^{2})}$ with $w : = ℏ / (2 σ)$ and $K^{2} = \sqrt{2 / π} σ / ℏ$ ; substituting $p = p_{0} + w u$ as in step 3.2 gives total mass $1$ , mean $p_{0}$ and variance $w^{2} = ℏ^{2} / (4 σ^{2})$ , and the density does not depend on $t$ ; this is clause (c).

5.1step 1.3step 2.1step 4.1F2F6algebra

[The momentum path solves the momentum equation] Put $q (p) : = p^{2} / (2 m ℏ)$ and $φ_{t} (p) : = e^{- i t q (p)} φ_{0} (p)$ . The Gaussian formula of step 2.1 gives $q φ_{0} \in L^{2} (d p)$ and $p^{2} φ_{0} \in L^{2} (d p)$ , so $φ_{t} \in D (B)$ for every $t$ . For fixed $t$ and $h \neq 0$ , $\frac{φ_{t + h} (p) - φ_{t} (p)}{h} = φ_{t} (p) \frac{e^{- i h q (p)} - 1}{h}$ tends pointwise to $- i q (p) φ_{t} (p) = - (i / ℏ) \frac{p^{2}}{2 m} φ_{t} (p)$ . Since $∣ e^{- i s} - 1 ∣ \leq ∣ s ∣$ , the difference between the quotient and its limit has squared modulus at most $4 ∣ q (p) φ_{0} (p) ∣^{2}$ , an integrable function; dominated convergence [F6] therefore gives convergence in $L^{2} (d p)$ . Also $φ_{t + h} \to φ_{t}$ and $B φ_{t + h} \to B φ_{t}$ in $L^{2} (d p)$ by pointwise convergence and the respective integrable square majorants $4 ∣ φ_{0} ∣^{2}$ and $4 ∣ B φ_{0} ∣^{2}$ . Thus $t \mapsto φ_{t}$ is continuously differentiable with $φ_{t}^{'} = - (i / ℏ) B φ_{t}$ . Consequently $ψ_{t} = U^{*} φ_{t}$ [step 4.1] is $C^{1}$ in $L^{2}$ , lies in $D (H_{0}) = U^{*} D (B)$ , and $\frac{d}{d t} ψ_{t} = U^{*} φ_{t}^{'} = - \frac{i}{ℏ} U^{*} B φ_{t} = - \frac{i}{ℏ} H_{0} U^{*} φ_{t} = - \frac{i}{ℏ} H_{0} ψ_{t}$ , using that $U^{*}$ is an isometry [F2] and $H_{0} U^{*} = U^{*} B$ from step 1.3; at $t = 0$ the formula of the Statement reduces to $ψ_{0}$ .

6.1step 1.1step 5.1F4F5F6algebra

[Uniqueness of the Cauchy problem] Let $V (t) : = e^{i t T}$ with $T : = - H_{0} / ℏ$ [F5]; since $ψ_{0} \in D (H_{0}) = D (T)$ [step 1.1, F4], one has $V (t) ψ_{0} \in D (T)$ for all $t$ and $t \mapsto V (t) ψ_{0}$ is $C^{1}$ with $\frac{d}{d t} V (t) ψ_{0} = i T V (t) ψ_{0} = - \frac{i}{ℏ} H_{0} V (t) ψ_{0}$ , because [F5] gives $\frac{1}{h} (V (t + h) - V (t)) ψ_{0} = V (t) \frac{1}{h} (V (h) ψ_{0} - ψ_{0}) \to V (t) (i T ψ_{0})$ as $ψ_{0} \in D (T)$ and $V (t)$ is bounded, and $T V (t) = V (t) T$ on $D (T)$ by [F5]; by step 5.1, $w_{t} : = ψ_{t} - V (t) ψ_{0}$ is $C^{1}$ with $w_{t} \in D (H_{0})$ and $w_{t}^{'} = - \frac{i}{ℏ} H_{0} w_{t}$ , and $w_{0} = 0$ since $V (0) = I$ and the Statement's formula at $t = 0$ is the initial packet; for the real function $t \mapsto ∥ w_{t} ∥^{2}$ one has $\frac{∥ w_{t + h} ∥^{2} - ∥ w_{t} ∥^{2}}{h} = 2 Re ⟨ w_{t}, \frac{w_{t + h} - w_{t}}{h} ⟩ + \frac{∥ w_{t + h} - w_{t} ∥^{2}}{h} \to 2 Re ⟨ w_{t}, w_{t}^{'} ⟩$ , the last term tending to $0$ ; now, since the inner product is conjugate-linear in its second argument, $⟨ w_{t}, w_{t}^{'} ⟩ = ⟨ w_{t}, - \frac{i}{ℏ} H_{0} w_{t} ⟩ = \frac{i}{ℏ} ⟨ w_{t}, H_{0} w_{t} ⟩$ . The number $⟨ w_{t}, H_{0} w_{t} ⟩ = ⟨ H_{0} w_{t}, w_{t} ⟩$ is real because $H_{0}$ is self-adjoint, hence symmetric, on $D (H_{0})$ [F4], so $Re ⟨ w_{t}, w_{t}^{'} ⟩ = 0$ and $∥ w_{t} ∥^{2}$ has vanishing derivative, hence is constant by the fundamental theorem [F6] and equals $∥ w_{0} ∥^{2} = 0$ ; thus $ψ_{t} = V (t) ψ_{0} = e^{- i t H_{0} / ℏ} ψ_{0}$ for every $t$ , which is clause (a).

7.1step 3.2step 4.2step 6.1F2F6algebra

[Operator standard deviations] For $ψ \in S (R)$ one has $U (P ψ) = M_{p} U ψ$ with $(M_{p} g) (p) = p g (p)$ , because $F_{2} (P ψ) (ξ) = \int_{R} (- i ℏ ψ^{'} (x)) e^{- 2 π i x ξ} d x = 2 π ℏ ξ F_{2} ψ (ξ)$ by integration by parts, the boundary terms $[ψ (x) e^{- 2 π i x ξ}]$ vanishing at $\pm \infty$ as $ψ \in S (R)$ [F6], and applying $D_{ℏ}$ with $p = 2 π ℏ ξ$ gives the claim [F2]; the unit vector $ψ_{t}$ of step 6.1 is the packet of steps 3.2 and 4.2, and for it this gives $b_{t} : = ⟨ P ψ_{t}, ψ_{t} ⟩ = ⟨ U P ψ_{t}, U ψ_{t} ⟩ = \int_{R} p ∣ U ψ_{t} (p) ∣^{2} d p = p_{0}$ , a real number, and $Δ P : = ∥ (P - b_{t}) ψ_{t} ∥_{2} = ∥ U (P - b_{t}) ψ_{t} ∥_{2} = (\int_{R} (p - p_{0})^{2} ∣ U ψ_{t} (p) ∣^{2} d p)^{1 / 2} = ℏ / (2 σ)$ by step 4.2, while $a_{t} : = ⟨ Q ψ_{t}, ψ_{t} ⟩ = \int_{R} x ∣ ψ_{t} (x) ∣^{2} d x = c_{t}$ and $Δ Q : = ∥ (Q - a_{t}) ψ_{t} ∥_{2} = (\int_{R} (x - c_{t})^{2} ∣ ψ_{t} (x) ∣^{2} d x)^{1 / 2} = σ \sqrt{1 + τ^{2}}$ by step 3.2; hence $Δ Q Δ P = \frac{ℏ}{2} \sqrt{1 + τ^{2}}$ .

8.1step 7.1F7algebra∎

[Robertson comparison and scope] By [F7] with $c = ℏ$ , applied to the unit Schwartz vector $ψ_{t}$ of step 7.1, one has $Δ Q Δ P \geq ℏ / 2$ ; the exact value of step 7.1 is $\frac{ℏ}{2} \sqrt{1 + τ^{2}}$ , which equals $ℏ / 2$ exactly when $τ = 0$ , i.e. exactly at $t = 0$ since $τ = ℏ t / (2 m σ^{2})$ , and is strictly larger otherwise; thus clause (d) holds, and clause (e) records that all of the above are computations inside the free model defined by $H_{0}$ , with no measurement interpretation.

TheoremStatement: AI-adaptedProof: AI-adaptedOpen item page →

Joint PVM construction for strongly commuting regular projection-valued measures

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $H$ be a nonzero complex Hilbert space (Hilbert space) and let $E, F$ be regular projection-valued measures on the Borel $σ$ -algebra of $R$ (Projection valued measure, The Borel sigma-algebra of a topological space) satisfying the strong commutation relation

$E (B) F (C) = F (C) E (B) for all Borel B, C \subseteq R .$

Then there is a unique regular projection-valued measure $G$ on the Borel $σ$ -algebra of $R^{2}$ with

$G (B \times C) = E (B) F (C) for every Borel rectangle B \times C,$

where rectangle means a measurable rectangle in the sense of Measurable rectangles in a product of measurable spaces. Its marginals are $G (B \times R) = E (B)$ and $G (R \times C) = F (C)$ .

For the scalar construction, let $A_{0}$ be the algebra of finite unions of Borel rectangles, and define $P_{0} (⨆_{i = 1}^{n} B_{i} \times C_{i}) : = \sum_{i = 1}^{n} E (B_{i}) F (C_{i})$ on disjoint rectangular decompositions. For every $ψ \in H$ , $μ_{ψ} (A) : = ⟨ P_{0} (A) ψ, ψ ⟩$ is a finite premeasure on $A_{0}$ and has a unique extension $G_{ψ} = ⟨ G (\cdot) ψ, ψ ⟩$ to the Borel $σ$ -algebra. For $x, y \in H$ , define the polarized measure $ν_{x, y} : = \frac{1}{4} \sum_{k = 0}^{3} i^{k} G_{x + i^{k} y}$ . Then every Borel $S \subseteq R^{2}$ has a unique bounded operator $G (S)$ satisfying $⟨ G (S) x, y ⟩ = ν_{x, y} (S)$ for all $x, y \in H$ , and $0 \leq G (S) \leq I$ .

Facts & Assumptions

Given: A nonzero complex Hilbert space $H$ ; regular PVMs $E, F$ on $B (R)$ with $E (B) F (C) = F (C) E (B)$ ; the algebra $A_{0}$ of finite unions of measurable rectangles in $R^{2}$ .

[F1]

A PVM is a map into orthogonal projections with $E (\emptyset) = 0$ , $E (R) = I$ , $E (B \cap C) = E (B) E (C)$ and strong countable additivity; regularity means every scalar measure $E_{x} (B) = ⟨ E (B) x, x ⟩$ is a regular Borel measure. For an orthogonal projection $P$ one has $⟨ P x, x ⟩ = ∥ P x ∥^{2} \geq 0$ (Projection valued measure, The Hilbert orthogonal projection onto a closed subspace, Hilbert projections are linear, self-adjoint and contractive).

[F2]

For a PVM, $E_{x} (B) = ⟨ E (B) x, x ⟩$ is a positive measure with $E_{x} (R) = ∥ x ∥^{2}$ ; the polarized $E_{x, y} (B) = ⟨ E (B) x, y ⟩$ is a finite complex measure with $E_{y, x} = \overline{E_{x, y}}$ , $∣ E_{x, y} (B) ∣ \leq ∥ x ∥ ∥ y ∥$ and the polarization identity $E_{x, y} = \frac{1}{4} \sum_{k = 0}^{3} i^{k} E_{x + i^{k} y}$ (Scalar and complex measures from a pvm).

[F3]

A premeasure on an algebra is finitely additive and countably additive on disjoint families whose union lies in the algebra; the outer set function it induces is an outer measure under ${A C}_{ω}$ , the Carathéodory $σ$ -algebra carries the extended measure, the source algebra consists of Carathéodory measurable sets, the extension agrees with the premeasure on the source algebra, and a finite premeasure has at most one extension to the generated $σ$ -algebra (Premeasures on algebras of sets, Assuming countable choice, the outer set function induced by a premeasure is an outer measure, Carathéodory's theorem: measurable sets form a sigma-algebra carrying a complete measure, The induced outer measure agrees with the premeasure on the source algebra, Assuming countable choice, every source-algebra set is measurable for the induced outer measure, A finite premeasure has at most one extension to its generated sigma-algebra).

[F4]

Finite positive measures agreeing on a $π$ -system that generates a $σ$ -algebra and having equal total mass agree everywhere on it (Finite measures agreeing on a generating pi-system and on the whole space are equal, Pi-systems).

[F5]

The $σ$ -algebra generated by the measurable rectangles of $B (R) \times B (R)$ is the product $σ$ -algebra, and $B (R) \otimes B (R) = B (R^{2})$ (The product sigma-algebra and its finite iterates, The Borel product of R^m and R^n is the Borel sigma-algebra of R^{m+n}).

[F6]

Every finite Borel measure on $R^{2}$ is regular once every open subset of $R^{2}$ is $σ$ -compact (Sigma-compact open sets make locally finite Borel measures regular). This topological hypothesis holds: for an open proper $U ⊊ R^{2}$ , the compact sets $K_{n} = {x \in [- n, n]^{2} : dist (x, R^{2} ∖ U) \geq 1 / n}$ cover $U$ ; for $U = R^{2}$ , the squares $[- n, n]^{2}$ cover it.

[F7]

For a bounded sesquilinear form on a Hilbert space there is a unique bounded operator representing it; equivalently, bounded linear functionals are represented by vectors (Riesz representation for Hilbert spaces).

[F8]

If $P$ is a $π$ -system and $D$ is a $λ$ -system containing $P$ , then $D$ contains the $σ$ -algebra generated by $P$ (Dynkin's pi-lambda theorem).

Proof

1.1givenF1algebra

[The finitely additive projection on the rectangle algebra] For a finite disjoint union $A = ⨆_{i = 1}^{n} (B_{i} \times C_{i})$ of measurable rectangles put $P (A) : = \sum_{i} E (B_{i}) F (C_{i})$ . This is well defined: two finite rectangle decompositions have a common refinement into disjoint rectangles (intersections of the pieces), and on a common refinement both sums agree by additivity of $E$ and $F$ ; in particular $P (\emptyset) = 0$ and $P (R^{2}) = I$ . Each $E (B_{i}) F (C_{i})$ is an orthogonal projection because $(E (B_{i}) F (C_{i}))^{2} = E (B_{i})^{2} F (C_{i})^{2} = E (B_{i}) F (C_{i})$ and $(E (B_{i}) F (C_{i}))^{*} = F (C_{i}) E (B_{i}) = E (B_{i}) F (C_{i})$ by [F1] and strong commutation; summands belonging to disjoint rectangles are orthogonal, since $E (B_{i}) F (C_{i}) E (B_{j}) F (C_{j}) = E (B_{i} \cap B_{j}) F (C_{i} \cap C_{j}) = 0$ . Hence $P (A)$ is a sum of pairwise orthogonal projections, so $P (A)^{2} = P (A) = P (A)^{*}$ ; and from the refinement description $P$ is finitely additive on disjoint unions, $P (A \cap A^{'}) = P (A) P (A^{'})$ and $P (A \cup A^{'}) = P (A) + P (A^{'})$ for disjoint $A, A^{'} \in A_{0}$ .

1.2F5F8algebra

[Uniqueness for finite complex measures] Let $μ, ν$ be finite complex measures on $B (R^{2})$ that agree on $A_{0}$ , which contains $R^{2}$ . Then $D : = {S \in B (R^{2}) : μ (S) = ν (S)}$ contains $R^{2}$ and the $π$ -system $A_{0}$ ; it is a $λ$ -system because equality on the whole space gives closure under complements and countable additivity gives closure under disjoint countable unions. By Dynkin's $π$ - $λ$ theorem [F8], $D$ contains $σ (A_{0}) = B (R^{2})$ by [F5], so $μ = ν$ on every Borel set.

2.1step 1.1F1F3algebra

[The scalar set functions are premeasures] For $ψ \in H$ define $μ_{ψ} (A) : = ⟨ P (A) ψ, ψ ⟩ = ∥ P (A) ψ ∥^{2}$ for $A \in A_{0}$ ; by step 1.1, $μ_{ψ} (\emptyset) = 0$ , $μ_{ψ}$ is finitely additive, and $μ_{ψ} \leq ∥ ψ ∥^{2}$ . To verify countable additivity, fix pairwise disjoint $A_{n} \in A_{0}$ with union $A \in A_{0}$ and put $R_{N} : = A ∖ ⋃_{n \leq N} A_{n} \in A_{0}$ . Finite additivity gives $μ_{ψ} (A) = \sum_{n \leq N} μ_{ψ} (A_{n}) + μ_{ψ} (R_{N})$ , and $R_{N} ↓ \emptyset$ , so continuity from above at $\emptyset$ implies $μ_{ψ} (A) = \sum_{n} μ_{ψ} (A_{n})$ . It remains to prove that continuity: for $A_{n} ↓ \emptyset$ in $A_{0}$ , show $μ_{ψ} (A_{n}) \to 0$ . Given $ε > 0$ , approximate each $A_{n}$ from inside by a compact union of rectangles: for a rectangle $B \times C$ one has $P (B \times C) - P (K_{B} \times K_{C}) = E (B ∖ K_{B}) F (C) + E (K_{B}) F (C ∖ K_{C})$ with the two summands orthogonal, so $μ_{ψ} (B \times C) - μ_{ψ} (K_{B} \times K_{C}) \leq E_{ψ} (B ∖ K_{B}) + F_{ψ} (C ∖ K_{C})$ for $E_{ψ} (B) = ⟨ E (B) ψ, ψ ⟩$ , $F_{ψ} (C) = ⟨ F (C) ψ, ψ ⟩$ ; by regularity [F1] compact $K_{B} \subseteq B$ , $K_{C} \subseteq C$ make this arbitrarily small, and finite unions are handled by summing. Choose compact $L_{n} \subseteq A_{n}$ with $μ_{ψ} (A_{n} ∖ L_{n}) < ε 2^{- n}$ . If $μ_{ψ} (A_{n}) \geq ε$ for all $n$ , then for every $N$ additivity and $L_{1} \cap \dots \cap L_{N} \supseteq A_{N} ∖ ⋃_{n \leq N} (A_{n} ∖ L_{n})$ give $μ_{ψ} (L_{1} \cap \dots \cap L_{N}) \geq μ_{ψ} (A_{N}) - \sum_{n \leq N} μ_{ψ} (A_{n} ∖ L_{n}) \geq ε - \sum_{n \leq N} ε 2^{- n} > 0$ ; hence every finite intersection $L_{1} \cap \dots \cap L_{N}$ is nonempty, and compactness gives a point of $⋂_{n} L_{n} \subseteq ⋂_{n} A_{n} = \emptyset$ , a contradiction. Thus $μ_{ψ} (A_{n}) \to 0$ , and $μ_{ψ}$ is a finite premeasure on $A_{0}$ .

3.1step 1.1step 2.1F2F3F4F5algebra

[Carathéodory extension and polarization] By steps 1.1 and 2.1 the algebra $A_{0}$ of finite unions of measurable rectangles is a $π$ -system containing $R^{2}$ and generates $B (R^{2})$ by [F5]; hence $μ_{ψ}$ extends to a finite measure $ν_{ψ}$ on $B (R^{2})$ , unique by [F3], and $ν_{ψ} (S) = μ_{ψ} (S)$ for $S \in A_{0}$ . Uniqueness also extends the scalar homogeneity $ν_{α x} = ∣ α ∣^{2} ν_{x}$ and the parallelogram identity from $A_{0}$ to every Borel set: the corresponding finite positive measures agree on $A_{0}$ and have the same total mass. For $x, y \in H$ define $ν_{x, y} (S) : = \frac{1}{4} \sum_{k = 0}^{3} i^{k} ν_{x + i^{k} y} (S)$ ; this is a finite complex measure and for $S \in A_{0}$ it coincides with $⟨ P (S) x, y ⟩$ by the polarization identity [F2]. The scalar homogeneity and parallelogram identity imply, by expansion of the four polarization terms, that $(x, y) \mapsto ν_{x, y} (S)$ is sesquilinear and Hermitian for every Borel $S$ , with $ν_{x, x} (S) = ν_{x} (S) \geq 0$ . Positivity gives Cauchy–Schwarz for this form (apply $0 \leq ν_{x + λ y} (S)$ for all $λ \in C$ ), so $∣ ν_{x, y} (S) ∣ \leq ν_{x} (S)^{1 / 2} ν_{y} (S)^{1 / 2} \leq ∥ x ∥ ∥ y ∥$ .

4.1step 3.1F7algebra

[The operators $G (S)$ ] Fix a Borel $S \subseteq R^{2}$ . For fixed $x$ , the map $y \mapsto \overline{ν_{x, y} (S)}$ is a bounded linear functional of norm at most $∥ x ∥$ by step 3.1, so by [F7] there is a unique $G (S) x \in H$ with $\overline{ν_{x, y} (S)} = ⟨ y, G (S) x ⟩$ , that is $ν_{x, y} (S) = ⟨ G (S) x, y ⟩$ , and $∥ G (S) x ∥ \leq ∥ x ∥$ . The assignment $x \mapsto G (S) x$ is linear because the form is linear in its first argument and the representative is unique, so $G (S) \in B (H)$ with $∥ G (S) ∥ \leq 1$ ; moreover $⟨ G (S) x, y ⟩ = ν_{x, y} (S) = \overline{ν_{y, x} (S)} = \overline{⟨ G (S) y, x ⟩} = ⟨ x, G (S) y ⟩$ , so $G (S)$ is self-adjoint, and $⟨ G (S) x, x ⟩ = ν_{x} (S) \in [0, ∥ x ∥^{2}]$ gives $0 \leq G (S) \leq I$ . For every $A \in A_{0}$ , step 3.1 gives $ν_{x, y} (A) = ⟨ P (A) x, y ⟩$ , so the defining pairings imply $G (A) = P (A)$ ; in particular $G (B \times C) = E (B) F (C)$ .

5.1step 1.1step 1.2step 4.1F1F2F6F8algebra

[Multiplicativity and the PVM axioms] First observe that for arbitrary Borel $S$ and rectangle $R = B^{'} \times C^{'}$ , for every $x, y$ the scalar set functions $S^{'} \mapsto ⟨ G (S^{'} \cap R) x, y ⟩ = ν_{x, y} (S^{'} \cap R)$ and $S^{'} \mapsto ⟨ G (S^{'}) G (R) x, y ⟩ = ν_{G (R) x, y} (S^{'})$ are finite complex measures; they agree on rectangles $S^{'} = B \times C$ because $G (S^{'} \cap R) = E (B \cap B^{'}) F (C \cap C^{'}) = E (B) E (B^{'}) F (C) F (C^{'}) = E (B) F (C) E (B^{'}) F (C^{'}) = G (S^{'}) G (R)$ by [F1] and the commutation hypothesis, and they agree on $R^{2}$ since each value is $G (R)$ . By step 1.2 [F8], these scalar measures agree on all Borel $S^{'}$ , so $G (S^{'} \cap R) = G (S^{'}) G (R)$ for all Borel $S^{'}$ . Now fix arbitrary Borel $S$ and compare, for every $x, y$ , the finite complex measures $T \mapsto ⟨ G (S \cap T) x, y ⟩ = ν_{x, y} (S \cap T)$ and $T \mapsto ⟨ G (S) G (T) x, y ⟩ = ν_{x, G (S) y} (T)$ ; by the previous result they agree on every rectangle $T$ , and by step 1.2 [F8] they agree for all Borel $T$ . Taking $S = T$ yields $G (S)^{2} = G (S)$ , so every $G (S)$ is an orthogonal projection; $G (\emptyset) = 0$ and $G (R^{2}) = I$ follow from the rectangle values. Strong countable additivity holds because for pairwise disjoint $S_{n}$ with union $S$ the scalar measures satisfy $ν_{x} (S) = \sum_{n} ν_{x} (S_{n})$ , so $\sum_{n} ∥ G (S_{n}) x ∥^{2} = \sum_{n} ν_{x} (S_{n}) = ν_{x} (S) \leq ∥ x ∥^{2} < \infty$ and the orthogonal series $\sum_{n} G (S_{n}) x$ converges in norm to a vector whose pairings with every $y$ equal $\sum_{n} ν_{x, y} (S_{n}) = ν_{x, y} (S) = ⟨ G (S) x, y ⟩$ , hence to $G (S) x$ . Finally the scalar measures of $G$ are the $ν_{ψ}$ , finite Borel measures on $R^{2}$ and hence regular by [F6]; so $G$ is a regular PVM with $G (B \times C) = E (B) F (C)$ .

6.1step 5.1step 1.2F1F2F8algebra∎

[Marginals and uniqueness] For the marginal, $G (B \times R) = E (B) F (R) = E (B)$ and $G (R \times C) = F (C)$ by [F1]. If $G^{'}$ is another regular PVM with $G^{'} (B \times C) = E (B) F (C)$ , then for every $x, y$ the finite complex measures $S \mapsto ⟨ G^{'} (S) x, y ⟩$ and $ν_{x, y}$ agree on rectangles, including $R^{2}$ , so step 1.2 [F8] gives their equality on $B (R^{2})$ . Taking $y = G^{'} (S) x - G (S) x$ gives $G^{'} (S) = G (S)$ for every Borel $S$ . Hence $G$ is unique, completing the proof.

LemmaStatement: AI-adaptedProof: AI-adaptedOpen item page →

Compatible Lüders repeatability

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $H$ be a nonzero complex Hilbert space, let $E, F$ be regular projection-valued measures on the Borel $σ$ -algebra of $R$ with $E (B) F (C) = F (C) E (B)$ for all Borel $B, C \subseteq R$ , and let $G$ be the joint projection-valued measure on the Borel $σ$ -algebra of $R^{2}$ with $G (B \times C) = E (B) F (C)$ and marginals $E, F$ (Joint PVM construction for strongly commuting regular projection-valued measures). Fix finite families of Borel sets $B_{1}, \dots, B_{n}$ with $B_{i} \cap B_{k} = \emptyset$ for $i \neq k$ and $C_{1}, \dots, C_{m}$ with $C_{j} \cap C_{l} = \emptyset$ for $j \neq l$ , put

$P_{i} : = E (B_{i}), Q_{j} : = F (C_{j}), R_{i j} : = P_{i} Q_{j},$

and let $ρ$ be a density operator on $H$ (Density operators and scalar spectral probability(a)). The listed projections are stipulated to be outcomes of ideal Lüders instruments, with outcome probabilities $tr (ρ P_{i})$ and post-measurement states $ρ_{i} : = P_{i} ρ P_{i} / tr (ρ P_{i})$ when $tr (ρ P_{i}) > 0$ , and likewise for $(Q_{j})$ . If either listed family fails to sum to the identity, include its complementary projection $I - \sum_{i} P_{i}$ or $I - \sum_{j} Q_{j}$ as an additional outcome; only probabilities for the listed outcomes are displayed below. A sequential run measures the first instrument and then the second on the resulting state. Then:

(a) $P_{i}, Q_{j}$ and $R_{i j}$ are orthogonal projections (The Hilbert orthogonal projection onto a closed subspace), $R_{i j} = Q_{j} P_{i} = G (B_{i} \times C_{j})$ , and $P_{i} P_{k} = 0$ for $i \neq k$ , $Q_{j} Q_{l} = 0$ for $j \neq l$ , $R_{i j} R_{k l} = 0$ whenever $(i, j) \neq (k, l)$ .

(b) The probability of the listed ordered outcome pair $(i then j)$ equals, when $tr (ρ P_{i}) > 0$ , $tr (P_{i} ρ P_{i} Q_{j}) / tr (ρ P_{i}) \cdot tr (ρ P_{i}) = tr (ρ R_{i j})$ , and is zero if the first outcome has zero probability; the same number is obtained in the reversed order, so the sequential order of the two compatible sharp measurements gives the same masses to all listed outcome pairs: $tr (P_{i} ρ P_{i} Q_{j}) = tr (Q_{j} ρ Q_{j} P_{i}) = tr (ρ R_{i j}) = tr (ρ G (B_{i} \times C_{j})) .$

(c) If $tr (ρ R_{i j}) > 0$ , the conditional state after the pair is order-independent: $\frac{Q_{j} P_{i} ρ P_{i} Q_{j}}{tr (ρ R_{i j})} = \frac{P_{i} Q_{j} ρ Q_{j} P_{i}}{tr (ρ R_{i j})} = \frac{R_{i j} ρ R_{i j}}{tr (ρ R_{i j})},$ and it is a density operator.

(d) Repeatability. Under the positive-probability hypothesis $tr (ρ R_{i j}) > 0$ of (c), with $σ : = R_{i j} ρ R_{i j} / tr (ρ R_{i j})$ one has $tr (σ R_{i j}) = 1$ and $tr (σ P_{i}) = tr (σ Q_{j}) = 1$ , and the Lüders update by $R_{i j}$ , $P_{i}$ or $Q_{j}$ leaves $σ$ unchanged: $R_{i j} σ R_{i j} / tr (σ R_{i j}) = P_{i} σ P_{i} / tr (σ P_{i}) = Q_{j} σ Q_{j} / tr (σ Q_{j}) = σ$ . Thus immediate repetition of the compatible pair, or of either single sharp measurement, reproduces the same outcomes with certainty.

(e) If $⋃_{j} C_{j} = R$ then $\sum_{j} tr (ρ R_{i j}) = tr (ρ P_{i})$ , and if $⋃_{i} B_{i} = R$ then $\sum_{i} tr (ρ R_{i j}) = tr (ρ Q_{j})$ : the joint law has the sharp-bin marginals.

(f) Scope. All of this is a conditional calculation for the stipulated ideal Lüders instrument and the joint PVM $G$ ; no claim is made that a laboratory apparatus realizes the Lüders instrument or the joint measurement, and no dynamical time development is described.

Facts & Assumptions

Given: A nonzero complex Hilbert space $H$ ; regular PVMs $E, F$ with $E (B) F (C) = F (C) E (B)$ ; the joint PVM $G$ ; Borel sets $B_{1}, \dots, B_{n}$ and $C_{1}, \dots, C_{m}$ pairwise disjoint within each family; a density operator $ρ$ ; the projections $P_{i}, Q_{j}, R_{i j}$ .

[F1]

An orthogonal projection on $H$ is a bounded operator $P$ with $P^{2} = P = P^{*}$ ; for a PVM $E$ every value $E (B)$ is an orthogonal projection, $E (\emptyset) = 0$ , $E (R) = I$ , $E (B \cap C) = E (B) E (C)$ , and $E$ is strongly countably additive (The Hilbert orthogonal projection onto a closed subspace, Joint PVM construction for strongly commuting regular projection-valued measures).

[F2]

$G$ is a regular PVM on the Borel $σ$ -algebra of $R^{2}$ with $G (B \times C) = E (B) F (C)$ for Borel rectangles and marginals $G (B \times R) = E (B)$ , $G (R \times C) = F (C)$ ; it is the unique such PVM and every value $G (S)$ is an orthogonal projection with $0 \leq G (S) \leq I$ (Joint PVM construction for strongly commuting regular projection-valued measures).

[F3]

A density operator is a self-adjoint positive trace-class $ρ$ with $tr ρ = 1$ ; for every PVM $E$ on a measurable space the set function $μ_{ρ} (S) : = tr (ρ E (S))$ is a probability measure, and in every expansion $ρ = \sum_{n} w_{n} T_{u_{n}, u_{n}}$ with orthonormal $(u_{n})$ and weights $w_{n} \geq 0$ , $\sum_{n} w_{n} = 1$ , one has $μ_{ρ} (S) = \sum_{n} w_{n} ∥ E (S) u_{n} ∥^{2}$ (Density operators and scalar spectral probability(a),(b)).

[F4]

For bounded $B$ and trace-class $T$ : $B T$ and $T B$ are trace class, $tr (B T) = tr (T B)$ , $tr$ is linear on trace-class operators, and $∣ tr (B T) ∣ \leq ∥ B ∥ ∥ T ∥_{1}$ (Nuclear bounded cyclicity for trace-class operators(b), Trace is absolutely convergent and basis independent).

[F5]

The Lüders instrument and the sequential protocol of the Statement are stipulated; the conditional state is defined only when the corresponding probability is positive.

Proof

1.1F1F2givenalgebra

[Sharp bins and their products are projections] By [F1], $P_{i} = E (B_{i})$ and $Q_{j} = F (C_{j})$ are orthogonal projections; $P_{i} P_{k} = E (B_{i}) E (B_{k}) = E (B_{i} \cap B_{k}) = E (\emptyset) = 0$ for $i \neq k$ and $Q_{j} Q_{l} = 0$ for $j \neq l$ ; the commutation hypothesis and [F2] give $P_{i} Q_{j} = E (B_{i}) F (C_{j}) = G (B_{i} \times C_{j}) = F (C_{j}) E (B_{i}) = Q_{j} P_{i}$ , so $R_{i j} = P_{i} Q_{j} = Q_{j} P_{i}$ is self-adjoint ( $R_{i j}^{*} = Q_{j}^{*} P_{i}^{*} = Q_{j} P_{i} = R_{i j}$ ) and idempotent ( $R_{i j}^{2} = P_{i} Q_{j} P_{i} Q_{j} = P_{i} (P_{i} Q_{j}) Q_{j} = P_{i}^{2} Q_{j}^{2} = P_{i} Q_{j}$ ), hence an orthogonal projection; also $R_{i j} R_{k l} = P_{i} Q_{j} P_{k} Q_{l} = P_{i} P_{k} Q_{j} Q_{l}$ using commutation, which vanishes unless $i = k$ and $j = l$ .

1.2F3F4algebra

[Trace and Born rules] For every bounded operator $B$ one has $tr (ρ B) = tr (B ρ)$ , and $tr (ρ \cdot)$ is linear in its bounded argument with $∣ tr (ρ B) ∣ \leq ∥ B ∥$ ; for every PVM and Borel set $S$ the number $tr (ρ E (S))$ lies in $[0, 1]$ and equals $\sum_{n} w_{n} ∥ E (S) u_{n} ∥^{2}$ in any expansion of [F3], so $tr (ρ R_{i j}) = tr (ρ G (B_{i} \times C_{j})) = \sum_{n} w_{n} ∥ G (B_{i} \times C_{j}) u_{n} ∥^{2}$ and $tr (ρ P_{i}) = \sum_{n} w_{n} ∥ P_{i} u_{n} ∥^{2}$ are nonnegative and at most $1$ .

2.1step 1.1step 1.2F3F4F5algebra

[Sequential Lüders probability] If $tr (ρ P_{i}) > 0$ , measuring the first instrument gives listed outcome $i$ with probability $tr (ρ P_{i})$ and state $ρ_{i} = P_{i} ρ P_{i} / tr (ρ P_{i})$ (stipulated, [F5]); measuring then the second gives conditional probability $tr (ρ_{i} Q_{j}) = tr (P_{i} ρ P_{i} Q_{j}) / tr (ρ P_{i})$ . Cyclicity and $P_{i} Q_{j} = Q_{j} P_{i}$ , $P_{i}^{2} = P_{i}$ give $tr (P_{i} ρ P_{i} Q_{j}) = tr (ρ P_{i} Q_{j} P_{i}) = tr (ρ P_{i} Q_{j})$ , so the probability of the listed pair is $tr (ρ P_{i} Q_{j}) = tr (ρ R_{i j})$ by step 1.1. When $tr (ρ Q_{j}) > 0$ , the reversed order gives conditional probability $tr (Q_{j} ρ Q_{j} P_{i}) / tr (ρ Q_{j}) = tr (ρ Q_{j} P_{i}) / tr (ρ Q_{j}) = tr (ρ R_{i j}) / tr (ρ Q_{j})$ ; multiplying by the probability of the first outcome gives the same joint probability. If a listed first outcome has zero probability, then $0 \leq R_{i j} \leq P_{i}, Q_{j}$ and the positive density expansion [F3] give $tr (ρ R_{i j}) = 0$ , so the listed pair has probability zero and no division is taken.

3.1step 1.1step 2.1F3F4algebra

[Order-independent conditional state] Assume $tr (ρ R_{i j}) > 0$ ; running $(P_{i})$ then $(Q_{j})$ produces the product of the two stipulated denominator-normalizations, i.e. the state $Q_{j} P_{i} ρ P_{i} Q_{j} / tr (ρ R_{i j})$ (by the computation of step 2.1), while the reversed order produces $P_{i} Q_{j} ρ Q_{j} P_{i} / tr (ρ R_{i j})$ ; by step 1.1 both numerators equal $R_{i j} ρ R_{i j}$ and the denominators are the same positive number, so the two conditional states coincide and equal $R_{i j} ρ R_{i j} / tr (ρ R_{i j})$ , which is a density operator because $R_{i j} ρ R_{i j}$ is positive trace class with trace $tr (ρ R_{i j}) > 0$ [F3, F4].

3.2step 1.1step 2.1F1F4algebra

[Marginals of the joint law] If $⋃_{j} C_{j} = R$ , the finite additivity of a PVM gives $\sum_{j} Q_{j} = F (⋃_{j} C_{j}) = F (R) = I$ [F1], so $\sum_{j} tr (ρ P_{i} Q_{j}) = tr (ρ P_{i} \sum_{j} Q_{j}) = tr (ρ P_{i})$ by linearity of $tr (ρ \cdot)$ and the ideal property [F4]; symmetrically $\sum_{i} tr (ρ P_{i} Q_{j}) = tr (ρ Q_{j})$ when $⋃_{i} B_{i} = R$ , so the joint law has the sharp-bin marginals.

4.1step 1.1step 3.1F4algebra

[Repeatability] Let $σ : = R_{i j} ρ R_{i j} / tr (ρ R_{i j})$ from step 3.1. Then $tr (σ R_{i j}) = tr (R_{i j} ρ R_{i j} R_{i j}) / tr (ρ R_{i j}) = tr (ρ R_{i j}) / tr (ρ R_{i j}) = 1$ since $R_{i j}^{2} = R_{i j}$ and $R_{i j} R_{i j} R_{i j} = R_{i j}$ [step 1.1, F4]; likewise $tr (σ P_{i}) = tr (P_{i} R_{i j} ρ R_{i j} P_{i}) / tr (ρ R_{i j}) = tr (R_{i j} ρ R_{i j}) / tr (ρ R_{i j}) = 1$ because $P_{i} R_{i j} = R_{i j} = R_{i j} P_{i}$ , and $tr (σ Q_{j}) = 1$ the same way. Hence the Lüders updates $R_{i j} σ R_{i j} / tr (σ R_{i j})$ , $P_{i} σ P_{i} / tr (σ P_{i})$ and $Q_{j} σ Q_{j} / tr (σ Q_{j})$ all return the same state: e.g. $R_{i j} σ R_{i j} = R_{i j}^{3} ρ R_{i j}^{3} / tr (ρ R_{i j}) = R_{i j} ρ R_{i j} / tr (ρ R_{i j}) = σ$ and the denominators equal $1$ , and similarly $P_{i} σ P_{i} = P_{i} R_{i j} ρ R_{i j} P_{i} / tr (ρ R_{i j}) = R_{i j} ρ R_{i j} / tr (ρ R_{i j}) = σ$ and $Q_{j} σ Q_{j} = σ$ . Thus an immediate repetition of the pair, or of either single sharp measurement, returns the same outcome with probability $1$ .

5.1step 4.1F2F5algebra∎

[Scope] Steps 2.1-4.1 are conditional computations inside the stipulated Lüders instrument and the joint PVM $G$ of [F2]; they assert joint probabilities, order independence and repeatability for that ideal model only, with no laboratory realization, no nonselective statement beyond step 3.2, and no time development.

5 · Examples, experiments and counterexamples

None yet.