Alphabeta Math

1 · Prerequisites

2 · Summary

The companion page to qm-math-composites records three explicit mathematical checks behind its tensor, density and instrument results. All three items are mathematics: they carry domain: mathematics, depend only on the mathematical items of the parent page and on mathematical suppliers, and make no physical or empirical claim.

Bell reduced state computes the reduced states of the unit vector $Ψ = (e_{+} \otimes e_{-} - e_{-} \otimes e_{+}) / \sqrt{2}$ in $C^{2} \otimes C^{2}$ : the coefficient matrix is $C = \frac{1}{\sqrt{2}} (\begin{array}{cc} 0 & 1 \\ - 1 & 0 \end{array})$ , both partial traces are $\frac{1}{2} I$ , the reduced purity is $\frac{1}{2}$ while $tr (T_{Ψ, Ψ}^{2}) = 1$ , the Schmidt coefficients are $s_{1} = s_{2} = 1 / \sqrt{2}$ , and $Ψ$ is not a product vector. Nonunique mixed ensemble shows that the maximally mixed operator $\frac{1}{2} I$ on $C^{2}$ has the two distinct orthonormal decompositions $\frac{1}{2} T_{e_{1}, e_{1}} + \frac{1}{2} T_{e_{2}, e_{2}}$ and $\frac{1}{2} T_{e_{+}, e_{+}} + \frac{1}{2} T_{e_{-}, e_{-}}$ , so this density operator has two distinct orthonormal convex decompositions into rank-one density operators. Selective-conditioning counterexample refutes the selective claim that local conditioning preserves the opposite reduced state: on the same Bell state, the Lüders outcome of the second factor with $M_{0} = T_{e_{-}, e_{-}}$ has probability $\frac{1}{2}$ and leaves the first factor in $T_{e_{+}, e_{+}}$ , which differs from the unconditional $\frac{1}{2} I$ , while the nonselective sum over both outcomes does return $\frac{1}{2} I$ as in Local trace-preserving Kraus maps do not change the opposite reduced state — the counterexample therefore separates the nonselective identity from its selective counterpart rather than contradicting the former.

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

None yet.

5 · Examples, experiments and counterexamples

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Bell reduced state

Example

Let $V = W = C^{2}$ with the standard inner product and orthonormal basis $e_{+} = (\binom{1}{0})$ , $e_{-} = (\binom{0}{1})$ (Hilbert tensor construction: inner product, completion and product bases), and let

$Ψ : = \frac{1}{\sqrt{2}} (e_{+} \otimes e_{-} - e_{-} \otimes e_{+}) \in V \otimes W .$

Here a product vector means an elementary tensor $v \otimes u$ with $v \in V$ and $u \in W$ ; $T_{Ψ, Ψ}$ is the rank-one density operator determined by the unit vector $Ψ$ .

Then:

(a) $∥ Ψ ∥ = 1$ , and the coefficient matrix of $Ψ$ in the bases $(e_{+}, e_{-})$ is $C = \frac{1}{\sqrt{2}} (\begin{array}{cc} 0 & 1 \\ - 1 & 0 \end{array})$ ;

(b) the partial traces satisfy ${Tr}_{W} T_{Ψ, Ψ} = C C^{*} = \frac{1}{2} I_{V}$ and ${Tr}_{V} T_{Ψ, Ψ} = C^{*} C = \frac{1}{2} I_{W}$ , so both reduced states are the maximally mixed matrix $\frac{1}{2} I$ ;

(c) the reduced density $ρ = \frac{1}{2} I$ has eigenvalues $\frac{1}{2}, \frac{1}{2}$ and trace-square $tr ρ^{2} = \frac{1}{2}$ , while $tr (T_{Ψ, Ψ}^{2}) = 1$ ; the Schmidt coefficients of $Ψ$ are $s_{1} = s_{2} = \frac{1}{\sqrt{2}}$ , and $Ψ$ is not a product vector.

Facts & Assumptions

Given: $V = W = C^{2}$ with the standard inner product; the orthonormal basis $(e_{+}, e_{-})$ ; the vector $Ψ$ above; the operators $T_{u, v} x = ⟨ x, v ⟩ u$ on the finite-dimensional spaces.

[A1]

On $V \otimes W$ the pairing is $⟨ u \otimes v, u^{'} \otimes v^{'} ⟩ = ⟨ u, u^{'} ⟩_{V} ⟨ v, v^{'} ⟩_{W}$ ; the four elementary tensors $e_{\pm} \otimes e_{\pm}$ form an orthonormal basis, each $Ψ \in V \otimes W$ has a unique expansion $\sum_{i j} c_{i j} e_{i} \otimes e_{j}$ with $∥ Ψ ∥^{2} = \sum_{i j} ∣ c_{i j} ∣^{2}$ , and the basis-free conventions (linearity in the first argument, conjugate symmetry) are those of Hilbert tensor construction: inner product, completion and product bases.

[A2]

For endomorphisms $R$ of $V \otimes W$ there is a unique ${Tr}_{W} R$ in $End (V)$ with $tr (({Tr}_{W} R) A) = tr (R (A \otimes I_{W}))$ for all $A$ ; it is linear and positive, ${Tr}_{W} T_{Ψ, Ψ} = C C^{*}$ where $C : W \to V$ is the coefficient map determined by $Ψ = \sum_{j} C f_{j} \otimes f_{j}$ , and ${Tr}_{V}$ is defined symmetrically; the Schmidt decomposition $Ψ = \sum_{r = 1}^{k} s_{r} v_{r} \otimes u_{r}$ has orthonormal families and positive $s_{1} \geq \dots \geq s_{k}$ equal to the positive singular values of $C$ , with $∥ Ψ ∥^{2} = \sum_{r} s_{r}^{2}$ , and for $∥ Ψ ∥ = 1$ the reduced density $ρ = {Tr}_{W} T_{Ψ, Ψ}$ satisfies $tr ρ^{2} = \sum_{r} s_{r}^{4}$ , which equals $1$ exactly when $k = 1$ (Finite partial trace, Schmidt decomposition and finite no-signalling(a),(b),(c)).

[A3]

$T_{u, v}$ is trace class with $tr (T_{u, v}) = ⟨ u, v ⟩$ , and $tr$ is linear (Rank-one operator calculus).

[A4]

In an orthonormal basis of a finite-dimensional inner product space the matrix of $T_{u, v}$ is the column-times-row matrix $u v^{*}$ , so $tr (T_{u, v}) = ⟨ u, v ⟩$ ; $T_{x, z} T_{w, y} = ⟨ w, z ⟩ T_{x, y}$ for rank-one operators.

Verification

1.1A1algebra

[Coefficient matrix and norm] By [A1], $Ψ = \frac{1}{\sqrt{2}} e_{+} \otimes e_{-} - \frac{1}{\sqrt{2}} e_{-} \otimes e_{+}$ is the expansion with $c_{+ -} = \frac{1}{\sqrt{2}}$ , $c_{- +} = - \frac{1}{\sqrt{2}}$ and the other coefficients $0$ , so the coefficient map $C : W \to V$ is the matrix $C = \frac{1}{\sqrt{2}} (\begin{array}{cc} 0 & 1 \\ - 1 & 0 \end{array})$ in the bases $(e_{+}, e_{-})$ (columns indexed by $W$ ), and $∥ Ψ ∥^{2} = ∣ c_{+ -} ∣^{2} + ∣ c_{- +} ∣^{2} = 1$ .

2.1step 1.1A2algebra

[Partial traces] Computing $C C^{*} = \frac{1}{2} (\begin{array}{cc} 0 & 1 \\ - 1 & 0 \end{array}) (\begin{array}{cc} 0 & - 1 \\ 1 & 0 \end{array}) = \frac{1}{2} I_{V}$ and $C^{*} C = \frac{1}{2} I_{W}$ , [A2] gives ${Tr}_{W} T_{Ψ, Ψ} = C C^{*} = \frac{1}{2} I_{V}$ and, by symmetry, ${Tr}_{V} T_{Ψ, Ψ} = C^{*} C = \frac{1}{2} I_{W}$ ; both are positive with trace $1$ , as they must be for the reduction of a unit vector.

3.1step 1.1step 2.1A2A3A4algebra∎

[Purity and Schmidt coefficients] The matrix $\frac{1}{2} I_{V}$ has the single eigenvalue $\frac{1}{2}$ with multiplicity $2$ (in any basis its matrix is diagonal with entries $\frac{1}{2}$ ), so $tr ρ^{2} = 2 \cdot (\frac{1}{2})^{2} = \frac{1}{2}$ for $ρ = {Tr}_{W} T_{Ψ, Ψ}$ ; the positive singular values of $C$ are the square roots of the eigenvalues of $C^{*} C = \frac{1}{2} I_{W}$ , namely $s_{1} = s_{2} = \frac{1}{\sqrt{2}}$ , so the Schmidt decomposition of $Ψ$ has $k = 2$ terms and $\sum_{r} s_{r}^{2} = 1$ , in agreement with the norm formula of [A2]; since $k = 2 > 1$ , $Ψ$ is not a product vector. Finally $T_{Ψ, Ψ}^{2} = T_{Ψ, Ψ}$ because $T_{Ψ, Ψ} T_{Ψ, Ψ} = ⟨ Ψ, Ψ ⟩ T_{Ψ, Ψ}$ by [A4] and $∥ Ψ ∥ = 1$ , so $tr (T_{Ψ, Ψ}^{2}) = tr (T_{Ψ, Ψ}) = ⟨ Ψ, Ψ ⟩ = 1$ by [A3].

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Nonunique mixed ensemble

Example

On $H = C^{2}$ with the standard inner product and orthonormal basis $e_{1} = (\binom{1}{0})$ , $e_{2} = (\binom{0}{1})$ , put $e_{\pm} : = \frac{1}{\sqrt{2}} (e_{1} \pm e_{2})$ (Density operators and scalar spectral probability). Then the maximally mixed density operator $\frac{1}{2} I$ admits the two orthonormal convex decompositions

$\frac{1}{2} I = \frac{1}{2} T_{e_{1}, e_{1}} + \frac{1}{2} T_{e_{2}, e_{2}} = \frac{1}{2} T_{e_{+}, e_{+}} + \frac{1}{2} T_{e_{-}, e_{-}},$

where $(e_{1}, e_{2})$ and $(e_{+}, e_{-})$ are distinct orthonormal bases of $H$ . Thus the same density operator has two distinct orthonormal convex decompositions into rank-one density operators.

Facts & Assumptions

Given: $H = C^{2}$ with the standard inner product; the orthonormal basis $(e_{1}, e_{2})$ ; the unit vectors $e_{\pm}$ ; the operators $T_{u, v} x = ⟨ x, v ⟩ u$ .

[A1]

A density operator is a self-adjoint positive trace-class operator with trace $1$ (Density operators and scalar spectral probability(a), definition only). For a finite family of unit vectors $u_{j} i n m a t h b b C^{2}$ and weights $w_{j} g e 0$ with $s u m_{j} w_{j} = 1$ , set $h o : = s u m_{j} w_{j} u_{j} u_{j}^{*}$ . This is a finite matrix and hence trace class; it is self-adjoint, $x^{*} h o x = s u m_{j} w_{j} ∣ u_{j}^{*} x ∣^{2} g e 0$ , and its finite trace is $s u m_{j} w_{j} s u m_{k = 1}^{2} ∣ (u_{j})_{k} ∣^{2} = s u m_{j} w_{j} = 1$ . Thus it is a density operator even without orthogonality. This is a local finite-matrix calculation, not an invocation of the supplier's Countable Choice expansion.

[A2]

$T_{u, v}$ is trace class with $tr (T_{u, v}) = ⟨ u, v ⟩$ and $∥ T_{u, v} ∥_{1} = ∥ u ∥ ∥ v ∥$ ; the trace is linear on trace-class operators (Rank-one operator calculus). In the present $H = C^{2}$ these facts follow directly from $T_{u, v} = u v^{*}$ : the diagonal sum is $\sum_{k} u_{k} \overline{v_{k}} = ⟨ u, v ⟩$ , and $(u v^{*})^{*} (u v^{*}) = ∥ u ∥^{2} v v^{*}$ has at most one nonzero eigenvalue, $∥ u ∥^{2} ∥ v ∥^{2}$ , so its singular-value sum is $∥ u ∥ ∥ v ∥$ . Finite diagonal sums are linear. No choice principle is used in these finite computations.

[A3]

In an orthonormal basis of $C^{2}$ the matrix of $T_{u, v}$ is the column-times-row matrix $u v^{*}$ ; $T_{e_{1}, e_{1}}$ and $T_{e_{2}, e_{2}}$ therefore have matrices $(\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array})$ and $(\begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array})$ .

Verification

1.1A1A2A3algebra

[Computational basis decomposition] By [A3], $\frac{1}{2} T_{e_{1}, e_{1}} + \frac{1}{2} T_{e_{2}, e_{2}}$ has matrix $\frac{1}{2} (\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array}) + \frac{1}{2} (\begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array}) = \frac{1}{2} I$ , and by [A2] it is a trace-class operator of trace $\frac{1}{2} + \frac{1}{2} = 1$ ; by [A1] it is a density operator coming from the orthonormal family $(e_{1}, e_{2})$ with weights $\frac{1}{2}, \frac{1}{2}$ .

2.1step 1.1A1A3algebra

[Fixed $\pm$ basis decomposition] The vectors $e_{\pm} = \frac{1}{\sqrt{2}} (e_{1} \pm e_{2})$ are unit and orthogonal since $⟨ e_{+}, e_{-} ⟩ = \frac{1}{2} (⟨ e_{1}, e_{1} ⟩ - ⟨ e_{1}, e_{2} ⟩ + ⟨ e_{2}, e_{1} ⟩ - ⟨ e_{2}, e_{2} ⟩) = 0$ ; their matrices are $T_{e_{+}, e_{+}} = \frac{1}{2} (\begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array})$ and $T_{e_{-}, e_{-}} = \frac{1}{2} (\begin{array}{cc} 1 & - 1 \\ - 1 & 1 \end{array})$ , so $\frac{1}{2} T_{e_{+}, e_{+}} + \frac{1}{2} T_{e_{-}, e_{-}} = \frac{1}{2} I$ again, with weights $\frac{1}{2}, \frac{1}{2}$ on an orthonormal family; thus the same operator $\frac{1}{2} I$ has a second decomposition.

3.1step 1.1step 2.1A1algebra∎

[The decompositions are different] The two orthonormal families are distinct: $∣ ⟨ e_{1}, e_{+} ⟩ ∣ = \frac{1}{\sqrt{2}} \notin {0, 1}$ , so $e_{1}$ is not a scalar multiple of $e_{\pm}$ and $T_{e_{1}, e_{1}}$ is not one of $T_{e_{\pm}, e_{\pm}}$ (their matrices of step 2.1 differ from $(\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array})$ ); hence the two lists of rank-one density operators are not the same up to reordering, while both convex combinations equal the single operator $\frac{1}{2} I$ by steps 1.1 and 2.1. Therefore this density operator has a nonunique orthonormal convex decomposition into rank-one density operators, consistent with the expansion in [A1].

Remarks

The operator identity is basis-independent. Both decompositions define the same matrix $ρ = \frac{1}{2} I$ , so $tr (ρ A) = \frac{1}{2} tr A$ for every endomorphism $A$ of $H$ .
Nonuniqueness needs no optimization. The two decompositions here are already orthonormal and vary only in the choice of basis; no operational preparation procedure is specified.

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Selective-conditioning counterexample

Statement refuted

False claim: for every density operator $ρ$ on a tensor product $H \otimes K$ of finite-dimensional complex inner product spaces and every finite local family $(M_{a})$ on $K$ with $\sum_{a} M_{a}^{*} M_{a} = I_{K}$ , the conditional reduced state on $H$ after the outcome $a$ is again ${Tr}_{K} ρ$ whenever its probability is positive.

The claim confuses the nonselective identity ${Tr}_{K} (\sum_{a} (I \otimes M_{a}) ρ (I \otimes M_{a}^{*})) = {Tr}_{K} ρ$ of Local trace-preserving Kraus maps do not change the opposite reduced state with the corresponding statement for one selected outcome. The displayed unit vector in $C^{2} \otimes C^{2}$ and one outcome of the explicitly defined projection instrument below refute the selective claim.

Facts & Assumptions

Given: $H = K = C^{2}$ with orthonormal bases $(e_{+}, e_{-})$ ; the unit vector $Ψ = \frac{1}{\sqrt{2}} (e_{+} \otimes e_{-} - e_{-} \otimes e_{+})$ ; the density operator $ρ = T_{Ψ, Ψ}$ with $T_{u, v} x = ⟨ x, v ⟩ u$ ; the rank-one projections $P_{0} : = T_{e_{-}, e_{-}}$ and $P_{1} : = T_{e_{+}, e_{+}} = I_{K} - P_{0}$ on $K$ .

[L1]

On $H \otimes K$ the pairing is $⟨ u \otimes v, u^{'} \otimes v^{'} ⟩ = ⟨ u, u^{'} ⟩_{H} ⟨ v, v^{'} ⟩_{K}$ ; the four tensors $e_{\pm} \otimes e_{\pm}$ form an orthonormal basis, $∥ Ψ ∥^{2} = 1$ , and $ρ = T_{Ψ, Ψ}$ is a density operator with $tr ρ = 1$ (Hilbert tensor construction: inner product, completion and product bases, Density operators and scalar spectral probability(a)); a partial trace ${Tr}_{K}$ is the unique linear map with $tr (({Tr}_{K} T) A) = tr (T (A \otimes I_{K}))$ , it is linear, and ${Tr}_{W} T_{Φ, Φ} = C C^{*}$ for the coefficient map $C$ of $Φ$ (Finite partial trace, Schmidt decomposition and finite no-signalling(a),(b)).

[L2]

${Tr}_{W}$ is linear, and for endomorphisms $M_{a}$ of $W$ with $\sum_{a} M_{a}^{*} M_{a} = I_{W}$ the nonselective sum satisfies ${Tr}_{W} (\sum_{a} (I \otimes M_{a}) R (I \otimes M_{a}^{*})) = {Tr}_{W} R$ for every endomorphism $R$ ; the lemma covers only the nonselective sum and says nothing about one selected summand (Finite partial trace, Schmidt decomposition and finite no-signalling(d), Local trace-preserving Kraus maps do not change the opposite reduced state).

[L3]

The refuted claim: for every unit $Ψ$ and every outcome $a$ of a local family $(M_{a})$ with $\sum_{a} M_{a}^{*} M_{a} = I$ , the conditional reduced state ${Tr}_{K} ((I \otimes M_{a}) ρ (I \otimes M_{a}^{*})) / tr ((I \otimes M_{a}) ρ (I \otimes M_{a}^{*}))$ equals ${Tr}_{K} ρ$ whenever the probability is positive.

Counterexample

1.1L1algebra

[The instrument] On $K$ take the self-adjoint projections $M_{0} : = P_{0} = T_{e_{-}, e_{-}}$ and $M_{1} : = P_{1} = T_{e_{+}, e_{+}}$ , so that $M_{a}^{*} M_{a} = P_{a}$ for $a = 0, 1$ and $\sum_{a} M_{a}^{*} M_{a} = P_{0} + P_{1} = I_{K}$ : this is the stipulated Lüders instrument for the sharp observable $P_{0}$ . Its outcome- $0$ non-normalized conditional operator on $H \otimes K$ is $(I \otimes P_{0}) ρ (I \otimes P_{0})$ .

2.1step 1.1L1algebra

[The selected outcome] Let $Ψ_{0} : = (I \otimes P_{0}) Ψ = \frac{1}{\sqrt{2}} e_{+} \otimes e_{-}$ . Its coefficient map in the chosen bases has matrix $C_{0} = \frac{1}{\sqrt{2}} (\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array})$ . The outcome probability is $tr ((I \otimes P_{0}) ρ (I \otimes P_{0})) = ∥ Ψ_{0} ∥^{2} = \frac{1}{2} > 0$ , and the non-normalized conditional operator is $T_{Ψ_{0}, Ψ_{0}} = \frac{1}{2} T_{e_{+} \otimes e_{-}, e_{+} \otimes e_{-}}$ . After normalization the conditional state is $T_{e_{+} \otimes e_{-}, e_{+} \otimes e_{-}}$ .

3.1step 2.1L1algebra

[The conditional reduced state differs] The coefficient map of the unit vector $e_{+} \otimes e_{-}$ has one nonzero matrix entry, equal to $1$ in the $e_{+}$ row and $e_{-}$ column, so its product with its adjoint is $T_{e_{+}, e_{+}}$ . By [L1], the conditional reduced state on $H$ is therefore ${Tr}_{K} T_{e_{+} \otimes e_{-}, e_{+} \otimes e_{-}} = T_{e_{+}, e_{+}}$ . The unconditional reduced state is ${Tr}_{K} ρ = C C^{*}$ for the coefficient matrix $C = \frac{1}{\sqrt{2}} (\begin{array}{cc} 0 & 1 \\ - 1 & 0 \end{array})$ of $Ψ$ , namely $\frac{1}{2} I_{H}$ . These differ, since $⟨ T_{e_{+}, e_{+}} e_{+}, e_{+} ⟩ = 1 \neq \frac{1}{2} = ⟨ \frac{1}{2} I_{H} e_{+}, e_{+} ⟩$ . Hence the selective conditional reduced state is not the unconditional one, and the claim of [L3] fails for this unit vector and instrument.

4.1step 3.1L1L2algebra∎

[The nonselective identity still holds] The outcome- $1$ summand is $(I \otimes P_{1}) ρ (I \otimes P_{1})$ , and $(I \otimes P_{1}) Ψ = - \frac{1}{\sqrt{2}} e_{-} \otimes e_{+}$ , so this summand equals $\frac{1}{2} T_{e_{-} \otimes e_{+}, e_{-} \otimes e_{+}}$ . The coefficient map of the unit vector $e_{-} \otimes e_{+}$ has one nonzero entry in the $e_{-}$ row, so its reduced state on $H$ is $T_{e_{-}, e_{-}}$ by [L1]. Adding both summands and using the linearity of ${Tr}_{K}$ [L2] gives ${Tr}_{K} (\sum_{a} (I \otimes M_{a}) ρ (I \otimes M_{a}^{*})) = \frac{1}{2} T_{e_{+}, e_{+}} + \frac{1}{2} T_{e_{-}, e_{-}} = \frac{1}{2} I_{H} = {Tr}_{K} ρ$ , in agreement with the nonselective identity of [L2]. The refuted claim is thus exactly the illegitimate replacement of that nonselective identity by its selective counterpart.

Qm Math Composites — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

5 · Examples, experiments and counterexamples

Bell reduced state

Example

Facts & Assumptions

Verification

Nonunique mixed ensemble

Example

Facts & Assumptions

Verification

Remarks

Selective-conditioning counterexample

Statement refuted

Facts & Assumptions

Counterexample

Sources