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Cutoff dipole relative bound

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 $α > 0$ , let $h$ be the transverse space and $H_{f} = ⨁_{n} T_{n}$ the free transverse Fock Hamiltonian of Transverse continuum free Fock Hamiltonian, and let $A_{\pm} (g)$ , $g \in h$ , be the maximal creation and annihilation operators on $F (h)$ of Closed adjoint sector domains. Fix a cutoff $Λ > 0$ and vectors $g_{1}, \dots, g_{J} \in h$ ( $J \geq 1$ ) with $supp g_{j} \subseteq {k : 0 < ε (k) \leq Λ} and \frac{g_{j}}{\sqrt{ε}} \in h, ε (k) = α ∣ k ∣,$ and define the symmetric operator $ϕ (g) : = A_{-} (g) + A_{+} (g), D (ϕ (g)) : = D (H_{f}) .$ Let $d \geq 1$ , let $H_{a}$ be a Hermitian $d \times d$ matrix, and let $D_{1}, \dots, D_{J}$ be Hermitian $d \times d$ matrices with operator norms $∥ D_{j} ∥$ (The operator norm as the least bound and as the unit-sphere or unit-ball supremum). Then:

(a) for every $Ψ \in D (H_{f})$ , $∥ A_{-} (g) Ψ ∥^{2} \leq ∥ \frac{g}{\sqrt{ε}} ∥^{2} ⟨ H_{f} Ψ, Ψ ⟩, ∥ A_{+} (g) Ψ ∥^{2} = ∥ A_{-} (g) Ψ ∥^{2} + ∥ g ∥^{2} ∥ Ψ ∥^{2};$ in particular $Ψ \in D (A_{+} (g)) \cap D (A_{-} (g))$ and for every $η > 0$ there is a finite $C_{η}$ with $∥ ϕ (g) Ψ ∥ \leq η ∥ H_{f} Ψ ∥ + C_{η} ∥ Ψ ∥;$

(b) $ϕ (g)$ is symmetric on $D (H_{f})$ and has $H_{f}$ -bound zero (Relative boundedness with respect to an operator, Symmetric, self-adjoint and essentially self-adjoint operators);

(c) the free part $H_{0} : = H_{a} \otimes I + I \otimes H_{f}$ with domain $C^{d} \otimes D (H_{f})$ is self-adjoint on $C^{d} \otimes F (h)$ (Hilbert space); for every $η_{0} \in R$ the operator $H : = H_{0} + η_{0} \sum_{j = 1}^{J} D_{j} \otimes ϕ (g_{j}), D (H) : = C^{d} \otimes D (H_{f}),$ is self-adjoint on $C^{d} \otimes F (h)$ and bounded below (Kato-Rellich theorem, Adjoint of a densely defined operator).

The product $D_{j} \otimes ϕ (g_{j})$ and the coupling $η_{0}$ carry the dimension of energy in a physical normalisation; this normalisation is a model choice and no physical statement about radiation or spontaneous emission is made here.

Facts & Assumptions

Given: the cutoff $Λ > 0$ and vectors $g_{j}$ with $supp g_{j} \subseteq {0 < ε \leq Λ}$ , the maps $A_{\pm} (g_{j})$ and $ϕ (g_{j})$ on $F (h)$ , the free Hamiltonian $H_{f}$ with sector multipliers $T_{n}$ and number projections $Π_{n}$ , Hermitian matrices $H_{a}, D_{j}$ on $C^{d}$ , and $Ψ, Φ \in D (H_{f})$ .

[F1]

$A_{\pm} (g)$ are closed operators with $A_{+} (g) = A_{-} (g)^{*}$ and $A_{-} (g) = A_{+} (g)^{*}$ ; on the core $F_{0}$ the sector formulas $(A_{+} (g) ψ)_{n} = \sqrt{n} P_{+}^{(n)} (g \otimes ψ_{n - 1})$ and $(A_{-} (g) ψ)_{n} = \sqrt{n + 1} Ψ_{g}^{(n)} (ψ_{n + 1})$ hold, where $Ψ_{g}^{(n)}$ is the contraction in the first tensor factor with coefficient $⟨ \cdot, g ⟩$ (Closed adjoint sector domains, Creation contraction bounds).

[F2]

$H_{f} = ⨁_{n} T_{n}$ is self-adjoint and positive with $T_{n} = ε_{n} ∣_{P_{+}^{(n)}}$ , $ε_{n} (k_{1}, \dots, k_{n}) = \sum_{i \leq n} ε (k_{i})$ , and $⟨ H_{f} Ψ, Ψ ⟩ = \sum_{n} ⟨ ε_{n} Ψ_{n}, Ψ_{n} ⟩ < \infty$ for $Ψ \in D (H_{f})$ (Transverse continuum free Fock Hamiltonian).

[F3]

Under the iterated Euclidean product unitary, $h^{\otimes m}$ is identified with a closed subspace of $L^{2} ((R^{3})^{m}; (C^{3})^{\otimes m})$ ; elements of $ran P_{+}^{(m)}$ correspond to tensor-valued functions $F$ invariant under simultaneous momentum and polarization-factor permutations, so $∥ F ∥$ is invariant under momentum permutations, and the energy multiplier is $ε_{m} (k_{1}, \dots, k_{m}) = \sum_{i = 1}^{m} ε (k_{i})$ . For $F$ in its operator domain, $⟨ ε_{m} F, F ⟩ = \sum_{i = 1}^{m} \int ε (k_{i}) ∥ F (k_{1}, \dots, k_{m}) ∥^{2}$ , and symmetry makes each integral equal to $m^{- 1} ⟨ ε_{m} F, F ⟩$ (Euclidean L² product unitary: L²(R^a)⊗̂L²(R^b) ≅ L²(R^{a+b}), Hilbert tensor construction: inner product, completion and product bases, Permutation sector projections on the n-fold tensor power).

[F4]

Cauchy-Schwarz for the weighted $L^{2}$ integrals in step 1.1 follows from Cauchy-Schwarz inequality for $L^{2}$ ; Cauchy-Schwarz in the Fock Hilbert space, used for $⟨ H_{f} Ψ, Ψ ⟩$ , follows from Cauchy–Schwarz: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ , with equality exactly for dependent pairs.

[F5]

For $Ψ \in F_{0}$ the identity $∥ A_{+} (g) Ψ ∥^{2} = ∥ A_{-} (g) Ψ ∥^{2} + ∥ g ∥^{2} ∥ Ψ ∥^{2}$ holds (Closed adjoint sector domains (d)).

[F6]

Kato-Rellich: if $A$ is self-adjoint and $B$ is symmetric with $D (A) \subseteq D (B)$ and $A$ -bound less than $1$ , then $A + B$ on $D (A)$ is self-adjoint and bounded below when $A$ is (Kato-Rellich theorem, Relative boundedness with respect to an operator).

[F7]

Tonelli applies to the nonnegative squared norms on the sigma-finite Lebesgue product spaces; Fubini applies to absolutely integrable scalar component pairings (Tonelli's theorem for nonnegative measurable functions on a sigma-finite product, Fubini's theorem for L^1 functions on a sigma-finite product).

Proof

1.1F1F2F3F4F7algebra

[Contraction estimate] Let $Ψ \in D (H_{f})$ and $n \geq 0$ ; represent $Ψ_{n + 1} \in ran P_{+}^{(n + 1)}$ by its symmetric function $F$ under [F3]. For $v \in C^{3}$ define the first-factor contraction $C_{v} : (C^{3})^{\otimes (n + 1)} \to (C^{3})^{\otimes n}$ by $C_{v} (a \otimes b) = ⟨ a, v ⟩ b$ , extended linearly; finite orthonormal coordinates give $∥ C_{v} ∥ \leq ∥ v ∥$ . The contraction is represented by the tensor-valued function $G (k^{'}) = \int C_{g (k_{1})} F (k_{1}, k^{'}) d k_{1}$ : unweighted Cauchy–Schwarz gives integrable slices almost everywhere, Fubini identifies the pairing first for elementary tensors and then by density, and the bounded contraction formula [F1] extends the identity. Weighted Cauchy–Schwarz, tested against unit vectors of the finite-dimensional target tensor space, gives $∥ G (k^{'}) ∥^{2} \leq (\int ∥ g (k_{1}) ∥^{2} / ε (k_{1}) d k_{1}) (\int ε (k_{1}) ∥ F (k_{1}, k^{'}) ∥^{2} d k_{1})$ ; the null point $k_{1} = 0$ is omitted. Integrating over $k^{'}$ and using the symmetry computation of [F3] gives $∥ Ψ_{g}^{(n)} Ψ_{n + 1} ∥^{2} \leq ∥ g / \sqrt{ε} ∥^{2} \cdot \frac{1}{n + 1} ⟨ ε_{n + 1} F, F ⟩ = \frac{1}{n + 1} ∥ g / \sqrt{ε} ∥^{2} ⟨ ε_{n + 1} Ψ_{n + 1}, Ψ_{n + 1} ⟩$ . Hence $∥ (A_{-} (g) Ψ)_{n} ∥^{2} = (n + 1) ∥ Ψ_{g}^{(n)} Ψ_{n + 1} ∥^{2} \leq ∥ g / \sqrt{ε} ∥^{2} ⟨ ε_{n + 1} Ψ_{n + 1}, Ψ_{n + 1} ⟩$ ; summing over $n$ and using [F2] gives $∥ A_{-} (g) Ψ ∥^{2} \leq ∥ g / \sqrt{ε} ∥^{2} ⟨ H_{f} Ψ, Ψ ⟩$ .

1.2F2F3F4F6algebra

[The free part] Fix the standard orthonormal basis $(u_{a})_{a = 1}^{d}$ of $C^{d}$ . The map $(Ψ_{a})_{a = 1}^{d} \mapsto \sum_{a} u_{a} \otimes Ψ_{a}$ is an isometry from the finite Hilbert sum $F (h)^{d}$ onto $C^{d} \otimes F (h)$ , since elementary tensors span a dense subspace and the finite sum is complete. Under this identification $T : = I \otimes H_{f}$ has dense domain $D (H_{f})^{d}$ and acts componentwise. It is symmetric by [F2]. If $Φ \in D (T^{*})$ with $T^{*} Φ = Z$ , testing the adjoint identity on a vector with only component $a$ nonzero gives $Φ_{a} \in D (H_{f}^{*}) = D (H_{f})$ and $Z_{a} = H_{f} Φ_{a}$ for each $a$ . Conversely symmetry gives $T \subseteq T^{*}$ , so $T = T^{*}$ and $T \geq 0$ . The matrix amplification $R : = H_{a} \otimes I$ is bounded and self-adjoint: its entries act by scalar multiples of the identity, its adjoint entries are the conjugate transposed entries, and finite-dimensional Cauchy–Schwarz bounds its norm. It has $T$ -bound zero; hence [F6] makes $H_{0} = T + R$ self-adjoint on $D (H_{f})^{d} = C^{d} \otimes D (H_{f})$ and bounded below by $- ∥ R ∥$ , also directly from $T \geq 0$ and $∣ ⟨ R Ψ, Ψ ⟩ ∣ \leq ∥ R ∥ ∥ Ψ ∥^{2}$ . This componentwise adjoint argument establishes the general amplification here without applying the fixed- $N$ kinetic-operator interface.

2.1F1F5step 1.1algebra

[The norm identity on $D (H_{f})$ ] Let $Ψ \in D (H_{f})$ and let $Ψ^{(N)}$ be its truncations to sectors $\leq N$ . Step 1.1 gives $A_{-} (g) Ψ^{(N)} \to A_{-} (g) Ψ$ and $Ψ^{(N)} \to Ψ$ . For $N > M$ , $χ : = Ψ^{(N)} - Ψ^{(M)}$ is in $F_{0}$ , so [F5] gives $∥ A_{+} (g) χ ∥^{2} = ∥ A_{-} (g) χ ∥^{2} + ∥ g ∥^{2} ∥ χ ∥^{2} \to 0$ as $M, N \to \infty$ . Thus $A_{+} (g) Ψ^{(N)}$ is Cauchy; closedness of $A_{+} (g)$ from [F1] implies $Ψ \in D (A_{+} (g))$ and $A_{+} (g) Ψ^{(N)} \to A_{+} (g) Ψ$ . Passing to the limit in [F5] for $Ψ^{(N)}$ now proves $∥ A_{+} (g) Ψ ∥^{2} = ∥ A_{-} (g) Ψ ∥^{2} + ∥ g ∥^{2} ∥ Ψ ∥^{2}$ .

3.1F4step 1.1step 2.1algebra

[The relative bound] From steps 1.1 and 2.1, $∥ ϕ (g) Ψ ∥ \leq ∥ A_{-} (g) Ψ ∥ + ∥ A_{+} (g) Ψ ∥ \leq 2 ∥ g / \sqrt{ε} ∥ \sqrt{⟨ H_{f} Ψ, Ψ ⟩} + ∥ g ∥ ∥ Ψ ∥$ for $Ψ \in D (H_{f})$ ; since $⟨ H_{f} Ψ, Ψ ⟩ \leq ∥ H_{f} Ψ ∥ ∥ Ψ ∥$ by [F4], the inequality $2 \sqrt{a b} \leq η a + η^{- 1} b$ with $a = ∥ H_{f} Ψ ∥$ , $b = ∥ g / \sqrt{ε} ∥^{2} ∥ Ψ ∥$ gives, for every $η > 0$ , $∥ ϕ (g) Ψ ∥ \leq η ∥ H_{f} Ψ ∥ + C_{η} ∥ Ψ ∥$ with $C_{η} = η^{- 1} ∥ g / \sqrt{ε} ∥^{2} + ∥ g ∥$ . This proves (a) and shows the $H_{f}$ -bound zero.

4.1F1step 1.1step 2.1step 3.1algebra

[Symmetry of $ϕ (g)$ ] For $Ψ, Φ \in D (H_{f})$ , steps 1.1 and 2.1 put both vectors in the domains of the two closed operators, and the mutual adjointness $A_{+} (g) = A_{-} (g)^{*}$ of [F1] gives $⟨ ϕ (g) Ψ, Φ ⟩ = ⟨ A_{-} (g) Ψ, Φ ⟩ + ⟨ A_{+} (g) Ψ, Φ ⟩ = ⟨ Ψ, A_{+} (g) Φ ⟩ + ⟨ Ψ, A_{-} (g) Φ ⟩ = ⟨ Ψ, ϕ (g) Φ ⟩$ . By step 3.1, $ϕ (g)$ is symmetric on the dense domain $D (H_{f})$ with $H_{f}$ -bound zero, which is (b).

5.1F6step 3.1step 4.1step 1.2algebra

[Kato-Rellich] Let $B : = η_{0} \sum_{j} D_{j} \otimes ϕ (g_{j})$ with domain $C^{d} \otimes D (H_{f})$ . Since the $D_{j}$ are bounded and each $ϕ (g_{j})$ is symmetric with $H_{f}$ -bound zero (step 4.1, step 3.1), $B$ is symmetric and for every $η > 0$ there are finite constants with $∥ B Ψ ∥ \leq η ∥ H_{f} Ψ ∥ + C_{η} ∥ Ψ ∥$ ; as $H_{0}$ differs from $I \otimes H_{f}$ by the bounded operator $H_{a} \otimes I$ , the same estimate holds with $H_{f}$ replaced by $H_{0}$ up to constants, so the $H_{0}$ -bound of $B$ is zero and in particular less than $1$ . As $H_{0}$ is self-adjoint (step 1.2), [F6] gives that $H = H_{0} + B$ is self-adjoint on $C^{d} \otimes D (H_{f})$ and bounded below. This proves (c).

6.1step 1.1step 2.1step 3.1step 4.1step 1.2step 5.1algebra∎

[Conclusion and scope] Steps 1.1 and 2.1 prove (a), step 4.1 proves (b), and steps 1.2 and 5.1 prove (c). The final paragraph of the statement records that the normalisation of the coupling and of the dipole matrices is a model choice, and no physical radiation or emission statement is asserted. This completes the proof.

Cutoff dipole relative bound

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

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