Physics theoremStatement: AI-adaptedProof: AI-generated

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Photon-number expansion of the diagonal bath correlator

PDF p. 12: Eqs. (S3.4)–(S3.5)

Statement

Assume the Axiom of Choice, Countable Choice and Dependent Choice, and adopt the finite-mode thermal-bath and real-polarization dipole-coupling postulates. Let the finite set of modes be indexed by $j = 1, \dots, m$ , with $m \geq 1$ , frequencies $ω_{j} > 0$ , real transverse polarization vectors $ϵ_{j}$ , and $β > 0$ . Write $P_{α j} = {\hat{e}}_{α} \cdot ϵ_{j}$ and $Q_{j} (t) = κ (ω_{j})^{*} a_{j}^{†} e^{i ω_{j} t} + κ (ω_{j}) a_{j} e^{- i ω_{j} t}$ . Here $a_{j}$ and $a_{j}^{†}$ are the mode annihilation and creation operators, ${\hat{e}}_{α}$ is a Cartesian unit vector, and $κ$ contains the selected finite-mode field normalization. For the bath operator $B_{α} = - d_{0} E_{α} / \sqrt{γ}$ , with $γ > 0$ , put $J_{α α} (t) = {Tr}_{B} [{\tilde{B}}_{α} (t) B_{α} ρ_{B}]$ .

Use the maximal mode operators in the complete occupation-basis realization, identified with the finite-mode symmetric Fock realization by matching occupation vectors. Set $D = ⋂_{j = 1}^{m} (D (a_{j}) \cap D (a_{j}^{†}))$ . In this calculation, $E_{α}$ and $B_{α}$ mean the extensions of the initial finite-particle field sums to $D$ , with $B_{α} = - γ^{- 1 / 2} \sum_{j} P_{α j} Q_{j} (0)$ there. The free evolution is $U (t) ∣ n ⟩ = e^{- i t E_{n}} ∣ n ⟩$ , and ${\tilde{B}}_{α} (t) = U (t)^{*} B_{α} U (t)$ has domain $U (t)^{*} D = D$ . The Gibbs-weighted products in the correlator are defined on the whole bath space and are trace class, as proved below.

Let $n = (n_{1}, \dots, n_{m}) \in N_{0}^{m}$ , $E_{n} = \sum_{j} n_{j} ω_{j}$ , and $Z = \prod_{j} (1 - e^{- β ω_{j}})^{- 1}$ . Then, for every real $t$ ,

$J_{α α} (t) = \frac{1}{γ Z} \sum_{n} ⟨ n ∣ \sum_{j, k = 1}^{m} P_{α j} P_{α k} Q_{j} (t) Q_{k} (0) e^{- β H_{B}} ∣ n ⟩ = \frac{1}{γ Z} \sum_{n} e^{- β E_{n}} ⟨ n ∣ \sum_{j, k = 1}^{m} P_{α j} P_{α k} Q_{j} (t) Q_{k} (0) ∣ n ⟩ .$

Both series converge absolutely. In particular, with $N_{j} = (e^{β ω_{j}} - 1)^{- 1}$ ,

$J_{α α} (t) = \frac{1}{γ} \sum_{j = 1}^{m} P_{α j}^{2} ∣ κ (ω_{j}) ∣^{2} [N_{j} e^{i ω_{j} t} + (N_{j} + 1) e^{- i ω_{j} t}] .$

The article writes the corresponding number-state expansion with continuum mode integrals in equations (S3.4)–(S3.5). Those integrals are formal here: no interchange with an infinite-volume thermal trace or limit of finite-mode expressions is claimed. The explicit factor $1 / γ$ above follows from the stated bath-operator normalization; it may be absorbed only by redefining the coupling coefficients or correlator consistently.

Facts & Assumptions

Given: The mode data, choice assumptions, thermal preparation and coupling postulates in the Statement.

[F1]

For a stationary bath density operator, the bath correlation matrix is defined when every two-time product ${\tilde{B}}_{α} (t) {\tilde{B}}_{η} (s) ρ_{B}$ has a well-defined trace and all resulting entries are invariant under a common time translation. Then $J_{α η} (t - s) = {Tr}_{B} [{\tilde{B}}_{α} (t) {\tilde{B}}_{η} (s) ρ_{B}]$ ; its diagonal at $s = 0$ is the correlator used here. The channel labels $α, η$ range independently over ${x, y, z}$ . Bath correlation matrix and Fourier convention

[F2]

The finite regulator supplies independent positive-frequency oscillator modes, the complete occupation basis, the maximal shifts $a_{j} ∣ n ⟩ = \sqrt{n_{j}} ∣ n - e_{j} ⟩$ (zero for $n_{j} = 0$ ) and $a_{j}^{†} ∣ n ⟩ = \sqrt{n_{j} + 1} ∣ n + e_{j} ⟩$ , their occupation-weight domains, and diagonal free evolution. Here $e_{j}$ is the unit occupation vector. A continuum thermal preparation and regulator-removal limit require separate specifications. Regularized transverse photon dipole bath model

[F3]

For the finite thermal bath, $H_{B} = \sum_{j} ω_{j} a_{j}^{†} a_{j}$ and $ρ_{B} = Z^{- 1} e^{- β H_{B}}$ , with the stated number basis and partition function. Thermal photon bath and partition function

[F4]

In the real-polarization convention, $B_{α} = - d_{0} E_{α} / \sqrt{γ}$ and the field coefficients are $P_{α j} κ (ω_{j})$ for annihilation and their conjugates for creation. Dimensionless dipole and ULE coupling identification

[F5]

In the finite-mode symmetric Fock realization, the occupation vectors form a complete orthonormal basis with the stated diagonal energies and Gibbs weights. For finitely many positive-frequency modes and positive $β$ , every ordered product of two creation or annihilation operators composed with the Gibbs density is defined on the entire bath space and is trace class. Their thermal moments are $⟨ a_{j}^{†} a_{k} ⟩ = δ_{j k} N_{j}$ , $⟨ a_{j} a_{k}^{†} ⟩ = δ_{j k} (N_{j} + 1)$ , and the moments of two creations or two annihilations vanish. Finite-mode bosonic Gibbs occupations and moments

[F6]

Under Countable Choice, the trace of a trace-class operator equals its absolutely convergent diagonal sum in every supplied Hilbert basis; finite linear combinations remain trace class and their trace is linear. Apply this with the complete occupation basis. To use the trace supplier's first-variable-linear pairing with the Fock second-variable-linear pairing, put $(u, v)_{1} = ⟨ v, u ⟩_{2}$ ; then $(T u, u)_{1} = ⟨ u, T u ⟩_{2}$ , with unchanged norms and operator domains. Trace is absolutely convergent and basis independent

[A1]

Assume the Axiom of Choice, Countable Choice and Dependent Choice for the cited constructions. The Axiom of Choice The Axiom of Countable Choice (AC_ω) The axiom of dependent choice: a relation in which every element is related to something admits an ℕ-indexed chain

Proof

Goal: Prove the two number-state expressions, their absolute convergence, and the evaluated finite-mode correlator with the stated field domains.

1.1F2F3F5A1algebra

Match the complete occupation bases in [F2] and [F5]. On finite linear combinations this map preserves the norm and has dense range, so it extends to a unitary identification. The ladder actions agree: the symmetric Fock occupation vectors in [F5] have the raising and lowering actions obtained from their normalized symmetric tensors, and [F2] gives the same coordinate shifts. The maximal domains are exactly the square-summability domains of these weighted coordinates; thus the identification intertwines the maximal mode operators and the diagonal Hamiltonians. Conjugating the first-variable-linear pairing of the tensor realization gives the second-variable-linear Fock convention without changing norms or domains. With $ℏ = 1$ , [F2] and [F3] give $U (t) ∣ n ⟩ = e^{- i t E_{n}} ∣ n ⟩$ . Its modulus-one phases preserve every mode domain. The coordinate shifts therefore give $U (t)^{*} a_{j} U (t) = e^{- i ω_{j} t} a_{j}$ and $U (t)^{*} a_{j}^{†} U (t) = e^{i ω_{j} t} a_{j}^{†}$ on their full maximal domains, including the zero lowering coordinate at $n_{j} = 0$ .

2.1F2F4step 1.1algebra

The finite field sum from [F4] extends to the common linear domain $D$ specified in the Statement and agrees with the initial field on finite-particle vectors. By step 1.1, $U (t)^{*} D = D$ , and conjugation gives ${\tilde{B}}_{α} (t) = - γ^{- 1 / 2} \sum_{j} P_{α j} Q_{j} (t)$ on precisely this domain.

3.1F3F5F6step 2.1algebra

Write $L$ for any one of the finitely many $a_{j}, a_{j}^{†}$ . By [F5], $L^{'} L ρ_{B}$ is defined on the whole space for every ordered pair $L^{'}, L$ . In particular, $ρ_{B} ψ$ is in every mode domain and $L ρ_{B} ψ$ is in every mode domain for every bath vector $ψ$ . For every Cartesian channel $η$ and real $s$ , step 2.1 expresses ${\tilde{B}}_{η} (s)$ as a finite linear combination of these mode operators on $D$ . Hence ${\tilde{B}}_{η} (s) ρ_{B} ψ \in D$ , and ${\tilde{B}}_{α} (t) {\tilde{B}}_{η} (s) ρ_{B}$ is an actual whole-space operator product for every $α, η \in {x, y, z}$ and $t, s \in R$ . Its finite expansion consists of the trace-class products in [F5], so [F6] makes it trace class. Define its trace to be $C_{α η} (t, s)$ . Substituting [F3] and step 2.1 gives $C_{α η} (t, s) = (γ Z)^{- 1} {Tr}_{B} [\sum_{j, k} P_{α j} P_{η k} Q_{j} (t) Q_{k} (s) e^{- β H_{B}}]$ .

4.1F1F3F5F6step 3.1algebra

Expand each $Q_{j} (t) Q_{k} (s)$ in step 3.1 into its four ladder-operator products. By [F5], the two equal-kind products have zero thermal trace, while the mixed products vanish unless $j = k$ . For $j = k$ , their respective coefficients and moments are $∣ κ (ω_{j}) ∣^{2} e^{i ω_{j} (t - s)} N_{j}$ and $∣ κ (ω_{j}) ∣^{2} e^{- i ω_{j} (t - s)} (N_{j} + 1)$ . Trace linearity therefore gives $C_{α η} (t, s) = γ^{- 1} \sum_{j = 1}^{m} P_{α j} P_{η j} ∣ κ (ω_{j}) ∣^{2} [N_{j} e^{i ω_{j} (t - s)} + (N_{j} + 1) e^{- i ω_{j} (t - s)}]$ for every channel pair. This depends only on $t - s$ , so $C_{α η} (t + u, s + u) = C_{α η} (t, s)$ for every real $u$ . The Gibbs preparation is stationary by [F3], and step 3.1 supplies all nine trace-class products; thus all conditions of [F1] hold for the three-channel matrix. Consequently $J_{α η} (t - s) = C_{α η} (t, s)$ , including the diagonal at $s = 0$ .

5.1F3F5F6step 3.1step 4.1algebra

Each summand in the finite double sum of step 3.1, with $η = α$ and $s = 0$ , is trace class by [F5] and $e^{- β H_{B}} = Z ρ_{B}$ , with $0 < Z < \infty$ from [F3]. Applying [F6] to the supplied occupation basis gives its absolutely convergent diagonal sum; trace linearity permits summing these finitely many series. Also $e^{- β H_{B}} ∣ n ⟩ = e^{- β E_{n}} ∣ n ⟩$ by [F3] and [F5]. These are exactly the two number-state expressions in the Statement. The finite sum of the absolute diagonal sums bounds the absolute sum of the combined series, proving absolute convergence of both expressions without any continuum interchange.

6.1step 3.1step 4.1step 5.1algebra∎

Set $η = α$ and $s = 0$ in step 4.1. The polarization factors become $P_{α j}^{2}$ , and the surviving terms are $∣ κ (ω_{j}) ∣^{2} e^{i ω_{j} t} N_{j}$ and $∣ κ (ω_{j}) ∣^{2} e^{- i ω_{j} t} (N_{j} + 1)$ , with the normalization $1 / γ$ from step 3.1. This is the final displayed formula, and step 5.1 supplies both absolutely convergent number-state representations. No continuum limit is used.

Source notes

Equations (S3.4)–(S3.5) of the article use continuum integrations and display the Gibbs operator before replacing it by its occupation-basis eigenvalue. The finite sums above are the regulated counterpart, not a proof that those continuum expressions exist. The article's preceding identification uses $B_{α} = - d_{0} E_{α} / \sqrt{γ}$ ; with the separately stated $κ = d_{0} E$ , that identification yields the displayed $1 / γ$ in the correlator. The conditional calculation makes this normalization explicit without assuming a value of $γ$ .

Depends on

Used by · 3

Dependency tree · 2 levels

Each result is drawn at its shortest distance from this item. Arrows run from prerequisites to the items that use them; this item has a heavier outline. External source nodes link directly to the cited text. The view shows up to two dependency steps; open a prerequisite to continue.

Sources

Monolithic Quantum Error Correction in the Presence of Distinguishability · PDF page 11, equation (S3.3) and the subsequent bath-operator identification; page 12, equations (S3.4)–(S3.5)