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Physics theoremStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)
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Jaynes–Cummings invariant blocks and Rabi frequency

Physical scope

An exact conditional calculation for the single-mode, two-level rotating-wave model on the full infinite-dimensional oscillator Fock space. The excitation blocks and self-adjoint direct-sum realization are model deductions. The usual weak-coupling rotating-wave scale condition is not uniform over all Fock states; no continuum spontaneous-emission law, detector observation, or laboratory identification is asserted.

Statement

Take the one-mode, two-level specialization of the finite-mode assignment and use its SI convention with ℏ>0 in joule-seconds. Let ω>0 be the selected angular frequency in seconds−1, let HA=C2 have orthonormal states ∣g⟩,∣e⟩, and choose ω0>0 in seconds−1, d∈R in coulomb-metres, and E∈R in volts per metre. Set σ+=∣e⟩⟨g∣, σ−=∣g⟩⟨e∣, σz=∣e⟩⟨e∣−∣g⟩⟨g∣, and σx=σ++σ−. Choose HA=(ℏω0/2)σz and D=dσx; then the dipole expression −ED⊗(a+a∗) has energy units. Define the real coupling g:=−Ed/ℏ in seconds−1. Retaining from the dipole expression just the excitation-exchange terms σ+⊗a and σ−⊗a∗ gives coefficient −Ed=ℏg. Thus the rotating-wave Hamiltonian below is a specified specialization of the model assignment, not an independently measured interaction.

Let Fs(C) be the infinite-dimensional one-mode bosonic Fock space with occupation basis ∣n⟩, n∈N0, and operators a∣n⟩=n∣n−1⟩, a∗∣n⟩=n+1∣n+1⟩ on the finite-particle core. Write IA and IF for the identity operators on the atomic and field factors, respectively. Define Δ=ω0−ω. For this rotating-wave model, whose usual scale ∣g∣n+1≪ω,ω0 applies only to the photon-number sectors under study and is not uniform over the full Fock space, set Hcore=ℏω IA⊗a∗a+ℏω02σz⊗IF+ℏg(σ+⊗a+σ−⊗a∗) on the finite-particle core of C2⊗^Fs(C). The Hamiltonian is the exact operator for the selected rotating-wave model; no error bound relative to the unapproximated dipole interaction is asserted.

The total excitation operator Nexc:=IA⊗a∗a+σ+σ−⊗IF is conserved on that core. The ground line C∣g,0⟩ is invariant. For each n∈N0, the subspace Kn=span⁡{∣e,n⟩,∣g,n+1⟩} is invariant and the restriction has matrix Hn=(ℏωn+ℏω02ℏgn+1ℏgn+1ℏω(n+1)−ℏω02). Its eigenvalue splitting is ℏΔ2+4g2(n+1). Define the full Hamiltonian to act on the ground line by −ℏω0/2 and as the orthogonal direct sum of the Hn on the Kn, with squared-block-norm domain D(HJC)={c∣g,0⟩⊕⨁n≥0ψn: c∈C, ψn∈Kn, ∑n≥0∥Hnψn∥2<∞}. This direct-sum operator is self-adjoint and extends Hcore.

Facts & Assumptions

Given: The finite one-mode model data ω,ω0>0, g∈R, ℏ>0, the product Hilbert space C2⊗^Fs(C), the finite-particle core, and the rotating-wave Hamiltonian specified in the Statement.

[A1]

The finite-mode light–matter assignment declares the one-mode frequency, two-level atom, SI-unit conventions, Hermitian dipole matrix, and rotating-wave option. The choices D=dσx and g=−Ed/ℏ make the exchange coefficient in the Statement exactly the retained part of the dipole expression. The RWA scale applies only on the stated low-excitation sectors; no approximation error or continuum-emission limit is supplied (Finite-mode dipole and detector assignment, Dimensions, SI unit conventions and quantity types of the framework).

[A2]

Carry AC and ACω as the finite-mode assignment and Fock/domain suppliers' mathematical assumptions; they are not physical laws. The self-adjointness proof uses ACω for the stated Fock/domain results (The Axiom of Choice, The Axiom of Countable Choice (ACω)).

[F1]

On the one-mode occupation basis, a∣n⟩=n∣n−1⟩, a∗∣n⟩=n+1∣n+1⟩, a∣0⟩=0, and ∣n⟩ are orthonormal (Fixed-mode number law).

[F2]

The one-mode Fock space is a Hilbert direct sum, the finite-particle subspace is dense, and its closed creation and annihilation operators have the stated domains and adjoint relations under ACω (Closed adjoint sector domains).

[F3]

The completed ordered Hilbert tensor product has the product orthonormal basis; atom and field operators act on separate factors and commute on the algebraic core (Hilbert tensor construction: inner product, completion and product bases).

Proof

1.1givenA1F1F3algebra

On the finite-particle core, [a∗a,a]=−a and [a∗a,a∗]=a∗ by [F1], while [σ+σ−,σ+]=σ+ and [σ+σ−,σ−]=−σ−. Therefore [Nexc,σ+⊗a]=σ+⊗a−σ+⊗a=0 and [Nexc,σ−⊗a∗]=−σ−⊗a∗+σ−⊗a∗=0. The free terms commute with Nexc, so [Nexc,Hcore]=0 on the core.

2.1A1F1F3step 1.1algebra

The vectors ∣g,0⟩ and ∣e,n⟩,∣g,n+1⟩ have excitation eigenvalues 0 and n+1, respectively. Applying the displayed ladder actions [F1] to each basis vector shows Hcore∣g,0⟩=−(ℏω0/2)∣g,0⟩, Hcore∣e,n⟩=(ℏωn+ℏω0/2)∣e,n⟩+ℏgn+1∣g,n+1⟩, and Hcore∣g,n+1⟩=ℏgn+1∣e,n⟩+(ℏω(n+1)−ℏω0/2)∣g,n+1⟩. The product occupation basis [F3] consists exactly of ∣g,0⟩ and these pairs: ∣e,m⟩ occurs in the pair indexed by m, while ∣g,m⟩ with m≥1 occurs in the pair indexed by m−1. They are mutually orthogonal and exhaust the space. Hence the displayed subspaces are invariant blocks with matrix Hn.

3.1step 2.1algebra

Write Hn=ℏω(n+12)I+ℏ(Δ/2gn+1gn+1−Δ/2). The traceless matrix has characteristic polynomial λ2−(Δ2/4+g2(n+1)), so its eigenvalues are ±12Δ2+4g2(n+1). Their difference gives the stated splitting.

4.1A2F2F3step 2.1algebra∎

Each block matrix and the ground-line scalar are Hermitian. Finite-support vectors are dense by [F2] and lie in the stated domain. If y∈D(HJC∗) with adjoint value z, testing against vectors supported in the ground line and each individual block gives zg=−(ℏω0/2)yg and zn=Hnyn. Since z is square-summable, ∑n∥Hnyn∥2<∞, so y∈D(HJC) and HJC∗y=HJCy. Conversely, for x,y∈D(HJC), Cauchy–Schwarz makes the block inner-product series absolutely convergent, and the Hermitian block identities give ⟨HJCx,y⟩=⟨x,HJCy⟩. Hence HJC⊆HJC∗ and equality holds. The direct-sum operator is self-adjoint, and its restriction to finite-support vectors is Hcore.

Remarks

The indexed blocks form an orthogonal decomposition by total excitation, not a finite-dimensional approximation of the full Fock space. The RWA inequality above is a model-validity scale for specified sectors, not a uniform bound on omitted counter-rotating terms.

diagram failed to render:
\begin{tikzpicture}[>=stealth,baseline=(current bounding box.center)]
\node[draw,minimum width=2.3cm,minimum height=1.1cm] (g) at (0,0) {$\mathbb C|g,0\rangle$};
\node at (1.8,0) {$\oplus$};
\node[draw,minimum width=3.2cm,minimum height=1.1cm] (k0) at (4.1,0) {$\mathcal K_0=\operatorname{span}\{|e,0\rangle,|g,1\rangle\}$};
\node at (6.2,0) {$\oplus$};
\node[draw,minimum width=3.2cm,minimum height=1.1cm] (k1) at (8.5,0) {$\mathcal K_1=\operatorname{span}\{|e,1\rangle,|g,2\rangle\}$};
\node at (10.5,0) {$\oplus\cdots$};
\end{tikzpicture}

The boxes label mutually orthogonal excitation sectors in C∣g,0⟩⊕⨁n≥0Kn. Each ⊕ denotes the Hilbert direct sum; no transition between blocks is represented. The proof of invariance and self-adjointness is the operator argument above and does not rely on the diagram.

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