5.2.1 Interaction Picture#

Worked solutions for the homework problems in the 5.2.1 Interaction Picture lecture. Each problem statement is reproduced verbatim, followed by a complete worked solution.

1. Expectations across pictures. Let \(\vert\psi\rangle_{\mathcal{S}}\) and \(\hat{O}\) be any state and observable in the Schrödinger picture. With \(\vert\psi\rangle_{\mathcal{I}}=\hat{U}_0^{\dagger}\vert\psi\rangle_{\mathcal{S}}\) and \(\hat{O}_{\mathcal{I}}=\hat{U}_0^{\dagger}\hat{O}\hat{U}_0\), show that \({}_{\mathcal{S}}\langle\psi\vert\hat{O}\vert\psi\rangle_{\mathcal{S}}={}_{\mathcal{I}}\langle\psi\vert\hat{O}_{\mathcal{I}}\vert\psi\rangle_{\mathcal{I}}\).

Solution.

The transformation to the interaction picture is implemented by the unperturbed propagator \(\hat{U}_0(t)=\mathrm{e}^{-\mathrm{i}\hat{H}_0t/\hbar}\), which is unitary: \(\hat{U}_0^{\dagger}\hat{U}_0=\hat{U}_0\hat{U}_0^{\dagger}=\hat{I}\). That single fact does all the work.

First find the interaction-picture bra. Taking the Hermitian adjoint of \(\vert\psi\rangle_{\mathcal{I}}=\hat{U}_0^{\dagger}\vert\psi\rangle_{\mathcal{S}}\),

\[ {}_{\mathcal{I}}\langle\psi\vert=\big(\hat{U}_0^{\dagger}\vert\psi\rangle_{\mathcal{S}}\big)^{\dagger}={}_{\mathcal{S}}\langle\psi\vert\,\hat{U}_0, \]

since \((\hat{U}_0^{\dagger})^{\dagger}=\hat{U}_0\). Now substitute this and the definition \(\hat{O}_{\mathcal{I}}=\hat{U}_0^{\dagger}\hat{O}\hat{U}_0\) into the interaction-picture expectation value:

\[\begin{split} \begin{split} {}_{\mathcal{I}}\langle\psi\vert\hat{O}_{\mathcal{I}}\vert\psi\rangle_{\mathcal{I}} &=\big({}_{\mathcal{S}}\langle\psi\vert\hat{U}_0\big)\,\big(\hat{U}_0^{\dagger}\hat{O}\hat{U}_0\big)\,\big(\hat{U}_0^{\dagger}\vert\psi\rangle_{\mathcal{S}}\big)\\ &={}_{\mathcal{S}}\langle\psi\vert\,(\hat{U}_0\hat{U}_0^{\dagger})\,\hat{O}\,(\hat{U}_0\hat{U}_0^{\dagger})\,\vert\psi\rangle_{\mathcal{S}}\\ &={}_{\mathcal{S}}\langle\psi\vert\,\hat{O}\,\vert\psi\rangle_{\mathcal{S}}. \end{split} \end{split}\]

The two \(\hat{U}_0\hat{U}_0^{\dagger}=\hat{I}\) collapses remove the transformation entirely. The expectation value is therefore picture-independent — provided the state and the operator are transformed together by the same \(\hat{U}_0\). This is the consistency requirement noted in the lecture: a picture change is a similarity transformation of the whole bracket, and unitarity guarantees it leaves every physical prediction (expectation values, and hence probabilities and measurement statistics) invariant. The identical argument with \(\hat{U}_0\) replaced by the full propagator \(\hat{U}\) gives the Schrödinger ↔ Heisenberg equivalence; the interaction picture is just the intermediate choice that transforms with the unperturbed propagator only.

2. Transition frequencies. The unperturbed Hamiltonian \(\hat{H}_0\) has eigenstates \(\{\vert n\rangle\}\) with energies \(\{E_n^{(0)}\}\).

(a) Show that in the interaction picture, the matrix elements of \(\hat{V}_{\mathcal{I}}(t)\) are \(\langle m\vert \hat{V}_{\mathcal{I}}(t)\vert n\rangle = V_{mn}(t)\,\mathrm{e}^{\mathrm{i}\omega_{mn}t}\) where \(\omega_{mn} = (E_m^{(0)} - E_n^{(0)})/\hbar\).

(b) Interpret the oscillatory factors: why do matrix elements between nearly degenerate states (\(\omega_{mn} \approx 0\)) evolve slowly, while those between widely separated states oscillate rapidly?

Solution.

(a) The unperturbed propagator acts diagonally on the eigenbasis of \(\hat{H}_0\). From \(\hat{U}_0(t)=\mathrm{e}^{-\mathrm{i}\hat{H}_0t/\hbar}=\sum_k\vert k\rangle\,\mathrm{e}^{-\mathrm{i}E_k^{(0)}t/\hbar}\,\langle k\vert\),

\[ \hat{U}_0(t)\vert n\rangle=\mathrm{e}^{-\mathrm{i}E_n^{(0)}t/\hbar}\vert n\rangle, \qquad \langle m\vert\hat{U}_0^{\dagger}(t)=\mathrm{e}^{+\mathrm{i}E_m^{(0)}t/\hbar}\langle m\vert, \]

the second relation being the adjoint of \(\hat{U}_0(t)\vert m\rangle=\mathrm{e}^{-\mathrm{i}E_m^{(0)}t/\hbar}\vert m\rangle\). Insert the definition \(\hat{V}_{\mathcal{I}}(t)=\hat{U}_0^{\dagger}(t)\,\hat{V}(t)\,\hat{U}_0(t)\) and let the propagators act on the surrounding bra and ket:

\[\begin{split} \begin{split} \langle m\vert\hat{V}_{\mathcal{I}}(t)\vert n\rangle &=\langle m\vert\,\hat{U}_0^{\dagger}(t)\,\hat{V}(t)\,\hat{U}_0(t)\,\vert n\rangle\\ &=\mathrm{e}^{+\mathrm{i}E_m^{(0)}t/\hbar}\,\langle m\vert\hat{V}(t)\vert n\rangle\,\mathrm{e}^{-\mathrm{i}E_n^{(0)}t/\hbar}\\ &=\mathrm{e}^{\mathrm{i}(E_m^{(0)}-E_n^{(0)})t/\hbar}\,\langle m\vert\hat{V}(t)\vert n\rangle\\ &=V_{mn}(t)\,\mathrm{e}^{\mathrm{i}\omega_{mn}t}, \end{split} \end{split}\]

where \(V_{mn}(t)=\langle m\vert\hat{V}(t)\vert n\rangle\) is the bare Schrödinger-picture matrix element and \(\omega_{mn}=(E_m^{(0)}-E_n^{(0)})/\hbar\) is the Bohr frequency of the \(n\to m\) transition. The picture change therefore does nothing to the size of a matrix element; it only multiplies it by a pure phase \(\mathrm{e}^{\mathrm{i}\omega_{mn}t}\) set by the energy gap.

(b) The phase \(\mathrm{e}^{\mathrm{i}\omega_{mn}t}\) winds at a rate fixed by the energy difference \(E_m^{(0)}-E_n^{(0)}\). Two nearly degenerate levels have a small gap, so \(\omega_{mn}\approx0\) and the phase barely advances over the timescales of interest — the interaction-picture matrix element is quasi-static. Two widely separated levels have a large gap, so the phase cycles rapidly. The physical consequence shows up the moment these elements drive transitions: the leading transition amplitude is the time integral \(\int_0^t\langle m\vert\hat{V}_{\mathcal{I}}(t')\vert n\rangle\,\mathrm{d}t'\). A slowly varying integrand accumulates coherently and builds a large amplitude, while a rapidly oscillating one repeatedly cancels itself and contributes almost nothing. The interaction picture thus makes energy conservation visible as a phase-matching condition: transitions are efficient between near-degenerate states — or, when a drive of frequency \(\omega\) is present, when the drive restores phase matching by \(\omega_{mn}-\omega\approx0\) (resonance, Problem 5) — and strongly suppressed otherwise. The background \(\hat{H}_0\) evolution was removed precisely so that what survives in \(\hat{V}_{\mathcal{I}}\) is exactly this relative phase winding that controls the interference.

3. Choosing a picture. Three quantum systems require calculations: (A) a bound particle in a slowly varying potential, (B) an atom driven by a monochromatic laser, (C) a free particle kicked by a brief impulse. For each, state which picture (Schrödinger, Heisenberg, or interaction) is most natural and explain your reasoning.

Solution.

(A) Bound particle in a slowly varying potential — Schrödinger picture. Here the time-dependent potential is the entire Hamiltonian; there is no weak perturbation to single out, and no observable with simple closed operator dynamics to exploit. The natural description tracks the state itself by integrating \(\mathrm{i}\hbar\,\partial_t\vert\psi\rangle=\hat{H}(t)\vert\psi\rangle\) directly. The “slowly varying” qualifier invites the adiabatic analysis — following the instantaneous eigenstates of \(\hat{H}(t)\) — which is itself a statement about how the Schrödinger-picture state tracks the spectrum, so the Schrödinger picture remains the home frame.

(B) Atom driven by a monochromatic laser — interaction picture. This is the textbook interaction-picture setting and matches its definition exactly: a solvable, time-independent reference Hamiltonian \(\hat{H}_0\) (the atom, with known eigenstates and energies) plus a weak, oscillatory perturbation \(\hat{V}(t)\) (the laser-atom coupling). Passing to the interaction picture strips the rapid free-atom phase evolution and leaves an equation generated by \(\hat{V}_{\mathcal{I}}(t)\) alone, whose matrix elements carry Bohr-frequency phases — exactly the structure that exposes resonance and the rotating-wave approximation (Problem 5).

(C) Free particle kicked by a brief impulse — Heisenberg picture. A free particle has trivially solvable operator dynamics: the Heisenberg equations give \(\hat{p}(t)=\hat{p}\) and \(\hat{x}(t)=\hat{x}+\hat{p}\,t/m\) in closed form. A brief impulse acts as a near-instantaneous translation of \(\hat{p}\), so the observables of interest evolve by simple operator equations punctuated by a discrete jump at the kick. The Heisenberg picture — whose operators carry the dynamics, and whose strength is precisely such operator-evolution and correlator computations — is the natural choice.

4. Operator-picture transformation. Define the interaction-picture operator \(\hat{O}_{\mathcal{I}}(t):=\hat{U}_0^{\dagger}(t)\,\hat{O}(t)\,\hat{U}_0(t)\) for any Schrödinger-picture observable \(\hat{O}(t)\) that may carry its own explicit time dependence.

(a) Differentiate the definition and use \(\mathrm{i}\hbar\,\partial_t\hat{U}_0=\hat{H}_0\hat{U}_0\) to show that

\[ \mathrm{i}\hbar\,\partial_t\hat{O}_{\mathcal{I}}(t)=[\hat{O}_{\mathcal{I}}(t),\hat{H}_0]+\mathrm{i}\hbar\,(\partial_t\hat{O})_{\mathcal{I}}(t). \]

(b) Contrast this with the Heisenberg equation of motion. Which Hamiltonian generates operator evolution in each picture, and what dynamics does the interaction-picture state equation \(\mathrm{i}\hbar\,\partial_t|\psi\rangle_{\mathcal{I}}=\hat{V}_{\mathcal{I}}(t)|\psi\rangle_{\mathcal{I}}\) track that the Heisenberg picture instead absorbs into the operators?

Solution.

(a) The definition \(\hat{O}_{\mathcal{I}}(t)=\hat{U}_0^{\dagger}(t)\,\hat{O}(t)\,\hat{U}_0(t)\) is a product of three time-dependent factors. The product rule gives

\[ \partial_t\hat{O}_{\mathcal{I}}=(\partial_t\hat{U}_0^{\dagger})\,\hat{O}\,\hat{U}_0+\hat{U}_0^{\dagger}\,(\partial_t\hat{O})\,\hat{U}_0+\hat{U}_0^{\dagger}\,\hat{O}\,(\partial_t\hat{U}_0). \]

Multiply through by \(\mathrm{i}\hbar\). The propagator equation \(\mathrm{i}\hbar\,\partial_t\hat{U}_0=\hat{H}_0\hat{U}_0\) supplies the last term; its Hermitian conjugate, using \(\hat{H}_0^{\dagger}=\hat{H}_0\), supplies the first:

\[ \mathrm{i}\hbar\,\partial_t\hat{U}_0=\hat{H}_0\hat{U}_0, \qquad \mathrm{i}\hbar\,\partial_t\hat{U}_0^{\dagger}=-\hat{U}_0^{\dagger}\hat{H}_0. \]

Substituting these into the three terms,

\[ \mathrm{i}\hbar\,\partial_t\hat{O}_{\mathcal{I}} =-\hat{U}_0^{\dagger}\hat{H}_0\,\hat{O}\,\hat{U}_0 +\mathrm{i}\hbar\,\hat{U}_0^{\dagger}(\partial_t\hat{O})\hat{U}_0 +\hat{U}_0^{\dagger}\hat{O}\,\hat{H}_0\hat{U}_0. \]

In the first and third terms insert the identity \(\hat{U}_0\hat{U}_0^{\dagger}=\hat{I}\) between \(\hat{H}_0\) and \(\hat{O}\), so each becomes a product of two interaction-picture operators:

\[\begin{split} \begin{split} \mathrm{i}\hbar\,\partial_t\hat{O}_{\mathcal{I}} &=-(\hat{U}_0^{\dagger}\hat{H}_0\hat{U}_0)(\hat{U}_0^{\dagger}\hat{O}\hat{U}_0) +(\hat{U}_0^{\dagger}\hat{O}\hat{U}_0)(\hat{U}_0^{\dagger}\hat{H}_0\hat{U}_0) +\mathrm{i}\hbar\,\hat{U}_0^{\dagger}(\partial_t\hat{O})\hat{U}_0\\ &=-(\hat{H}_0)_{\mathcal{I}}\,\hat{O}_{\mathcal{I}} +\hat{O}_{\mathcal{I}}\,(\hat{H}_0)_{\mathcal{I}} +\mathrm{i}\hbar\,(\partial_t\hat{O})_{\mathcal{I}}. \end{split} \end{split}\]

The reference Hamiltonian is invariant under its own interaction-picture transformation, because \(\hat{U}_0=\mathrm{e}^{-\mathrm{i}\hat{H}_0t/\hbar}\) commutes with \(\hat{H}_0\):

\[ (\hat{H}_0)_{\mathcal{I}}=\hat{U}_0^{\dagger}\hat{H}_0\hat{U}_0=\hat{H}_0\,\hat{U}_0^{\dagger}\hat{U}_0=\hat{H}_0. \]

Hence the first two terms collapse to a commutator, \(-\hat{H}_0\hat{O}_{\mathcal{I}}+\hat{O}_{\mathcal{I}}\hat{H}_0=[\hat{O}_{\mathcal{I}},\hat{H}_0]\), and

\[ \mathrm{i}\hbar\,\partial_t\hat{O}_{\mathcal{I}}(t)=[\hat{O}_{\mathcal{I}}(t),\hat{H}_0]+\mathrm{i}\hbar\,(\partial_t\hat{O})_{\mathcal{I}}(t), \]

as claimed. The final term \((\partial_t\hat{O})_{\mathcal{I}}\) is the interaction-picture transform of any explicit time dependence the Schrödinger operator already carried; it is present only when \(\hat{O}\) depends on \(t\) by itself (e.g. an externally modulated observable).

(b) The Heisenberg-picture operator \(\hat{O}_{\mathcal{H}}(t)=\hat{U}^{\dagger}(t)\,\hat{O}(t)\,\hat{U}(t)\) is built from the full propagator \(\hat{U}\), and the identical derivation — with \(\hat{U}\) and \(\hat{H}=\hat{H}_0+\hat{V}\) in place of \(\hat{U}_0\) and \(\hat{H}_0\) — yields

\[ \mathrm{i}\hbar\,\partial_t\hat{O}_{\mathcal{H}}(t)=[\hat{O}_{\mathcal{H}}(t),\hat{H}]+\mathrm{i}\hbar\,(\partial_t\hat{O})_{\mathcal{H}}(t). \]

The two equations have the same shape but different generators. Summarizing which Hamiltonian moves the operators in each picture:

Schrödinger picture: operators do not evolve at all (apart from any explicit \(\partial_t\hat{O}\)); the entire dynamics, generated by the full \(\hat{H}\), is carried by the state.
Heisenberg picture: operators evolve under the full \(\hat{H}=\hat{H}_0+\hat{V}\); the state is frozen.
Interaction picture: operators evolve under \(\hat{H}_0\) only (part (a)); the residual \(\hat{V}\)-driven motion is handed back to the state.

The interaction picture therefore splits the labor. The solvable background \(\hat{H}_0\) goes into the operators via the \(\hat{U}_0\) transformation — precisely part (a) — while the perturbation goes into the state through the interaction-picture state equation \(\mathrm{i}\hbar\,\partial_t\vert\psi\rangle_{\mathcal{I}}=\hat{V}_{\mathcal{I}}\vert\psi\rangle_{\mathcal{I}}\). That state equation tracks exactly the \(\hat{V}\)-driven evolution: the transitions and redistribution of probability among the unperturbed levels that constitute the physical effect of the perturbation. The Heisenberg picture has no moving state to track this — it absorbs all of the dynamics, the \(\hat{V}\) part included, into the operators, leaving the state stationary. In short, the interaction-picture state equation isolates what \(\hat{V}\) does; the Heisenberg picture buries that same content inside the operator evolution.

5. Two-level system under monochromatic drive. Consider \(\hat{H}_0=\tfrac{\hbar\omega_0}{2}\hat{\sigma}^z\) with eigenstates \(\vert\uparrow\rangle\) (energy \(+\hbar\omega_0/2\)) and \(\vert\downarrow\rangle\) (energy \(-\hbar\omega_0/2\)), perturbed by

\[ \hat{V}(t)=\hbar\Omega\cos(\omega t)\,\hat{\sigma}^x. \]

(a) Rewrite \(\hat{V}(t)\) in the eigenbasis of \(\hat{H}_0\) and identify the Bohr frequency \(\omega_{\uparrow\downarrow}=(E_\uparrow-E_\downarrow)/\hbar\).

(b) Compute \(\langle\uparrow\vert\hat{V}_{\mathcal{I}}(t)\vert\downarrow\rangle\) in the interaction picture and show that it splits into two exponentials at frequencies \(\omega_0\pm\omega\).

(c) The rotating-wave approximation keeps only the slowly oscillating contribution. State the resonance condition on \(\omega\) that selects the slow term, and write down the matrix element kept by the approximation.

Solution.

(a) The eigenbasis of \(\hat{H}_0=\tfrac{\hbar\omega_0}{2}\hat{\sigma}^z\) is \(\{\vert\uparrow\rangle,\vert\downarrow\rangle\}\) — the \(\hat{\sigma}^z\) eigenstates. In that basis \(\hat{\sigma}^x\) is the pure spin-flip operator, \(\hat{\sigma}^x=\vert\uparrow\rangle\langle\downarrow\vert+\vert\downarrow\rangle\langle\uparrow\vert\), so

\[ \hat{V}(t)=\hbar\Omega\cos(\omega t)\,\big(\vert\uparrow\rangle\langle\downarrow\vert+\vert\downarrow\rangle\langle\uparrow\vert\big). \]

The perturbation is purely off-diagonal: it connects \(\vert\uparrow\rangle\) and \(\vert\downarrow\rangle\) and has no diagonal part. The unperturbed energies are \(E_\uparrow=+\hbar\omega_0/2\) and \(E_\downarrow=-\hbar\omega_0/2\), so the Bohr frequency of the up-from-down transition is

\[ \omega_{\uparrow\downarrow}=\frac{E_\uparrow-E_\downarrow}{\hbar}=\frac{(\hbar\omega_0/2)-(-\hbar\omega_0/2)}{\hbar}=\omega_0. \]

(b) Apply the matrix-element rule from Problem 2, \(\langle m\vert\hat{V}_{\mathcal{I}}(t)\vert n\rangle=V_{mn}(t)\,\mathrm{e}^{\mathrm{i}\omega_{mn}t}\), with \(m=\uparrow\), \(n=\downarrow\). The bare matrix element is \(V_{\uparrow\downarrow}(t)=\langle\uparrow\vert\hat{V}(t)\vert\downarrow\rangle=\hbar\Omega\cos(\omega t)\) and \(\omega_{\uparrow\downarrow}=\omega_0\), so

\[ \langle\uparrow\vert\hat{V}_{\mathcal{I}}(t)\vert\downarrow\rangle=\hbar\Omega\cos(\omega t)\,\mathrm{e}^{\mathrm{i}\omega_0 t}. \]

Write the cosine as a sum of phasors, \(\cos(\omega t)=\tfrac12(\mathrm{e}^{\mathrm{i}\omega t}+\mathrm{e}^{-\mathrm{i}\omega t})\), and combine the exponentials:

\[\begin{split} \begin{split} \langle\uparrow\vert\hat{V}_{\mathcal{I}}(t)\vert\downarrow\rangle &=\frac{\hbar\Omega}{2}\big(\mathrm{e}^{\mathrm{i}\omega t}+\mathrm{e}^{-\mathrm{i}\omega t}\big)\,\mathrm{e}^{\mathrm{i}\omega_0 t}\\ &=\frac{\hbar\Omega}{2}\,\mathrm{e}^{\mathrm{i}(\omega_0+\omega)t}+\frac{\hbar\Omega}{2}\,\mathrm{e}^{\mathrm{i}(\omega_0-\omega)t}. \end{split} \end{split}\]

The interaction-picture matrix element is thus a sum of two terms: one oscillating at the sum frequency \(\omega_0+\omega\) (the counter-rotating term) and one at the difference frequency \(\omega_0-\omega\) (the co-rotating term).

(c) Near resonance the drive frequency is tuned close to the transition frequency, \(\omega\approx\omega_0\). Then the difference-frequency term oscillates slowly, \(\omega_0-\omega\approx0\), while the sum-frequency term oscillates fast, \(\omega_0+\omega\approx2\omega_0\). Over any interval long compared with \(1/\omega_0\) the fast term integrates to nearly zero, whereas the slow term accumulates coherently and dominates the transition amplitude (the phase-matching argument of Problem 2b). The rotating-wave approximation keeps only the slow co-rotating term and discards the fast counter-rotating one:

\[ \langle\uparrow\vert\hat{V}_{\mathcal{I}}(t)\vert\downarrow\rangle\;\xrightarrow{\ \text{RWA}\ }\;\frac{\hbar\Omega}{2}\,\mathrm{e}^{\mathrm{i}(\omega_0-\omega)t}=\frac{\hbar\Omega}{2}\,\mathrm{e}^{-\mathrm{i}\delta t},\qquad\delta\equiv\omega-\omega_0, \]

with \(\delta\) the detuning. The resonance condition that selects this term is

\[ \omega\approx\omega_0\qquad(\text{equivalently }\delta\to0). \]

Exactly on resonance (\(\omega=\omega_0\)) the kept matrix element becomes time-independent and equal to \(\hbar\Omega/2\); a constant off-diagonal coupling of this size drives full Rabi oscillations between \(\vert\uparrow\rangle\) and \(\vert\downarrow\rangle\) at the Rabi frequency \(\Omega\). (The companion element \(\langle\downarrow\vert\hat{V}_{\mathcal{I}}(t)\vert\uparrow\rangle\) is the complex conjugate; the RWA keeps its slow part \(\tfrac{\hbar\Omega}{2}\,\mathrm{e}^{+\mathrm{i}\delta t}\), so the truncated \(\hat{V}_{\mathcal{I}}\) remains Hermitian.)

6. Misconception check. One might argue: In the interaction picture, the perturbation \(\hat{V}\) has been absorbed into the change of frame and no longer affects the dynamics. Identify what is correct and what is wrong in this claim. In two or three sentences, explain in what sense \(\hat{V}\) survives the transformation and what is actually removed by passing to the comoving frame.

Solution.

What is correct. A change of frame has taken place, and it has absorbed something into the picture’s definitions. Transforming by \(\hat{U}_0(t)=\mathrm{e}^{-\mathrm{i}\hat{H}_0t/\hbar}\) removes the background evolution generated by \(\hat{H}_0\): the interaction-picture state no longer winds under \(\hat{H}_0\), and that piece of the dynamics has been moved into the defining transformation (and into the explicit phases carried by \(\hat{V}_{\mathcal{I}}\) and by interaction-picture operators). So the intuition that “something has been absorbed into the frame change” is right.

What is wrong. It is \(\hat{H}_0\) — not \(\hat{V}\) — that is removed from the explicit equation of motion. The interaction-picture state equation is \(\mathrm{i}\hbar\,\partial_t\vert\psi\rangle_{\mathcal{I}}=\hat{V}_{\mathcal{I}}(t)\vert\psi\rangle_{\mathcal{I}}\): its generator is precisely the transformed perturbation. Far from disappearing, \(\hat{V}\) is the entire content of the interaction-picture dynamics — it survives as \(\hat{V}_{\mathcal{I}}(t)=\hat{U}_0^{\dagger}\hat{V}\hat{U}_0\), the same perturbation merely dressed with Bohr-frequency phases \(\mathrm{e}^{\mathrm{i}\omega_{mn}t}\). The decisive check: if \(\hat{V}=0\) then \(\hat{V}_{\mathcal{I}}=0\), \(\hat{U}_{\mathcal{I}}=\hat{I}\), and the interaction-picture state is frozen — every bit of nontrivial motion left in this picture is due to \(\hat{V}\).

In short (two to three sentences): the comoving frame removes the trivial, known background phase winding generated by \(\hat{H}_0\), not the perturbation. The perturbation survives the transformation as \(\hat{V}_{\mathcal{I}}=\hat{U}_0^{\dagger}\hat{V}\hat{U}_0\) and is exactly the generator of the interaction-picture equation of motion. What the change of frame removes is the part of the evolution we already know how to solve; what it isolates and keeps front and center is the dynamics driven by \(\hat{V}\).