# 5.2.1 Interaction Picture
Worked solutions for the homework problems in the [5.2.1 Interaction Picture](../ch5_perturbation-theory/5-2-1-interaction-picture) lecture. Each problem statement is reproduced verbatim, followed by a complete worked solution.

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**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:

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

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.

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**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:

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

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.

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**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.

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**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:

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

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.

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**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:

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

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.)

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**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 $\hat{V}$.