Interaction Picture for Perturbation Theory

The interaction picture is the standard workbench for time-dependent perturbation theory. It removes the evolution generated by a chosen solvable Hamiltonian $H_0$ and leaves the residual interaction to drive the state. In a useful split, the rapid phases associated with known energy gaps become explicit, so resonance, off-resonant cancellation, degeneracy, and secular growth can be read directly from matrix elements.

This page owns that calculation setup. The exact picture transformation is canonical at Interaction Picture, the general frame-generator formula is derived at Rotating Frames, and formal time-ordered evolution belongs to Dyson Expansion as Formal Evolution. Here the question is how to choose and use the split before truncating a transition amplitude.

Choose the Hamiltonian Split

Write

$$H(t)=H_0+\lambda V(t),$$

where $H_0$ is time independent for now and

$$H_0\lvert n\rangle = E_n\lvert n\rangle.$$

The bookkeeping parameter $\lambda$ tracks perturbative order. A useful split should satisfy four practical conditions:

1. the propagator generated by $H_0$ is known or efficiently computable;
2. the initial state and desired final channels have simple descriptions in the $H_0$ basis;
3. the transformed interaction has identifiable small matrix elements or rapidly oscillating phases;
4. important degeneracies, large diagonal shifts, and near-resonant couplings are not hidden in an allegedly small remainder.

The split is not unique. Moving a solvable term from $V$ into $H_0$ changes the interaction picture and reorganizes the perturbation series without changing the exact physics. Good bookkeeping places the dominant, exactly manageable evolution in $H_0$.

What “small” means here

An instantaneous operator coefficient can be small while its effect accumulates for a long time. For one channel, the interaction-picture diagnostic is

$$\epsilon_{fi}(T) = \frac{\lvert\lambda\rvert}{\hbar} \left\lvert \int_{t_0}^{t_0+T}dt\, \langle f\rvert V_I(t)\lvert i\rangle \right\rvert.$$

Off resonance, oscillatory cancellation can keep $\epsilon_{fi}$ small. On resonance or inside a degenerate subspace, the same matrix element may accumulate approximately linearly with $T$. Small coupling must therefore be assessed together with detuning and observation time.

Exact Transformation

The reference propagator is

$$U_0(t,t_0) = e^{-iH_0(t-t_0)/\hbar}.$$

Define

$$\lvert\psi_I(t)\rangle = U_0^\dagger(t,t_0) \lvert\psi_S(t)\rangle$$

and, for a Schrödinger-picture operator $A_S(t)$,

$$A_I(t) = U_0^\dagger(t,t_0) A_S(t) U_0(t,t_0).$$

The interaction Hamiltonian is

$$V_I(t) = U_0^\dagger(t,t_0) V(t) U_0(t,t_0).$$

Substituting the transformation into the Schrödinger equation gives the exact interaction-picture equation

$$i\hbar \frac{d}{dt}\lvert\psi_I(t)\rangle = \lambda V_I(t) \lvert\psi_I(t)\rangle.$$

No approximation has been made. Perturbation theory begins only when the solution of this equation is expanded and truncated in powers of $\lambda$.

Propagator factorization

Let $U(t,t_0)$ be the full propagator. Define

$$U_I(t,t_0) = U_0^\dagger(t,t_0)U(t,t_0).$$

Then

$$U(t,t_0) = U_0(t,t_0)U_I(t,t_0),$$

and

$$i\hbar \frac{\partial U_I(t,t_0)}{\partial t} = \lambda V_I(t)U_I(t,t_0),$$

with

$$U_I(t_0,t_0)=I.$$

The same reference time $t_0$ must be used in $U_0$, $U_I$, state transformations, and phase factors. Changing it is allowed, but doing so in only part of a calculation introduces spurious phases.

Matrix Elements Expose the Phases

For the time-independent $H_0$ basis,

$$U_0(t,t_0)\lvert n\rangle = e^{-iE_n(t-t_0)/\hbar} \lvert n\rangle.$$

Therefore

$$\langle m\rvert V_I(t)\lvert n\rangle = e^{i\omega_{mn}(t-t_0)}V_{mn}(t),$$

where

$$\omega_{mn} = \frac{E_m-E_n}{\hbar}$$

and

$$V_{mn}(t) = \langle m\rvert V(t)\lvert n\rangle.$$

Equivalently, the operator itself is

$$V_I(t) = \sum_{m,n} e^{i\omega_{mn}(t-t_0)} V_{mn}(t) \lvert m\rangle\langle n\rvert.$$

This formula is the practical reason for using the interaction picture. It separates two sources of time dependence:

• $e^{i\omega_{mn}(t-t_0)}$ comes from the known reference spectrum;
• $V_{mn}(t)$ comes from the applied drive or residual interaction.

Terms whose phases rotate rapidly tend to cancel under time integration. Terms with stationary or slowly varying phase can accumulate coherently. The comparison is quantitative only after the pulse duration and matrix-element scale are specified.

Diagonal and off-diagonal terms

For $m=n$,

$$\omega_{nn}=0,$$

so a diagonal matrix element does not acquire a fast interaction-picture phase. It commonly produces a phase or energy shift rather than population transfer. If that shift becomes large over the time of interest, leaving it in $V$ creates secular phase terms. Absorbing the diagonal part into a redefined $H_0$ often gives a better expansion.

For $m\ne n$, the phase compares the drive spectrum with the Bohr frequency. Off-diagonal does not automatically mean small, and diagonal does not automatically mean irrelevant.

Coupled Coefficient Equations

Expand

$$\lvert\psi_I(t)\rangle = \sum_n c_n(t)\lvert n\rangle.$$

Projecting the exact equation onto $\langle m\rvert$ gives

$$i\hbar\dot c_m(t) = \lambda \sum_n e^{i\omega_{mn}(t-t_0)} V_{mn}(t)c_n(t).$$

These equations are exact. The common first-order approximation starts from $c_n(t_0)=\delta_{ni}$ and replaces the coefficients on the right by their zeroth-order values:

$$i\hbar\dot c_f^{(1)}(t) = e^{i\omega_{fi}(t-t_0)}V_{fi}(t).$$

Integration yields

$$c_f^{(1)}(t) = -\frac{i}{\hbar} \int_{t_0}^{t}dt'\, e^{i\omega_{fi}(t'-t_0)} V_{fi}(t').$$

That amplitude is developed at First-Order Transition Probability. The interaction-picture page’s task is to make every phase and matrix element entering it unambiguous.

Schrödinger- and interaction-picture amplitudes

For an $H_0$ eigenstate $\lvert f\rangle$,

$$\langle f\vert\psi_I(t)\rangle = e^{iE_f(t-t_0)/\hbar} \langle f\vert\psi_S(t)\rangle.$$

The coefficients differ by a known phase, so

$$\left\lvert \langle f\vert\psi_I(t)\rangle \right\rvert^2 = \left\lvert \langle f\vert\psi_S(t)\rangle \right\rvert^2.$$

Picture changes do not change physical probabilities. They change where the time dependence is stored.

Degenerate and Nearly Degenerate Subspaces

If $E_m=E_n$, then

$$\omega_{mn}=0.$$

The interaction-picture matrix element inside that degenerate subspace has no reference oscillation to average it away:

$$\langle m\rvert V_I(t)\lvert n\rangle = V_{mn}(t).$$

A constant or slowly varying perturbation can therefore mix the subspace coherently on a time scale

$$\tau_{\mathrm{mix}} \sim \frac{\hbar}{\lvert\lambda V_{mn}\rvert}.$$

If the observation time is not short compared with this scale, treating each basis vector as an independent unperturbed channel is ineffective. Natural options are:

1. diagonalize the important interaction inside the degenerate subspace and include it in the reference problem;
2. retain the whole subspace and solve its coupled coefficient equations together;
3. construct an effective Hamiltonian for the subspace when distant states contribute virtually.

For a near-degeneracy with detuning

$$\delta_{mn}=\omega_{mn}-\omega_{\mathrm{drive}},$$

the relevant comparison is not simply $\lambda\ll1$ but

$$\frac{\lvert\lambda V_{mn}\rvert}{\hbar\lvert\delta_{mn}\rvert}.$$

When this ratio is not small, a rotating-frame or quasi-degenerate treatment is usually more faithful than a fixed-order transition amplitude.

Example: Transversely Driven Two-Level System

Take

$$H_0 = \frac{\hbar\omega_0}{2}\sigma_z$$

and

$$V(t) = \frac{\hbar\Omega}{2} \cos(\omega t)\sigma_x.$$

Set $t_0=0$ and define

$$\sigma_\pm = \frac12\left(\sigma_x\pm i\sigma_y\right).$$

The reference evolution gives

$$U_0^\dagger(t)\sigma_\pm U_0(t) = e^{\pm i\omega_0t}\sigma_\pm.$$

Since

$$\sigma_x=\sigma_++\sigma_-,$$

the interaction-picture perturbation is

$$\begin{aligned} V_I(t) = \frac{\hbar\Omega}{4} \big[{} &\sigma_+e^{i(\omega_0-\omega)t} \\ &+\sigma_-e^{-i(\omega_0-\omega)t} \\ &+\sigma_+e^{i(\omega_0+\omega)t} \\ &+\sigma_-e^{-i(\omega_0+\omega)t} \big]. \end{aligned}$$

The transformation has sorted the drive into two frequency scales:

$$\Delta=\omega_0-\omega$$

and

$$\Sigma=\omega_0+\omega.$$

Near resonance, $\lvert\Delta\rvert\ll\Sigma$. The terms oscillating with $\Delta$ can accumulate, whereas the terms oscillating with $\Sigma$ often average strongly. Dropping the latter is the rotating-wave approximation, not part of the exact interaction-picture transformation. Its assumptions and corrections belong to Rotating-Wave Approximation.

At sufficiently short time or weak drive, first-order perturbation theory describes the emerging excited-state amplitude. Once substantial population is transferred, Rabi Oscillations: First Encounter supplies the coherent nonperturbative solution.

Example: Driven Harmonic Oscillator

For

$$H_0 = \hbar\omega_0 \left(a^\dagger a+\frac12\right),$$

the ladder operators transform as

$$a_I(t)=a e^{-i\omega_0(t-t_0)},$$ $$a_I^\dagger(t) = a^\dagger e^{i\omega_0(t-t_0)}.$$

Let a force couple through position:

$$V(t)=-F(t)x, \qquad x=x_0(a+a^\dagger),$$

with

$$x_0=\sqrt{\frac{\hbar}{2m\omega_0}}.$$

Then

$$\begin{aligned} V_I(t) = -F(t)x_0 \big[{} &a e^{-i\omega_0(t-t_0)} \\ &+a^\dagger e^{i\omega_0(t-t_0)} \big]. \end{aligned}$$

The operator structure immediately gives

$$\Delta n=\pm1.$$

The phase structure then decides which of those allowed matrix elements is enhanced by the frequency content of $F(t)$. Selection rules and resonance are separate filters: the first asks whether the matrix element vanishes, and the second asks whether its phase accumulates coherently.

Density Operators and Observables

For a mixed state,

$$\rho_I(t) = U_0^\dagger(t,t_0) \rho_S(t) U_0(t,t_0).$$

It obeys

$$i\hbar\dot\rho_I(t) = \lambda \left[V_I(t),\rho_I(t)\right].$$

An observable transforms as

$$A_I(t) = U_0^\dagger(t,t_0) A_S(t) U_0(t,t_0),$$

and expectation values are picture invariant:

$$\operatorname{tr}\!\left[\rho_S(t)A_S(t)\right] = \operatorname{tr}\!\left[\rho_I(t)A_I(t)\right].$$

For the projector onto one nondegenerate $H_0$ eigenstate,

$$P_f=\lvert f\rangle\langle f\rvert,$$

one has $[P_f,H_0]=0$, so $P_{f,I}=P_f$. This is why interaction-picture coefficient magnitudes can be read directly as $H_0$-basis populations. A general observable need not remain unchanged.

Time-Dependent Reference Hamiltonians

The method is not restricted to a constant $H_0$. Suppose

$$H(t)=H_0(t)+\lambda V(t),$$

and the reference propagator is known from

$$i\hbar \frac{\partial U_0(t,t_0)}{\partial t} = H_0(t)U_0(t,t_0).$$

Define the same transformed state and interaction:

$$\lvert\psi_I(t)\rangle = U_0^\dagger(t,t_0) \lvert\psi_S(t)\rangle,$$ $$V_I(t) = U_0^\dagger(t,t_0) V(t)U_0(t,t_0).$$

The interaction-picture equation remains

$$i\hbar \frac{d}{dt}\lvert\psi_I(t)\rangle = \lambda V_I(t) \lvert\psi_I(t)\rangle.$$

What changes is that $U_0$ may itself require time ordering. It cannot be replaced by $e^{-i\int H_0dt/\hbar}$ unless the relevant reference Hamiltonians commute at different times or another exact construction is available.

This generalized split is useful when a strong drive or slowly varying control is solvable and a smaller residual term is the true perturbation. The picture should follow the dominant tractable dynamics, not an arbitrary declaration that one term is “free.”

Interaction Picture Versus Rotating Frame

Every interaction picture is a time-dependent unitary frame, but not every rotating frame is presented as a free-plus-interaction split.

Construction	Unitary chosen from	Transformed Hamiltonian
Interaction picture	Propagator $U_0$ of a reference Hamiltonian	$\lambda U_0^\dagger VU_0$
General rotating frame	Any differentiable unitary $R(t)$	$R^\dagger HR-i\hbar R^\dagger\dot R$
Heisenberg picture	Full exact propagator	State is constant for closed exact evolution

For $R=U_0$, the frame-generator term cancels $H_0$, leaving precisely $V_I$. A later rotating transformation may remove a drive frequency or make near-resonant terms slowly varying. That second transformation is a new frame choice and should not be silently merged with the first.

A Practical Setup Checklist

1. Specify the split and time interval. Write $H_0$, $V(t)$, $\lambda$, $t_0$, and the switching protocol.
2. Solve the reference evolution. Obtain $U_0(t,t_0)$ and a convenient basis or set of channels.
3. Transform the entire interaction. Compute $V_I=U_0^\dagger VU_0$, including any explicit time dependence already present in $V$.
4. Write matrix elements before integrating. Separate reference phases, drive frequencies, envelopes, and symmetry zeros.
5. Inspect degeneracies and slow phases. Enlarge the model space or reorganize $H_0$ when a nominal perturbation accumulates coherently.
6. Choose the required order. Use the Dyson expansion or coupled coefficient equations consistently.
7. Check depletion and secular terms. A small instantaneous coupling can fail after long resonant evolution.
8. Return to observables. Transform operators when necessary and state which basis population or measurement probability is being reported.

Common Mistakes

• Calling the interaction picture an approximation. The unitary transformation is exact; truncation of $U_I$ or the coefficient equations is the approximation.
• Transforming the state but not the interaction. The equation of motion contains $V_I$, not the original $V$.
• Forgetting explicit time dependence in $V(t)$. The interaction-picture conjugation does not replace the drive envelope or drive phase.
• Using inconsistent reference times. A mismatch in $t_0$ changes phase conventions and can corrupt interference terms.
• Assuming every oscillatory term is negligible. The integral depends on amplitude, detuning, pulse shape, and duration; a visual “fast” label is not an error estimate.
• Leaving a large diagonal term in the perturbation. A secular phase may indicate that the term belongs in a redefined $H_0$.
• Ignoring degeneracy. Matrix elements inside a degenerate subspace do not acquire suppressing Bohr phases.
• Applying the rotating-wave approximation during the transformation. The exact $V_I$ contains both rotating and counter-rotating terms; dropping terms is a separate approximation.
• Reading every coefficient as a physical amplitude without defining endpoints. Interaction- and Schrödinger-picture coefficients can differ by phases, while probabilities agree.
• Writing an ordinary exponential for time-dependent $H_0(t)$. Time ordering is generally required when $[H_0(t),H_0(t')]\ne0$.

Exercises

Derive the transformed equation

Starting from $\lvert\psi_I\rangle=U_0^\dagger\lvert\psi_S\rangle$, derive the interaction-picture Schrödinger equation for $H=H_0+\lambda V(t)$.

Solution

Differentiate the state:

$$\frac{d}{dt}\lvert\psi_I\rangle = \dot U_0^\dagger\lvert\psi_S\rangle +U_0^\dagger\lvert\dot\psi_S\rangle.$$

The reference propagator obeys

$$i\hbar\dot U_0=H_0U_0,$$

so

$$i\hbar\dot U_0^\dagger = -U_0^\dagger H_0.$$

Using

$$i\hbar\lvert\dot\psi_S\rangle = (H_0+\lambda V)\lvert\psi_S\rangle,$$

the two transformed $H_0$ terms cancel. The result is

$$i\hbar \frac{d}{dt}\lvert\psi_I\rangle = \lambda U_0^\dagger VU_0 \lvert\psi_I\rangle = \lambda V_I\lvert\psi_I\rangle.$$

Recover the Bohr phase

For time-independent $H_0$, show that

$$\langle m\rvert V_I(t)\lvert n\rangle = e^{i(E_m-E_n)(t-t_0)/\hbar}V_{mn}(t).$$

Solution

Insert $V_I=U_0^\dagger VU_0$ and use

$$U_0\lvert n\rangle = e^{-iE_n(t-t_0)/\hbar}\lvert n\rangle.$$

On the bra side,

$$\langle m\rvert U_0^\dagger = e^{iE_m(t-t_0)/\hbar}\langle m\rvert.$$

Multiplying the phases gives

$$e^{i(E_m-E_n)(t-t_0)/\hbar},$$

and the remaining matrix element is $V_{mn}(t)$.

Diagnose a commuting perturbation

Suppose $V$ is time independent and $[H_0,V]=0$. What becomes of $V_I(t)$? What can still happen inside a degenerate eigenspace of $H_0$?

Solution

Commutation implies

$$V_I(t) = e^{iH_0(t-t_0)/\hbar} V e^{-iH_0(t-t_0)/\hbar} = V.$$

One may choose a basis that diagonalizes $H_0$ and $V$ simultaneously. In that common basis, $V$ changes phases but not populations.

Inside a degenerate eigenspace, an arbitrary $H_0$ eigenbasis need not diagonalize $V$. Coefficients in that arbitrary basis can mix even though no transfer occurs between different $H_0$ energies. Diagonalizing $V$ within the degenerate subspace identifies the stationary combinations of the full Hamiltonian.

Transform a transverse two-level drive

For

$$H_0=\frac{\hbar\omega_0}{2}\sigma_z,$$

show that $U_0^\dagger\sigma_+U_0=e^{i\omega_0t}\sigma_+$ when $t_0=0$. Use this to identify the two frequencies in a drive proportional to $\cos(\omega t)\sigma_x$.

Solution

Since

$$[\sigma_z,\sigma_+]=2\sigma_+,$$

the adjoint action gives

$$e^{i\omega_0t\sigma_z/2} \sigma_+ e^{-i\omega_0t\sigma_z/2} = e^{i\omega_0t}\sigma_+.$$

Writing

$$\cos(\omega t) = \frac12\left(e^{i\omega t}+e^{-i\omega t}\right)$$

and $\sigma_x=\sigma_++\sigma_-$ produces phases at

$$\omega_0-\omega$$

and

$$\omega_0+\omega,$$

together with their negatives. Near resonance, the difference-frequency terms are slow and the sum-frequency terms are counter-rotating.

Compare detuning with coupling

Two states are coupled by a constant interaction matrix element $v$. In a frame where the pair has detuning $\delta$, estimate the first-order amplitude scale for times $T\gg1/\lvert\delta\rvert$. When is fixed-order perturbation theory suspect?

Solution

The relevant integral is

$$\int_0^Tdt\,e^{i\delta t} = \frac{e^{i\delta T}-1}{i\delta}.$$

For $T\gg1/\lvert\delta\rvert$, its magnitude is bounded by a number of order $1/\lvert\delta\rvert$. Thus

$$\lvert c_f^{(1)}\rvert \sim \frac{\lvert\lambda v\rvert}{\hbar\lvert\delta\rvert}.$$

When this ratio is not small, the two states should be retained and diagonalized or evolved together. At exact resonance, the first-order amplitude instead grows as $\lvert\lambda v\rvert T/\hbar$ until depletion invalidates first order.

Generalize to a time-dependent reference problem

Let $H_0(t)$ have known propagator $U_0(t,t_0)$. Show that the transformed equation still contains only $V_I(t)$, and state why an ordinary exponential for $U_0$ is generally invalid.

Solution

The derivation uses only

$$i\hbar\dot U_0(t,t_0) = H_0(t)U_0(t,t_0),$$

not time independence. Differentiating $U_0^\dagger\lvert\psi_S\rangle$ again cancels the two $H_0(t)$ terms, leaving

$$i\hbar\frac{d}{dt}\lvert\psi_I\rangle = \lambda U_0^\dagger VU_0 \lvert\psi_I\rangle.$$

If

$$[H_0(t),H_0(t')]\ne0,$$

then factors generated at different times do not commute. The reference propagator requires time ordering or another exact solution; in general it is not

$$\exp\!\left[ -\frac{i}{\hbar}\int_{t_0}^{t}H_0(t')\,dt' \right].$$

Cross-Links

• Time-Dependent Perturbation Theory and Transitions
• Interaction Picture
• Dyson Expansion as Formal Evolution
• Dyson Expansion for Transition Amplitudes
• Interaction-Picture Evolution Formula Card
• Time-Dependent Hamiltonians
• First-Order Transition Probability
• Harmonic Perturbations
• Rotating Frames
• Rotating-Wave Approximation
• Rabi Oscillations: First Encounter

References

• P. A. M. Dirac, The Principles of Quantum Mechanics, 4th ed., Oxford University Press, 1958.
• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Vol. 2, Wiley, 1977.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• A. Messiah, Quantum Mechanics, Vol. 2, Dover, 1999.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• N. V. Vitanov, T. Halfmann, B. W. Shore, and K. Bergmann, “Laser-induced population transfer by adiabatic passage techniques,” Annual Review of Physical Chemistry 52, 763–809, 2001.