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.

Required background. Interaction Picture gives the exact change of picture; Time-Dependent Hamiltonians explains reference propagators. Here we use those tools to choose a Hamiltonian split and identify which transition amplitudes can be truncated.

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. Check the sum $\sum_{f\ne i}\epsilon_{fi}^2$, not only individual channels. Even a vanishing first-order integral does not bound higher orders: an off-resonant drive can produce a cumulative second-order phase while transferring little population.

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

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

The physical first-order contribution is $\lambda c_f^{(1)}$. Its finite-time line shape is developed at First-Order Transition Probability.

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

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

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. See Rotating-Wave Approximation for the scale separation and corrections. With the drive convention used here, the resonant rotating-wave Hamiltonian is $\lambda\hbar\Omega\sigma_x/4$, so the resonant population oscillation has Rabi angular frequency $\lvert\lambda\Omega\rvert/2$.

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

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

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 any normalized $H_0$ 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 $\lambda V_I$. 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.

Exercises

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 any $T$ and nonzero $\delta$, its magnitude is at most $2/\lvert\delta\rvert$. Thus

$$\lvert\lambda c_f^{(1)}\rvert \le \frac{2\lvert\lambda v\rvert}{\hbar\lvert\delta\rvert}.$$

This is an upper bound, not a nonzero late-time limit: the amplitude has oscillatory zeros. When the coupling-to-detuning 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.

References

• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Vol. 2, Wiley, 1977.
• P. A. M. Dirac, The Principles of Quantum Mechanics, 4th ed., Oxford University Press, 1958.
• A. Messiah, Quantum Mechanics, Vol. 2, Dover, 1999.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• 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.