Interaction-Picture Evolution

Purpose

The interaction picture separates a chosen solvable evolution from the remaining dynamics. Write

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

and let $U_0(t,t_0)$ be the exact propagator generated by $H_0(t)$. The interaction-picture state and interaction are

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

The transformed state obeys

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

This transformation is exact. Perturbation theory begins only when the resulting evolution operator or state is expanded and truncated in powers of $\lambda$.

The exact picture is defined at Interaction Picture. Its use as a perturbative workbench is developed at Interaction Picture for Perturbation Theory.

At a glance

Task	Formula
Split the Hamiltonian	$H=H_0+\lambda V$
Transform a state	$\lvert\psi_I\rangle=U_0^\dagger\lvert\psi_S\rangle$
Transform an observable	$A_I=U_0^\dagger A_SU_0$
Transform the interaction	$V_I=U_0^\dagger VU_0$
Factor the full propagator	$U=U_0U_I$
Define interaction evolution	$U_I=U_0^\dagger U$
Evolve the state	$i\hbar\partial_t\lvert\psi_I\rangle=\lambda V_I\lvert\psi_I\rangle$
Evolve a density operator	$i\hbar\dot\rho_I=\lambda[V_I,\rho_I]$
Solve formally	$U_I=\mathcal T\exp[-(i\lambda/\hbar)\int V_I(t')dt']$
First-order transition amplitude	$c_{fi}^{(1)}=-(i\lambda/\hbar)\int dt\,\langle f\rvert V_I(t)\lvert i\rangle$

Every $U_0$, $U_I$, state transformation, and phase must use the same reference time $t_0$.

Reference evolution

The reference propagator satisfies

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

If $H_0$ is time independent,

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

If $H_0(t)$ is time dependent, then generally

$$U_0(t,t_0) = \mathcal T \exp\!\left[ -\frac{i}{\hbar} \int_{t_0}^{t}H_0(s)\,ds \right],$$

with the time-ordered exponential understood through its defining evolution equation. An ordinary exponential of the integral is valid only when the reference Hamiltonians commute at different times or another special reduction applies.

The reference split is not unique. Moving a solvable term between $H_0$ and $V$ changes the picture and reorganizes a truncated perturbation series, but does not change exact predictions.

States, operators, and density operators

At the reference time,

$$U_0(t_0,t_0)=I,$$

so interaction- and Schrödinger-picture objects agree:

$$\lvert\psi_I(t_0)\rangle = \lvert\psi_S(t_0)\rangle.$$

A Schrödinger-picture observable becomes

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

A density operator becomes

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

Expectation values are picture invariant:

$$\begin{aligned} \langle\psi_S\rvert A_S\lvert\psi_S\rangle &= \langle\psi_I\rvert A_I\lvert\psi_I\rangle,\\ \operatorname{Tr}(\rho_SA_S) &= \operatorname{Tr}(\rho_IA_I). \end{aligned}$$

Thus a picture change redistributes time dependence; it does not alter measurement probabilities.

Deriving the state equation

Differentiate the transformed 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 and full equations imply

$$i\hbar\dot U_0^\dagger = -U_0^\dagger H_0,$$ $$i\hbar\lvert\dot\psi_S\rangle = (H_0+\lambda V)\lvert\psi_S\rangle.$$

The two $H_0$ terms cancel, leaving

$$i\hbar\lvert\dot\psi_I\rangle = \lambda U_0^\dagger VU_0 \lvert\psi_I\rangle = \lambda V_I\lvert\psi_I\rangle.$$

The cancellation works for time-dependent $H_0(t)$ as long as $U_0$ solves its exact reference evolution equation.

For a density operator, the same transformation gives

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

This equation preserves trace, Hermiticity, positivity, and purity because it is still closed unitary evolution.

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

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

The order is essential: $U_I$ acts first on a ket at $t_0$, and $U_0$ then restores the reference motion. Differentiation gives

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

with

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

Hence

$$U_I(t,t_0) = \mathcal T \exp\!\left[ -\frac{i\lambda}{\hbar} \int_{t_0}^{t}V_I(s)\,ds \right].$$

If

$$[V_I(t_1),V_I(t_2)]=0$$

throughout the interval, time ordering is unnecessary and the expression is an ordinary operator exponential.

Dyson expansion

The first terms are

$$\begin{aligned} U_I(t,t_0) &= I -\frac{i\lambda}{\hbar} \int_{t_0}^{t}dt_1\,V_I(t_1)\\ &\quad- \frac{\lambda^2}{\hbar^2} \int_{t_0}^{t}dt_1 \int_{t_0}^{t_1}dt_2\, V_I(t_1)V_I(t_2)\\ &\quad+O(\lambda^3). \end{aligned}$$

The later-time operator appears to the left. The same result can be written with full-square integrals and a factor $1/2!$ only if an explicit time-ordering operator is retained.

The expansion and its order-by-order checks are collected at Dyson Series. The interaction picture supplies its most common quantum-mechanical generator, $\lambda V_I$.

Bohr-frequency phases

For time-independent $H_0$, choose eigenstates

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

Then

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

and

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

where

$$\omega_{mn} = \frac{E_m-E_n}{\hbar}, \qquad V_{mn}(t) = \langle m\rvert V(t)\lvert n\rangle.$$

Equivalently,

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

This formula separates reference-spectrum phases from explicit drive time dependence. Rapid phases can cancel after integration, while nearly stationary phases can accumulate. That statement is a diagnostic, not an automatic approximation: amplitudes, pulse duration, envelopes, and detuning all matter.

Inside a degenerate eigenspace, $\omega_{mn}=0$. The interaction then has no rapid reference phase to suppress mixing, so degenerate or quasi-degenerate treatment may be required.

Coefficient equations and first order

Expand

$$\lvert\psi_I(t)\rangle = \sum_nc_n(t)\lvert n\rangle.$$

The exact coefficients obey

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

For an initial state $\lvert i\rangle$, the first-order amplitude to a different reference eigenstate $\lvert f\rangle$ is

$$\begin{aligned} c_{fi}^{(1)}(t) &= -\frac{i\lambda}{\hbar} \int_{t_0}^{t}dt'\, \langle f\rvert V_I(t')\lvert i\rangle\\ &= -\frac{i\lambda}{\hbar} \int_{t_0}^{t}dt'\, e^{i\omega_{fi}(t'-t_0)}V_{fi}(t'). \end{aligned}$$

The transition probability begins at order $\lambda^2$ when the zeroth-order amplitude vanishes:

$$P_{i\to f}(t) = \lvert c_{fi}^{(1)}(t)\rvert^2 +O(\lambda^3), \qquad f\neq i.$$

For channels already populated at zeroth order, probabilities contain interference between different amplitude orders and must be expanded more carefully. The canonical transition calculation is at First-Order Transition Probability.

Commuting interaction

If $H_0$ and a time-independent $V$ commute,

$$[H_0,V]=0,$$

then

$$V_I(t)=V.$$

The interaction evolution is exactly

$$U_I(t,t_0) = e^{-i\lambda V(t-t_0)/\hbar},$$

and

$$U(t,t_0) = e^{-iH_0(t-t_0)/\hbar} e^{-i\lambda V(t-t_0)/\hbar}.$$

In a common eigenbasis, the perturbation changes phases and energies but does not transfer population. Within a degenerate $H_0$ subspace, an arbitrary $H_0$ basis may still mix under $V$ until $V$ is diagonalized in that subspace.

Two-level drive example

Take

$$H_0 = \frac{\hbar\omega_0}{2}\sigma_z,$$ $$V(t) = \frac{\hbar\Omega}{2} \cos(\omega t)\sigma_x, \qquad t_0=0.$$

Using

$$\sigma_\pm = \frac12(\sigma_x\pm i\sigma_y),$$

the reference transformation gives

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

Therefore

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

The interaction picture has exposed difference- and sum-frequency terms. It has not discarded either. Removing the rapidly rotating pair near resonance is the separate rotating-wave approximation, whose validity requires a scale analysis.

Choosing a useful split

A productive $H_0+\lambda V$ split usually has these features:

• $U_0$ is known exactly or can be computed reliably.
• Initial and target states have a simple representation under $H_0$.
• The residual $V_I$ has controlled matrix elements or useful oscillatory structure.
• Large diagonal shifts and strongly coupled degenerate subspaces are included in the reference problem when possible.
• The bookkeeping parameter corresponds to an actual small scale over the relevant time interval.

A small instantaneous coupling can still produce an order-one effect after a long resonant time. A useful channel diagnostic is

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

When this is not small, fixed first order is not controlled for that channel even if $\lambda$ is numerically small.

Units and checks

$H_0$, $V$, and $V_I$ have units of energy. Therefore

$$\frac{\lambda}{\hbar} \int dt\,V_I(t)$$

is dimensionless when $\lambda$ is dimensionless. The transformed interaction has the same spectrum and operator norm as $V(t)$ at each instant because unitary conjugation preserves both.

Useful exact checks are:

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

and picture invariance of expectation values. In a numerical calculation, compare direct Schrödinger evolution with $U_0U_I$ before trusting a perturbative truncation.

Common mistakes

• Calling the interaction picture approximate by definition.
• Transforming the state but using the untransformed $V$ in its equation.
• Forgetting explicit time dependence already present in $V(t)$.
• Reversing the factorization to $U_IU_0$.
• Using inconsistent reference times in phases and propagators.
• Dropping time ordering when $V_I$ fails to commute at different times.
• Writing an ordinary exponential for a noncommuting $H_0(t)$.
• Assuming every rapidly oscillating term is negligible without an error scale.
• Ignoring degenerate or nearly degenerate subspaces.
• Leaving a large secular diagonal term in the nominal perturbation.
• Applying the rotating-wave approximation during the exact picture change.
• Comparing coefficients across pictures without accounting for known phases.

Related formulas

• Time-Evolution Operator
• Dyson Series
• Heisenberg Equation
• Time Ordering
• Picture Transformations
• Density Operators in Different Pictures
• Rotating Frames
• Rotating-Wave Approximation

Exercises

1. Derive the transformed equation

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

Solution

Differentiate:

$$\lvert\dot\psi_I\rangle = \dot U_0^\dagger\lvert\psi_S\rangle +U_0^\dagger\lvert\dot\psi_S\rangle.$$

Using

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

and

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

the transformed $H_0$ terms cancel. Since $\lvert\psi_S\rangle=U_0\lvert\psi_I\rangle$,

$$i\hbar\lvert\dot\psi_I\rangle = \lambda U_0^\dagger VU_0 \lvert\psi_I\rangle = \lambda V_I\lvert\psi_I\rangle.$$

No assumption that $H_0$ is time independent was used.

2. Recover the Bohr phase

For time-independent $H_0$, derive the matrix-element formula for $V_I(t)$ between two $H_0$ eigenstates.

Solution

The reference propagator acts as

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

while

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

Therefore

$$\begin{aligned} \langle m\rvert V_I(t)\lvert n\rangle &= e^{i(E_m-E_n)(t-t_0)/\hbar}\\ &\quad\times \langle m\rvert V(t)\lvert n\rangle\\ &= e^{i\omega_{mn}(t-t_0)}V_{mn}(t). \end{aligned}$$

3. Commuting perturbation

Suppose $V$ is time independent and $[H_0,V]=0$. Find $V_I$, $U_I$, and the full propagator.

Solution

Commutation makes the conjugation trivial:

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

Hence

$$U_I(t,t_0) = e^{-i\lambda V(t-t_0)/\hbar}.$$

The full propagator is

$$U(t,t_0) = e^{-iH_0(t-t_0)/\hbar} e^{-i\lambda V(t-t_0)/\hbar}.$$

Because the factors commute, they may also be combined into $e^{-i(H_0+\lambda V)(t-t_0)/\hbar}$.

4. First-order constant pulse

Let $V_{fi}(t)=v$ for $t_0\leq t\leq t_0+T$ and zero otherwise. Compute the first-order amplitude for $f\neq i$ and discuss the resonant limit.

Solution

The amplitude is

$$\begin{aligned} c_{fi}^{(1)} &= -\frac{i\lambda v}{\hbar} \int_0^Tds\,e^{i\omega_{fi}s}\\ &= -\frac{i\lambda v}{\hbar} e^{i\omega_{fi}T/2} \frac{2\sin(\omega_{fi}T/2)}{\omega_{fi}}. \end{aligned}$$

As $\omega_{fi}\to0$,

$$\frac{2\sin(\omega_{fi}T/2)}{\omega_{fi}} \longrightarrow T,$$

so

$$c_{fi}^{(1)} \longrightarrow -\frac{i\lambda vT}{\hbar}.$$

The linear growth signals coherent accumulation. Once $\lvert\lambda v\rvert T/\hbar$ is not small, a fixed first-order treatment is no longer controlled.

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.
• 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.