Interaction Picture

The interaction picture is a special time-dependent unitary frame selected by a Hamiltonian split. The general exact transformation and frame-generator term are derived in Picture Transformations. Here the Hamiltonian is split into a solvable part and an interaction:

$$H(t)=H_0+V(t).$$

Here $H_0$ is time independent. Its motion is absorbed into operators, while states evolve under the transformed interaction.

For a compact Core Formalism comparison with the Schrödinger and Heisenberg pictures, see Pictures of Motion Overview.

Why Split the Hamiltonian

The split is useful when $H_0$ is exactly solvable and $V(t)$ is the part responsible for transitions. This is the natural language of time-dependent perturbation theory, light-matter interactions, and field-theory perturbation expansions.

The split is a choice. It should be made to simplify the problem.

Definitions

Let

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

The interaction-picture state is

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

An operator becomes

$$A_I(t)=U_0^\dagger(t,t_0)A_SU_0(t,t_0).$$

Interaction Hamiltonian

The transformed interaction is

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

The interaction-picture state satisfies

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

Thus the exact problem has been rewritten so that only the interaction drives the state.

Interaction-Picture Evolution

Define $U_I(t,t_0)$ by

$$\lvert\psi_I(t)\rangle =U_I(t,t_0)\lvert\psi_I(t_0)\rangle.$$

Because $U_0(t_0,t_0)=I$, the full propagator factors as $U=U_0U_I$, or $U_I=U_0^\dagger U$. For $t\ge t_0$,

$$U_I(t,t_0) = \mathcal T \exp\left[ -\frac{i}{\hbar} \int_{t_0}^{t}V_I(t')\,dt' \right].$$

Expanding this expression gives the Dyson series.

Joining intervals in one interaction picture

The reference time used to define a picture need not be the starting time of each evolution interval. Keep one reference time $\tau$ fixed and write $\lvert\psi_I^{[\tau]}(t)\rangle=U_0^\dagger(t,\tau)\lvert\psi_S(t)\rangle$. Applying the two-endpoint picture transformation to evolution from $s$ to $t$ gives

$$U_I^{[\tau]}(t,s) =U_0^\dagger(t,\tau)U(t,s)U_0(s,\tau).$$

The adjacent frame factors then cancel when intervals are composed:

$$U_I^{[\tau]}(t_2,t_1)U_I^{[\tau]}(t_1,t_0) =U_I^{[\tau]}(t_2,t_0).$$

The simpler expression $U_0^\dagger(t,s)U(t,s)$ instead chooses $s$ as the reference time. Multiplying several such expressions without translating them into one frame generally fails, even if $H_0$ and $V$ are constant. For the time-independent $H_0$ used here, the translation is

$$U_I^{[\tau]}(t,s) =U_0^\dagger(s,\tau) \bigl[U_0^\dagger(t,s)U(t,s)\bigr]U_0(s,\tau).$$

This is the same fixed-picture composition rule used in Tong’s treatment of interaction-picture evolution. The exercise below tests it with a finite, exactly solvable spin rotation.

A static perturbation becomes a rotating interaction

Take $t_0=0$, $H_0=\hbar\omega\sigma_z/2$, and $V=\hbar g\sigma_x/2$. Direct matrix multiplication gives

$$V_I(t)=\frac{\hbar g}{2} \left(\sigma_x\cos\omega t-\sigma_y\sin\omega t\right).$$

Although $V$ is static, its interaction-picture matrix elements oscillate at the unperturbed energy gap. In an energy eigenbasis the general formula is

$$(V_I)_{mn}(t)=e^{i(E_m-E_n)t/\hbar}V_{mn}(t).$$

These phases control interference between transition amplitudes; they are retained in interaction-picture perturbation theory. No small-$g$ assumption was used to obtain them.

Common Mistakes

• Thinking the interaction picture is approximate by definition.
• Forgetting to transform $V(t)$ into $V_I(t)$.
• Choosing an $H_0$ that does not simplify the problem.
• Mixing Schrödinger-picture and interaction-picture operators in the same formula.
• Dropping time ordering when $V_I(t)$ does not commute with itself at different times.
• Resetting the interaction-picture reference time between intervals without transforming the intermediate state.

References

• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Wiley, 1977.
• M. E. Peskin and D. V. Schroeder, An Introduction to Quantum Field Theory, Addison-Wesley, 1995.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• D. Tong, Quantum Field Theory, University of Cambridge lecture notes, 2006, §3.1, “The Interaction Picture”.

Exercises

1. If $V(t)=0$, what happens to $\lvert\psi_I(t)\rangle$?

Solution

When $V(t)=0$, $V_I(t)=0$. The interaction-picture equation becomes

$$i\hbar\frac{d}{dt}\lvert\psi_I(t)\rangle=0.$$

So $\lvert\psi_I(t)\rangle$ is constant. All $H_0$ motion has been moved into the operators.

2. In the spin example above set $\omega=g=\nu>0$, choose the fixed reference $\tau=0$, and take two intervals of length $\delta=\pi/(\sqrt2\nu)$. Starting in $\lvert0\rangle$, the $+1$ eigenstate of $\sigma_z$, find the probability of $\lvert1\rangle$ at $2\delta$. Compare correct composition with the incorrect product $W^2$, where $W=U_0^\dagger(\delta,0)U(\delta,0)$ is used for both intervals.

Solution

Since $H=\hbar\nu(\sigma_x+\sigma_z)/2$,

$$U(\delta,0)=-\frac{i}{\sqrt2}(\sigma_x+\sigma_z), \qquad U(2\delta,0)=-I.$$

The exact final state is $-\lvert0\rangle$, so the transition probability is zero. The correct interaction-picture product is $U_I^{[0]}(2\delta,\delta)U_I^{[0]}(\delta,0) =-U_0^\dagger(2\delta,0)$, which is diagonal in the $\sigma_z$ basis. The measured projector $\lvert1\rangle\langle1\rvert$ commutes with $U_0$, so it has the same matrix in both pictures.

For the reset-frame product, put $\phi=\nu\delta/2=\pi/(2\sqrt2)$. Then

$$W=-\frac{i}{\sqrt2}e^{i\phi\sigma_z}(\sigma_x+\sigma_z), \qquad \langle1\rvert W^2\lvert0\rangle=-i e^{-i\phi}\sin\phi.$$

The incorrect transition probability is therefore $\sin^2\phi\simeq0.803$. Both factors are unitary, so checking norm conservation would not detect the error. The second factor must act in the same interaction picture as the state produced by the first.