The interaction picture

Intuition

When the Hamiltonian is a solved part plus a perturbation that changes in time, most of what the state does is already known: every level turns its phase at its own frequency. That motion carries no news. The interaction picture takes it away, by turning every state back at the rate the solved part alone would turn it. What is left changes only because of the perturbation, and changes slowly when the perturbation is weak. The solved part’s motion is handed to the operators instead, so the picture sits between Schrödinger’s, where the state does all the moving, and Heisenberg’s, where the operators do.

To see how a gust of wind disturbs the hands of a clock, watch the clock from a frame that turns with its hands at their normal rate. In that frame an undisturbed hand stands still, and every movement you see is the gust’s doing.

plane diagram: interaction-picture-envelope

The real part of the amplitude of one level while a weak perturbation slowly empties it. In the Schrödinger picture it turns at the level’s own frequency $E/\hbar$ inside a slow envelope; in the interaction picture that turning is removed, and what is left is the envelope, the slow change the perturbation causes.

Split the Hamiltonian as $\hat{H}(t)=\hat{H}_{0}+\hat{V}(t)$ , with solved. Undo in the state the evolution alone would cause, and give it to the operators.

\begin{aligned}\ket{\psi_{I}(t)}&=e^{i\hat{H}_{0}t/\hbar}\ket{\psi(t)}\\ \hat{V}_{I}(t)&=e^{i\hat{H}_{0}t/\hbar}\,\hat{V}(t)\,e^{-i\hat{H}_{0}t/\hbar}\\ i\hbar\frac{d}{dt}\ket{\psi_{I}(t)}&=\hat{V}_{I}(t)\ket{\psi_{I}(t)}\end{aligned}

Properties

With $\hat{V}=0$ the state $\ket{\psi_{I}}$ stands still: everything does has been taken out.

The equation of the interaction picture

i\hbar\frac{d}{dt}\ket{\psi_{I}(t)}=\hat{V}_{I}(t)\ket{\psi_{I}(t)}

Differentiate the definition by the product rule. The derivative of the exponential brings down $-\hat{H}_{0}$ , which cancels the $\hat{H}_{0}$ of Schrödinger’s equation, and what is left is the perturbation between the same two exponentials.

Proof steps

\ket{\psi_{I}(t)}=e^{i\hat{H}_{0}t/\hbar}\ket{\psi(t)}

The definition.

\begin{aligned}i\hbar\frac{d}{dt}\ket{\psi_{I}}&=-\hat{H}_{0}e^{i\hat{H}_{0}t/\hbar}\ket{\psi}\\ &\quad+e^{i\hat{H}_{0}t/\hbar}\,i\hbar\frac{d}{dt}\ket{\psi}\end{aligned}

The product rule: the exponential’s derivative brings down $i\hat{H}_{0}/\hbar$ .

i\hbar\frac{d}{dt}\ket{\psi}=\bigl(\hat{H}_{0}+\hat{V}\bigr)\ket{\psi}

Schrödinger’s equation for the whole Hamiltonian.

i\hbar\frac{d}{dt}\ket{\psi_{I}}=e^{i\hat{H}_{0}t/\hbar}\,\hat{V}\ket{\psi}

$\hat{H}_{0}$ commutes with its own exponential, so the two $\hat{H}_{0}$ terms cancel.

\begin{aligned}i\hbar\frac{d}{dt}\ket{\psi_{I}}&=e^{i\hat{H}_{0}t/\hbar}\,\hat{V}e^{-i\hat{H}_{0}t/\hbar}\ket{\psi_{I}}\\ &=\hat{V}_{I}\ket{\psi_{I}}\end{aligned}

Write $\ket{\psi}=e^{-i\hat{H}_{0}t/\hbar}\ket{\psi_{I}}$ .

Applications

Practice

Removing the Known Motion

The solved part of the Hamiltonian turns each level’s phase at its own frequency. The interaction picture undoes that turning in the state, so that the state changes only because of the perturbation.

\ket{\psi_{I}(t)}=e^{i\hat{H}_{0}t/\hbar}\ket{\psi(t)}

Try it

What does passing to the interaction picture remove from the time dependence of the state?