The Interaction Picture

This is introduced in order to facilitate the solution of the Schrödinger equation

$$\frac{d}{dt}\vert\Psi(t)\rangle = -i H(t)\vert\Psi(t)\rangle %%\nonumber\\$$ (3.2)

We define

$$\vert\Psi(t)\rangle_I\equiv e^{iH_0 t} \vert\Psi(t)\rangle$$ (3.3)

and derive the new Schrödinger equation for $\vert\Psi(t)\rangle_I$,

$$\frac{d}{dt}\vert\Psi_I(t)\rangle = iH_0 \vert\Psi_I(t)\rangle + e^{iH_0 t} \frac{d}{dt}\vert\Psi(t)\rangle$$

$$= iH_0 \vert\Psi_I(t)\rangle -i e^{iH_0 t}\left[ H_0 + V(t)\right] e^{-iH_0 t} \vert\Psi_I(t)\rangle$$

$$= -i e^{iH_0 t} V(t) e^{-iH_0 t} \vert\Psi_I(t)\rangle \equiv -i V_I(t) \vert\Psi_I(t)\rangle.$$ (3.4)

The Schrödinger equation therefore is transformed into the interaction picture

$$\frac{d}{dt}\vert\Psi(t)\rangle = -i H(t)\vert\Psi(t)\rangle \leftrightarrow \frac{d}{dt}\vert\Psi_I(t)\rangle=-i V_I(t) \vert\Psi_I(t)\rangle$$

$$\vert\Psi(t)\rangle_I \equiv e^{iH_0 t} \vert\Psi(t)\rangle,\quad V_I(t)\equiv e^{iH_0 t} V(t) e^{-iH_0 t}.$$ (3.5)