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Consider a system with Hamiltonian $\hat{H}=\hat{H_0}+\hat{V}$. We define the interaction picture kets $|\psi(t)\rangle _I$ by $$\tag{1} |\psi(t)\rangle _I=\exp\left(\frac{i}{\hbar}\hat{H}_0(t-t_0)\right)|\psi(t)\rangle$$ and interaction picture operators $\hat{O}_I$ by $$\tag{2} \hat{O}_I= \exp\left(\frac{i}{\hbar}\hat{H}_0(t-t_0)\right)\hat{O}\exp\left(-\frac{i}{\hbar}\hat{H}_0(t-t_0)\right).$$

From the ordinary Schrödinger equation, we obtain the equation of motion for the state: $$\tag{3} i\hbar\frac{d}{dt}|\psi(t)\rangle _I = \hat{V}_I|\psi(t)\rangle _I.$$ As is done e.g. in Sakurai's book, we also define the time-evolution operator $\hat{U}(t,t_0)_I$ by $$\tag{4} |\psi(t)\rangle _I = \hat{U}(t,t_0)_I |\psi(t_0)\rangle _I.$$ Differentiating (4) w.r.t $t$ yields $$\tag{5} \begin{align} i\hbar\frac{d}{dt}|\psi(t)\rangle _I &= i\hbar\frac{d}{dt}\left(\hat{U}(t,t_0)_I\right) |\psi(t_0)\rangle \\ &=i\hbar\frac{d}{dt}\left(\hat{U}(t,t_0)_I\right)\hat{U}(t,t_0)_I^{-1} |\psi(t)\rangle. \end{align}$$ Comparison of (3) and (5) shows that we must have the following equation of motion for $\hat{U}(t,t_0)_I$: $$\tag{6} i\hbar\frac{d}{dt}\hat{U}(t,t_0)_I = \hat{V}_I\hat{U}(t,t_0)_I.$$ However, if we take $\hat{O}=\hat{U}(t,t_0)$ in equation (2), then equation (2) would yield $$\tag{7} \begin{align} &i\hbar\frac{d}{dt}\hat{U}(t,t_0)_I\\ &= \exp\left(\frac{i}{\hbar}\hat{H}_0(t-t_0)\right)(-\hat{H}_0\hat{U}(t,t_0)+\hat{U}(t,t_0)\hat{H}_0)\exp\left(-\frac{i}{\hbar}\hat{H}_0(t-t_0)\right)\phantom{abc}\\&+\exp\left(\frac{i}{\hbar}\hat{H}_0(t-t_0)\right)\hat{H}\hat{U}(t,t_0)\exp\left(-\frac{i}{\hbar}\hat{H}_0(t-t_0)\right)\\ &= \hat{V}_I\hat{U}(t,t_0)_I+\hat{U}(t,t_0)_I\hat{H}_0. \end{align}$$

Evidently, equations (6) and (7) do not agree. Am I misunderstanding something here?

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1 Answer 1

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The relation between the time evolution operator $U_\mathrm I$ in the interaction picture and the time evolution operator $U_\mathrm S$ in the Schrödinger picture is not given by equation $(2)$ in the OP. In other words, the operator obtained from transforming $U_\mathrm S$ into the interaction picture is not equal to the time evolution operator in the interaction picture.

In fact, the correct relation reads $$U_\mathrm I (t,t^\prime) = e^{iH_0t}\, U_\mathrm S (t,t^\prime) \,e^{-iH_0t^\prime } \tag{1} \quad , $$ which follows from $$|\Psi(t)\rangle_\mathrm I = e^{iH_0t}\,U_\mathrm S(t,t^\prime) \, |\Psi(t^\prime)\rangle_\mathrm S = e^{iH_0t}\,U_\mathrm S(t,t^\prime)\, e^{-iH_0t^\prime} |\Psi(t^\prime)\rangle_\mathrm I \overset{!}{=} U_\mathrm I (t,t^\prime)\, |\Psi(t^\prime)\rangle_\mathrm I\quad . $$


By using the correct relation between these evolution operators, we obtain the correct differential equation for $U_\mathrm I$. Indeed, differentiating equation $(1)$ with respect to $t$ shows that

\begin{align} i \frac{\mathrm d U_\mathrm I (t,t^\prime)}{\mathrm d t} &= - e^{iH_0t}\,H_0\, U_\mathrm S(t,t^\prime) \, e^{-iH_0t^\prime} + e^{iH_0t} \,\overbrace{i\frac{\mathrm d U_\mathrm S (t,t^\prime)}{\mathrm dt}}^{=H\,U_\mathrm S{(t,t^\prime)}}\, e^{-iH_0t^\prime} \\ &= -e^{iH_0t}\,H_0\, U_\mathrm S(t,t^\prime) \, e^{-iH_0t^\prime} + e^{iH_0t} \,(H_0 +V)\,U_\mathrm S (t,t^\prime)\, e^{-iH_0t^\prime} \\ &= e^{iH_0t}\, V \, U_{\mathrm S}(t,t^\prime) \, e^{-iH_0t^\prime} \\ &= e^{iH_0t}\, V \, e^{-iH_0t}\, e^{iH_0t} \,U_{\mathrm S}(t,t^\prime) \, e^{-iH_0t^\prime} \\ &= V_\mathrm I\, U_\mathrm I(t,t^\prime) \quad . \end{align}

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