Consider a system with Hamiltonian $\hat{H}=\hat{H_0}+\hat{V}$. We define the interaction picture kets $|\psi(t)\rangle _I$ by $$\tag{1} |\psi(t)\rangle _I=\exp\left(\frac{i}{\hbar}\hat{H}_0(t-t_0)\right)|\psi(t)\rangle_{S}$$ 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 $$\tag{5} \begin{align} i\hbar\frac{d}{dt}|\psi(t)\rangle _I &= i\hbar\frac{d}{dt}\left(\hat{U}(t,t_0)_I\right) |\psi(t_0)\rangle_{I} \\ &=i\hbar\frac{d}{dt}\left(\hat{U}(t,t_0)_I\right)\hat{U}(t,t_0)_I^{-1} |\psi(t)\rangle_{I}. \end{align}$$ 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?


1 Answer 1


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}

  • $\begingroup$ Note: We've set, implicitly, $t_0=0$ (the time where the Schrödinger and Dirac picture coincide, i.e. the initial condition of the time-evolution). But one can easily generalize $(1)$ following the same line of thoughts. $\endgroup$ Commented Oct 4, 2023 at 21:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.