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Consider a Hamiltonian $H(t) = H_0 +V(t)$. Let $\hat{\psi}=e^{iH_0 t} \psi$ be the state in the interaction picture where $\psi$ is that in the Schrodinger picture, i.e., $i\psi'=H(t)\psi$. Then we know that (formally) $$ \hat{\psi}(t)=S(t)\psi_0 $$ where $$ S(t)=T \exp\left( -i\int_0^t\hat{V}\right), \quad \hat{V}=e^{iH_0 t}Ve^{-iH_0 t} $$ However, this seems to lead to a paradox. More specifically, consider the case where the interaction $V$ is time-independent, i.e.., $V(t) =V$ so that $H(t)=H$ would also be time independent. Then we see that $\psi =e^{-iHt} \psi_0$ and that $S(t) = e^{-i\hat{V}t}$. Hence, \begin{align} e^{-i\hat{V}t} \psi_0 &= e^{iH_0 t} e^{-iHt}\psi_0 \\ e^{-iVt} &= e^{-iHt} e^{iH_0 t}\\ \end{align} Now obviously, this equality cannot hold in general by the BCH equality, so what is wrong? The most obvious justification would be that the interaction picture requires that $V(t)$ disappear at $t\to \pm \infty$, but I feel that this is rather weak since the argument (at least formally) does not rely upon this fact.

To sum up, is the interaction picture rigorous in any sense? If so, please provide a reference for such proof. If not, why is it a good approximation considering the above-mentioned paradox?

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Just because $V$ is $t$-independent doesn't mean $\hat{V}$ is. We know that $\hat{V}(t)=e^{-iH_0 t}Ve^{iH_0t}$. Therefore, you cannot generally write $S(t)=e^{i\hat V t}$, but must use the time-ordered exponential. The only time you can use the non-time-ordered exponential is when $[V,H]=0$, in which case your "paradox" disappears.

(Also: The interaction picture is an exact identity, and does not require any level of approximation)

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  • $\begingroup$ Your answer is better than mine... +1. $\endgroup$ Commented Feb 7, 2021 at 23:54

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