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Consider a Hamiltonian $H$ of two components, a time-independent $H_0$ and a time-dependent $V(t)$,

$$H(t)=H_0+V(t)\,.$$

In the Schrödinger picture of quantum mechanics, the state $\left|\psi(t)\right\rangle_{\mathrm{S}}$ evolves according to:

$$\mathrm{i}\hbar\frac{\partial}{\partial t}\left|\psi(t)\right\rangle_{\mathrm{S}}=H(t)\left|\psi(t)\right\rangle_{\mathrm{S}}\,.$$

This equation admits a formal solution given by, $$\left|\psi(t)\right\rangle_{\mathrm{S}}=U(t,t_0)\left|\psi(t_0)\right\rangle_{\mathrm{S}}\,.$$

$U(t,t_0)$ is the evolution operator corresponding to $H(t)$, defined by $$U(t,t_0)=\mathcal{T}\exp\left(-\frac{\mathrm{i}}{\hbar}\int\limits_{t_0}^{t}\mathrm{d}t'H(t')\right)\,,$$ where $\mathcal{T}$ is the time-ordering operator.

One define the states and operators in the interaction picture as $$\left|\psi(t)\right\rangle_{\mathrm{I}}=\exp\left(\frac{\mathrm{i}}{\hbar}H_{0}t\right)\left|\psi(t)\right\rangle_{\mathrm{S}}\,,$$ $$A_{\mathrm{I}}(t)=\exp\left(\frac{\mathrm{i}}{\hbar}H_{0}t\right)A_{\mathrm{S}}(t)\exp\left(-\frac{\mathrm{i}}{\hbar}H_{0}t\right)\,.$$

$A_{\mathrm{S}}(t)$ is the operator in the Schrödinger picture, such as $V(t)$, which in general could depend on time.

One could prove that $$\mathrm{i}\hbar\frac{\partial}{\partial t}\left|\psi(t)\right\rangle_{\mathrm{I}}=V_{\mathrm{I}}(t)\left|\psi(t)\right\rangle_{\mathrm{I}}\,.$$

This resolves into $$\left|\psi(t)\right\rangle_{\mathrm{I}}=U'(t,t_0)\left|\psi(t_0)\right\rangle_{\mathrm{I}}\,,$$

where $$U'(t,t_0)=\mathcal{T}\exp\left(-\frac{\mathrm{i}}{\hbar}\int\limits_{t_0}^{t}\mathrm{d}t'V_{\mathrm{I}}(t')\right)\,.$$

I know that $U$ and $U'$ are related to each other by $$U'(t,t_0)=\exp\left(\frac{\mathrm{i}}{\hbar}H_{0}t\right)U(t,t_0)\exp\left(-\frac{\mathrm{i}}{\hbar}H_{0}t_0\right)\,.$$

This can be proved by a differentiation procedure (and essentially this is how the interaction picture is derived from the Schrödinger one). But how can one prove this based on the expansion in Neumann's series of the evolution operators only?

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It is not correct that $$U'(t,t_0)=\exp\left(\frac{\mathrm{i}}{\hbar}H_{0}t\right)U(t,t_0)\exp\left(-\frac{\mathrm{i}}{\hbar}H_{0}t_0\right)\,.$$ You can simply see this when setting $V\equiv 0$. We then have that $$ U(t,t_0) = \exp\left[-i (t-t_0) H_0\right]$$ while $$U'(t,t_0)= 1 \neq\exp\left(\frac{\mathrm{i}}{\hbar}H_{0}t\right)U(t,t_0)\exp\left(-\frac{\mathrm{i}}{\hbar}H_{0}t_0\right)\,.$$

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  • $\begingroup$ Respectfully disagree (I wanted to say Holy Cow!). In the case you considered, couldn't you just split $U(t,t_0)$ into $\exp(-\frac{\mathrm{i}}{\hbar}H_0t)\times\exp(\frac{\mathrm{i}}{\hbar}H_0t_0)$. $H_0$ commutes with itself! $\endgroup$ – Willi Tschau Jan 22 at 15:51

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