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How would an operator A in the Schrodinger $A_s$ and interaction $A_I$ picture be related if the commutation relation

$$[A_s,H_0] = 0$$ holds where $H_0$ is a solved hamiltoniain.

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It is fairly simple to see that they would be same. $$\begin{aligned} A_{s}(t) & = \text{exp}\left(-\imath H_{0} t\right)A_{s}\text{exp}\left(\imath H_{0} t\right) \\ & = \sum_{n,m}\left(-\imath\right)^{n}\left(\imath\right)^{m}\frac{1}{n!m!}H_{0}^{n}A_{s}H_{0}^{m} \\ & = \sum_{n,m}\left(-\imath\right)^{n}\imath^{m}\frac{1}{n!m!}A_{s}H_{0}^{n+m} \\ & \left(\text{Since, } \left[A_{s},H_{0}\right]=0\right)\\ & = A_{s} \end{aligned}$$

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  • $\begingroup$ im not sure though how that show the interaction and schrodinger picture are the same though $\endgroup$ – DJA Apr 5 at 21:20

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