Let K be any Matrix, not necessarily the hamitonian. Is $$e^{-Kt}\left|\psi\right>$$ equal to $$e^{-K\left|\psi\right>t}$$ even if it is not the the eigenvector of K?(adsbygoogle = window.adsbygoogle || []).push({});

I think so as i just taylor expand the $$e^{-Kt}$$ out but I want to confirm.

In that case can i say that $$\left<\psi\right|e^{-Kt}\left|\psi\right>$$ = $$e^{-\left<\psi\right|K\left|\psi\right>t}$$?

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# B Matrix exponential and applying it a random state

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