B Matrix exponential and applying it a random state

[td21]

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?

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

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

 

[DrClaude]

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?

I'm not sure this is formally correct, as it looks weird. The product of an operator on a ket should give back a ket, not the exponential of a ket (whatever that is).

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

This is definitely not correct. Considering that you can expand $|\psi\rangle$ in terms of the eigenstates of $\hat{K}$,
$$
|\psi\rangle = \sum_k c_k |k \rangle
$$
with
$$
\hat{K} |k \rangle = k |k \rangle
$$
then
$$
e^{-\hat{K}t} |\psi\rangle = \sum_k c_k |k \rangle e^{-k t}
$$
so
$$
\langle \psi | e^{-\hat{K}t} |\psi\rangle = \sum_k \left|c_k\right|^2 e^{-k t} \neq e^{-\langle \psi | \hat{K} |\psi\rangle t}
$$