# B Matrix exponential and applying it a random state

1. Jan 11, 2017

### 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 $$\left<\psi\right|e^{-Kt}\left|\psi\right>$$ = $$e^{-\left<\psi\right|K\left|\psi\right>t}$$?

Last edited: Jan 11, 2017
2. Jan 11, 2017

### Staff: Mentor

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).

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
$$\langle \psi | e^{-\hat{K}t} |\psi\rangle = \sum_k \left|c_k\right|^2 e^{-k t} \neq e^{-\langle \psi | \hat{K} |\psi\rangle t}$$