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B Matrix exponential and applying it a random state

  1. Jan 11, 2017 #1

    td21

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    Gold Member

    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. jcsd
  3. Jan 11, 2017 #2

    DrClaude

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