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Probability Amplitude and Time Evolution Operator

  1. Feb 9, 2009 #1
    Hello,

    I am fairly new to quantum physics (I'm actually an engineer, not a physicist). I think I am getting a decent grasp on things, but I have a question.

    Suppose you have two time dependent states: [tex]\Psi_{1}[/tex] and [tex]\Psi_{2}[/tex].

    Also, suppose that we have a constant potential, represented in our Hamiltonian as V.

    Our Hamiltonian is thus represented as: [tex]\hat{H}=1/(2m)\partial^2/x^2+V(x)[/tex]

    Now suppose that we want to find the probability amplitude to go from state 1 to the 2nd state in a time interval t1 to t2.

    It is my understanding that the following operation is used:

    [tex]<\Psi_{2}|exp(-i\hat{H}(t2-t1))|\Psi_{1}>[/tex]

    This is of course equal to

    [tex]\int\Psi_{2}*exp(-i\hat{H}(t2-t1))\Psi_{1}dx[/tex]

    However, how do I evaluate the following operation?

    [tex]exp(\hat{H})\Psi_{1}[/tex]

    Many thanks.
     
  2. jcsd
  3. Feb 10, 2009 #2

    nicksauce

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    The general idea of evaluating a function of an operator is to expand the function...

    exp(H) = 1 + H + H^2/2 + H^3/6 + ...

    If Psi is an eigenfunction of H with eigenvalue E, then it is not hard to show that exp(H) Psi = exp(E) Psi.
     
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