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Propagator operator

  1. Jan 21, 2012 #1
    So in the attachment, in fact (6), the formula for the propagator rectangled in red...

    is the Hamiltonian ACTING on (t-t')?

    is the Hamiltonian a function of (t-t')?

    or should it be (this is what I think), to be more clear

    U(t,t') = exp((-i/h)(t-t')H), so that when acting on a state |psi>, we have

    U(t,t')|psi> = exp((-i/h)(t-t')H|psi>)

    where H operates on whatever comes next. The last one makes most sense to me because U is overall an operator, so the RHS should also be an operator.

    Unless it means like this:

    U(t,t')|psi> = exp((-i/h)H[(t-t')|psi>]) ?

    I'm really not sure which one...

    Attached Files:

    Last edited: Jan 21, 2012
  2. jcsd
  3. Jan 22, 2012 #2
    It is just the Hamiltonian multiplied by (t-t'), so it is what you think it is. Here, you don't have to be careful with the ordering of the operator and the time-dependence, because the Hamiltonian never operates on time. Time and the Hamiltonian always commute.

    Also keep in mind that this derivation applies only to time-independent Hamiltonians.
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