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Path integrals and propagators

  1. Jun 28, 2007 #1
    we know that for the SE equation we find the propagator

    [tex] (i\hbar \partial _{t} - \hbar ^{2} \nabla +V(x,y,z) )K(x,x')=\delta (x-x') [/tex]

    with m=1/2 for simplicity

    then we know that the propagator K(x,x') may be obtained from the evaluation of the Path integral.

    [tex] K(x,x')=C \int \mathcal D[x] e^{iS[x]/\hbar} [/tex] (sum over all path X(t) )

    my question is, since we can't know the evaluation of the path integral exactly, but give a WKB approach of this if we name the result of the path integral by [tex] K_{WKB}(x,x') [/tex].

    then my question is if at least as an approximation this function satisfies.

    [tex] (i\hbar \partial _{t} - \hbar ^{2} \nabla +V(x,y,z) )K_{WKB}(x,x')=\delta (x-x') [/tex]

    the notation WKB means that we have evaluated the propagator and so on in a semiclassical way.
  2. jcsd
  3. Jun 28, 2007 #2


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    you seem to have lost your x and x' dependence... at least on the RHS of the above equation
  4. Jun 28, 2007 #3


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    twice you said "my question is" but you never asked a question.
  5. Jun 28, 2007 #4
    Isn't that K(x,x') a Green's function? I think the propagator is the 'operator inverse' of the Green's funciton.
  6. Jun 28, 2007 #5


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    no. "propagator" and "green's function" are synonymous.
  7. Jun 28, 2007 #6
    Yeah I remember now, propagator = greens function = operator inverse of the field equation operator
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