- #1
Klaus_Hoffmann
- 86
- 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.
[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.