Path integrals and propagators

[Klaus_Hoffmann]

we know that for the SE equation we find the propagator

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

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.

K(x,x')=C \int \mathcal D[x] e^{iS[x]/\hbar} (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 K_{WKB}(x,x').

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

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

the notation WKB means that we have evaluated the propagator and so on in a semiclassical way.

 

[olgranpappy]

we know that for the SE equation we find the propagator

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

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.

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

you seem to have lost your x and x' dependence... at least on the RHS of the above equation

 

[olgranpappy]

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 K_{WKB}(x,x').

then my question is...

twice you said "my question is" but you never asked a question.

 

[smallphi]

we know that for the SE equation we find the propagator

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

with m=1/2 for simplicity

Isn't that K(x,x') a Green's function? I think the propagator is the 'operator inverse' of the Green's funciton.

 

[olgranpappy]

no. "propagator" and "green's function" are synonymous.

 

[smallphi]

Yeah I remember now, propagator = greens function = operator inverse of the field equation operator