# Path integrals and propagators

1. Jun 28, 2007

### 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.

2. Jun 28, 2007

### olgranpappy

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

3. Jun 28, 2007

### olgranpappy

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

4. Jun 28, 2007

### smallphi

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

5. Jun 28, 2007

### olgranpappy

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

6. Jun 28, 2007

### smallphi

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