Green's Function in Time-Dep. Schrodinger Eqn.

[kreil]

Show, from it's definition,

$$\psi(x,t) = \int dx' G(x,t;x',t_0) \psi(x',t_0)$$

$$G(x,t;x',t_0)= \langle x | U(t,t_0) | x' \rangle$$

that the propagator G(x,t;x',t') is the Green Function of the Time-Dependent Schrödinger Equation,

$$\left ( H_x - i \hbar \frac{\partial}{\partial t} \right ) G(x,t;x',t_0) = -i \hbar \delta(x-x') \delta(t-t')$$

where H is the Hamiltonian expressed as a differential operator in the x representation. Calculate the propagator for a free particle by this method.Is the best way to do this to substitute that form of the propagator into the Schrödinger eqn and take the derivatives? If so, how are the delta functions defined?

 

[gabbagabbahey]

If I were you, I'd multupily both sides of your final equation by $\psi(x',t_0)$ and integrate over all values of $x'$ and $t_0$...integrating the delta function terms should be straightforward once you look up the defining properties of the Durac Delta distribution.

 

[kreil]

I am aware of the usual properties, but thought for some reason they may have a definition in terms of bras/kets in the context of this problem.

So the right hand side integration fixes x'=x and t0=t, so that the RHS is $-i \hbar \psi(x,t)$

How should I deal with the derivatives on the left? i.e.

$$\left ( H_x - i \hbar \frac{\partial}{\partial t} \right ) \int \int G(x,t;x,t_0) \psi(x',t_0) dx' dt_0= \left ( -H_x- i \hbar \frac{\partial}{\partial t} \right ) \int \psi(x,t) dt_0=-i\hbar \psi(x,t)$$

The form on the RHS of the equation seems to suggest that only the time derivative contributes. If so, why are the other terms zero?

 

[kreil]

Edit: I think the part that most confuses me about this is that the time dependent Schrödinger equation is,

$$\left ( H_x - i \hbar \frac{\partial}{\partial t}\right ) \psi =0$$

In other Green's function problems I have done, there is a function on the RHS of the differential equation.

In this case, the following confuses me:

$$\left ( H_x - i \hbar \frac{\partial}{\partial t}\right ) \psi = \left ( H_x - i \hbar \frac{\partial}{\partial t}\right ) \int dx' G(x,t;x',t_0) \psi(x',t_0) = \int \left [ \left ( H_x - i \hbar \frac{\partial}{\partial t}\right ) G(x,t;x',t_0) \right ] \psi(x',t_0) dx' = -i \hbar \int \psi(x', t_0) \delta(x-x') \delta(t-t')dx'$$

$$=-i \hbar \psi(x,t_0) \delta(t-t_0)=0$$

$$-i \hbar \int \psi(x,t_0) \delta(t-t_0) dt_0 = \int 0 dt_0$$

$$-i \hbar \psi(x,t) = constant$$

 

[kreil]

Nobody has any thoughts on how to do this? I'm still stuck.

 

[fzero]

You have to use the expression

$$G(x,t;x',t_0)= \langle x | U(t,t_0) | x' \rangle$$

for the Green function. Act with

$$\left ( H_x - i \hbar \frac{\partial}{\partial t} \right )$$

and use the definition of $$U(t,t_0)$$.

 

[kreil]

I can take the differential operator inside and let it act directly on U? I thought so, but I got zero as an answer when I did this:

$$\langle x | \left [ \frac{-\hbar^2}{2m} \frac{\partial^2 U(t,t_0)}{\partial x^2} + W(x)U(t,t_0) - i\hbar \frac{\partial U(t,t_0)}{\partial t}\right ] | x' \rangle$$

$$\langle x | \left [ W(x)U(t,t_0) - H_x U(t,t_0)| x' \rangle = \langle x | W(x)U(t,t_0) - W(x)U(t,t_0) \right ] | x' \rangle =0$$

since

$$\frac{\partial U}{\partial x}=0$$

$$\frac{\partial U}{\partial t} = \frac{-i H_x}{\hbar}U(t,t_0)$$

 

[fzero]

You have to be more careful. The states are in the position basis, so you can't just push the x-derivatives past the position states. It might be a good idea to insert a convenient complete set of states into the matrix element. Note that the Green function will satisfy Schrödinger's equation, so you'll get 0 almost everywhere. You will also need to consider the behavior as $$t\rightarrow t_0$$.

 

[kreil]

I arrived at the expression

$$\left ( H_x - i \hbar \frac{\partial}{\partial t} \right ) G(x,t;x',t') =...=(2\pi \hbar)^{-1} \left ( H_x - i \hbar \frac{\partial}{\partial t} \right ) e^{-iV(x') \Delta t/\hbar} \int e^{-ip^2 \Delta t/2m \hbar} e^{ip(x-x')/\hbar} dp$$

At this point should I distribute the derivatives and then integrate, or integrate and then take the derivatives? I must have some delta functions in the answer, which I assume are defined by a Fourier transform?

 

[fzero]

I would have inserted energy eigenstates. The problem with the momentum eigenstates is that the momentum does not commute with the Hamiltonian in the presence of a non-trivial potential, so $$<x|V|p>$$ is not proportional to a plane wave. Energy eigenstates would be better, since $$<x|n>$$ always satisfies the Schrödinger equation. You'll get one delta function from noting that the Green function has a discontinuity at $$t=t'$$. The other can be seen by looking at the $$t\rightarrow t'$$ limit.

 

[kreil]

Thanks, I think I got it