This is a continuation of the previous post on Green’s Functions.

When you have a hammer, every problem looks like a nail.

Take the Schrodinger equation: $\hat{H} \psi(\vec{x},t) = i h \frac{\partial\psi(\vec{x},t)}{\partial t}$

Writing it as $(-\frac{h^2}{2 m}\nabla^2 - i h \frac{\partial}{\partial t}) \psi(\vec{x},t)=-V(\vec{x},t)\psi(\vec{x},t)$, we can try and come up with a green’s function for it.

What we’d like to be able to do is start with the wavefunction and potential at some point in time, and have it tell us the wavefunction for all times.

The standard QM thing to do would be to write the wavefunction as the sum of eigenstates of the Hamiltonian, them time evolve each of them. But instead, let’s see how far Mr. Green can take us.

Here’s the defining relation for the green function in omega space:

$(-\frac{h^2}{2 m}\nabla^2 - i h \frac{\partial}{\partial t}) G=\frac{h}{i} \delta^3(\vec{x-x_0})\delta(t-t_0)$

(where a factor of $\frac{h}{i}$ has been introduced for later convenience)

$\rightarrow (\frac{h^2}{2 m}\nabla^2 + h \omega) G_\omega = -\frac{h}{i} \delta^3(\vec{x-x_0})$

$\rightarrow (\nabla^2 + k^2)G_{\omega}=-\frac{h}{i} \frac{2 m}{h^2}\delta^3(\vec{x-x_0})$ where $k=\sqrt{\frac{2 m \omega}{h}}$

This is just our old friend the Helmholtz Equation, solved in the previous post, with a constant in front. The solution is: $G_{\omega}(\vec{x,x_0},\omega)=\frac{h}{i} \frac{2 m}{h^2}\frac{1}{4\pi r}e^{\pm i k r}$

To get the space-time propagator, we fourier transform back.

$G(\vec{x,x_0},t,t_0)=\frac{h}{i} \frac{2m}{h^2}\frac{1}{4\pi r}\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{\pm i k r}e^{-i \omega (t-t_0)}d\omega$

giving $G(\vec{x,x_0},t,t_0)=(\frac{m}{2 \pi i h (t-t_0)})^{\frac{3}{2}}e^{\frac{i m r^2}{2 h (t-t_0)}}$

This is the Green’s function for the Schrodinger equation in 3 dimensions. In D dimensions, $\frac{3}{2} \rightarrow \frac{D}{2}$

The integral form of the Schrodinger equation is then:
$\psi(\vec{x},t)=\psi_0(\vec{x},t)-\frac{i}{h}\int_{R^3} \int_{-\infty}^{t} G(\vec{x,x_0},t,t_0) V(\vec{x_0},t_0) \psi(\vec{x_0},t_0) d t_0 d^3 x_0$

The equation above is useful in the study of quantum scattering, particularly in the Lipmann-Schwinger equation.

The propagator G that we derived can also be derived from usual quantum mechanics. It’s the same thing as the transition amplitude $\langle \vec{x},t | \vec{x_0},t_0 \rangle$ ! So it’s actually the quantum mechanical propagator (or the transition amplitude) to go from $(\vec{x_0},t_0) \rightarrow (\vec{x},t)$

It usually shows up in QM texts on the way to deriving the path integral formulation. The reason for that is that the path integral formulation breaks up a path between two points x0 and x into N segments at x1,x2, .. ,xn-1.

Figure from A modern approach to Quantum Mechanics, John Townsend (p. 222)

So to calculate the total transition amplitude, we need N propagators multiplied together (one for each step such as x1 -> x2, x2->x3,and so on). To take into account the sum over all possible paths between the two endpoints, this product needs to be integrated over all possible intermediary values of x1,x2,.. xn-1. Then take the limit $n \rightarrow \infty$ and that’s it, you have Feynman’s crazy idea!

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