Free-Particle Schrödinger Green's Function

The Green's function satisfies

$\psi (x,t) = \int G(x,y,t) \psi(y,0) dy,$

where $\psi(x,t)$ is a solution to the time-dependent Schrodinger equation. The Green's equation therefore propagates the solution at a single point forward in time.

Note as given in the hint that the general solution can be written out in a superposition of eigenfunctions of the Hamiltonian:

$\psi(x,t) = \sum_n c_n \phi_n (x) e^{-i E_n t / \hbar},$

where the coefficients $c_n$ are defined by:

$c_n = \int dy \phi^{\dagger}_n (y) \psi (y,0);$

the overlap of the eigenfunctions with the solution at time zero.

Plugging into the general solution, the general solution can be represented as:

$\psi(x,t) = \int dy \sum_n \phi^{\dagger}_n (y) \phi_n (x) e^{-iE_n t / \hbar} \psi(y,0),$

so the Green's function can be read off as:

$G(x,y,t) = \sum_n \phi^{\dagger}_n (y) \phi_n (x) e^{-iE_n t / \hbar}.$

For the free particle, there is actually a continuous spectrum of eigenfunctions of the Hamiltonian, so the sum should be converted to an integral over the energies (note that $n$ above is just a proxy for the energy in the discrete case; we have $E = \frac{p^2}{2m} = \frac{\hbar^2 k^2}{2m}$ ):

$G(x,y,t) = \int_0^{\infty} dE \phi^{\dagger}_E (y) \phi_E (x) e^{-iE_n t / \hbar}.$

The (non-normalized) eigenfunctions of momentum (and energy) take the form $e^{ikx} = e^{i\sqrt{2mE}x / \hbar}$ . A careful change of variables, normalizing the eigenfunctions, allows one to rewrite the above integral as an integral over momentum:

$G(x,y,t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} dk e^{ik (x-y)} e^{\frac{\hbar k^2 t}{2im}}.$

Using the formula for the Gaussian integrals by completing the square:

$\int_{-\infty}^{\infty} dk e^{-c_1 k^2 + c_2k} = \sqrt{\frac{\pi}{c_1}} e^{c_2^2 / (4c_1)},$

for $c_1$ and $c_2$ constants, this integral can be evaluated directly as:

$G(x,y,t) =\sqrt{\frac{m}{2\pi i \hbar t}} \exp \left(\frac{-m(x-y)^2}{2i \hbar t} \right),$

which is the answer as claimed.