Evolution of Delta Function in Free Particle Hamiltonian

Question

I was curious what would happen if we had a free particle hamiltonian and saw what would happen if your initial wavefunction was a delta function and time evolve it using Schrodinger's Equation (say in one dimension). Here were my thoughts:
$$-frac{hbar^2}{2m}partial^2_xpsi = ihbar partial_t psi$$
Taking the Fourier transform, which I call $F(cdot)$, and $F(psi) = tilde{psi}$:
$$partial_t tilde{psi} = frac{hbar i}{2m} (2pi i k)^2 tilde{psi}$$
$$implies tilde{psi} = tilde{psi}(t=0)exp(frac{-2pi^2hbar k^2 it}{m})$$
If we our initial wavefunction is a delta function (say centered at $x=0)$:
$$tilde{psi}(t=0) = F(delta(x)) = int delta(x) exp(-2pi i xk) dx= 1$$
$$implies tilde{psi}(k, t) = exp(frac{-2pi^2hbar k^2 it}{m})$$
Thus, in position space:
$$psi(x, t) = F(tilde{psi}) = int exp(frac{-2pi^2hbar k^2 it}{m}) exp(2pi ixk)dk$$
I believe this is correct, and (if it is) my main question is how do you interpret it?

Gec · Answer

You've found the integral representation of Green's function for the free particle Schrodinger equation. Delta-function is not a square-integrable function, but in quantum mechanics we often consider $delta(x)$ as the eigenfunction of the operator $hat{x}$ in coordinate representation:
$$
hat{x}|x'rangle = x'|x'rangle, quad langle x| x'rangle = delta(x-x').
$$
From your question formulation follows equality
$$
psi(x,t) = langle x | e^{-frac{i}hbar hat{H} t} | x' = 0 rangle qquad (*)
$$
Free particle problem possess translational invariance, so
$$
 psi(x,t) = langle x+x_0 | e^{-frac{i}hbar hat{H} t} | x_0 rangle
$$
for any $x_0$. As $psi(x,t)$ is the matrix element of the evolution operator, it is straightforward to express the solution to the non-stationary Schrodinger equation in the following form:
$$
phi(x,t) = int psi(x-x',t) phi(x',0) dx'
$$

Evolution of Delta Function in Free Particle Hamiltonian

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