Long term dispersion of the wavepacket

Question

Consider a one-dimensional wavepacket $\psi(x,t)$ that evolves according to the free particle Schrödinger's equation. It is initially localized around $x=0$ with a certain width $\Delta x$. This initial wavepacket corresponds to a momentum-space wavepacket $\phi(p,0)$.

We let this wavepacket evolve for a very long time $t$. Intuitively, a position measurement at $t$, combined with the knowledge of the particle's initial position at approximately $x=0$ can be used to calculate the particle's speed of travel and thus measure its initial momentum (of course, a certain level of uncertainty always applies).

This implies that, for very large $t$, approximately: $$|\psi(x,t)|^2\propto\left|\phi\left(m\frac{x}{t},0\right)\right|^2$$

Somewhat analogous to the way the diffraction pattern of a plane wave through slit, at large distances, is given by the Fourier Transform of the slit's aperture, as seen here.

Is this right? If it is, how can it be proved more rigorously?

Answer 1

It’s actually exactly like optics. This is because the paraxial Helmholtz equation is the same as the time dependent Schrödinger equation. There are different ways to get it.

One method is to use the real space propagator to infer the large time dynamics. Normalizing the equation of motion to: $$ i\partial_t\psi=\frac{1}{2}\partial_x^2\psi $$ The kernel is: $$ \psi(x,t)=\int K(x-y,t)\psi_0(y)dy\\ K(x,t)=\frac{1}{\sqrt{2\pi it}}e^{-x^2/2it} $$ Using convention for the Fourier transform: $$ \mathcal F\psi(k)=\int \psi(x)e^{-ikx}dx $$ By expanding the square, you recognize a Fourier transform: $$ \psi(x,t)= \frac{1}{\sqrt{2\pi it}}e^{-x^2/2it}\mathcal F \left[e^{-y^2/2it}\psi_0(y)\right]\left(\frac{x}{t}\right) $$ For large times, you can neglect the quadratic exponential in the Fourier transform and get: $$ \psi(x,t)= \frac{e^{-x^2/2it}}{\sqrt{2\pi it}}\mathcal F \psi_0\left(\frac{x}{t}\right) $$

Another way is to use the kernel in Fourier space: $$ \psi(x,t) = \int e^{-itk^2/2+ikx}\phi_0(k)\frac{dk}{2\pi} $$ and use the stationary point approximation since $t\to \infty$ assuming $x/t$ is constant. The stationary point is at: $$ k=x/t $$ Intuitively, you are selecting the wavepacket travelling at the group velocity $x/t$. This again gives you (doing the Gaussian integral): $$ \psi(x,t)= \frac{e^{it(x/t)^2/2}}{\sqrt{2\pi it}}\phi_0\left(\frac{x}{t}\right) $$

Hope this helps.