How to Determine a Photon's Wavefunction After it Collapses?

[Flamel]

TL;DR Summary

How would one determine the spread of photon wavefunction after it has been collapsed?

Suppose one measures the position of a photon without destroying it. From my understanding, the wavefunction of the photon should collapse, and will return to a more spread out state over time. How would one calculate this, specifically the rate at which the wavefunction spreads out from the center?

 

[Demystifier]

It depends on the exact shape of the wave function $\psi({\bf x}, t_0)$ at the time $t_0$ of collapse. Assuming that you know it, you first compute its Fourier transform $\tilde{\psi}({\bf q})$ to write this wave function as
$$\psi({\bf x}, t_0)=\int d^3q\, \tilde{\psi}({\bf q}) e^{i{\bf q}\cdot{\bf x}}$$
Then the wave function at later times $t$ is given by
$$\psi({\bf x}, t)=\int d^3q\, \tilde{\psi}({\bf q}) e^{-i\omega({\bf q})(t-t_0)} e^{i{\bf q}\cdot{\bf x}}$$
where $\omega({\bf q})=\sqrt{{\bf q}^2}$ and I use units $\hbar=c=1$.

 

[Flamel]

It depends on the exact shape of the wave function $\psi({\bf x}, t_0)$ at the time $t_0$ of collapse. Assuming that you know it, you first compute its Fourier transform $\tilde{\psi}({\bf q})$ to write this wave function as
$$\psi({\bf x}, t_0)=\int d^3q\, \tilde{\psi}({\bf q}) e^{i{\bf q}\cdot{\bf x}}$$
Then the wave function at later times $t$ is given by
$$\psi({\bf x}, t)=\int d^3q\, \tilde{\psi}({\bf q}) e^{-i\omega({\bf q})(t-t_0)} e^{i{\bf q}\cdot{\bf x}}$$
where $\omega({\bf q})=\sqrt{{\bf q}^2}$ and I use units $\hbar=c=1$.

Thanks. If I understand this correctly, I would need to Fourier transform the wavefunction, then plug it into the bottom equation at $\tilde{\psi}({\bf q})$, correct? How might the equations change if $\hbar$ and c are not equal to 1 and $\omega({\bf q})$ is not equal to $\sqrt{{\bf q}^2}$?

Also how would I find the wavefunction after it collapses? I think I would need to use one of Maxwell's equations, but I'm not sure which one or how to go about it.

 

[Demystifier]

Thanks. If I understand this correctly, I would need to Fourier transform the wavefunction, then plug it into the bottom equation at $\tilde{\psi}({\bf q})$, correct?

Yes.

How might the equations change if $\hbar$ and c are not equal to 1 and $\omega({\bf q})$ is not equal to $\sqrt{{\bf q}^2}$?

That's left as an exercise for the reader.

Also how would I find the wavefunction after it collapses? I think I would need to use one of Maxwell's equations, but I'm not sure which one or how to go about it.

There is no simple way to find this, because it depends on details of the measuring apparatus. Note also that most photon detectors do destroy the photon, so your initial task makes more sense for electrons. For practical purposes, you can model your wave function after the collapse as a narrow Gaussian.

 

[Heikki Tuuri]

You can calculate the wave using classical electromagnetic waves. A photon is an energy quantum of that classical wave.

An example: a flux of coherent photons from a laser meets a narrow slit. The slit corresponds to measuring the x coordinate of a photon quite precisely.

The diffraction pattern of the electromagnetic wave tells you how the photon proceeds. Its velocity in the x direction has afterwards a large uncertainty and the diffraction pattern is very wide.