# Photon wavefunction and speed of light

• thermonuclear

#### thermonuclear

Consider a photon which is sent towards a detector. The instant before the photon hits the detector, let's say one mm-light (the time the light travels one millimeter), the photon should be located at a position of one mm far away from the detector.

But since the photon has an associated wavefunction, the photon might be theoretically in any point of space with some probability. This means that with some probability different from zero, the photon might have hit the detector already.

The question: does this contradict special relativity, i.e. the principle that there is nothing which travels faster than c?

Have a nice day.

You are correct that in the theory of quantum electrodynamics, which deals with the interaction of light and matter, there can be non-zero contributions from solutions that involve light traveling faster than or slower than c.

You are also correct in that special relativity is violated by light not traveling at c. The resolution is simple: special relativity is not a quantum theory. At macroscopic scales, the contributions from solutions where light does not travel at c rapidly become (extremely) negligible, and special relativity is recovered.

- Warren

Hmm Ambi,

Am I wrong in my last post? Or is there something I'm missing?

- Warren

Originally posted by thermonuclear
But since the photon has an associated wavefunction, ...
I think it doesn't. There is no such thing as a single-photon wavefunction.

Originally posted by chroot
Am I wrong in my last post? Or is there something I'm missing?

To perturbatively calculate amplitudes in QFT, you sum over virtual processes, and there is no requirement for those virtual processes to be "on-shell": virtual particles can propagate at any speed. But the amplitude you get in the end, which represents something real and measurable, does not carry information from real events outside the past lightcone, so the speed of light is not violated.

Originally posted by Ambitwistor
To perturbatively calculate amplitudes in QFT, you sum over virtual processes, and there is no requirement for those virtual processes to be "on-shell": virtual particles can propagate at any speed. But the amplitude you get in the end, which represents something real and measurable, does not carry information from real events outside the past lightcone, so the speed of light is not violated.
Right, but why? Is it because the off-shell virtual processes' amplitudes are negligible? What happens to the "information from real events outside the past lightcone?"

I'm familiar with the fact that the perturbation series always sums to an "on-shell" solution, but I guess I just don't understand the mechanism that always makes that so.

- Warren

What do you mean when you say there is no such thing as a single-photon wavefunction? Are you saying that there are no 1-photon states in the Fock space of QED? Or is this just a terminological issue over the meaning of the word "wavefunction", as in:

http://groups.google.com/groups?selm=795dho$9ji$1@pravda.ucr.edu

I'm sorry, I don't know the proof. I'm not sure if there is a proof starting from the perturbative sum over virtual processes, or whether they just prove it directly from the axioms of QFT.

Originally posted by Ambitwistor
What do you mean when you say there is no such thing as a single-photon wavefunction?

I mean it in the following sense: "It's not possible to define any photon wavefunction from which a probability amplitude for spatial localization can be calculated". (Landau-Lifschitz IV, chapter 1, §4.)

IOW: Since a photon has sharp momentum and is massless, it is completely unlocalized via the uncertainty principle &delta;p&delta;q > h.

"the photon might have hit the detector already" is a statement about spatial localization and as such makes no sense.

Originally posted by arcnets
I mean it in the following sense: "It's not possible to define any photon wavefunction from which a probability amplitude for spatial localization can be calculated". (Landau-Lifschitz IV, chapter 1, §4.)

Yes, you cannot perform Newton-Wigner localization on photons.

"the photon might have hit the detector already" is a statement about spatial localization and as such makes no sense.

I'm not sure what you're getting at here. Whether and when a detector measures a photon is a physically meaningful statement.

Originally posted by Ambitwistor
Whether and when a detector measures a photon is a physically meaningful statement.
Yes, it is.

"the photon might have hit the detector already" is a quote from the original post. Let me re-phrase this:

"At a given time, before the observation is made, there is a finite chance that the photon is localized inside the detector".

I think this makes no sense since it implies the localization of a photon from an uncollapsed wavefunction.

Hmm, okay, after re-reading I think I see what you are getting at. You're saying that if the detector hasn't measured a photon, we can't speak of whether it's inside the detector (but not yet registered), because it cannot be localized ... correct?

Exactly.

chroot-

QFT is consistent with special relativity. special relativity is not an approximation. you are guaranteed that no propagation happens from outside the lightcone, because when you quantise the field, you choose canonical commutation relations that vanish for spacelike separated seperations, which implies the lack of a causal connection between the two operators.

Originally posted by lethe
you are guaranteed that no propagation happens from outside the lightcone, because when you quantise the field, you choose canonical commutation relations that vanish for spacelike separated seperations, which implies the lack of a causal connection between the two operators.

Well ... see here: