# Photon Momentum varies along waveform?

1. Aug 26, 2014

### jmatejka

An M.I.T. Physics lecture mentioned B and E crossing over the x axis as lowest energy state/position of the wave(s).

Along the x axis, path of photon, is photon momentum here the lowest also?

Does momentum vary?

For a tiny target, if a photon collides at this exact point is there no momentum transfer?

2. Aug 26, 2014

### Staff: Mentor

Can you give a link? I'm not sure what you're referring to here; the description you're giving doesn't seem correct.

3. Aug 26, 2014

### jmatejka

4. Aug 26, 2014

### Staff: Mentor

A photon does not change momentum as it propagates.

Be very cautious about "mixing" the classical description of electromagnetic radiation in terms of waves of oscillating E and B fields on the one hand, and the quantum description in terms of photons on the other hand. Photons are not tiny "bullets" or localized bundles of classical E and B fields.

As far as I know, the only "safe" way to connect the two pictures at the introductory or intermediate physics level (i.e. without going into the mathematical details of quantum electrodynamics) is by way of the total energy of the radiation. If the classical E and B fields in some volume of space have a certain amplitude, you can calculate the energy contained in that volume; and if you know the frequency, you can use that energy to calculate the number of photons. You can also go the other way if you have a very large number of photons. However, if you have only a few photons, I don't think it's meaningful to talk about classical E and B fields.

5. Aug 26, 2014

### jmatejka

Thanks Guys!

6. Aug 26, 2014

### WannabeNewton

The Poynting vector $\vec{S}$ represents the momentum density of the EM field. To get the total momentum $P_{EM}$ of the EM you have to integrate over a volume and clearly the momentum of an EM wave is not going to vanish in an entire finite volume even if the density vanishes at a single point $(t,\vec{x})$ where $\vec{E} = \vec{B} = 0$ i.e. $P_{EM} \neq 0$ for the radiation field since it is integrated over a finite volume and the radiation field certainly has to carry momentum through this volume.

Furthermore you are trying to use a classical field variable to describe the quantum mechanical concept of a photon. You have to first quantize the EM momentum to get a momentum operator $\hat{P}_{EM}$ for the EM field that acts on single photon states from which you will find that the single photon states are momentum eigenstates of the EM momentum operator: $\hat{P}_{EM}|k \rangle = \vec{k}|k \rangle$ where $|k \rangle$ are single photon states. In other words the single photon states are states of definite momentum $\vec{k}$ of the EM field, through $\hat{P}_{EM}$.

Since photons are basically "excitations" or "vibrations" of the EM field, much like the vibrational normal modes in a crystal, you can think of this as saying that the momentum (not momentum density!) of the EM field is composed of the momenta of the photons that are excitations of it. Anyways, the upshot is you cannot have $\vec{k} = 0$ i.e. photon states of zero momentum do not exist, simply by virtue of their dispersion relation.