# Electric & magnetic field of a photon in QM

1. Nov 29, 2014

### dsoodak

Generally, all that matters is the amplitude of the wavefunction. You usually can't even MEASURE the phase unless you are making a particle interfere with itself.

With photons, however, the real & complex components seem to appear as electric and magnetic fields.

Is there anything equivalent to a static electric or magnetic field except with another particle such as an electron? I think someone once told me they read an article about this but when I asked them later they didn't remember.

It also seems like there should be a way to measure the phase of an electron beam, even if it involves (as it does with photons) sacrificing a certain percentage of them.

Dustin Soodak

2. Nov 29, 2014

### Staff: Mentor

As far as I understand, the "phase of an electron" only exists in the math involved with calculating the wavefunction. Once you measure an electron the wavefunction collapses and there is no more phase. I don't know if an electron beam has a phase like a light beam does since the wavefunction isn't a measureable entity like the electric and magnetic field vectors are in an EM wave. Hopefully someone with far more knowledge in this area than I have can give a good answer.

3. Nov 29, 2014

### Staff: Mentor

As far as I know, the electric and magnetic fields of a classical electromagnetic wave "emerge" from the QED description of a collection of a large number of photons. I don't think it's meaningful to talk about the electric and magnetic fields of a single photon.

4. Nov 29, 2014

Staff Emeritus
It's not. If you're in an eigenstate of photon number you are not in an eigenstate of E or B field.

5. Nov 30, 2014

### BruceW

this is a tricky issue.. are you talking about the polarization of light? There is some connection between polarization of photons and polarization of a beam of photons. But I'm not sure how direct this connection is. I'll come back to this thread later, when I have some time. It's pretty interesting.

6. Nov 30, 2014

### Staff: Mentor

Or are you trying to make a connection between the perpendicular real and imaginary axes in the common graphical representation (Argand diagram) for a complex number, and the perpendicular directions of the $\vec E$ and $\vec B$ fields in a classical electromagnetic wave? If so, these have no connection as far as I know.

7. Dec 1, 2014

### dsoodak

One of my QM books started with the classical EM wave equation, then showed various polarizations as superpositions of different sets of polarization base states (like circular polarization as superposition of horizontal+i*vertical). I could be wrong, but this section seemed to imply a 1-1 mapping between classical & QM for photons (except for quantization, of course). Also, every "derivation" of Schrodinger's equation I've seen started with classical EM and expanded it using SR to include particles with a rest mass.
Even if you just think of EM fields as emergent phenomena, they are still measurable quantities that seem to correlate with phase of photons.
Of course, you still need coherent light like laser or radio transmissions in order to MAKE this measurement and I'm not aware of a method of creating coherent electron beams...or, for that matter, if there is an equivalent property of electrons to measure.

8. Dec 1, 2014

### vanhees71

Forget photons at this level of understanding. First learn non-relativistic quantum theory. There (and even in quantum optics) you can go a very long way with the "semi-classical approximation", i.e., treating the electromagnetic fields as classical ("external") fields and the particles (atomic nuclei and electrons in atomic, molecular, and solid-state physics) with non-relativistic quantum theory. Particularly this includes the understanding of the photoelectric effect which is often wrongly used as an example for the need of the quantized electromagnetic field, as photons.

When you master this material, you can go further and study many-body theory ("second quantization") and then finally relativistic quantum theory, which should be treated as a quantum field theory from the very beginning. What's called "relativistic quantum mechanics" is an oldfashioned attempt to formulate relativistic quantum theory as single-particle quantum mechanics in analogy to non-relativistic wave mechanics, but this doesn't work, because as soon as it comes to relativistic energies in collisions of particles (i.e., you have particles smashing together with a kinetic energy that's not very small compared to its rest mass times $c^2$), there is some probability to create and/or destroy particles. Thus a many-body treatment of relativistic quantum theory is unavoidable anyway, and the most simple and elegant formalism for this is quantum field theory.

9. Dec 1, 2014

### naima

i do not find the E and B operators. Thanks.

10. Dec 1, 2014

### BruceW

single particle relativistic quantum mechanics is surely useful for things like neutrinos. i.e. you can calculate how long it takes to undergo flavour oscillation. But I agree that to understand how classical electromagnetism emerges from quantum mechanics, it is very necessary to use second quantization.

So... for a single photon, (I am a novice to quantum field theory, so anyone please correct me if I get it wrong). There is no classical electromagnetic field associated with it. We can only get a classical electromagnetic field if we consider a system with a non-fixed number of photons. BUT, on the other hand, a single photon does have a polarization. So for example, if we sent a single high-energy photon through a vacuum, someone else could in principle measure the polarization of the photon which we sent to him/her ?

Also, the phase of the polarization of a single photon is not a measurable thing, since it is just an unimportant phase factor of the state. But, presumably, the phase of the polarization of a classical EM wave is a measurable thing. (i.e. at any given time and place, the electric field will be pointing in a specific direction). So it seems to me that it is necessary to understand second quantization to be able to understand how the phase of polarization of a classical EM wave emerges, given that a single photon does not have a measurable phase of polarization.

11. Dec 1, 2014

### vanhees71

Sure, a single photon has a polarization-degree of freedom. As with any other observable in quantum theory, it depends on the state it is prepared in, whether it has a determined polarization or not. I don't understand, what you mean by "phase of the polarization". A single-photon state has a very indetermined phase. In some sense phase and photon number are complementary properties. A coherent state, where the phase is well defined, has an undetermined photon number. The photon number is Poisson distributed in this case. A coherent state of high average photon number is, by the way, the closest representant of a classical electromagnetic field.

Neutrino oscillation is a total different topic, and it's very messy in the literature. I cannot point to a single paper, where this issue is really treated in a way completely satisfactory to me :-(.

12. Dec 7, 2014

### BruceW

yeah, sorry I don't know what I meant by phase of polarization. I think I just meant the overall phase factor of the state of a photon, which is not a measurable thing. I was hoping that maybe if each photon has a different phase, then somehow it might be possible for the expectation value of the spin of the entire system to be zero. But this is not correct, is it?

I did have a nice book on quantum field theory, but I had to give it back to the library. So I've been looking at this wikipedia page: http://en.wikipedia.org/wiki/Quantization_of_the_electromagnetic_field
If we have a Fock state (in the Coulomb gauge), with just one photon, then it is an eigenstate of number of photons. So the expectation value of the number of photons is just one. And also the wiki page says the spin operator is just:
$$\Sigma_\mu \ \mu N(\mu )$$
where $\mu$ can be +-1 and $N(\mu )$ is the number operator for photons with spin value $\mu$. So it seems that our single-photon Fock state would be an eigenstate of the spin operator, so that the expectation value of spin would simply be the spin of the photon... But of course, the expectation value of the electric field is zero for this state... So the spin they are talking about is a classical quantity, but it's not related to the classical electric field. Maybe this is not so weird. After all, a Fock state will have nonzero classical energy and momentum, even though the classical electric field is zero.

also, I found this paper: http://arxiv.org/abs/arXiv:1006.3876 where they mention that the "separation of the total angular momentum into orbital and spin parts has restricted physical meaning." So maybe this confirms that the spin doesn't fully correspond to our usual intuitive idea of what spin is.

13. Dec 8, 2014

### vanhees71

In relativistic quantum theory there's no physically well-defined definition of a separate orbital and spin angular-momentum part. The reason is that, contrary to the issue in the Galilei group, Lorentz boosts don't commute and the composition of two Lorentz boosts with velocities in non-collinear direction always involve a rotation (the famous Wigner rotation). In relativistic QT the spin of a massive particle is defined by the representation of rotations in the restframe of the particle. For massless particles the issue is somewhat more complicated, there any particle with spin 1/2 and large has only two polarization degrees of freedom (helicities $\pm s$).

On the other hand, of course, photon polarizations are measurable. You can use polarization foils to let through only photons in a certain polarization direction (of course, that's idealized; in reality such polarization foils have a finite probability to let through also photons with perpendicular polarization, but they can be made very effective).

14. Dec 12, 2014

### BruceW

I see. Yeah, I guess it's a subtle issue. I wonder if the OP'er is still around? ah well, I learned something at least :)
p.s. also It looks like ( Experimental generation of multi-photon Fock states
Merlin Cooper, Laura J. Wright, Christoph Söller, and Brian J. Smith ) this group has experimentally created a Fock state of up to 3 photons. You could hardly call this a beam, but still it would imply it is possible to have a beam of many photons in a Fock state. And this is quite novel, since as you were saying, they could be made with a specific polarization, and the beam can be blocked by a polarizing filter, even though there is zero classical electric field ! Interesting stuff.

15. Dec 12, 2014

### vanhees71

For these issues, I'd recommend a quantum optics book. I'm not an expert, but my favorite is the book by Scully and Zubairy.