# Magnetic field and operators

1. Nov 30, 2014

### naima

I know that the electric field can be expressed in term of creation and annihilation operators; Is it the same for the magnetic field B ?

2. Nov 30, 2014

### Matterwave

The electric field and the magnetic field are simply different manifestations of each other, so we refer to them both as the "electro-magnetic field", which we quantize in QED.

3. Dec 2, 2014

4. Dec 2, 2014

### Matterwave

Did you have a follow up question, or were you just providing some light reading? :D

5. Dec 2, 2014

### naima

I realize (for the first time) that E and B do not commute. have you read that in a text book?
I had something in mind: You can easily see wigner functions of photons. They depend on position and momentum. I thought that the phase space could be related to the E B plane but i did not see how.
This comes from proportional coefficients!

6. Dec 2, 2014

### Matterwave

Most of the time the Electro-magnetic field is quantized in QFT using path integral formalisms. This is far simpler I hear than with canonical quanization methods. As such, I never read about E and B commuting or not. However, for specific instances, like the E&M field inside a cavity, it might be easier to quantize it via canonical quantization as was done in your link. In that case, E and B are conjugates of each other and do not commute. However, canonical quantization makes the Lorentz invariance of the theory clouded and harder to see than in path integral formalisms. The fact that E can change into B under Lorentz transformations is harder to see, and often not discussed in those models since it's almost always simpler to simply stick with the rest frame of the cavity.

7. Dec 3, 2014

### Skilroyishere

To all, I'm new to this site. It excites me to now that minds are so interested in the field of fields. however, I'm curious if any of all of the above have any experience in super conductors. This may answer your question. Specifically by modifying the electromagnetic field that encompasses it by modulating the fields integrity with a harmonic pulse, by matching the harmonic with the shape of the object core.

8. Dec 4, 2014

### f95toli

The derivation in the link is pretty much the "standard" way of deriving the quantization in both textbooks on quantum optics and cavity- and circuit-QED (the latter is as it happens the field the authors work in). Path integral formalism is virtually never used in CQED (at least I've never seen in done) and it is not something you would come across in a normal textbook.
Note also that these field are quite different than e.g. particle and sub-atomic physics in that you have a fairly obvious connection to the macroscopic word (the cavity) meaning it is often quite convenient to be able to switch from "classical" EM theory in the many photon limit, to 2nd quantization when looking at pure quantum effects.
I do this all the time.

9. Dec 4, 2014

### Matterwave

Yeah, when I was talking about the path integral formalism, I was talking about just regular QED in free space. Most of the QED texts I've seen e.g. Peskin, Ryder, etc., seem to quantize it using path integral formalisms.

10. Dec 5, 2014

### vanhees71

The reason for the dominance of the path-integral approach presenting QED in most modern textbooks is that it is much simpler than the operator formalism, concerning how to deal with gauge invariance. Particularly when it comes to non-Abelian gauge theories, the Faddeev-Popov formalism is way more convenient than the covariant operator formalism. Of course, finally the result of both methods is the same (at least in the sense of perturbative QFT; neither the one or the other formalism is strictly formulated in a mathematically strict sense).

It's also a bit of the applications you have in mind, when presenting the material. Quantum-optics texts usually deal a lot with free photons, and there the most simple way is to fix the gauge completely using the radiation gauge for free photons, i.e., $A^0=0$ and $\vec{\nabla} \cdot \vec{A}=0$ and then quantizing the theory with only physical (spacelike transverse) field modes canonically.

For HEP applications that's a bit inconvenient, because you get Feynman rules that are not manifestly covariant. That's why one would use the Gupta-Bleuler formalism in Lorenz (covariant) gauge, but that's a bit complicated. With the path-integral formalism it's easy to derive everything in any gauge you like. Of course, at the end you get out the same physics.

In quantum optics, there's usually no reason to develop a manifestly Lorentz-covariant formalism, because it doesn't make the calculations more simple in any way. Of course, at the end there's only one successful quantum theory of the electromagnetic interaction, and that's QED. It's only used in many different fields (HEP, quantum optics, nano physics, condensed-matter physics), and in each field different choices of gauge and manifestly Lorentz invariant or non-invariant formalisms are more or less convenient. That's all.

11. Dec 15, 2014

### naima

I come back to this question
$E_x(z,t) = E_0 (a + a^\dagger) sin (kz)$
$B_y (z,t) = -i B_0(a - a^\dagger) cos(kz)$
I used $[a a^\dagger] = I$ to conclude that E and B do not commute.
I wonder now if in E and B formulas the $a$ and $a^\dagger$ are not related to different modes so we would have $[a_l, a^\dagger_m] = 0$ and that E and could commute.