Elementary QFT question, part 2

[snoopies622]

I'm still having trouble understanding the connection between a static electric field and harmonic oscillators.

I understand that a static electric field can be expressed as a scalar (potential) field, and that through Fourier analysis this scalar field can in turn be expressed as the sum of a countably infinite set of static sine waves, each with a characteristic wavelength and amplitude, and - on the other side - that every harmonic oscillator has a characteristic frequency, and that a quantum harmonic oscillator of a given frequency has a set of discrete energy levels.

But since the sine waves are static they have no frequency, and unless it's in motion an oscillator has no wavelength to speak of. So what is the connection?

Thanks.

 

[tom.stoer]

You should study QED in the A° = 0 and div A = 0 gauge. THis gauge is special as it can be shown that "static charges do not radiate". You will see that the Hamiltonian has different terms, one is the Coulomb potential

$$\sim e^2 \int d^3x\,d^3y\,\frac{\rho(x)\,\rho(y)}{|x-y|}$$

and then there is a second term which contains the physical polarizations of the photons. The latter one is derived via Fourier expansion of the physical photon field.

So in this special gauge it becomes clear that the Fourier coefficients and the Coulomb potential are something different. This remains valid in other physical gauges and even in non-abelian gauge theories like QCD

 

[juvenile1]

You mean static waves have no frequency? Think about the De Brogile's matter wave

 

[tom.stoer]

I mean that in QED formulated in the Coulomb gauge the Coulomb potential is static and does not "consist" of photons or waves. That means that "photon" is not a gauge invariant concept but it depends how you formulate the theory. Therefore you should not pay too much attention what a photon "is" but focus on the physical effects that can be described.

 

[snoopies622]

..."static charges do not radiate".
...the Fourier coefficients and the Coulomb potential are something different.

So then energy discontinuity (quantization) does not manifest itself in a static field?

 

[tom.stoer]

So then energy discontinuity (quantization) does not manifest itself in a static field?

Sorry, I don't understand what you mean.

 

[snoopies622]

Well, the term you gave as the Coulomb potential looks continuous, and if static charges don't radiate then they don't emit photons, so in what sense is the field quantized, that is - in what way is it not like a classical electrical field?

 

[tom.stoer]

I said "... the Hamiltonian has different terms, one is the Coulomb potential" and I wrote down only the Coulomb potential.

You are right, the denominator of thr Coulomb potential is a classical entity, but the densities in the numerator are bilinear in fermionic field operators. In addition the other terms of the Hamiltonian contain gauge field operators. So quantization is hidden in the operators, not in the denominator of the potential.

I am afraid you have to learn some basics of quantum field theory ...

 

[DrFaustus]

The connection between harmonic oscillators and quantum fields only holds for free fields, independently on whether it is a scalar, spinor or gauge field. And the reason is more mathematical than physical, meaning that you can see that the Lagrangian of a free field is really a collection of Lagrangians of harmonic oscillators. And then you draw the analogies from here. But again, strictly speaking this is only true for free fields. And since the idea of harmonic oscillators is "nice", people like to talk about it for interacting fields as well.

A static charge is a source of a real life electro(magnetic) field, and hence not a free field. Because of that, if you shoot an electron at your static charge, it will interact with it by emitting and absorbing photons.

The "field quantization" is really just a mathematical construction wherein you replace classical fields by "field operators" that (will hopefully) satisfy the canonical commutation relations. Then you realize that you can actually use this machinery for real life physics.

 

[tom.stoer]

In canonical quantization the field operators in momentum space can be translated into creation and annihilation operators. For compact space (e.g. interval of length L) k-space becomes discrete and formally these operators will look like $$a_n$$, $$a^\dagger_n$$ with

$$[a_m, a_n] = [a^\dagger_m, a^\dagger_n] = 0$$
$$[a_m, a^\dagger_n] = \delta_{mn}$$

Formally these are infinitly many harmonic oscillators.

These quantization rules are not restricted to free fields; why do you say that

... strictly speaking this is only true for free fields.

 

[DrFaustus]

tom.stoer -> Because it is :) What I'm really referring here to is the rigorous (mathematical) construction of interacting fields. But even at a formal level, you cannot extend the harmonic oscillator analogy to an interacting field. Write down the interacting equations of motion, just like you said on a compact interval and in momentum space, and see that you cannot write it's solution in terms of anything that resembles a harmonic oscillator. Simply because the equation of motion is not a collection of harmonic oscillators anymore...

 

[tom.stoer]

Agreed. Of course for interacting fields the solution has nothing to do with an harmonic oscillator solution, but the basic operators remain the same. But I think we should stop this private discussion as it is of no relevance for snoopies622 ...

 

[snoopies622]

I am afraid you have to learn some basics of quantum field theory ...

Indeed! I was hoping that going from ordinary quantum mechanics (if that's the appropriate phrase for what I have in mind) to quantization of the EM field would mean only a few additional hypotheses. There's wishful thinking for you.

 

[tom.stoer]

The quantization of the EM field is tricky due to gauge invariance. You should try to understand the quantization of a scalar (Klein-Gordon) field first; this is sometimes discussed in QM II textbooks.