What is the relationship between Maxwell's equations and quantum field theory?

[DrDu]

If I where a QFT guy I would certainly be able to mock up a confining interaction where the field carrying the interaction depends on c and then would discuss the limit c to infinity.

 

[samalkhaiat]

If I where a QFT guy I would certainly be able to mock up a confining interaction where the field carrying the interaction depends on c and then would discuss the limit c to infinity.

If you are not a “QFT guy”, shouldn’t you listen to someone who makes his living from QFT?

 

[bhobba]

This is a very interesting thread I am greatly enjoying.

My concern is the OP said he had not done much QM and very likely no QFT. So I thought the following paper explaining how gauge symmetry and the vector potential naturally come out in QM will help his understanding:
http://www.niser.ac.in/~sbasak/p303_2010/23.11.pdf

I want to also assure the OP that this gauge view is absolutely essential to the standard model which of course QED is part of. Its part of what's called Yang-Mills theory:
https://en.wikipedia.org/wiki/Yang–Mills_theory

Thanks
Bill

 

[DrDu]

If you are not a “QFT guy”, shouldn’t you listen to someone who makes his living from QFT?

I am certainly eager to do so! I did some reading. Apparently, in the Coulomb gauge, the Gauss constraint holds at operator level. Doesn't this clearly show that the extensions to the Maxwell equations for A which appear in other gauges, are artifacts of working with A instead of F?

 

[martinbn]

Isn't the "law of local interactions" in field theory rather a tautology as you can transform any non-local interaction into a local one introducing sufficient auxiliary fields?

Is it still local if you need an infinite number of auxiliary fields?

 

[samalkhaiat]

Apparently, in the Coulomb gauge, the Gauss constraint holds at operator level.

All Maxwell's equations hold as operator equations in the Coulomb gauge.

Doesn't this clearly show that the extensions to the Maxwell equations for A which appear in other gauges, are artifacts of working with A instead of F?

I don’t understand what you are saying here. In this thread I stressed the points which represent our current understanding of QFT:

1) After 90 years, we now almost certain that QED cannot be formulated without introducing the vector potential as primary field in the theory.

2) Suppose that (1) is false. That is a "smart person" came along and managed to formulate QED using only $F_{\mu\nu}(x), J^{\mu}(x)$ and the electron field $\psi (x)$ (i.e., with no vector potential). Then the theorem in #25 forces the “smart person” to realize that his formalism make sense only if the Maxwell’s equations are abandoned as operator equations.

3) In the usual formulation of QED, i.e., the one based on $A_{\mu}(x) , J^{\mu}(x)$ and $\psi (x)$, you have two options:
i) Give up locality and have the Maxwell equations as operator equations. This happens in the Coulomb gauge quantization.
ii) Keep locality and covariance but modify the Maxwell equations. This is the widely use Gupta-Bleuler covariant quantization.

 

[Phylosopher]

This is a very interesting thread I am greatly enjoying.

My concern is the OP said he had not done much QM and very likely no QFT. So I thought the following paper explaining how gauge symmetry and the vector potential naturally come out in QM will help his understanding:
http://www.niser.ac.in/~sbasak/p303_2010/23.11.pdf

I want to also assure the OP that this gauge view is absolutely essential to the standard model which of course QED is part of. Its part of what's called Yang-Mills theory:
https://en.wikipedia.org/wiki/Yang–Mills_theory

Thanks
Bill

Thank you sir. I am enjoying the discussion too! This would be my future reference after studying QFT ^^".
It seems that QFT theories from what I am seeing from the discussion were not a direct implications of the previously famous theory of ElectroMagnetism/Maxwell's equations, or at least the view of the last was lost in the process. Otherwise, there won't be this much disagreement between the members of the forum. I actually had the same discussion with one of my professors and the results were the same as here.

 

[bhobba]

Thank you sir. I am enjoying the discussion too! This would be my future reference after studying QFT
It seems that QFT theories from what I am seeing from the discussion were not a direct implications of the previously famous theory of ElectroMagnetism/Maxwell's equations, or at least the view of the last was lost in the process. Otherwise, there won't be this much disagreement between the members of the forum. I actually had the same discussion with one of my professors and the results were the same as here.

Well we have a few people whose knowledge of such things is extremely advanced - Samalkhaiat and Dr Neumaier are among them. It's unusual for any of those to disagree, but it happens occasionally and we all learn something. We also have professors that are particle physicists and some that work at Cern - a very broad range of people. I am not - just a retired guy that studied applied math and is now interested in physics.

If you want to go into this in a bit more detail start by looking into spontaneous emission. Its actually caused by the electron not being in a stationary state because its coupled to Quantum EM Field that exists everywhere. This is the precise reason you need Quantum Field Theory. It's all explained in this interesting paper that hopefully your QM is advanced enough to understand:
http://www.physics.usu.edu/torre/3700_Spring_2015/What_is_a_photon.pdf

Its your first step into the beautiful world of Quantum Field Theory. If you want to go further I suggest the following books in no particular order:
https://www.amazon.com/dp/0984513957/?tag=pfamazon01-20
https://www.amazon.com/dp/019969933X/?tag=pfamazon01-20

If you want to go further still, write back for how to progress to THE book on QFT - Weinberg:
https://www.amazon.com/dp/0521670535/?tag=pfamazon01-20

Thanks
Bill

 

[DrDu]

3) In the usual formulation of QED, i.e., the one based on $A_{\mu}(x) , J^{\mu}(x)$ and $\psi (x)$, you have two options:
i) Give up locality and have the Maxwell equations as operator equations. This happens in the Coulomb gauge quantization.
ii) Keep locality and covariance but modify the Maxwell equations. This is the widely use Gupta-Bleuler covariant quantization.

This is quite interesting. Apparently, different gauges live in different spaces, i.e. they are unitarily inequivalent. Isn't this precisely the definition of a broken symmetry?

 

[Demystifier]

I think this thread can be reduced to the question: What is QED really about?

The problem is that this is a philosophical question, not a mathematical one, which is why it is hard to give an answer on which everybody will agree. Here are some possible answers to the question, neither of which is completely wrong.

1) QED is really about electrons, positrons and photons, because they are particles that we really observe.

2) QED is really about cross sections in particle detectors, not about particles as such, because we only measure the cross sections.

3) QED is really about fields, not about particles or cross sections, because the former are fundamental in QED while the later are derived.

3a) QED is really about the field $F_{\mu\nu}$, not about $A_{\mu}$, because QED is a gauge invariant theory and $A_{\mu}$ is not gauge invariant.

3b) QED is really about the field $A_{\mu}$, not about $F_{\mu\nu}$, because $F_{\mu\nu}$ cannot be expressed without $A_{\mu}$ so $A_{\mu}$ is more fundamental.

 

[samalkhaiat]

This is quite interesting.

This is just a repeat of what I have been saying.

Apparently, different gauges live in different spaces,

A gauge choice defines a slice in the gauge group orbits. Did you mean this?

i.e. they are unitarily inequivalent.

Why is that so? In many cases, a unitary transformation does exist between two gauge choices.

Isn't this precisely the definition of a broken symmetry?

No. That is precisely the definition of broken understanding of QED.

 

[A. Neumaier]

Both classically and in quantum mechanics, the free Maxwell equations (for electromagnetic fields without sources) are linear, hence free electromagnetic waves and free photons don't interact.

No. The free Maxwell equations are not about the vector potential but about the electromagnetic field, and these hold also on the operator level!

QED is not about the Maxwell equations but about a bigger system of equations involving a fermionic field not known before 1925. Such a field does not figure in Maxwell's equations. Neither do Maxwell's equations demand a derivation from an action principle; they stand for themselves.

The free Maxwell equation $\partial^{\mu}F_{\mu\nu}=0$ does not hold as operator equation in the usual covariant quantization of the em-field that one can find in almost all usual textbooks. And that is a complete answer to the remarks raised in #1.

This just means that one has to avoid the usual covariant quantization of the em-field that one can find in almost all usual textbooks.

What is constructed is a redundant reducible representation of the free massless spin 1 field on a vector space of unphysical wave functions that not even forms a Hilbert space, thus violating one of the basic principles of quantum mechanics. This is the reason why spurious terms are present in your equations. The physically relevant representation is the restriction to the wave functions defined by transverse momenta, and on this (irreducible) physical representation space, the free Maxwell equations hold in their classical form.

Which you confirm:

The space $\mathcal{H}_{ph}$ carries one good news for you. Indeed, the Maxwell equations do hold as operator equations in [itex]\mathcal{H}_{ph}[However, this comes with heavy price, because all elements of $\mathcal{H}_{ph}$ have zero electric charge, and $[J^{0}(f)]$ generates trivial automorphism, i.e., $\mathcal{H}_{ph}$ has no room for electrons.

Maxwell's equations as stated everywhere do not contain physical charged fields but only the electromagnetic field strength and a charge current. These are gauge invariant fields, hence operate in the quantum case on a physical Hilbert space with a positive definite metric.

I
The theorem say that “one cannot hope to formulate QED in term of $F_{\mu\nu}$, $j^{\mu}$ and local charged fields without essentially going to the Gupta-Bleuler formulation.

But nothing in the theorem you cite prevents one from formulating QED in terms of a Lagrangian with nonlocal fields, without introducing an indefinite metric. For scalar electrons this was done in Section 3 of the 1962 paper Quantum electrodynamics without potentials by S Mandelstam. The spinor case (i.e., full perturbative QED) was treated in a subsequent 1964 paper by Levi. The interacting Maxwell equations hold on the operator level.

we have a few people whose knowledge of such things is extremely advanced - Samalkhaiat and Dr Neumaier are among them. It's unusual for any of those to disagree, but it happens occasionally and we all learn something.

We don't disagree on the fact that the standard treatment of QED is based on a gauge theory and usually represented by field operators on an indefinite metric space. We only disagree on the claim that the gauged representation and the indefinite metric are unavoidable. The fact is that they are not; the theorem cited by samalkhaiat only holds under assumptions made for convenience, not for physical necessity.