Does Charge Conjugation change the electromagnetic field?

[gjj]

Is there a difference between the meaning of charge conjugation in Relativistic Quantum Mechanics and its meaning in Quantum Field Theory?

In chapter 4.7.5 of "Thomson Modern Particle Physics" the charge conjugation operator is derived without changing the electromagnetic field A^μ. This approach is also used in other books on relativistic quantum mechanics. Wachter "Relativistic Quantum Mechanics", for instance, defines an "Extended Charge Conjugation" that sets A^μ --> -A^μ, but that's in addition to the usual charge conjugation that makes no change to the fields.

On the other hand when I look at Wikipedia C-symmetry, the statement is made that under charge conjugation every charge is reversed and therefore so are the electromagnetic fields (A^μ --> -A^μ). Wikipedia says "the dynamics would preserve the same form" which I interpret to mean that the motion of the particle would be unaffected by charge conjugation since we've flipped both the charge and the field.

Is charge conjugation defined differently in relativistic quantum mechanics and quantum field theory?

 

[vanhees71]

I don't know, how one cannot make $A^{\mu} \rightarrow -A^{\mu}$ under Charge conjugation, if one describes standard electrodynamics, be it classical or quantum. After all electrodynamics is $C$-invariant (it's also $P$ and $T$ invariant and thus also $CP$ invariant; it's of course $CPT$ invariant anyway, bcause of the Pauli-Lüders theorem). Checking some reasonable textbooks, that's also treated in this way: Weinberg, QT of Fields Vol. 1, Schwartz, QFT and the SM, Duncan, Conceptual framework of QFT, Itzykson/Zuber QFT.

I can imagine how some authors come to the wrong conclusion concerning the C operation on the em. field. This could happen easily when errorneously taking the socalled "relativistic quantum mechanics" (or "first-quantization approach") carelessly literally. Having then defined the C-symmetry operation of Dirac spinor fields (1st quantization!) in the usual way and applying it to the curren $\bar{\psi} \gamma^{\mu} \psi$ leads to the result that the current seems not to change under C and thus to keep C a symmetry of the pseudotheory where electrons and positrons are first-quantized and the electromagnetic field stays unquantized (which in fact very often is a good approximation in non-relativistic QT, the socaled semiclassical approximation), the em. field shouldn't change under C too.

That's of course against our intuition. After all, C is misnomed as "charge conjugation", because its proper definition is that any particle is exchanged by its anti-particle and an antiparticle should sign-flip all its charge-like quantum numbers, including electric charge. That's indeed what comes out of the only consistent relativistic quantum theory we have today, i.e., relativistic QFT! The Dirac field is necessarily to be quantized as fermions, because it's an half-integer field ($s=1/2$), and consequently the proper normal-ordered current operator flips sign under C as it must be. Consequently, to keep QED C-symmetric, also $A^{\mu}$ must flip sign under C. Taking C as charge conjugation, that's also clear from classical electrodynamics since if you flip the sign of all charges, this implies $\rho \rightarrow -\rho$ and $\vec{j} \rightarrow -\vec{j}$. The em. field is given from these charge-current disribution by the retarded potentials (or Jefimenko equations to directly get the physical fields), which are linear in $\rho$ and $\vec{j}$ and thus also flip sign.

Also if you look on the very fundamental definition of electric charge and the electromagnetic field (in classical physics) via the action of the fields on a test charge, C invariance implies that the field must flip under the C transformation. If you'd only flip the sign of the charge, the Lorentz force $\vec{F}=q \vec{E} + q \vec{v} \times \vec{B}$ (in SI units) would flip its direction either, and you'd get a different trajectory of the charged particle starting with the same initial condition, which would mean C is not a symmetry of electrodynamics.

So there's no good reason, not to flip the em. field under C but it rather contradicts the very well established symmetry of the em. interaction under C. Only the weak interaction violates C (as well as P, C, and CP), as is now proven experimentally for all these types of discrete symmetries. So far, however, synmmetry under CPT still holds (within the accuracy of all experiments), keeping this beautiful prediction of relativistic local microcausal QFT valid.

 

[gjj]

Thank you for your answer. I am now clear that in QFT charge conjugation means flipping all the spins and thus also flipping the electric and magnetic fields. Unfortunately, I'm looking at the Feynman-Stueckelberg approach to QED which I realize is not the best approach to QED.

In your second paragraph (the one that starts with "I can imagine ...") you point out how some authors, using this RQM (Feynman-Stueckelberg) approach, could apply charge conjugation and not flip the E&M fields. The reasoning here, as I understand it, is that the charge conjugation operator is applied to a particular Dirac spinor (I think that's what you mean by "C-symmetry operation of Dirac spinor fields" ) rather than to all electrons. Since the electrons that create the electric and magnetic fields are left alone, so are the corresponding fields. This is what I thought those authors who explore the 1st quantization, RQM, Feynman-Stueckelberg approach (Bjorken and Drell, Greiner, Wachter) were doing and if I understand you (although you prefer QFT) this is also your opinion.

In the RQM, Feynman-Stueckelberg approach, the current remains the same after charge conjugation because the charge of the given electron flips as does its 3-momentum. You find this unintuitive, but it sounds something like saying that in a given electric field a positive charge going to the left is like negative charge going to the right.

 

[vanhees71]

I don't know, what you mean by "Feynman-Stueckelberg approach". I only know this expression as precisely describing the modern approach to relativistic QT, i.e., using QFT. The "Feynman-Stueckelberg trick" then simply boils down to write a creation operator in front of the negative-frequence eigenmodes of the free fields (used in perturbation theory using the interaction picture, ignoring Haag's theorem with good success ;-)). It's precisely the way one should present relativistic QT from the very beginning. There is no 1st-quantized consistent treatment of interacting relativistic particles known. So why should one study it?

The only way to get along with the 1st-quantization approach is Dirac's hand-wavy "hole argument", introducing the Dirac sea. This, however, has the disadvantage that it provides a wrong picture to the student, because it says that there is an infinite number of electrons occupying the "negative-energy states", but then where is this infinite amount of negative charge in our observations? Of course, it's not there, and Dirac just has to renormalize the charge, counting his fully-occupied Dirac sea as the new vacuum with 0 energy and 0 charge. That's cumbersome and confusing. Using QFT right away, using the very plausible assumptions of microcausality and existence of a stable ground state (i.e., that the energy spectrum has to be bounded from below) leads directly to the correct free-field/particle vacuum with renormalized energy, momentum, angular momentum, and charge, simply achieved by introduction of normal ordering. Then it's clear from the very beginning that we deal with a Fock space, which is necessarily needed since at relativistic energies we always deal with creation and annihilation processes.