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I seem to have troubles finding definitions of the charge conjugation operator that are independant of the theory considered.

Weinberg defined it as the operator mapping particle types to antiparticles :

$$\operatorname C \Psi^{\pm}_{p_1 \sigma_1 n_1;p_2 \sigma_2 n_2; ...} = \xi_{n_1} \xi_{n_2} ... \Psi^{\pm}_{p_1 \sigma_1 n_1^c;p_2 \sigma_2 n_2^c; ...}$$

He does not really seem to specify what he means by "antiparticles" around there, but I'm guessing this is the one-particle state that is conjugate to this one. This assumes that it is possible to decompose everything into one-particle states.

Wightman seems to go with $C \gamma^\mu C^{-1} = \bar \gamma^\mu$, which isn't terribly satisfying and also only works for spinor fields.

I've seen thrown around that the $C$ conjugation corresponds roughly to the notion of complex conjugation on the wavefunction but never really expanded upon.

Is there a generic definition of charge conjugation that does not depend on how the theory is constructed? The CPT theorem in AQFT indeed seems to not have any of those extraneous constructions, but the action of the different symmetries is a bit hidden as

$$(\Psi_0, \phi(x_1) ... \phi(x_n) \Psi_0) = (\Psi_0, \phi(-x_n) ... \phi(-x_1) \Psi_0)$$

Is the action of $C$ symmetry $\Psi' = C \Psi$ just a state such that for any operator $A$,

$$(\Psi, A \Psi) = (\Psi', A^\dagger \Psi')$$

or something to that effect? From some parts seems like it may just be $C \phi C^{-1} = \phi^*$.

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  • $\begingroup$ Wightman (1-47) defines the action of $C$ on a two-component spinor. A field in an arbitrary representation of Loretnz can always be understood as a tensor with several (dotted and undotted) spinor indices, or direct sums thereof. Therefore, Wightman's definition works for a field of arbitrary spin. Just act on its spinor indices as (1-47) indicates. $\endgroup$
    – AccidentalFourierTransform
    Commented Mar 5, 2018 at 21:40
  • $\begingroup$ What about the case of a scalar field? $\endgroup$
    – Slereah
    Commented Mar 5, 2018 at 21:45
  • $\begingroup$ Well, no indices, no transformation (up to a phase) :-P $\endgroup$
    – AccidentalFourierTransform
    Commented Mar 5, 2018 at 21:47
  • $\begingroup$ Except according to him later on it's $\phi \to \phi^*$! $\endgroup$
    – Slereah
    Commented Mar 5, 2018 at 21:47

2 Answers 2

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There is no natural definition of charge conjugation that works for all QFTs. Rather, you should understand the CPT theorem instead as a combination of reflection-positivity and Wick rotation. See this paper, Appendix A.2.

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All of your fields naturally lie in some representation of the group of all symmetries (these include gauge symmetries, global gauge transformations and global Lorentz transformations). Charge conjugation is simply passing to the conjugate representation of that group.

E.g. complex scalars are 1d irreps of $U(1)$, and the conjugate object is $\phi^{*}$. The same logic also works for spinors, gauge fields, etc.

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  • $\begingroup$ What about symmetries that don't act on the fields? This idea can only work in some very limited scope. $\endgroup$
    – Ryan Thorngren
    Commented Mar 6, 2018 at 0:28
  • $\begingroup$ @RyanThorngren for that symmetries, fields lie in the trivial representation. Why do you think that the scope is limited? $\endgroup$
    – Prof. Legolasov
    Commented Mar 6, 2018 at 0:29
  • $\begingroup$ Now ask yourself what happens when there is a dual set of fields. You would define a different charge conjugation if you did things this way. Further, sometimes your procedure is not defined. For example there can be fields valued in representations without a real or quaternionic structure (eg. quarks in a triplet of SU(3)), then the dual representation is really a different representation, and there is no symmetry-preserving map between them. You would apply charge conjugation and end up with a different theory, so you don't get an operator on the Hilbert space. $\endgroup$
    – Ryan Thorngren
    Commented Mar 6, 2018 at 0:35
  • $\begingroup$ @RyanThorngren how does your definition of charge conjugation work in this latter case then? I don't see any plausible definition. $\endgroup$
    – Prof. Legolasov
    Commented Mar 6, 2018 at 1:01
  • $\begingroup$ I guess to have a kinetic term such theories need to include the anti-particles as separate fields and C can just switch them. That's how it goes in QCD anyway. I think what you described can work in any theory near a Gaussian point because you always need something to pair with. I don't think there is a charge conjugation in a general qft. Take some weird TQFT for instance... what does it mean? $\endgroup$
    – Ryan Thorngren
    Commented Mar 6, 2018 at 1:45

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