# What feature of QFT requires the C in the CPT theorem?

Classical tensor field theories have a PT theorem, so what changes in a QFT to require charge conjugation to be a part of the theorem? Charge conjugation seems a bit unrelated to space-time, but is an integral part of the theorem.

I have a suspicion this has to do with the Grassmann algebra of fermions, if this is the case, then would a purely bosonic QFT have a PT theorem?

EDIT: Robin gives a counter-example of this idea below, so it must be another aspect of QFT.

• Isn't the whole point of the $CPT$ theorem that while none of $C$ or $P$ or $T$ or $CP$ or $PT$ or $CT$ have to be conserved, $CPT$ is always conserved. Charge conjugation is supposed to interchange particles and antiparticles; I wasn't sure one could think of this classically although after a google search this suggests replacing $\tau\to-\tau$ for proper time would be the action of $C$ classically. – snulty Oct 10 '16 at 8:57

I am not very familiar with details of the proof of the $CPT$ theorem, but could it be that $T$ is anti-unitary? For example consider a bosonic QFT with a Klein-Gordon field $\phi$ and a vector field $A^\mu$, and take the interaction Lagrangian $$\mathcal L_\text{int} = \frac{1}{M^2} \epsilon^{\mu\nu\sigma\rho} (\partial_\nu A_\mu) (\partial_\rho \phi) (\partial_\sigma \phi^\dagger).$$
Under $PT$, $\epsilon^{\mu\nu\sigma\rho}$ is unchanged, but $\phi \leftrightarrow \phi^\dagger$ because $PT$ is anti-linear. Thus we need the anti-linear $C$, which also switches $\phi \leftrightarrow \phi^\dagger$, to make $\mathcal L_\text{int}$ invariant.
• Well, that's sort of what an anti-unitary operator does. E.g. expand $\phi$ and $\phi^\dagger$ in creation and annihilation operators and you will see that $T$ switches particles and anti-particles. – Robin Ekman Oct 12 '16 at 0:04
• This is not correct---charge conjugation is always linear. It may look like it is anti-linear because of how it acts on fields, but it sends $i\mapsto +i$. Haag's book has a relatively good (if rather formal) discussion of this. – user3521569 Aug 4 '19 at 16:46