I am now studying QFT using Schwartz's book, and I am going through the part discussing about how the charge conjugation, parity, and time-reversal operator acting on various object should look like. He starts with how charge conjugation operator on the spinors act: $$C : \psi \rightarrow -i \gamma_{2} \psi^{*}$$

and then enlarges the action of $C$ on the other objects so that the Lorentz invariance is attained.

However, I find this logic to be somewhat strange. For example, suppose we have a Lagrangian $$\mathcal{L} = \phi^{*}(\Box + m^{2}) \phi + \frac{\lambda}{3!} \phi^{3}.$$ Then after imposing a transformation such that $\phi \rightarrow - \phi$, we see that $\mathcal{L}$ is not invariant under this. Then we just simply say that this Lagrangian does not have this symmetry and go on.

But the argument on the action of $C$, $P$ and $T$ doesn't go this way. We first declare that the Lagrangian we are interested in must be invariant under $C$, $P$ and $T$, and adjust the action of them on various objects to make that happen. For example, Schwartz says that the QED interaction term $eA_{\mu} \bar{\psi} \gamma^{\mu} \psi$ should be invariant under Lorentz transformations, and since $$C : \bar{\psi} \gamma^{\mu} \psi \rightarrow - \bar{\psi} \gamma^{\mu} \psi,$$ he declares that $C: A_{\mu} \rightarrow -A_{\mu}$. (Schwartz, Quantum Field Theory and the Standard Model, p.194~195))

Why can we do this for $C$, $P$ and $T$? Why do we not do this for others?

For example, for $$\mathcal{L} = \phi (\Box + m^{2} ) \phi + \frac{\lambda}{3!} \phi^{3},$$ and the transformation $\phi \rightarrow -\phi,$ why not we extend the transformation to $\lambda \rightarrow -\lambda$, so that the Lagrangian $\mathcal{L}$ is invariant under it?

  • $\begingroup$ CPT symmetry follows from relativity, doesn't it? One can surely construct Lagrangians that are not obeying this symmetry, but then they are not relativistic. I would imagine that solid state physics has a lot of those systems. These are still very important emergent systems, they are just not considered "fundamental" like theories of the vacuum are. $\endgroup$ Commented May 4, 2023 at 18:36
  • $\begingroup$ @FlatterMann CPT is fundamental, but the question is about C, P, T separately (which are violated eg. in standard model) $\endgroup$ Commented May 4, 2023 at 20:37
  • $\begingroup$ @QCD_IS_GOOD In that case I simply misunderstood the question... exactly for the reason you mentioned. CP violation has been experimentally observed, after all. The only fundamental symmetry that is "different" because we don't get a choice if we want to keep relativity alive is CPT. At least that is my limited understanding as an experimentalist. Not that a small violation of relativity is completely out of the question... I would keep an open mind there as well. $\endgroup$ Commented May 5, 2023 at 1:57

1 Answer 1


When folks talk about symmetry transformation, usually it's the fields (and in some cases, coordinates as well) that are transformed, e.g. $\psi$, $\phi$, and $A_\mu$.

And in your case of $$ \mathcal{L} = \phi (\Box + m^{2} ) \phi + \frac{\lambda}{3!} \phi^{3} $$ where $\lambda$ is NOT a field, thus can NOT be transformed.

One workaround is to promote $\lambda$ to be a field that can transform properly, then you regain symmetry. One historical example: Steven Weinberg promoted the mass parameter $m$ of standard model fermion to the Higgs field (more precisely Higgs field multiplied by the Yukawa constant) to accommodate the electroweak symmetry.

  • $\begingroup$ (just to add on, promoting parameters to preserve symmetries is usually called spurion analysis) $\endgroup$ Commented May 22, 2023 at 17:30

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