CPT invariance in Dirac Bilinears

Question

I am currently working on exercise 3.7c of Peskin and Schroeder's An Introduction to Quantum Field Theory. The problem statement is:

Show that any Hermition, Lorentz scalar local operator built from $\psi(x)$ and $\phi(x)$ and their conjugates have $CPT = + 1$

The solution manual states:

" Any Lorentz-scalar hermitian local operator $O(x)$ constructed from $\psi(x)$ or $\phi(x)$ can be decomposed into groups, each of which is a Lorentz-tensor hermitian operator and contains either $\psi(x)$ or $\phi(x)$ only"

How do I prove this statement? To me, this statement seems false. Why cannot I have something that is a mixture of $\psi(x)$ or $\phi(x)$ (i.e. contains both $\psi(x)$ or $\phi(x)$)?

Next the solution states:

" and for operators constructed from $\phi(x)$, we note that all such operators can be decomposed further into a product (including Lorentz inner product) of operators of the form $(\partial_{\mu_1}· · · \partial_{\mu_m}\phi^\dagger)(\partial_{\mu_1}· · · \partial_{\mu_n}\phi) + c.c$"

As before, why is this true? Why can it not contain products of $\phi$'s and $\phi^\dagger$'s? Is Lorentz invariance broken or something, and if so why?

Answer 1

Firstly, I think you misunderstand what they're saying a little - they're saying a general operator is a sum of products of simpler operators, and asking what building blocks you can take these simpler operators to be. For example, $$\psi^\dagger \gamma^\mu \psi (\phi^\dagger \partial_\mu \phi + \phi \partial_\mu \phi^\dagger) $$ is the product of $\psi^\dagger \gamma^\mu \psi$ and $\phi^\dagger \partial_\mu \phi + \phi \partial_\mu \phi^\dagger$, each of which is a Lorentz tensor (specifically vector and covector) involving only one of $\psi, \phi$. Further, note that in this case, the latter is exactly of the form they suggest general $\phi$ operators can be decomposed into - in particular it involves products of terms containing both $\phi, \phi^\dagger$, so I don't understand exactly why you think such things aren't allowed.

However, the second claim is in fact wrong in general. The most trivial counterexample is $\phi + \phi^\dagger$. This is a Hermitian scalar, but it cannot be written as a product of terms of the suggested form. So your manual is a bit wide of the mark.

As for proving the first statement, well, it follows from the fact that all the Lorentz scalars you can build are contractions of products all the basic tensors you can build using derivatives, gamma matrices and the fields, plus a slight twist to show that you can take each term individually to be Hermitian. This follows from the identity $$AB + \overline{AB} = \frac 1 2 (A+\bar{A})(B + \bar{B}) - \frac 1 2 i(A-\bar{A})i(B - \bar{B})$$ for example.

Then to draw the conclusion for the $\phi$ part of the argument, we can instead note that $CPT$ acts on any Hermitian tensor built of $\phi, \phi^\dagger, \partial_\mu$ by swapping the operator with its conjugate, complex conjugation of numbers, and flipping the sign of all derivatives. The first two operations together leave the expression invariant as it is Hermitian. The last operation makes the expression transform with the overall sign given by the parity of the tensor's rank.

Then one checks that all fermion expressions also transform with this sign, and the result follows.