I have a series of exercises regarding C, P and T symmetry but I am not really sure how to start with the problems. If anyone could help me with one of the problems, or show me a few example problems with full solutions, I would be very grateful. Then I can hopefully solve the remaining problems myself... As an example, we can consider this problem:

Given the Lagrangian: $$L = \bar{\Psi}(i\gamma_\mu\partial^\mu - m)\Psi - \frac{1}{2}\partial_\mu\phi\partial^\mu\phi - \frac{1}{2}M^2\phi^2+ig\phi\bar{\Psi}\gamma_5\Psi $$

How should $\phi(x)$ transform under C, P and T such that these are all symmetries of the theory?

Should I work directly on the Lagrangian, or should I consider the action? If I find one solution, how do I know it is the sole solution?

  • $\begingroup$ @ChiralAnomaly the only information given by the problem is what I have written, so yes I am starting from scratch. If I were given a transform I guess I would just need to put it in the Lagrangian and see if it is unchanged or not... $\endgroup$ – a20 May 25 '20 at 9:38

You must inspect how the last piece of the Lagrangian transforms, the rest of them are invariant. For example, let's do P. Dirac fields transform as:

$$\psi \xrightarrow{\mathcal{P}} \gamma^0 \psi,$$ $$\overline{\psi} \xrightarrow{\mathcal{P}} \overline{\psi}\gamma^0.$$ So the quatity $\overline{\psi}\gamma_5\psi$ transform as

$$\overline{\psi}\gamma_5\psi \xrightarrow{\mathcal{P}} \overline{\psi}\gamma^0\gamma_5\gamma^0\psi=-\overline{\psi}\gamma_5\psi.$$

Then if you want the Lagrangian to persevere the symmetry you can impose
$$\phi \xrightarrow{\mathcal{P}} -\phi.$$ So it's a pseudoscalar field.

You can find more information on how to perform these discrete transformations in the section 3.6 of Peskin and Schroeder.

  • $\begingroup$ Thanks! So, the parity operators are distributive like this? $P^{-1}\phi\bar{\psi}\gamma_5\psi P = P^{-1}\phi PP^{-1} \bar{\psi} PP^{-1} \gamma_5PP^{-1}\psi P $? Is this because $PP^{-1}$ is unity? $\endgroup$ – a20 May 26 '20 at 18:05
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    $\begingroup$ Yes, you are right. C and T transformations, are a little trickier but is the same idea. And the full calculation for each transformation is in the Peskin. $\endgroup$ – Susy1312 May 26 '20 at 19:01
  • $\begingroup$ what do you mean by trickier? Is the same method for C and T not applicable? For T I get: $T^{-1}\phi \bar{\psi} \gamma_5 \psi T = T^{-1}\phi TT^{-1} \bar{\psi} \gamma_5 \psi T = T^{-1}\phi T (-\bar{\psi} \gamma_5 \psi) $. Where the last step is from an identity in Srednicki. Similarily for C I get: $C^{-1}\phi \bar{\psi} \gamma_5 \psi C = C^{-1}\phi CC^{-1} \bar{\psi} \gamma_5 \psi C = C^{-1}\phi C (+\bar{\psi} \gamma_5 \psi) $. $\endgroup$ – a20 May 26 '20 at 21:12
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    $\begingroup$ I mean, the way C and T acts on field is more complicated than P, but you can do It in the same way. What you said is correct but you haven't proved how $\overline{\psi}\gamma^5\psi$ transforms. $\endgroup$ – Susy1312 May 27 '20 at 5:52

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