Charge Conjugation and Internal Symmetry Representations

[erccarls]

Hi All,
I am trying to work through a QFT problem for independent study and I can't quite get my head around it. It is 5.16 from Tom Bank's book (http://www.nucleares.unam.mx/~Alberto/apuntes/banks.pdf) which goes as follows:

"Show that charge conjugation symmetry implies that the representation of the internal symmetry group G is real or pseudo-real." (I think we only need to deal with scalar fields here but I don't know that it matters.)

The book linked to above has more details on pages 50-52. I am am pretty confused but I think that I need to show that the representation R_S is not complex if C-symmetry exists. That is to show that R_s^\dagger=U^\dagger R_s U. (i.e. unitary equivalence) is implied by C-symmetry on scalar fields.

Thanks in advance for any help.

-Eric

 

[azntoon]

To show that charge conjugation symmetry implies the representation of the internal symmetry group G is real or pseudo-real, you can use the following steps: 1. First, recall that charge conjugation (C) symmetry is a symmetry of the Lagrangian which implies that the transition matrix elements of antiparticles and particles must be the same. 2. This implies that under a C-transformation, a scalar field should transform as follows: \phi(x) \rightarrow \phi^*(-x).3. Then, consider the representation of G on the scalar field. That is, for any element g in G,\phi(x) \rightarrow R_S(g)\phi(x).4. Using the fact that G is an internal symmetry group, if we apply the C-symmetry transformation to the scalar field, it should also obey the group transformation i.e. \phi^*(-x) \rightarrow R_S(g)\phi^*(-x).5. But this can only be true if R_S(g) is real (or pseudo-real).Therefore, we have shown that charge conjugation symmetry implies that the representation of the internal symmetry group G is real or pseudo-real.