# [Srednicki] Charge Conjugation of Dirac Spinor

1. Jan 15, 2014

### cedricyu803

1. The problem statement, all variables and given/known data
I am reading Srednicki's QFT up to CPT symmetries of Spinors
In eq. 40.42 of
http://web.physics.ucsb.edu/~mark/ms-qft-DRAFT.pdf
I attempted to get the 2nd equation:
$$C^{-1}\bar{\Psi}C=\Psi^{T}C$$

from the first one:
$$C^{-1}\Psi C=\bar{\Psi}^{T}C$$

2. Relevant equations

$$\bar{\Psi}=\Psi^\dagger \beta$$
where numerically $$\beta=\gamma^0$$

$$C^\dagger=C^{-1}=C^T=-C$$

3. The attempt at a solution

h.c. of the first equation:
$$C^{-1}\Psi^\dagger C=(C^{-1}\Psi C)^\dagger=(C\bar{\Psi}^T)^\dagger =(C(\Psi^\dagger \beta)^T)^\dagger=(C\beta\Psi^\ast)^\dagger=\Psi^T\beta C^\dagger=\Psi^T C\beta$$

So

$$C^{-1}\bar{\Psi}C=C^{-1}\Psi^\dagger \beta C=-C^{-1}\Psi^\dagger C \beta=-(\Psi^T C\beta) \beta=-\Psi^{T}C$$

I got an extra minus sign.

However, if I start from takingg transpose of the first equation I got the equation correctly.

What have I done wrong?

Also, for eq. 40.43
A is some general combination of gamma matrices.
Should it not be
$$C^{-1}\bar{\Psi}A\Psi C=\Psi^TA\bar{\Psi}^T$$
?

Why are there C's wedging A??

Thanks a lot

2. Jan 16, 2014

### strangerep

You seem to be using the same "C" on both sides on the equations. But they should be, e.g.,
$$C^{-1}\Psi C ~=~ \bar{\Psi}^{T} {\mathcal C} ~.$$(The "C" on the rhs is in caligraphic font.)

${\mathcal C}$ is a matrix. $C$ is a unitary representation of the charge-conjugation operator on this Hilbert space.

Maybe a review of ch23 would help to clarify this distinction?

3. Jan 16, 2014

### cedricyu803

Oh right. I wasn't aware of the difference between the operator and the matrix.
So the charge conjugation of $$\beta$$ does nothing to it. That's why I got a minus sign.

Thanks very much!

4. Jan 20, 2014

### russelljbarry15

There is a book out by partick labelle. It is a DeMystified book. But this one is really good. Called Supersymmetry. Chapters 2-4 cover Weyl, Majorana and Dirac Spinors. The book does it in a way that it writes out all the components of the matrices even the C one so you can really see what it going on. All the problems have solutions in the back. If you are that far in Srednicki's book you could probably read the three chapters and understand them in about a day. The book is cheap and those sections make it worth the price, since Srednicki will use it more through out the book. I found this book after Srednicki and I wish I had found it first. Then later if you want to do Supersymmetry go ahead. But it cover those topics like no other book I have ever seen, and those three chapters only focus on that subject. I think the level you are at you might want a really good understanding of those topics. Good Luck.

5. Jan 22, 2014

### cedricyu803

Thanks very much for your recommendation.

I will look for this book in a public library.

I am lucky to have been recommended Srednicki's book by my prof as well, after I told him I was having a hard time Peskin & Schroeder. Hopefully I can finish the whole book by the end of this semester.

Cheers