cedricyu803
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Homework Statement
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
Homework Equations
\bar{\Psi}=\Psi^\dagger \beta
where numerically \beta=\gamma^0
C^\dagger=C^{-1}=C^T=-C
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<br /> =(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