Charge conjugation in Peskin and Schroeder

  • #1
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Hey there

I'm trying to reconstruct the entire table of all Dirac bilinears under C, P, T and CPT transformations of page 71 and hit a wall on charge conjugation.

It's a computational problem, really. Here's a specific problem:
Equation 3.145 we have
$$-i\gamma ^2 \left( \psi ^{\dagger }\right) ^T =-i\left( \bar{\psi}\gamma ^0 \gamma ^2 \right) ^T$$

If I understand what is going on, we are taking the transpose of ##\gamma ^2##, which should have changed the sign of the whole expression. This is in Weyl basis. All my calculations following give wrong results but this and this is the first step that shows a problem, so it might be the root. What am I doing wrong?

BTW, I was going to post this in homework/coursework but I feel it doesn't fit the pre-made layout very well.
 

Answers and Replies

  • #2
PeterDonis
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If I understand what is going on, we are taking the transpose of ##\gamma ^2##
Not the way I'm reading the equation you posted. On the LHS, you're taking the transpose of ##\psi^\dagger##, then multiplying that on the left with ##\gamma^2##. On the RHS, you're multiplying togethr ##\bar{\psi}##, ##\gamma^0##, and ##\gamma^2##, and then taking the transpose of the result.
 
  • #3
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But how does that equality hold? Working from right to left, step by step, and neglecting that -i:
$$(\bar{\psi} \gamma ^0 \gamma^2)^T$$
$$(\psi ^\dagger \gamma^0 \gamma ^0 \gamma^2)^T$$
$$(\psi ^\dagger \gamma^2)^T$$
$$ \gamma^{2T} (\psi ^\dagger)^T$$
$$ -\gamma^2 (\psi ^\dagger)^T$$

I can almost sense it. There is always a dumb mistake and I will slap my forehead. Any moment now.
 
  • #4
strangerep
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I can almost sense it. There is always a dumb mistake and I will slap my forehead. Any moment now.
Check eq(3.25). Then check the Pauli matrices.

In a little more detail: $$(\sigma^2)^T ~=~ - \sigma^2 ~~.$$ Therefore, (in Weyl rep), $$(\gamma^2)^T ~=~ \gamma^2 ~~.$$
 
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  • #5
170
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Aaaaand there it is. Right on schedule. slaps forehead

This was a rock solid information in my brain so I never bothered to check. That the transpose of ##\gamma^2## is its negative. The confusion is, of course, because I'm just looking at the 2x2 short-hand version and forgot it's actually 4x4.

Edit: Thank you! =)
 

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