Primary calculation involving the Dirac gama matrices

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Homework Help Overview

The discussion revolves around a calculation involving the Dirac gamma matrices as part of exercise 3.2 from Peskin's Quantum Field Theory. Participants are exploring the treatment of momentum within the context of the Dirac equation.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to understand a specific step in the solution related to the treatment of momentum as an operator in the Dirac equation. Some participants clarify that momentum is indeed treated as an operator, leading to questions about the absence of a minus sign in front of momentum in the solution.

Discussion Status

Participants are actively engaging in clarifying the treatment of momentum in the Dirac equation, with some guidance provided regarding the sign conventions used in the metric. There is an ongoing exploration of the implications of these conventions on the interpretation of the momentum operator.

Contextual Notes

There is a mention of a specific sign convention related to the metric used, which may influence the understanding of the momentum operator's representation in the Dirac equation.

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Homework Statement
How to work out a calculation involving properties of gama matrices and the dirac equation.
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When working on the exercise 3.2 of Peskin's QFT, I find one of the calculating steps confused for me. I read the solution, which is showed in the picture. I just don't understand the boxed part.

I know it involved the Dirac equation, and the solution seems to treat the momentum as a operator, because only in this way can I relate the momentum in the equation with the partial derivative in the Dirac Equation. But I don't think the momentum in the solution of Dirac field serve as an operator.
 

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Momentum in Dirac equation indeed is an operator, in fact: ##p_\mu = i\partial_\mu##. So if that's the only problem, there's your answer.

Edit: Momentum in solutions of Dirac equation is eigenvalue of momentum operator, though they're usually denoted with the same letter.
 
Last edited:
Thank you for your answer, but why no minus sign in front of p?
 
It's the sign convention where metric is given by ##diag(1, -1, -1, -1)##. So in that convention the energy operator is given by ##p_0 = i\partial_t## as it should be, and 3-momentum operator is given by ##\textbf{p} = -i\nabla## because ##p^i = -p_i## for spatial indices.
 
Last edited:
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