Primary calculation involving the Dirac gama matrices

Click For Summary
SUMMARY

The discussion centers on the calculation steps in exercise 3.2 of Peskin's Quantum Field Theory (QFT), specifically regarding the treatment of momentum in the Dirac equation. The momentum is treated as an operator, represented as ##p_\mu = i\partial_\mu##, which connects momentum to the partial derivative in the Dirac equation. The confusion arises from the sign convention used, where the metric is defined as ##diag(1, -1, -1, -1)##, leading to the energy operator being ##p_0 = i\partial_t## and the spatial momentum operator as ##\textbf{p} = -i\nabla##.

PREREQUISITES
  • Understanding of the Dirac equation
  • Familiarity with quantum field theory concepts
  • Knowledge of operator notation in quantum mechanics
  • Grasp of metric conventions in relativistic physics
NEXT STEPS
  • Study the implications of the Dirac equation in quantum mechanics
  • Learn about the role of operators in quantum field theory
  • Research different metric conventions used in relativistic physics
  • Explore the relationship between eigenvalues and operators in quantum mechanics
USEFUL FOR

Students and researchers in theoretical physics, particularly those focusing on quantum field theory and the Dirac equation, will benefit from this discussion.

YSM
Messages
4
Reaction score
0
Homework Statement
How to work out a calculation involving properties of gama matrices and the dirac equation.
Relevant Equations
showed below
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.
 

Attachments

  • the calculation.png
    the calculation.png
    25.4 KB · Views: 271
Physics news on Phys.org
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:
  • Like
Likes   Reactions: vanhees71

Similar threads

  • · Replies 0 ·
Replies
0
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
1
Views
2K
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K