Make the Dirac Equation Consistent with Relativity

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SUMMARY

The discussion focuses on the consistency of the Dirac equation with relativity, specifically the requirement that the Dirac gamma matrices satisfy the relation ##[\gamma ^\mu ,\gamma ^\nu]=2g^{\mu \nu}##. Participants clarify that the Dirac Hamiltonian, expressed as ##\vec {\alpha} \cdot \vec p +\beta m##, leads to the conclusion that ##\beta^2=1## and that the anticommutator relation ##\alpha^i\alpha^j + \alpha^j\alpha^i = 2\delta^{ij}## is essential for consistency. The discussion emphasizes the importance of expanding ##(\alpha \cdot \mathbf{p})^2## to derive the necessary relationships between the gamma matrices and the momentum operator.

PREREQUISITES
  • Understanding of the Dirac equation and its components
  • Familiarity with gamma matrices and their properties
  • Knowledge of the energy-momentum relation in natural units
  • Basic concepts of quantum mechanics and relativistic physics
NEXT STEPS
  • Study the properties of Dirac gamma matrices in detail
  • Learn about the derivation of the anticommutator relation for gamma matrices
  • Explore the implications of the Dirac equation in quantum field theory
  • Investigate the role of the metric in relativistic quantum mechanics
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Physicists, graduate students in theoretical physics, and anyone studying quantum mechanics and relativistic equations will benefit from this discussion.

sk1105
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Homework Statement


The free Dirac equation is given by ##(i\gamma ^\mu \partial _\mu -m)\psi = 0## where ##m## is the particle's mass and ##\gamma ^\mu## are the Dirac gamma matrices. Show that for the equation to be consistent with Relativity, the gamma matrices must satisfy ##[\gamma ^\mu ,\gamma ^\nu]=2g^{\mu \nu}##.

Homework Equations


Dirac equation
Gamma matrices
##E^2=|\vec p|^2 + m^2## in natural units
We use the ##+---## metric.

The Attempt at a Solution


I know that the Dirac Hamiltonian is ##\vec {\alpha} \cdot \vec p +\beta m##, so I have equated it to the energy-momentum relation as follows:

##E^2 = (\vec {\alpha} \cdot \vec p + \beta m)^2 = (\vec {\alpha} \cdot \vec p)^2 + 2\beta m\vec {\alpha} \cdot \vec p + \beta^2m^2 = |\vec p|^2 + m^2##

It's clear that this leads to ##\beta^2=1##. I also know that we get ##\alpha^i\beta + \beta\alpha^i = 0##, although I'm less sure why. The bit that really puzzles me though is that we're supposed to get ##\alpha^i\alpha^j + \alpha^j\alpha^i = 2\delta^{ij}##, and I can't see how it follows from the equation above.

Then even with those three pieces I'm not sure how to arrive at the required commutator. Any help is much appreciated.
 
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sk1105 said:
##[\gamma ^\mu ,\gamma ^\nu]=2g^{\mu \nu}##.
That should be an anticommutator.
sk1105 said:
The bit that really puzzles me though is that we're supposed to get αiαj+αjαi=2δij\alpha^i\alpha^j + \alpha^j\alpha^i = 2\delta^{ij}, and I can't see how it follows from the equation above.
Expand ##(\alpha \cdot \mathbf{p})^2## and make it equal to ##|\mathbf{p}|^2##

sk1105 said:
##2\beta m\vec {\alpha} \cdot \vec p##
By the way you shouldn't write it like that since you know that ##\beta## does not commute with ##\alpha_i##.
 
blue_leaf77 said:
That should be an antimcommutator.

Yes you're right, my bad.

blue_leaf77 said:
Expand ##(α⋅p)2(\alpha \cdot \mathbf{p})^2## and make it equal to |p|2

That makes sense, I was just getting muddled with all the vector components. Thanks for your help.
 

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