Question about Lorentz Invariance and Gamma Matrices

[SheikYerbouti]

This is a pretty basic question, but I haven't seen it dealt with in the texts that I have used. In the proof where it is shown that the product of a spinor and its Dirac conjugate is Lorentz invariant, it is assumed that the gamma matrix \gamma^0 is invariant under a Lorentz transformation. I have generally seen that each of the gamma matrices are treated as Lorentz invariant, but I have never seen the justification for this. Why are the gamma matrices Lorentz invariant?

 

[dextercioby]

The gamma matrices are just made up of 16 numbers, not of 16 functions. So they are constant, they don't vary when one switches between different inertial reference frames.

 

[stevendaryl]

Apparently, the answer is a little complicated. A brief digression: If you remember in non-relativistic quantum mechanics, there are two different ways to do things, the "Schrödinger picture" and the "Heisenberg picture". In the Schrödinger picture, the operators H, \vec{p}, \vec{L}, \vec{S} are time-independent, while the wave function \psi evolves with time. In the Heisenberg picture, those operators are functions of time, and the wave function \psi is constant. These two ways of doing things are exactly equivalent, mathematically, although people prefer one or the other for intuitive or calculational reasons. the combination of wave functions and operators \psi^\dagger O \psi has the same value in either picture.

When you get to the Dirac equation, there is a similar choice that can be made. You can either view the gamma matrices \gamma^\mu as constants, invariant under Lorentz transformations and view the Dirac spinor \Psi to transform as a spinor under Lorentz transformations, or you can view \gamma^\mu as a matrix-valued 4-vector, which transforms as a vector under Lorentz transformations, and view \Psi as a set of 4 Lorentz scalars. The two approaches are mathematically equivalent. Almost all treatments of the Dirac equation view \Psi as a Lorentz spinor and \gamma^\mu as 4 constant matrices. But I have read that for applying the Dirac equation in curved spacetime, the other way of doing it is more convenient. The combination \bar{\Psi} \gamma^\mu \Psi is the same in either way of doing it.

 

[Demystifier]

Apparently, the answer is a little complicated. A brief digression: If you remember in non-relativistic quantum mechanics, there are two different ways to do things, the "Schrödinger picture" and the "Heisenberg picture". In the Schrödinger picture, the operators H, \vec{p}, \vec{L}, \vec{S} are time-independent, while the wave function \psi evolves with time. In the Heisenberg picture, those operators are functions of time, and the wave function \psi is constant. These two ways of doing things are exactly equivalent, mathematically, although people prefer one or the other for intuitive or calculational reasons. the combination of wave functions and operators \psi^\dagger O \psi has the same value in either picture.

When you get to the Dirac equation, there is a similar choice that can be made. You can either view the gamma matrices \gamma^\mu as constants, invariant under Lorentz transformations and view the Dirac spinor \Psi to transform as a spinor under Lorentz transformations, or you can view \gamma^\mu as a matrix-valued 4-vector, which transforms as a vector under Lorentz transformations, and view \Psi as a set of 4 Lorentz scalars. The two approaches are mathematically equivalent. Almost all treatments of the Dirac equation view \Psi as a Lorentz spinor and \gamma^\mu as 4 constant matrices. But I have read that for applying the Dirac equation in curved spacetime, the other way of doing it is more convenient. The combination \bar{\Psi} \gamma^\mu \Psi is the same in either way of doing it.

Are you referring to this paper?
http://lanl.arxiv.org/abs/1309.7070 [Eur. J. Phys. 35, 035003 (2014)]

 

[stevendaryl]

Are you referring to this paper?
http://lanl.arxiv.org/abs/1309.7070 [Eur. J. Phys. 35, 035003 (2014)]

Yes, but I didn't remember the reference. Thanks.