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 "Schrodinger picture" and the "Heisenberg picture". In the Schrodinger 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 "Schrodinger picture" and the "Heisenberg picture". In the Schrodinger 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.