- #1

DuckAmuck

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- TL;DR Summary
- What happens for non-unitary gauge transformations when it comes to fermion factors?

You see in the literature that the vector potentials in a gauge covariant derivative transform like:

[tex] A_\mu \rightarrow T A_\mu T^{-1} + i(\partial_\mu T) T^{-1} [/tex]

Where T is not necessarily unitary. (In the case that it is ##T^{-1} = T^\dagger##)

My question is then if T is not unitary, how is ##\bar{\psi}## transforming?

Since

[tex]\psi \rightarrow T\psi[/tex]

[tex]\bar{\psi} \rightarrow (T\psi)^\dagger \gamma_0 = \psi^\dagger \gamma_0 \gamma_0 T^\dagger \gamma_0 = \bar{\psi} \gamma_0 T^\dagger \gamma_0[/tex]

This seems to necessitate that [tex]\gamma_0 T^\dagger \gamma_0 = T^{-1}[/tex]

This can only be the case if ##T^\dagger = T^{-1}##, or if ##T^\dagger## is a 4x4 matrix that multiplies with ##\gamma_0## matrices to give the inverse.

Otherwise, in general, you are left with something like:

[tex]\bar{\psi} (\partial_\mu + i A_\mu) \psi \rightarrow \bar{\psi} \gamma_0 T^\dagger \gamma_0 (\partial_\mu + i T A_\mu T^{-1} - (\partial_\mu T)T^{-1} ) T\psi[/tex]

[tex]=\bar{\psi} (\gamma_0 T^\dagger \gamma_0 T) (\partial_\mu + i A_\mu) \psi[/tex]

So to be gauge invariant, this object [tex]\gamma_0 T^\dagger \gamma_0 T = 1[/tex] but that is not the case in general.

Can this be simplified more, maybe for cases where T is a 2x2 matrix that commutes with ##\gamma_0##?

[tex] A_\mu \rightarrow T A_\mu T^{-1} + i(\partial_\mu T) T^{-1} [/tex]

Where T is not necessarily unitary. (In the case that it is ##T^{-1} = T^\dagger##)

My question is then if T is not unitary, how is ##\bar{\psi}## transforming?

Since

[tex]\psi \rightarrow T\psi[/tex]

[tex]\bar{\psi} \rightarrow (T\psi)^\dagger \gamma_0 = \psi^\dagger \gamma_0 \gamma_0 T^\dagger \gamma_0 = \bar{\psi} \gamma_0 T^\dagger \gamma_0[/tex]

This seems to necessitate that [tex]\gamma_0 T^\dagger \gamma_0 = T^{-1}[/tex]

This can only be the case if ##T^\dagger = T^{-1}##, or if ##T^\dagger## is a 4x4 matrix that multiplies with ##\gamma_0## matrices to give the inverse.

Otherwise, in general, you are left with something like:

[tex]\bar{\psi} (\partial_\mu + i A_\mu) \psi \rightarrow \bar{\psi} \gamma_0 T^\dagger \gamma_0 (\partial_\mu + i T A_\mu T^{-1} - (\partial_\mu T)T^{-1} ) T\psi[/tex]

[tex]=\bar{\psi} (\gamma_0 T^\dagger \gamma_0 T) (\partial_\mu + i A_\mu) \psi[/tex]

So to be gauge invariant, this object [tex]\gamma_0 T^\dagger \gamma_0 T = 1[/tex] but that is not the case in general.

Can this be simplified more, maybe for cases where T is a 2x2 matrix that commutes with ##\gamma_0##?