# Non-unitary gauge transformation

• A
• DuckAmuck
In summary, the literature states that the vector potentials in a gauge covariant derivative transform according to ##A_\mu \rightarrow T A_\mu T^{-1} + i(\partial_\mu T) T^{-1}##, where T is not necessarily unitary. However, in the case of ##\bar{\psi}## transforming, it is not clear why T is assumed to be non-unitary as ##\bar{\psi}## transforms by ##T^\dagger## and not ##T^{-1}##. This can lead to non-invariant transformations for the dirac field, but this can be remedied by choosing a specific representation for the field.

#### DuckAmuck

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:
$$A_\mu \rightarrow T A_\mu T^{-1} + i(\partial_\mu T) T^{-1}$$
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
$$\psi \rightarrow T\psi$$
$$\bar{\psi} \rightarrow (T\psi)^\dagger \gamma_0 = \psi^\dagger \gamma_0 \gamma_0 T^\dagger \gamma_0 = \bar{\psi} \gamma_0 T^\dagger \gamma_0$$
This seems to necessitate that $$\gamma_0 T^\dagger \gamma_0 = T^{-1}$$
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:
$$\bar{\psi} (\partial_\mu + i A_\mu) \psi \rightarrow \bar{\psi} \gamma_0 T^\dagger \gamma_0 (\partial_\mu + i T A_\mu T^{-1} - (\partial_\mu T)T^{-1} ) T\psi$$
$$=\bar{\psi} (\gamma_0 T^\dagger \gamma_0 T) (\partial_\mu + i A_\mu) \psi$$
So to be gauge invariant, this object $$\gamma_0 T^\dagger \gamma_0 T = 1$$ 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##?

• ohwilleke
What is $\psi$? A Dirac spinor?

The $\gamma_0$ and $T$ are matrices in different spaces, the $\gamma_0$ acts on spinor index and $T$ on group "space" index. So you do not need to worry about the generator matrix for the group commute with $\gamma_0$ or not.

What group do you have in mind? SO(N)? G_2 ?

malawi_glenn said:
What is $\psi$? A Dirac spinor?

The $\gamma_0$ and $T$ are matrices in different spaces, the $\gamma_0$ acts on spinor index and $T$ on group "space" index. So you do not need to worry about the generator matrix for the group commute with $\gamma_0$ or not.

What group do you have in mind? SO(N)? G_2 ?
Yes, of course they are acting on different spaces in most cases. Was trying to keep things very generalized in an attempt to "rescue" invariance, but I think that may be overkill.
And, Psi is indeed a dirac spinor.
My question still remains on what to do about T being non-unitary.
As ##A_\mu \rightarrow T A_\mu T^{-1} +i (\partial_\mu T) T^{-1}##
But ##\bar{\psi} \rightarrow \bar{\psi} T^\dagger##
So it is not clear why literature asserts that the transformation on A is for non-unitary transformations, because ##\bar{\psi}## tranforms by ##T^\dagger## not ##T^{-1}##.
The transformation is something like:
##\bar{\psi} \gamma^\mu (\partial_\mu + iA_\mu)\psi \rightarrow \bar{\psi} \gamma^\mu T^\dagger T(\partial_\mu + iA_\mu)\psi## which is not invariant if T is not unitary. So what can remedy this if anything?

But you have to choose which representation the dirac field must transform under. And most (compact) Lie Groups have complex representations, but not all of them. This is why ##E_8## is never considered as a model for GUT's because it has no complex representations and you can not couple them to dirac fermions.