# The form of gauge invariance

When imposing local gauge invariance for a simple lagrangian describing a free Dirac fermion field: $$\mathcal{L}=\bar{\psi}\left(i\gamma^{\mu}\partial_{\mu}-m\right)\psi$$

Is there a particular reason for choosing the local gauge transformations to be: $$\psi\rightarrow \psi'=e^{-i\alpha(x)}\psi'$$ $$\bar{\psi}\rightarrow \bar{\psi'}=\bar{\psi'}e^{i\alpha(x)}$$ and not: $$\psi\rightarrow \psi'=e^{i\alpha(x)}\psi'$$ $$\bar{\psi}\rightarrow \bar{\psi'}=\bar{\psi'}e^{-i\alpha(x)}$$

I encountered the first convention more often, and I am wondering whether there is a physical or mathematical reason for that, or just a pure convention. Since it is trivial to see that no matter what convention we use, the symmetry will not be broken. $$e^{-i\alpha(x)}\cdot e^{i\alpha(x)} = e^{i\alpha(x)}\cdot e^{-i\alpha(x)} =1$$

## Edit:

• Sorry for not being clear. I was not suggesting that this lagrangian is gauge invariant. In order to maintain the gauge symmetry, the ordinary derivative has to be replaced by the covariant derivative $$D_{\mu}$$. $$D_{\mu}:=\partial_{\mu}+iqA_{\mu}$$

This leads to the introduction of the field $$A_{\mu}$$, and the new lagrangian will have by construction an interaction term between the Dirac field (that was initially considered) and the scalar field $$A_{\mu}$$.

I was really just thinking only at the form of the gauge transformations (the $$\pm$$ convention). I could as well picked the global invariance case, where there is no need of introducing the covariant derivative, because the lagrangian already has global gauge symmetry.

• It must be convention. You're just adding a phase term. Jun 15, 2020 at 14:47
• Comment to the post (v2): Why the usual gauge-covariant derivative is absent from your Lagrangian? How do you then maintain gauge symmetry? Jun 15, 2020 at 15:16

The free Dirac field has no such gauge symmetry - the position dependence of the $$\alpha(x)$$ means you cannot simply move it through the derivative to cancel the exponential with its conjugate. If you are introducing a gauge covariant derivative here, then any sign changes have to be reflected in your expression for the gauge covariant derivative.

In general, all transformations $$\psi\mapsto \mathrm{e}^{\mathrm{i}n\alpha(x)}, n\in\mathbb{Z}$$ are possible transformation behaviours for the field $$\psi$$. These correspond to all possible irreducible representations of the symmetry group $$\mathrm{U}(1)$$.

The gauge covariant derivative acts on a field in a given representation by convention as $$\partial_\mu - \mathrm{i}nA_\mu$$. If you flip the sign of this convention, then you have to flip the sign of the $$n$$ in the exponential, too. This is the purely "mathematical" part of the convention. Let's keep it fixed as $$-\mathrm{i}nA_\mu$$.

The $$n$$ is the charge of the field $$\psi$$, compared to a smallest charge corresponding to $$n=1$$ - usually either the charge of the electron or the charge of a quark in theories similar to the Standard Model. You can see this by deriving the Noether current for the global version of the symmetry. If $$\psi$$ is the only field in the theory, then the choice of $$n$$ is essentially arbitrary. So the choice of $$n$$ here is not merely mathematical convention once we have fixed the form of the gauge covariant derivative - it is the physical convention of which charges are called positive and which negative, and what the "base unit" for the charge is.

That's just a matter of convention. What is not a matter of convention is the position of the $$e^{\pm i\alpha(x)}$$ with respect to the spinor. Your transformations are wrongly written since the Dirac adjoint transforms with the exponential on the right $$\psi\to e^{-i\alpha(x)}\psi\qquad \bar\psi\to\bar\psi e^{i\alpha(x)}$$

But other that that, yes, it's not important whether $$\psi$$ transforms with the $$+$$ or the $$-$$, you'll get an overall sign difference in the subsequent derivation of the covariant derivative and so, which will make the same exact theory.

Which form is correct depends on which gauge transformation you apply to the potential. The minimal coupling is $$p^\mu \rightarrow p^\mu - eA^\mu ~.$$ Note that the minus sign here is not a convention. The transformation $$A^\mu \rightarrow A^\mu + \partial^\mu \alpha$$, according to gauge invariance, therefore requires $$\psi \rightarrow e^{ie\alpha/\hbar} \psi ~.$$ The transformation $$A^\mu \rightarrow A^\mu - \partial^\mu \alpha$$ requires $$\psi \rightarrow e^{-ie\alpha/\hbar} \psi ~.$$