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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:

  • Dirac adjoint transformation fixed
  • 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.

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

3 Answers 3

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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.

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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.

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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 $$\psi \rightarrow e^{-ie\alpha/\hbar} \psi ~.$$

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