I'm trying to better understand the argument that $U(1)$ local gauge invariance implies a coupling of EM and Dirac fields. I understand the math, but I'm not sure about the chain of logic.

You start with $\mathcal{L}_\mathrm{Dirac} = \bar\psi ( i\hbar \partial_\mu - m)\psi $ and the following transformations:

$$ \psi \rightarrow \psi' = e^{i\theta}\psi \qquad \left( \bar\psi \rightarrow \bar\psi' = \bar\psi e^{-i\theta} \right) \qquad (1)$$ $$ A_\mu \rightarrow A'_\mu + \partial_\mu \chi \qquad (2)$$

You often by applying (1) to $\mathcal{L}_\mathrm{Dirac}$, and note that to keep the form invariant, you have to change the partial derivative to the covariant derivative: $\partial_\mu \rightsquigarrow D_\mu = \partial_\mu - \frac{i}{e} A_\mu$. You do the calculation again, and then find

$$\mathcal{L}' = \bar \psi\left( i \hbar \gamma^\mu (D_\mu - i \partial_\mu\theta + i\partial_\mu\chi) - m\right)\psi$$

Thus, to make $\mathcal{L}$ invariant under the transformations, $\theta = \chi$ (up to factors of $e$ and $\hbar$ that I missed). The phase of the electron wavefunction must in a way absorb the change of the four-potential given by the gauge transformation, or vice versa.

I understand (2) comes from classical gauge freedom in electrodynamics. You are free to add a constant potential offset $\Phi_0$ to $\Phi$, or to add a divergence of a scalar field $\vec\nabla \chi$ to $\vec A$. Since $A_\mu = (\Phi, \vec A)$, it follows that (2) leaves the physics invariant.

But where does (1) come from? Is this just the regular freedom of phase change you have with any wavefunction? If so, why doesn't the $U(1)$ symmetry show up everywhere? Or does this come from postulating a $U(1)$ symmetry for $\psi$? If so, how is this motivated (other than "oh, it gives the right result")?

For example, if I think (2) is better motivated, is it possible to start with only (2) and possibly the covariant derivative, and get to (1)? In general, what assumptions can I put into the argument, and what can I get out of it?

  • 1
    $\begingroup$ The global U(1) symmetry is association with a conservation law: conservation of charge (or particle number, which is the same in this case). This charge is the (integrated) zeroth component of the Noether current associated with the U(1) symmetry. So the fact that charge is conserved goes hand in hand with the U(1) symmetry -- you demand one, and get the other for free. $\endgroup$
    – Olaf
    Commented Jul 22, 2013 at 11:30
  • 1
    $\begingroup$ If you want, you can call it a postulate. I see it as "model building". If you demand that particle number is conserved, then a U(1) symmetry is pretty much automatic. If you want to the system to couple to a gauge field, then you 'demand' local gauge invariance and introduce a minimal coupling. Again, a form of model building. $\endgroup$
    – Olaf
    Commented Jul 22, 2013 at 11:37
  • $\begingroup$ @Olaf: Thanks, I forgot that you can put in charge conservation. That also explains where the factors of $e$ come from! ... There are several statements, and we can use some as assumptions, and one as a conclusion. I guess I'm trying to figure out what the most pedagogical or intuitive set of inputs to the argument is. $\endgroup$
    – jdm
    Commented Jul 22, 2013 at 11:48
  • $\begingroup$ The answer to the question in bold face is, no. You can't start with transformation properties of the gauge potential and get the phase transformation of $\psi$. $\endgroup$
    – QuantumDot
    Commented Jul 6, 2016 at 15:09

1 Answer 1


Here is a different way to see this.

The massless gauge boson field $A_\mu$ is not a true Lorentz vector. In fact, the little group for massless particles is $ISO(2)$, that is the euclidean group $E(2)$, with 2 "translations" and 1 rotation ($SO(2)$). Under the "translations", $A_\mu$ transform as :

$$A_\mu \rightarrow A_\mu + \partial_\mu \Phi$$ where $\phi$ could be complex (one of the "translations" corresponds to the real part, and the other "translation" corresponds to the imaginary part).

Clearly, it is not the usual transformation for a Lorentz vector, under a Lorentz transformation.

So, this is a problem, and we can solve this problem, by making an interaction between $A_\mu$ and a conserved current $j_\mu$, so that, under the "translations" ,we have the transformation :

$$\int d^4x ~j^\mu A_\mu \rightarrow \int d^4x ~(j^\mu A_\mu + j^\mu \partial_u \phi) = \int d^4x ~(j^\mu A_\mu + \partial_u ( j^\mu\phi))$$
the last term could be transformed, in the action, in a surface integral which can be set to zero if fields decrease sufficiently fast at infinity. So, now, the term $\int d^4x ~ j^\mu A_\mu$ is a real Lorentz invariant action (while $A_\mu$ is not a Lorentz vector).

The conserved current $j^\mu$ is here the electromagnetic current $\bar \psi \gamma^\mu \psi$, we see that this current is invariant if we multiply $\psi$ by a local or global phase.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.