For a given gauge transformation, say, the electromagnetic field, where observable quantities aren't affected by transformations of the form $$\mathbf{A}' = \mathbf{A} + \nabla \chi,$$ $$\phi' = \phi - \frac{\partial \chi}{\partial t},$$ $$\Psi' = \Psi \cdot \exp(\frac{iq\chi}{\hbar}).$$

What exactly does the $U(1)$ symmetry Lie group have to do with these gauge transformation?

  • 2
    $\begingroup$ Group has nothing to do with them. They form a group! Very simple fact -- if you do one gauge transformation and from that one another one => the resulting transformation (a product) will be again a gauge transformation. So the product of two transformations is again a gauge transformation. That is it. They form a group. $\endgroup$
    – Asphir Dom
    Jul 26, 2013 at 21:36
  • $\begingroup$ @AsphirDom there is more than closure to a group $\endgroup$
    – Yossarian
    Feb 4, 2016 at 9:11

1 Answer 1


The group $\mathrm U(1)$ can be described as the set of complex numbers of unit modulus with the group multiplication given by multiplication of complex numbers. Given this characterization, notice that the transformation $$ \Psi \to e^{iq\chi/\hbar} \Psi $$ constitutes an action of $\mathrm U(1)$ on the Dirac field $\Psi$. If $\chi$ is then promoted to a function of spacetime position, in other words if we consider a local action of $\mathrm U(1)$ on the fields, and if we introduce the gauge field $A_\mu$ into the theory, then the theory becomes a $\mathrm U(1)$ gauge theory in the sense that the action of the theory is invariant under the local $\mathrm U(1)$ transformation.

  • $\begingroup$ So, U(1) is just the set of all possible values by which $\Psi$ is multiplied due to the gauge field $\chi$? $\endgroup$
    – abhishek
    Jul 25, 2013 at 3:25
  • 2
    $\begingroup$ @abhishek Pretty much. $\mathrm U(1)$ is the set of all values that $\Psi$ can be multiplied by when we perform a global transformation, namely one for which the $\chi$ doesn't depend on spacetime position. Once we take it to depend on spacetime position, we would say that we're performing a $\mathrm U(1)$ gauge transformation. In that case, it is necessary to introduce $A_\mu$ as a sort of auxiliary field in order for the action of the theory to be invariant under this gauged version of the $\mathrm U(1)$, and it is this new field $A_\mu$ that is typically called the gauge field. $\endgroup$ Jul 25, 2013 at 3:32

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