attempted some preliminary computations, but was unable to determine the answer to this query. I am inquiring into the properties of an action of the form: $$ \int dx^4 \: \left(D_\mu \psi (D^\mu\psi)^* - m^2 \psi \psi^* + A^\mu A_\mu \right)$$ Under the Gauge transformation $\psi \to \psi e^{i \Lambda(x^\mu)}$, where $D_\mu = \partial_\mu - ie A_\mu$ and $A_\mu$ is the electromagnetic 4-potential and $\psi$ is complex scalar field. While it is well know that the first two terms are gauge invariant. I am not sure how to approach the third term. I figured I could gain some insight by considering the transformation $A_\mu \to A_\mu + \partial_\mu \phi$, this did not really provide me with any insight as I am not sure how to relate this to the original transformation, nor do I think that this is appropriate to prove/disprove the gauge invariance of the action.

  • $\begingroup$ Comment to the post (v2): Does the action also have a $F_{\mu\nu}F^{\mu\nu}$ term? $\endgroup$
    – Qmechanic
    Commented Nov 1, 2017 at 8:01
  • $\begingroup$ Sure, but would that term be necessary to check U(1) Gauge invariance? What if one where to simply consider an action term of the form: $\int dx^4 A_\mu A^\mu $? $\endgroup$ Commented Nov 1, 2017 at 22:17
  • $\begingroup$ It is important for context to give the whole action. Are you following some references? $\endgroup$
    – Qmechanic
    Commented Nov 1, 2017 at 22:57

1 Answer 1


Your Lagrangian does not look gauge invariant. You should check the source where you got it from. There was probably an error there. Instead of $A^\mu A_\mu$ there should be $F^{\mu\nu} F_{\mu \nu}$ there, maybe with a constant factor.

EDIT (11/2/2017): According to the OP's comment, it looks like there is no reference or rationale for the Lagrangian. It is obvious that the Lagrangian is not gauge invariant, as all its terms but $A^\mu A_\mu$ are gauge invariant and the term $A^\mu A_\mu$ is not: for example, if $A^\mu=0$, then $A^\mu A_\mu=0$ and generally $(A^\mu+\partial^\mu \phi) (A_\mu+\partial_\mu \phi)=\partial^\mu \phi\partial_\mu \phi\neq 0$.

  • $\begingroup$ How do you determine whether an action looks gauge invariant? This problem is more for the sake of knowing, not from any book etc, so no-one is claiming that this is gauge invariant, but I would like to know how to find out. $\endgroup$ Commented Nov 2, 2017 at 5:29
  • $\begingroup$ @PTIYL : please see the EDIT to the answer. $\endgroup$
    – akhmeteli
    Commented Nov 2, 2017 at 7:10

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