To get rid of the extra term in the QED Lagrangian we need to redefine the electromagnetic four-vector:

$A^{\mu} \rightarrow A^{\mu} - \frac{1}{c} \partial_{\mu} a(x)$ where $a(x)$ is the function that defines the local U(1) symmetry, i.e. sending $\psi(x) \rightarrow e^{ia(x)}\psi(x)$ leaves the QED Lagrangian invariant if we redefine the field as stated above.

To still satisfy the Lorenz gauge ($\partial \cdot A = 0$) we thus need to impose $\partial^2 a(x) = 0$.

In the textbook I'm following the choice for the function a is: $a(x) = c e^{-iqx}$ where q and x are four-vectors and the multiplication means Minkowski scalar product.

I don't see how $\partial^2 \cdot c e^{iqx} = 0$, unless we impose a further condition on $q$, i.e. $q^2$ = 0. That condition would sort of make sense, since the e.m.-field corresponds to massless photons for which that condition is satisfied. Is that what's happening here?

  • $\begingroup$ That is not a redefinition of the gauge field, it is its natural transformation behaviour under a gauge transformation. $\endgroup$
    – ACuriousMind
    Jan 15, 2015 at 14:17
  • $\begingroup$ I don't know about the correct wording, but in my course this is what we want the em field to transform like to get rid of the extra term arising from the local transformation of the spinor field $\psi(x)$. It's natural, but it's still sort of a redefintion in that sense, no? $\endgroup$
    – user17574
    Jan 15, 2015 at 14:22
  • $\begingroup$ You don't impose $\partial^2 a(x) = 0$ to satisfy the Lorenz gauge. You impose that when you start from the Lorenz gauge and want to remain in the Lorenz gauge. Without more info about what your book is saying, my best guess would be that the $a(x)$ you give solved the equation $\partial_\mu A_\mu' = \partial_\mu A_\mu - \frac1c \partial^2a(x) = 0$. $\endgroup$ Jan 19, 2015 at 9:09
  • $\begingroup$ Yes and $\partial_{\mu} A^{\mu} - \frac{1}{c} \partial_{\mu}^2 a(x)$ only equals zero if $a(x) = 0$ for each component of $\partial_{\mu}^2$. Right? That would also not be the case for the choice of $a(x) = e^{iqx}$ unless each of the components of q squared is zero... $\endgroup$
    – user17574
    Jan 19, 2015 at 10:50

1 Answer 1


As you probably know, an equation of the form $$\partial^2 f=0$$ is called a wave equation. Your text has chosen a plane-wave gauge function $a=ce^{-iqx}.$ For this to solve the wave equation, the wave vector $q$ must be null. This is just the standard condition that must be imposed for a plane wave to satisfy the wave equation. I don't think there is anything too profound going on here because the gauge function is not physical.

  • $\begingroup$ Can we say this is a gauge function representing on-shell photons? f has a non-trivial solution which could be wave function of on-shell photons? $\endgroup$
    – aQuestion
    Jan 15, 2015 at 22:22
  • $\begingroup$ @Major_Tom: No. Photons do not, in any way, correspond to gauge transformations, they are the quanta of $A^\mu$, and nothing else. $\endgroup$
    – ACuriousMind
    Jan 15, 2015 at 22:30
  • $\begingroup$ @Ocelo7: What do you mean exactly by $q$ must be null? That $q \equiv 0$? $\endgroup$
    – user17574
    Jan 16, 2015 at 7:50
  • $\begingroup$ Let $\langle.,.\rangle$ be an inner product of indefinite signature (Minkowski obviously falls in this category). A vector $X$ is said to be null if $\langle X,X\rangle=0$. $\endgroup$
    – Ryan Unger
    Jan 16, 2015 at 10:11
  • $\begingroup$ q is a 4-vector (in Minkowski-metric) with components like $q=(q_0,\bf{q})=(\omega/c, \bf{k})$. $q^2=q_0^2- \bf{q}^2=(\omega/c)^2-\bf{k}^2=0$ because $\bf{k} =\omega/c$. $\endgroup$ Mar 5, 2015 at 13:16

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