How exactly does one come to this identity $$\int\mathcal{D}G(A^\alpha)\delta(G(A^\alpha)) = \int\mathcal{D}\alpha(x)\delta(G(A^\alpha)) \mathrm{det}\left(\frac{\delta G(A^\alpha)}{\delta\alpha}\right).$$

Let $A_\mu$ be the massless vector field appearing in the Maxwell equations, defined by the Lagragian $$\mathcal{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}.$$ Define now $$\begin{align*} A^\alpha_\mu (x) &:= A_\mu(x) +\frac{1}{e} \partial_\mu \alpha(x),\\ G(A^\alpha) &:= \partial_\mu A^\mu +\frac{1}{e}\partial^2\alpha(x)-\omega(x), \end{align*}$$ which implies setting $G(A^\alpha)=0$ fixes our gauge. According to my professor the following identity now holds: $$\mathbb{I}=\int\mathcal{D}G(A^\alpha)\delta(G(A^\alpha)) = \int\mathcal{D}\alpha(x)\delta(G(A^\alpha)) \mathrm{det}\left(\frac{\delta G(A^\alpha)}{\delta\alpha}\right).$$

Can somebody explain to me how we derive this? Especially the second equality is unclear to me. The first one seems to be just the generalization of $1=\int dx\, \delta(x)$, but for the second one I'm kind of lost where the determinant comes from and why we now can integrate over all possible $\alpha$. What part on the rhs guarantees me that I'm still in the same gauge as in the lhs?

After the suggestion in the comments I took a closer look at the linked Wikipedia page. Presumably the relevant equation from this page is ($\varphi:U\to\mathbb{R}^n$): $${\displaystyle \int _{\varphi (U)}f(\mathbf {v} )\,d\mathbf {v} =\int _{U}f(\varphi (\mathbf {u} ))\left|\det(D\varphi )(\mathbf {u} )\right|\,d\mathbf {u} .}$$ Now comparing this equation to the identitiy from my question I think $G$ plays the role of $\varphi$ and the $\delta$-function in my original expression plays the role of $f$. I will assume that the generalization is sensible for the determinant aka that ${\rm det}(D\varphi) \to {\rm det}(\frac{\delta G}{\delta \alpha})$ but how can it be that the $\delta$-function has the same argument on both sides of the equation? Shouldn't that argument be transformed as well? This links to my above question about the fixed gauge on the lhs and the rhs.

  • 4
    $\begingroup$ The determinant is simply the Jacobian you get under the change of integration variables. A quick read on how change of variables works in multivariate integration should clarify that the generalization to functional integration is straightforward. $\endgroup$
    – JF132
    Mar 3, 2020 at 19:19
  • 1
    $\begingroup$ @JF132 I think I would agree with you if the $\alpha$ wasn't hidden behind a $\partial^2$.. Could you maybe show me how to deal with that? $\endgroup$
    – Sito
    Mar 3, 2020 at 20:20
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    $\begingroup$ This is just an infinite-dimensional version of integration by substitution. $\endgroup$
    – Qmechanic
    Mar 3, 2020 at 20:25
  • $\begingroup$ @Qmechanic I edited my question to add what I don't understand in the infinite-dimensional version. Can you give me a hint on this? $\endgroup$
    – Sito
    Mar 4, 2020 at 13:32
  • $\begingroup$ @JF132 I added what exactly I don't understand in the generalization you mention. $\endgroup$
    – Sito
    Mar 4, 2020 at 13:33

1 Answer 1


It seems like part of your confusion is due to the "abuse" of notations.

On the left hand side, you have a functional integral over $G$. The LHS reads as: $$\int \mathcal{D}G \delta (G)$$

Here $G$ is an integration variable. We need to integrate over all possible values. You can forget about its connection to $\alpha$. Same thing in a path integral for a field, the field symbol is merely an integration variable, not the field we get from solving the equation of motion.

Now on the RHS, we now want to express it as a functional integral over $\alpha$. We can make a change of variable via $G=G(A^\alpha)$. To be explicit, the RHS reads as: $$\int \mathcal{D}\alpha\; \delta \big(\partial_\mu A^\mu+\frac{1}{e}\partial^2 \alpha-\omega\big) \Delta$$

where $\Delta$ is the Jacobian from the change of variable as mentioned.


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