Faddeev–Popov Quantization (Abelian), Part 2: The Faddeev–Popov Identity and the Photon Propagator

In Part 1, we saw that the path integral for a $U(1)$ gauge field is ill-defined: the kinetic operator $-\Box\,\delta_{\mu\nu} + \partial_\mu\partial_\nu$ has zero modes along gauge directions, and the integral overcounts by an infinite factor (the volume of the gauge orbit). We need to restrict the integral to one representative per orbit, but naively inserting a delta function misses a Jacobian.

Now we fix this properly. Honestly, the whole Faddeev–Popov trick is just inserting a very clever " $1$ " into the path integral.

The identity we need#

Our gauge-fixing condition is $\partial_\mu A_\mu^\chi = G(x)$ , where $A_\mu^\chi = A_\mu + \partial_\mu\chi$ is the gauge-transformed field. From Part 1, we know this requires $\Box\chi = G - \partial_\mu A_\mu$ , which has a unique solution.

What we want to prove is the following identity:

\det(\Box)\int\mathcal{D}\chi\;\delta\!\left(\partial_\mu A_\mu^\chi - G\right) = 1

If this is true, we can insert it into the path integral for free (it’s just multiplying by $1$ ). But it’s not obvious why it’s true, so let’s build up to it.

Warm-up: the discrete version#

Before tackling the functional case, let’s see why this identity works for ordinary integrals. Going from one dimension, to $n$ , to a functional integral is a clean way to see where the determinant comes from.

One dimension#

Consider a function $f(x)$ with a single zero at $x = x_0$ , so $f(x_0) = 0$ . A standard identity for the delta function gives:

\int dx\;\delta(f(x)) = \frac{1}{|f'(x_0)|}

Rearranging:

|f'(x_0)|\int dx\;\delta(f(x)) = 1

The factor $|f'(x_0)|$ is the Jacobian. It measures how fast $f$ changes at the zero. If you forget it, the integral doesn’t equal $1$ .

$n$ dimensions#

Now let $\vec{f}(\vec{x})$ be a vector-valued function of $n$ variables with a single zero at $\vec{x}_0$ . The generalization is:

\det\!\left(\frac{\partial f_i}{\partial x_j}\right)\!\Bigg|_{\vec{x}_0}\int d^n x\;\delta^{(n)}(\vec{f}(\vec{x})) = 1

The single derivative $|f'|$ has become a determinant of the Jacobian matrix, exactly what you’d expect from a multi-dimensional change of variables.

The functional version#

Now go to infinite dimensions. Replace $\vec{x} \to \chi(x)$ (a function, not a vector), replace $\vec{f} \to \partial_\mu A_\mu^\chi - G$ (the gauge condition applied to the transformed field), and replace the Jacobian matrix with the functional derivative:

\frac{\delta(\partial_\mu A_\mu^\chi)}{\delta\chi} = \Box

The determinant of this operator is $\det(\Box)$ , and the identity becomes:

\det(\Box)\int\mathcal{D}\chi\;\delta\!\left(\partial_\mu A_\mu^\chi - G\right) = 1

This is the Faddeev–Popov identity. It’s the functional integral version of the same change-of-variables formula you learned in multivariable calculus, just promoted to infinite dimensions.

From delta function to Lagrangian term#

We have our identity, but the delta function $\delta(\partial_\mu A_\mu^\chi - G)$ isn’t convenient to work with inside a path integral. We’d much rather have an exponential term in the action, and there’s a standard trick for this.

Recall the Gaussian integral identity: for any function $G(x)$ ,

N_\xi \int\mathcal{D}c\;e^{-\frac{1}{2\xi}\int d^4x\;c(x)^2}\;\delta(c - G) = e^{-\frac{1}{2\xi}\int d^4x\;G(x)^2}

where $N_\xi$ is a normalization constant. This is just “the delta function picks out $c = G$ .”

Now the trick: start with the Gaussian integral without the delta function evaluated:

N_\xi\int\mathcal{D}c\;e^{-\frac{1}{2\xi}\int c^2}\;\det(\Box)\int\mathcal{D}\chi\;\delta(\partial_\mu A_\mu^\chi - c) = 1

This equals $1$ because for each value of $c$ , the FP identity gives $1$ , and the Gaussian integral with $N_\xi$ is normalized.

Now use the delta function to perform the $c$ integral, which sets $c = \partial_\mu A_\mu^\chi$ :

N_\xi\;\det(\Box)\int\mathcal{D}\chi\;e^{-\frac{1}{2\xi}\int(\partial_\mu A_\mu^\chi)^2} = 1

This is still equal to $1$ , and it’s still an identity we can insert into the path integral. But now the delta function is gone, replaced by a Gaussian weight, which is just an exponential term in the action. Exactly what we wanted.

Inserting into the path integral#

We insert this identity into the partition function:

Z = \int\mathcal{D}A_\mu\;e^{-S[A]} \times \underbrace{N_\xi\;\det(\Box)\int\mathcal{D}\chi\;e^{-\frac{1}{2\xi}\int(\partial_\mu A_\mu^\chi)^2}}_{= 1}

Now three steps.

Step 1: Change variables $A_\mu \to A_\mu^\chi$ #

Inside the path integral over $A_\mu$ , shift to the gauge-transformed variable $A_\mu^\chi = A_\mu + \partial_\mu\chi$ . Three things happen:

The action is gauge-invariant: $S[A^\chi] = S[A]$ . This is why we built a gauge theory in the first place.
The measure is gauge-invariant: $\mathcal{D}A^\chi = \mathcal{D}A$ . For the photon field alone this is easy: $A_\mu \to A_\mu + \partial_\mu\chi$ is just a shift of the integration variable by something independent of $A$ . Once charged matter is included, invariance of the full measure (including the fermions) is a genuine assumption, and there exist theories where it fails. Those are called anomalous theories, where a symmetry of the classical action is broken at the quantum level. QED is not anomalous, so we’re safe here.
The gauge-fixing term simplifies: $\partial_\mu A_\mu^\chi$ becomes just $\partial_\mu A_\mu$ after the shift (since we’ve relabeled the integration variable).

After this change of variables, the $\chi$ dependence has completely dropped out of the integrand:

Z = N_\xi\;\det(\Box)\int\mathcal{D}\chi\int\mathcal{D}A_\mu\;e^{-S[A]\,-\,\frac{1}{2\xi}\int(\partial_\mu A_\mu)^2}

Step 2: Factor out the gauge volume#

Since nothing in the $A_\mu$ integral depends on $\chi$ anymore, the $\chi$ integral is just a constant, the volume of the gauge group:

\int\mathcal{D}\chi = V_{\text{gauge}}

This is the infinite overcounting factor from Part 1. It multiplies $Z$ overall, but it cancels in any ratio $\langle\mathcal{O}\rangle = Z^{-1}\int\mathcal{D}A\;\mathcal{O}(A)\,e^{-S_{\text{eff}}}$ . So expectation values of gauge-invariant operators are perfectly well-defined.

Step 3: The gauge-fixed action#

Dropping the constant prefactors, we arrive at:

\boxed{Z = \int\mathcal{D}A_\mu\;e^{-S_{\text{eff}}[A]}, \qquad S_{\text{eff}}[A] = S[A] + \frac{1}{2\xi}\int d^4x\;(\partial_\mu A_\mu)^2}

So the entire Faddeev–Popov procedure for $U(1)$ boils down to adding the term $\frac{1}{2\xi}(\partial_\mu A_\mu)^2$ to the action. You could have just written this down by hand and said “I’m gauge fixing” (and many textbooks do exactly that). What we’ve shown is that this isn’t an ad hoc modification of the theory. It follows from the geometry of gauge orbits and the correct treatment of the Jacobian.

Let me stress one thing. The reason this was so clean is that $\det(\Box)$ does not depend on $A_\mu$ . The operator $\Box$ is just the d’Alembertian and knows nothing about the gauge field. So it’s a constant that gets absorbed into the normalization. This is the exact point where the non-abelian story diverges. In non-abelian theories, the analogous determinant $\det(M[A])$ depends on $A$ , cannot be pulled out, and must be represented as a path integral over new fields, the Faddeev–Popov ghosts.

The photon propagator#

Now we can finally invert the kinetic operator. The gauge-fixed action is:

S_{\text{eff}} = \frac{1}{2}\int d^4x\;A_\mu\!\left(-\Box\,\delta_{\mu\nu} + \partial_\mu\partial_\nu - \frac{1}{\xi}\partial_\mu\partial_\nu\right)\!A_\nu = \frac{1}{2}\int d^4x\;A_\mu\!\left(-\Box\,\delta_{\mu\nu} + \left(1 - \frac{1}{\xi}\right)\partial_\mu\partial_\nu\right)\!A_\nu

Go to momentum space ( $\partial_\mu \to ik_\mu$ , $\Box \to -k^2$ ). The operator becomes:

\widetilde{\mathcal{O}}_{\mu\nu}(k) = k^2\delta_{\mu\nu} - \left(1 - \frac{1}{\xi}\right)k_\mu k_\nu

Before gauge fixing, the operator was $k^2\delta_{\mu\nu} - k_\mu k_\nu$ , which kills any vector proportional to $k_\mu$ . Now the $k_\mu k_\nu$ term has a modified coefficient, and you can check that the zero mode is gone (unless $\xi \to \infty$ , which undoes the gauge fixing).

Inverting this (a nice exercise in projecting onto transverse and longitudinal components, try it yourself) gives the photon propagator in Euclidean momentum space:

\boxed{G_{\mu\nu}(k) = \frac{1}{k^2}\left(\delta_{\mu\nu} - (1-\xi)\frac{k_\mu k_\nu}{k^2}\right)}

This depends on the gauge parameter $\xi$ , but any physical observable (computed from gauge-invariant operators) will be $\xi$ -independent.

Feynman gauge#

The simplest choice is $\xi = 1$ , called Feynman gauge. The propagator collapses to:

G_{\mu\nu}(k) = \frac{\delta_{\mu\nu}}{k^2}

This is as simple as a propagator can get. Every component of $A_\mu$ propagates the same way, just like a massless scalar. This is why Feynman gauge is the default choice for most QED calculations.

Other common choices:

$\xi = 0$ : Landau gauge. The propagator becomes purely transverse: $G_{\mu\nu} = \frac{1}{k^2}(\delta_{\mu\nu} - k_\mu k_\nu/k^2)$ . This enforces $\partial_\mu A_\mu = 0$ strictly.
$\xi \to \infty$ : The gauge-fixing term vanishes and we’re back to the original non-invertible operator. No propagator, as expected.

What counts as a valid gauge-fixing term?#

We derived the specific term $\frac{1}{2\xi}(\partial_\mu A_\mu)^2$ , but nothing in the Faddeev–Popov logic required the gauge-fixing function to be linear. The procedure works for any function $F(\partial_\mu A_\mu)$ , and you’d add $\frac{1}{2\xi}[F(\partial_\mu A_\mu)]^2$ to the action instead.

For example, choosing $F(\partial_\mu A_\mu) = (\partial_\mu A_\mu)^2$ gives a quartic gauge-fixing term in the Euclidean action:

+\frac{1}{2\xi}(\partial_\mu A_\mu)^4

(positive, so that $e^{-S}$ stays damped). This is perfectly valid, with one caveat about the determinant. For a nonlinear $F$ , the Jacobian is $\det\!\left(\delta F/\delta\chi\right) = \det\!\left(2(\partial_\mu A_\mu)\,\Box\right)$ , which now depends on $A$ . The extra factor $\det(2\,\partial_\mu A_\mu)$ is ultralocal (no derivatives act on anything), so it only contributes $\delta^{(4)}(0)$ -type terms, which vanish in dimensional regularization. You can derive it through the same FP procedure: start with the path integral plus an auxiliary field, shift by $\pi \to \pi + \frac{1}{\xi}\partial_\mu A_\mu$ , and perform a gauge transformation. The result only depends on the longitudinal mode of $A_\mu$ (the gauge degree of freedom), so it correctly picks a slice through each gauge orbit without touching the physical, transverse degrees of freedom. In the limit $\xi \to 0$ it enforces $\partial_\mu A_\mu = 0$ , and in the limit $\xi \to \infty$ it disappears. Correlation functions of gauge-invariant operators remain unchanged.

A term that does not work#

What about adding $\xi A_\mu^2$ to the action? Under a gauge transformation $A_\mu \to A_\mu + \partial_\mu\chi$ , this transforms as:

\xi A_\mu^2 \to \xi(A_\mu + \partial_\mu\chi)^2

This term is not of the form $f(\partial_\mu A_\mu)$ . It depends on the full gauge field, not just the longitudinal part. Physically, $A_\mu^2$ is a mass term for the photon. It modifies the transverse, physical degrees of freedom, not just the gauge redundancy.

In the limit $\xi \to \infty$ , it doesn’t select a gauge slice. It forces $A_\mu = 0$ everywhere, killing the gauge field entirely. The photon is gone, and with it the physics of the theory. Correlation functions of gauge-invariant operators do change. This is not gauge fixing, it’s a different theory.

So a valid gauge-fixing term must be a function of the gauge condition (like $\partial_\mu A_\mu$ ), which lives purely in the longitudinal sector. If it touches the transverse modes, it changes the physics.

Conceptual summary#

The whole argument of the two posts, in one list:

The path integral overcounts by a factor of the gauge group volume, and the kinetic operator is non-invertible.
The FP identity, $\det(\Box)\int\mathcal{D}\chi\;\delta(\partial_\mu A_\mu^\chi - G) = 1$ , is the functional version of the change-of-variables Jacobian from calculus.
A Gaussian trick converts the delta function into the exponential gauge-fixing term $\frac{1}{2\xi}(\partial_\mu A_\mu)^2$ in the action.
After a change of variables and using gauge invariance of the action and measure, the gauge volume factors out and we’re left with a well-defined, gauge-fixed path integral.
The operator can now be inverted, giving the photon propagator $G_{\mu\nu}(k) = \frac{1}{k^2}(\delta_{\mu\nu} - (1-\xi)k_\mu k_\nu/k^2)$ .
The whole procedure worked cleanly because $\det(\Box)$ is independent of $A_\mu$ , which is special to the abelian case.
Valid gauge-fixing terms must only constrain the longitudinal (gauge) sector. A photon mass term $A_\mu^2$ fails this criterion.

What’s next#

The natural continuation is the non-abelian case: $SU(2)$ Yang–Mills theory. There, the covariant derivative, field strength, and Lagrangian need to be checked for gauge invariance from scratch, since the transformations are more involved because the generators don’t commute. And the Faddeev–Popov determinant becomes field-dependent, forcing us to introduce ghost fields. That’s the subject of the next series.

This completes the abelian Faddeev–Popov series. Next up, the non-abelian series, starting with the classical machinery of $SU(2)$ Yang–Mills.

Source Material#

This post was inspired by my own course material that I developed for **Heidelberg University Advanced QFT Plenary Tutorials. You can download the original notes:

📄 Tutorial 6: Faddeev-Popov Quantization (PDF)