In Peskin & Schroeder chapter 16.2 the authors go through the computation of the non-Abelian gauge boson propagator using the Faddeev-Popov procedure as is done for the QED case. The difference here, is that our functional determinant is not just a constant which we can cancel out. It turns out that this functional determinant gives an extra contribution to the Feynman rules of the theory. My question here is the following:

During the derivation we arrive at $$ \begin{aligned}\int{\mathcal{D}Ae^{iS[A]}}&=\int\mathcal{D}Ae^{iS[A]}\int{\mathcal{D}\alpha(x)\delta(G(A^\alpha))\det\left(\frac{\delta G(A^\alpha)}{\delta\alpha}\right)}\\ &=\int\mathcal{D}\alpha(x)\int{\mathcal{D}Ae^{iS[A]}}\delta(G(A^\alpha))\det\left(\frac{\delta G(A^\alpha)}{\delta\alpha}\right). \end{aligned} $$ Now from here we use the fact that $\mathcal{D}A^\alpha=\mathcal{D}A$ and $S[A^\alpha]=S[A]$ where $A^\alpha$ is the gauge transform of $A$. And now we can rewrite the functional integral as $$ \left(\int{\mathcal{D}\alpha}\right)\int{\mathcal{D}A^\alpha e^{iS[A^\alpha]}}\delta(G(A^\alpha))\det\left(\frac{\delta G(A^\alpha)}{\delta\alpha}\right)=\left(\int{\mathcal{D}\alpha}\right)\int{\mathcal{D}A e^{iS[A]}}\delta(G(A))\det\left(\frac{\delta G(A^\alpha)}{\delta\alpha}\right) $$ where we just relabel the integration variable $A^\alpha$ to $A$. What about the part where we multiply with the determinant? Since that depends on $A^a_\mu$ shouldn't we relabel that? But if we do that then the derivation doesn't work because we need the gauge transformed determinant.

Would this be the case because initially, the determinant was in the integral of $\int{\mathcal{D}\alpha}$ instead of the integral of $\int{\mathcal{D}A\;e^{iS[A]}}$?

  • $\begingroup$ What do you mean? In P&S eq. (16.27) there is integrated over both $\alpha$ and $A$. $\endgroup$
    – Qmechanic
    May 27, 2022 at 18:56
  • $\begingroup$ Yes but we relabel the variable for $A$. And we turn everything into $A$s except the determinant, which we leave as a functional of $A^\alpha$. In the relabeling of the integral variable shouldn't we change the determinant to $A$ instead of $A^\alpha$? $\endgroup$ May 27, 2022 at 21:19
  • $\begingroup$ The functional determinant is also gauge invariant. $\endgroup$
    – Valac
    May 28, 2022 at 9:17

2 Answers 2


In the Lagrangian formalism of path integral quantization, one invokes the formula $$\int d\theta\,\delta(f(\theta))=\int df\det\Bigg|\frac{d\theta}{df}\Bigg|\delta(f(\theta))=\det\Bigg|\frac{d\theta}{df}\Bigg|_{f=0}. \tag{$\ast$}$$

Now consider the naive partition function $$\mathcal{Z}=\int\mathcal{D}[A]e^{-S[A]}. \tag{1}$$

The action $S[A]$ and the functional integral measure $\mathcal{D}[A]$ must be gauge invariant under an arbitrary gauge transformation $$A[U]=U^{-1}dU+U^{-1}AU\quad\mathrm{or}\quad\delta_{U}A=d\,\delta U+[A,\delta U], \tag{$\star$}$$

and $\mathcal{D}[A]\equiv\prod_{\mu,a,x}dA^{a}_{\mu}(x)$.

Next, consider a "haar measure" on the (infinite dimensional) Lie group of gauge transformations ($\star$), i.e $$\mathcal{D}[U]=\mathcal{D}[UU^{\prime}];\quad\mathrm{and}\quad\mathcal{D}[U]\equiv\prod_{x}dU(x),$$

and define the Faddeev-Popov determinant via the following functional integral identity $$1=\Delta[A]\int\mathcal{D}[U]\,\delta[\mathcal{F}(A[U])], \tag{2}$$

where $\mathcal{F}(A[U])$ is known as the gauge fixing condition to be specified.

Claim: The Faddeev-Popov determinant $\Delta[A]$ is gauge invariant.
proof: It can be shown as follows $$\Delta[A]^{-1}=\int\mathcal{D}[U]\,\delta[\mathcal{F}(A[U])],\quad\mathrm{hence}\quad\Delta[A[U]]^{-1}=\int\mathcal{D}[U^{\prime}]\,\delta[\mathcal{F}(A[UU^{\prime}])].$$ But since $\mathcal{D}U$ is gauge invariant, one has $$\Delta[A[U]]^{-1}=\int\mathcal{D}[UU^{\prime}]\,\delta[\mathcal{F}(A[UU^{\prime}])]=\Delta[A]^{-1}.$$

Now insert the functional integral identity (2) into the partition function (1), one has \begin{align} \mathcal{Z}&=\int\mathcal{D}[A]e^{-S[A]} \\ &=\int\mathcal{D}[A]\Delta[A]\int\mathcal{D}[U]\,\delta[\mathcal{F}(A[U])]e^{-S[A]} \\ &=\int\mathcal{D}[U]\int\mathcal{D}[A]\left(\Delta[A]\delta[\mathcal{F}(A[U])]e^{-S[A]}\right). \end{align}

Now observe that the integrand $$\int\mathcal{D}[A]\left(\Delta[A]\delta[\mathcal{F}(A[U])]e^{-S[A]}\right)$$

under the integral $\int\mathcal{D}[U]$ is actually independent of $U$, therefore it can be replaced by $$\int\mathcal{D}[A[U]]\left(\Delta[A[U]]\delta[\mathcal{F}(A[U])]e^{-S[A[U]]}\right)=\int\mathcal{D}[A]\left(\Delta[A]\delta[\mathcal{F}(A)]e^{-S[A]}\right).$$

Thus, the partition function can be written as $$\mathcal{Z}=\int\mathcal{D}[U]\int\mathcal{D}[A]\left(\Delta[A]\delta[\mathcal{F}(A)]e^{-S[A]}\right). \tag{3}$$

Next, use the functional version of the formula ($\ast$), one has \begin{align} \Delta[A]^{-1}&=\int\mathcal{D}[\mathcal{F}]\mathrm{Det}\Bigg|\frac{\delta U}{\delta\mathcal{F}(A[U])}\Bigg|\delta[\mathcal{F}] \\ &=\int\mathcal{D}[\mathcal{F}]\mathrm{Det}\Bigg|\frac{\delta\mathcal{F}(A[U])}{\delta U}\Bigg|^{-1}\delta[\mathcal{F}] \\ &=\mathrm{Det}\Bigg|\frac{\delta\mathcal{F}(A[U])}{\delta U}\Bigg|^{-1}\Bigg|_{\mathcal{F}(A[U])=0}, \end{align}

i.e $$\Delta[A]=\mathrm{Det}\Bigg|\frac{\delta\mathcal{F}(A[U])}{\delta U}\Bigg|_{\mathcal{F}(A[U])=0}.$$

One can pick a gauge fixing condition $\mathcal{F}(A[U])$ such that $\mathcal{F}(A[U])=0$ at $U=\mathrm{id}$.

Since from the above expression one finds that only infinitesimal gauge transformations are relevant in calculation, one can safely assume $$U(x)=\exp\left\{i\sum_{a}T^{a}\Lambda_{a}(x)\right\}, \tag{4}$$

where $\left\{T_{a}\right\}_{a=1,\cdots,N}$ is a basis of the Lie algebra of the gauge group. Then, the Fadeev-Popov determinant can be chosen such that $$\Delta[A]=\mathrm{Det}\Bigg|\frac{\delta\mathcal{F}(A[U])}{\delta U}\Bigg|_{U=\mathrm{id}}$$

Then, using the functional chain-rule and linearity of determinant, one has $$\frac{\delta\mathcal{F}(A[U](x))}{\delta U(y)}\Bigg|_{U=\mathrm{id}}=\sum_{a}\int d^{4}z\frac{\delta\mathcal{F}(A[U](x))}{\delta\Lambda_{a}(z)}\Bigg|_{\Lambda=0}\frac{\delta\Lambda_{a}(z)}{\delta U(y)}\Bigg|_{\Lambda=0}.$$

Notice that the factor on the right is the inverse of $$\left(\frac{\delta U(y)}{\delta\Lambda_{a}(z)}\right)\Bigg|_{\Lambda=0}=iT^{a}\delta(y-z),$$

which is a Lie algebra-valued constant. The above equation is an (infinite-dimensional) linear transformation on $$\mathcal{M}\equiv\frac{\delta\mathcal{F}(A[U])}{\delta\Lambda}.$$

Thus, up to some infinite constant, one can replace $\Delta[A]$ by $\mathrm{Det}\mathcal{M}$ in the partition function.

More precisely, one has $$\Delta[A]=\mathrm{Det}\Bigg|\frac{\delta\mathcal{F}(A[U])}{\delta U}\Bigg|_{U=\mathrm{id}}=\mathrm{Det}(\mathcal{M}\cdot𝟙^{-1})=\mathrm{Det}\mathcal{M},$$

where \begin{align} (\mathcal{M}\cdot𝟙^{-1})(x,y)&=\int d^{4}z\frac{\delta\mathcal{F}(A[U](x))}{\delta\Lambda(z)}\Bigg|_{\Lambda=0}\cdot\frac{\delta\Lambda(z)}{\delta U(y)}\Bigg|_{\Lambda=0} \\ &=\int d^{4}z\mathcal{M}(x,z)𝟙^{-1}(z,y). \end{align}

Using the functional chain-rule, one obtains \begin{align} \mathcal{M}^{ab}(x,y)&\equiv\int d^{4}z\left(\frac{\delta\mathcal{F}^{a}(A[U](x))}{\delta A[U]^{c}_{\mu}(z)}\frac{\delta A[U]^{c}_{\mu}(z)}{\delta\Lambda_{b}(y)}\right)\Bigg|_{\Lambda=0} \\ &=\int d^{4}z\frac{\delta\mathcal{F}^{a}(A(x))}{\delta A^{c}_{\mu}(z)}\left(\frac{\partial}{\partial z_{\mu}}\delta^{cb}+\sum_{d}f^{cbd}A_{d\mu}(z)\right)\delta(z-y). \tag{5} \end{align}

Also notice that under the change of variables, $$\mathcal{D}[U]=\mathrm{Det}\left(\frac{\delta U}{\delta\Lambda}\right)\mathcal{D}[\Lambda],$$

the Jacobian factor is an infinite constant (independent of gauge fields $A_{\mu}^{a}(x)$), which can be omitted in the functional integral.

Plugging (5) back into (3), one has $$\mathcal{Z}=\int\mathcal{D}[\Lambda]\int\mathcal{D}[A]\left(\mathrm{Det}\mathcal{M}\cdot\delta[\mathcal{F}(A)]\cdot e^{-S[A]}\right).$$

But the integrand $$\int\mathcal{D}[A]\left(\mathrm{Det}\mathcal{M}\cdot\delta[\mathcal{F}(A)]\cdot e^{-S[A]}\right)$$

under the integral $\int\mathcal{D}[\Lambda]$ is independent of $\Lambda$, thus the integral $\int\mathcal{D}[\Lambda]$ of gauge orbits only produces an infinite constant factor, which can be omitted.

Finally, one obtains the gauge-fixed partition function $$\mathcal{Z}=\int\mathcal{D}[A]\left(\mathrm{Det}\mathcal{M}\cdot\delta[\mathcal{F}(A)]\cdot e^{-S[A]}\right).$$

In your case, one can pick up the Lorenz gauge $$\mathcal{F}(A)=\partial_{\mu}A^{\mu}=0$$

in QED, and the Faddeev-Popov matrix is $$\mathcal{M}(x,y)=\Box\,\delta(x-y)$$

which is a constant.

More generally, in the non-Abelian case, since the $A^{a}_{\mu}(z)$ appears in the Faddeev-Popov determinant, the determinant cannot be factored out in the functional integral, and produces non-trivial interactions between photons and ghost fermions.

The gauge fixing procedure is even clearer in the canonical approach of the path-integral. TO BE CONTINUED

  • $\begingroup$ Such a comprehensive answer! Ofcourse I upvoted but could you show me how the determinant is gauge invariant in the specific case? i.e when it's defined as in the book ? $\endgroup$ May 28, 2022 at 12:23
  • $\begingroup$ @twistedmanifold You can check the claim after equation (2). That's the most general argument. A specific case is shown in the end, where you can see that in QED the determinant for the Lorenz gauge is just a d'Alembertian operator, which is just a constant. $\endgroup$
    – Valac
    May 28, 2022 at 12:27
  • $\begingroup$ @twistedmanifold I didn't read Peskin's book. But the reason why the Faddeev-Popov determinant is gauge invariant is that it is defined in that way. One can only insert gauge invariant stuff into the path-integral. When written as a determinant, it is not so obvious why it is gauge invariant. Equation (2) is the more fundamental definition of the determinant. $\endgroup$
    – Valac
    May 28, 2022 at 12:32
  • $\begingroup$ Thank you for your remarks. It's just that here, I can't show that the Fadeev Popov in this case is gauge invariant, which as you show should be true. Is there something that I'm missing? $\endgroup$ May 28, 2022 at 12:35

As in the Abelian case, the path integral becomes \begin{align} \int \mathcal{D} A \; \int \mathcal{D}\alpha \; \delta [ G(A^\alpha)] \det \left( \frac{\delta G(A^\alpha)}{\delta \alpha} \right) \exp\left[ i \int d^4x\; \left( -\frac{1}{4} \big( F(A)_{\mu\nu}^a\big)^2\right)\right] \end{align} and we can change the order of integration $A' = A^\alpha$: \begin{align} \int \mathcal{D}\alpha \; \int \mathcal{D} A \; \delta [ G(A')] \det \left( \frac{\delta G(A')}{\delta \alpha} \right) \exp\left[ i \int d^4x\; \left( -\frac{1}{4} \big( F(A)_{\mu\nu}^a\big)^2\right)\right] \end{align} Contrary to the Abelian case the determinant is not independent of the gauge fields and we cannot take it outside of the gauge field integration. As in QED we use the fact that the Lagrangian is invariant under gauge transformations and we can just replace $F(A)$ by $F(A')$ in the action. In QED we can also change the measure $\mathcal{D}A$ by $\mathcal{D}A'$ as the change in integration variables was just linear shift. In this case the change in integration variables is a bit more complicated and given by \begin{align} \big( A^\alpha \big) ^a_{\mu} = A_\mu^a +\frac{1}{g} \partial_\mu \alpha^a + f^{abc} A^b_\mu\alpha^c = A_\mu^a + \frac{1}{g}D_\mu\alpha^a \end{align}

It consist of a linear shift followed by a rotation in internal space. Both these operations preserve the measure and so we have $\mathcal{D}A=\mathcal{D}A'$ as well. We can thus write the path integral as \begin{align} \int \mathcal{D}\alpha \; \int \mathcal{D} A' \; \delta [ G(A')] \det \left( \frac{\delta G(A')}{\delta \alpha} \right) \exp\left[ i \int d^4x\; \left( -\frac{1}{4} \big( F(A')_{\mu\nu}^a\big)^2\right)\right] \end{align} and can now simply replace the integration variable $A'$ by $A$ to obtain \begin{align} \int \mathcal{D}\alpha \; \int \mathcal{D} A \; \delta [ G(A)] \det \left( \frac{\delta G(A)}{\delta \alpha} \right) \exp\left[ i \int d^4x\; \left( -\frac{1}{4} \big( F(A)_{\mu\nu}^a\big)^2\right)\right] \end{align}

In the Abelian case, the determinant is independent of the gauge field $A$ and can be taken out of the gauge field integration. It contributes to an overall normalisation constant and we can simply ignore it, as it is factored out in correlation functions and scattering amplitudes. In the non-Abelian case we cannot just ignore this determinant as it depends on the gauge field. We have \begin{align} \frac{\delta G (A^\alpha)}{\delta \alpha} &= \frac{\delta }{\delta \alpha} \left[ \partial^\mu \big( A^\alpha \big)^a_\mu(x) - \omega^a(x) \right] \nonumber\\ &= \partial^\mu \frac{\delta }{\delta \alpha} \big( A^\alpha \big)^a_\mu(x) =\partial^\mu \frac{\delta }{\delta \alpha} \big[ A^a_\mu + \frac{1}{g} D_\mu \alpha \big]= \frac{1}{g}\partial^\mu D_\mu \end{align} In the Abelian case the covariant derivative equals the ordinary derivative and we recover, of course, the QED result.

We chose to rewrite this determinant as a functional integral over anti-commuting fields $c$ and $\bar{c}$. \begin{equation} { \det \left( \frac{1}{g} \partial^\mu D_\mu \right) = \int \mathcal{D} c\, \mathcal{D}\bar{c}\, e^{i\int d^4x\, \bar{c}\big( - \partial^\mu D_\mu \big) c} } \label{eq:ymhhgwpfdg} \end{equation}

  • $\begingroup$ In the Abelian case, the covariant derivative does not equal to ordinary derivatives. In the Abelian case, the covariant derivative is $D=d+A$. The only difference is that the curvature is $F=dA$ instead of $F=dA+A\wedge A$ for the non-Abelian case. In the Abelian case, one obtains an ordinary derivative instead of a covariant derivative. $\endgroup$
    – Valac
    May 28, 2022 at 9:15
  • $\begingroup$ Look at your determinant. It's $\det\left(\frac{\delta G(A)}{\delta\alpha}\right)$ but in fact we require it to be $\det\left(\frac{\delta G(A^\alpha)}{\delta\alpha}\right)$. How do we reconcile that? $\endgroup$ May 28, 2022 at 11:43
  • $\begingroup$ The simplest explanation is that there is a typo in P&S. $\endgroup$
    – mike stone
    May 28, 2022 at 12:07
  • $\begingroup$ @mikestone no in fact the determinant is also gauge invariant so we can choose to multiply with the gauge transformed one. Libertarians answer is amazing and includes that part. Though I have some more questions. $\endgroup$ May 28, 2022 at 12:18

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