# Understanding the Gaussian weight and the parameter $\xi$ when quantizing gauge theories

In section 9.4 of Peskin & Schroeder's textbook on quantum field theory, when applying the Faddeev Popov procedure to quantize an Abelian gauge theory, they obtain the following functional integral: $$\int \mathcal{D}A e^{iS[A]} = \det\Big(\frac{1}{e}\partial^2\Big)\Big( \int \mathcal{D}\alpha\Big) \int\mathcal{D}Ae^{iS[A]}\delta\big(\partial^\mu A_\mu - \omega(x)\big).\tag{p.296}$$ Here $$A$$ is the gauge field. They then say:

This equality holds for any $$\omega(x)$$, so it will also hold if we replace the right-hand side with any properly normalized linear combination involving different functions $$\omega(x)$$. For our final trick, we will integrate over all $$\omega(x)$$, with a Gaussian weighting function centered on $$\omega = 0$$. The above expression is thus equal to $$N(\xi)\int D\omega \exp\Big[-i\int d^4x \frac{\omega^2}{2\xi} \Big] \det\Big(\frac{1}{e}\partial^2\Big) \Big(\int \mathcal{D}\alpha\Big) \int \mathcal{D} A e^{iS[A]} \delta(\partial^\mu A_\mu - \omega(x)) \\ = N(\xi)\det \Big(\frac{1}{e}\partial^2\Big)\Big(\int \mathcal{D}\alpha \Big) \int \mathcal{D}A e^{iS[A]} \exp\Big[-i\int d^4x \frac{1}{2\xi} (\partial^\mu A_\mu)^2\Big], \tag{9.56}$$ where $$N(\xi)$$ is an unimportant normalization constant and we have used the delta function to perform the integral over $$\omega$$. We can choose $$\xi$$ to be any finite constant. Effectively, we have added a new term $$-(\partial^\mu A_\mu)^2/2\xi$$ to the Lagrangian.

I have understood everything about this derivation except for the Gaussian weighting function and the role of the parameter $$\xi$$.

Regarding the Gaussian weighting function: How is this a Gaussian weight? A Gaussian integral is usually defined with the spatial integral being outside the exponential, but here the integral is inside the exponential. Also, if it is a Gaussian, why is the argument of the exponential negative?

Regarding $$\xi$$: What is the purpose of it and why do we introduce it as part of the Gaussian weight? Is it supposed to be interpreted as a variance? Second, one often chooses a value for $$\xi$$ and this is known as fixing a gauge. For example, $$\xi = 0$$ is the Landau gauge and $$\xi = 1$$ is the Feynman gauge. Does this gauge fixing have any relation to the gauge fixing of $$A$$ (e.g. the Lorenz gauge $$\partial_\mu A^\mu = 0$$)?

• It's a Gaussian integral because the integral is over $\omega$ and the integrand is of the form $e^{- \frac{1}{2\xi} \omega^2}$. Commented Jun 13 at 7:04
• @Prahar It doesn't matter that the integrand has a spatial integral, i.e. $\exp^{-i\int d^4x \frac{\omega^2}{2\xi}}$, instead of just $e^{- \frac{1}{2\xi} \omega^2}$? Commented Jun 13 at 18:00
• No. We deal with those type integrals by analytically continuing, or by allowing the integration contour for $\omega$ to have a small imaginary part. Commented Jun 14 at 4:06

1. Note that functional integral is independent of the gauge-fixing function $$G~=~\chi-\omega, \qquad \chi~=~d_{\mu}A^{\mu}, \tag{9.55}$$ cf. e.g. this Phys.SE post. In particular it does not actually depend on $$\omega$$.

We can therefore use an $$\omega$$-averaged functional integral. Refs. 1-2 use the simplest average: a Wick-rotated$$^1$$ Gaussian average, i.e. with a Gaussian weight function $$N(\xi)~=~\int\!{\cal D}\omega~e^{- \frac{i}{2\xi}\int\!d^4x~\omega^2}.$$

After Refs. 1-2 perform the Gaussian $$\omega$$-integration, the gauge-fixing action term becomes quadratic in the gauge-fixing (GF) function $$\chi$$, $$S_{GF}~=~-\int\!d^4x~\frac{\chi^2}{2\xi}.$$

2. Hence the gauge $$(\chi,\xi)$$ consists of a gauge-fixing function $$\chi$$ and a gauge parameter $$\xi$$, cf. e.g. my Phys.SE answer here.

3. Alternatively instead of the $$\omega$$-trick of Refs. 1-2, the same path integral (up to an overall normalization constant) in the gauge $$(\chi,\xi)$$ can be derived via the BRST formulation using a Lautrup-Nakanishi (LN) auxiliary field $$B$$, cf. e.g. my Phys.SE answers here and here.

References:

1. M.E. Peskin & D.V. Schroeder, An Intro to QFT, 1995; Section 9.4. p. 296 eq. (9.55).

2. M. Srednicki, QFT, 2007; Chapter 71 p. 433 eq. (71.21). A prepublication draft PDF file is available here.

$$^1$$ The imaginary unit $$i$$ in front of the action term can in principle be removed via a Wick-rotation to the Euclidean formulation, cf. e.g. this Phys.SE post.

• Thank you for your answer. I think I have now understood everything except the gauge parameter $\xi$. I understand why we need to fix a gauge $\chi$ when computing the path integral, but why do we also need to introduce a gauge parameter $\xi$? Is there any physics behind this or is it a computational tool? Commented Jun 13 at 18:05
• Different gauges are convenient for different calculations. Commented Jun 14 at 9:48