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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 $\partial_\mu A^\mu = 0$)?

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1 Answers1

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  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.

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