# Delta function conversion into gauge-fixing Lagrangian in the path integral

So, I am at the moment working on gauge-fixing a path integral. The procedure involves adding a delta function $\delta g$ to the path integral (together with the faddeev-popov determinant, but that is not important for this discussion). Once the delta function has been added, we make the argument that the gauge condition is arbitrary, and we are free to add any function $\omega$ to it as long as it does not affect the Faddeev-Popov deteminant, which it will not as it does not depend on the gauge variable.

Now comes the part where I find my understanding to be lacking. In the literature, we say that we average over all the arbitrary functions around $\omega = 0$, invoking a gauge weight. This argument seems a bit arbitrary for me; We should be free to use any weighting function so far as I am aware, since $\omega$ is arbitrary.

It looks like this: $$Z = \int DA_\mu\, \delta(g-\omega) e^{iS}\,,$$ $$Z = N(\xi)\int DA_\mu D\omega\, e^{-i\int dx\, \omega^2/2\xi} \delta(g-\omega) e^{iS}\,,$$ $$Z = N(\xi)\int DA_\mu\, e^{-i\int dx\, g^2/2\xi} e^{iS}\,.$$

I find no specific argument for why the exponential that gets added has the form that it has. It sure is convenient since it gives us the gauge condition in quadratic form, so that it integrates nicely with the Lagrangian we needed to gauge-fix in the first place. However, my current understanding is that this is an arbitrary choice; We might have just used a cubic weighing function, and ended up with a gauge-fixing Lagrangian of a completely different form.

Please tell me if my understanding is correct, and if it is not, please tell me the argument for the use of Gaussian weights.

Thanks.

• Yes, your understanding is correct. The choice of a quadratic weight is arbitrary but convenient. Other choices are valid too, – AccidentalFourierTransform May 14 '17 at 11:06
• Can you please post the whole procedure? I saw once a procedure in which the delta function was introduced by means of an auxiliary field $\omega$, which appeared in the exponent in the combination $\int d^4x\left[\omega^2/2-\omega g\right]$ in such a way that its equations of motion we exactly $\omega=g$; that would explain why a quadratic expression and not any other. Also, please take into account that a simil-Gaussian integral $\int{-\infty}^\infty e^{-x^{2n+1}}$ would not converge, so likewise won’t allow a discretization procedure for the path integral for arbitrary powers of $\omega$. – Francesco Bernardini Jan 18 at 2:51

1. OP is considering a Faddeev-Popov (FP) partition function $$Z_{\omega}~=~\int \! {\cal D}\phi~ \exp\left(\frac{i}{\hbar}S[\phi]\right)\Delta_{FP} ~\delta(\omega - G), \tag{1}$$ where $$\delta(\omega \!-\! G)~:=~\int \! {\cal D}B~ \exp\left(\frac{i}{\hbar}\int \! d^nx~B(\omega \!-\! G) \right) ; \tag{2}$$ $G$ is a gauge-fixing function that depends on the fields $\phi$, the function $\omega$ only depends on the spacetime point $x$; $\Delta_{FP}$ is the Faddeev-Popov determinant, which depends on $G$ but not $\omega$.
2. Now the first point is that the partition function (1) does not depend on $G$ nor $\omega$.
3. Therefore the $\omega$-average $$\langle Z_{\omega} \rangle_f ~:=~N(f) \int \! {\cal D}\omega ~Z_{\omega}\exp\left(\frac{i}{\hbar}\int \! d^nx~f(\omega) \right)$$ $$~\stackrel{(1)}{=}~N(f)\int \! {\cal D}\phi~ \exp\left(\frac{i}{\hbar}S[\phi]\right)\Delta_{FP}\exp\left(\frac{i}{\hbar}\int \! d^nx~f(G) \right), \tag{3}$$ does not depend on the function $f$, as long as the integrals are convergent. Here we have defined the normalization factor $$N(f)~:=~\frac{1}{\int \! {\cal D}\omega \exp\left(\frac{i}{\hbar}\int \! d^nx~f(\omega) \right)}.\tag{4}$$
4. Similarly, one may show that the partition function $$Z_f~:=~ \int \! {\cal D}\phi~ \exp\left(\frac{i}{\hbar}S[\phi]\right)\Delta_{FP}\int \! {\cal D}B~ \exp\left(\frac{i}{\hbar}\int \! d^nx~B(f(B) \!-\! G) \right)\tag{5}$$
does not depend on the function $f$.