I want to compute the following path integral $$Z[w] = \frac{1}{(2\pi)^{n/2}}\int d^n x \: \prod_{i=1}^{n}d\overline{\theta}_id\theta \: \exp{\left(-\overline{\theta}_i \partial_j w_i(x)\theta_j -\frac{1}{2}w_i(x)w_i(x)\right)}.\tag{1}$$ Here $w_i(x)$ are functions of the $n$ real variables $x_i$ and $\theta_i$ and $\overline{\theta}_i$ are $n$ independent Grassmann variables.

The first step seems easy: computation of the $\theta$ and $\overline{\theta}$ integrals give $$Z[w] = \frac{1}{(2\pi)^{n/2}}\int d^n x \: \det(\partial_j w_i(x)) \exp{\left(-\frac{1}{2}w_i(x)w_i(x)\right)}.\tag{2}$$

From here, I tried using that $$\det(\partial_j w_i (x)) = \det\left(\partial_j w_i \left(\frac{d}{db}\right)\right) \exp\left(b_i x_i\right)\bigg\vert_{b=0}.$$ But I don't seem to be able to apply this step.

  • $\begingroup$ Is this from some reference? Which page? Which context? $\endgroup$
    – Qmechanic
    May 13 at 13:08
  • $\begingroup$ @Qmechanic not specifically, just a small problem I wanted to solve $\endgroup$ May 13 at 13:09

1 Answer 1

  1. This resembles the Faddeev-Popov (FP) construction, where $\theta^i,\bar{\theta}_j$ are the FP ghost & antighost, respectively. In this answer we will assume that this is indeed the intention, possibly tweaking OP's coefficients to simplify and make it work.

  2. The point is that OP's partition function $Z[w]$ then does not depend on the gauge-fixing function $w^j$ as long as the FP determinant $\det(\partial_i w^j)$ is non-zero and the $x$-integral is convergent, cf. e.g. this Phys.SE post.

  3. A simple gauge choice is $w^j(x)=x^j$. Then the FP determinant becomes 1 and the $x$-integral becomes Gaussian.

  4. It is perhaps easiest to see the independence of the gauge-fixing choice $w$ in the BRST formalism$^1$ $$ Z[w] ~=~ \int \! d^n x ~d^n \theta ~d^n\bar{\theta} ~d^nB~e^{iS} ,\tag{A} $$ where we have introduced a Lautrup-Nakanishi (LN) auxiliary field $B$. The underlying "gauge" symmetry is $x$-translations. The BRST transformations are $$\begin{align} {\bf s}x^i~=~&\theta^i, \cr {\bf s}\theta^i~=~&0, \cr {\bf s}\bar{\theta}_i~=~&-B_i, \cr {\bf s}B_i~=~&0. \end{align}\tag{B}$$ The action is BRST-exact $$\begin{align} S~=~~~&-{\bf s}\psi \cr ~\stackrel{(B)+(D)}{=}&~ \bar{\theta}_i \partial_j w^i \theta^j+B_i\left(\frac{\xi}{2}B_i+w^i\right) \cr ~\stackrel{\text{int. out} B}{\longrightarrow}&~~~\bar{\theta}_i \partial_j w^i \theta^j-\frac{w^iw^i}{2\xi},\end{align}\tag{C}$$ where $$\psi ~=~ \bar{\theta}_i \left(\frac{\xi}{2}B_i+w^i\right)\tag{D}$$ is the gauge-fixing fermion. One can now show that $Z[w]$ does not depend on $\psi$, cf. e.g. my Phys.SE answer here. $\Box$


$^1$ Various $i\epsilon$ prescriptions are implicitly assumed when needed for convergence.

  • $\begingroup$ Thanks for the hint! Is there a way to further solve this integral? Looking online I only seem to find explanations why the determinant is introduced, not how we can compute it further $\endgroup$ May 13 at 13:43
  • $\begingroup$ Looking online where? Which page? $\endgroup$
    – Qmechanic
    May 13 at 13:49
  • $\begingroup$ Ok, that already brings me a big step closer! Is there a way to prove that it is independent of the gauge fixing function $w_i$? Or how can I compute the $x_i$ integrals further? $\endgroup$ May 13 at 13:58
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    May 13 at 14:20
  • $\begingroup$ Again, thanks for the updated answer. I am however still confused on the integral. I get that it is independent of $w$, that you have shown in detail, but I'm wondering if there is an explicit solution to this integral. $\endgroup$ May 14 at 9:10

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