# Completing the square for Grassmann variables

When working with path integrals of both bosonic and fermionic field variables, I'm a bit unsure of how to do the usual complete the square trick when an interaction between the two is concerned. Say you have a generic partition function like $$Z=\int D\phi D\bar\psi D\psi \,e^{iS[\phi,\bar\psi,\psi]}$$ where the action has your standard quadratic part, but also an interaction term like $\mathcal{L}_{int}\sim \bar\psi \phi\psi$. If you wanted to integrate out the Bose fields first, you would complete the square but that would involve making a substitution like $\phi\rightarrow \phi'= \phi-\bar\psi\psi$ and integrating over $\phi'$ (I may be wrong here). I'm confused on what it means to:

1. Subtract a product of Grassmann numbers from an ordinary number (they aren't the same kind of number)
2. Integrate over the new Bose field, which is technically a function of the Grassmann variables (are there any subtleties with that process?)

Any help would be greatly appreciated.

The idea is correct (it's called the Hubbard-Stratonovich transformation), but I can't say more without the details of the action. It is discussed in any good textbook on quantum field theory for condensed matter.

Concerning you questions (if you're okay with the physicist's approach to Grassman numbers) : the product of two Grassmann numbers commutes with any c-number (bosonic' number) or Grassmann number, so it can be considered for most purposes as a c-number. By the way, that's why the action of fermions is also a c-number, as you would expect.

Therefore,

1- you can safely shift $\phi$ by $\bar\psi\psi$, that's just a change of variables.

2- when you shift $\phi$, you do that at $\bar\psi\psi$ constant (because you do the integral over $\phi$ first), so there's no problem with the integration either.

I'm sure mathematicians would find plenty of subtleties (well, they don't even agree that functional integrals exist...), but as a physicist, you're ready to go !

• Could you please explain this or give a reference: "the product of two Grassman numbers commutes with any c-number (bosonic' number) or Grassman number, so it can be considered as a c-number. By the way, that's why the action of fermions is also a c-number, as you would expect."? My problem is if $\theta_i$ and $\theta_j$ are Grassmann variables, I would not say that $\kappa=\theta_i\theta_j$ is a c-number, as $\theta_i\kappa=0$. – akhmeteli Dec 3 '13 at 4:33
• @akhmeteli: Yes sure. What I meant is that $\kappa$ commutes with all Grassmann variables (including $\theta_i$ and $\theta_j$) or c-numbers, so it's 'almost' a c-number in that respect. – Adam Dec 3 '13 at 4:49

It seems that OP is pondering about the notion of supernumbers, and the generalization of Fubini and Tonnelli's theorems for integration over superdomains and supermanifolds. See e.g. this Phys.SE post and the references listed therein for details.

Example: Consider the integral

$$\tag{1} \int\!\mathrm{d}\theta\int\!\mathrm{d}\overline{\theta}\int_{\mathbb{R}}\! \mathrm{d}x~ e^{S(x,\theta,\overline{\theta})}~=~b\sqrt{2\pi},$$

where $x$ is a Grassmann-even variable, and $\theta,\overline{\theta}$ are Grassmann-odd variables, where $$\tag{2} S(x,\theta,\overline{\theta}) ~:=~-\frac{1}{2}x^2 +(ax+b)\overline{\theta}\theta ~=~-\frac{1}{2}(x-a\overline{\theta}\theta)^2 +b\overline{\theta}\theta,$$

and where $a,b\in\mathbb{C}$. We have completed the square wrt. $x$ on the rhs. of eq. (2).

It is an easy exercise to check that the integral (1) does not depend on whether we first integrate over the Grassmann-even variable $x$, and then the Grassmann-odd variables $\theta,\overline{\theta}$; or vice-versa.

Heuristically, the $x$-integration may be view as an integration along a contour $\gamma$ in the space $\mathbb{R}_c$ of Grassmann-even supernumbers. It doesn't matter which integration contour $\gamma$ is used as long as it starts and ends at the same points.

• Correction to the answer (v2): The symbol $\mathbb{R}_c$ should be $\mathbb{C}_c$. – Qmechanic Dec 11 '13 at 15:37

I don't think there are miracles here.

With an interaction term like $\lambda \bar \psi \psi \phi$, ou may always write something like :

$Z(j,\eta, \bar \eta) \sim e^{\large i\int d^4x ~\lambda~ {\frac{\delta}{\delta \eta(x)}\frac{\delta}{\delta \bar \eta(x)}\frac{\delta}{\delta J(x)}}} \int \mathcal{D\phi}\mathcal{D\psi}\mathcal{D \bar \psi} e^{i \int d^4x(\bar\psi S^{-1}\psi+ \phi D^{-1}\phi+\bar \eta \psi+\bar \psi\eta+J\phi)}$

and then taking this expression for $J=\eta=\bar\eta=0$

(Here $S^{-1}$ and $D^{-1}$ are the inverse propagators for fermions and scalars).

Now, you may develop this expression $Z(j,\eta, \bar \eta)$ by powers of $\lambda$, then for each power-of-$\lambda$-term, after having calculate it, make $J=\eta=\bar\eta=0$