# Contradictory results for Berezin integral

Say $$u = (u_1, \dots, u_{2n}) = (\xi_1, \eta_1, \dots \xi_n, \eta_n)\tag{1}$$ is a vector of Grassmann variables. For an antisymmetric bosonic matrix $$A$$ we know that $$\int e^{\frac{1}{2} \sum_{a,b} A_{ab}(\xi_a\eta_b - \eta_a\xi_b)} \prod_{i=1}^n d\xi_i d\eta_i = \det(A) \ .\tag{2}$$

Now, I need to calculate $$\mathbb E\left[e^{\langle t,u\rangle}\right]$$, with $$t = (t_1,\dots,t_{2n})$$ is bosonic; that is, $$\mathbb E\left[e^{\langle t,u\rangle}\right] = \int e^{\langle t,u\rangle} \ e^{\frac{1}{2} \sum_{a,b} A_{ab}(\xi_a\eta_b - \eta_a\xi_b)} \prod_{i=1}^n d\xi_i d\eta_i \ .\tag{3}$$

On the one hand, we know $$e^{\langle t,u\rangle} = 1 + \langle t,u\rangle,\tag{4}$$ in which case $$\mathbb E\left[e^{\langle t,u\rangle}\right] = \int e^{\frac{1}{2} \sum_{a,b} A_{ab}(\xi_a\eta_b - \eta_a\xi_b)} \prod_{i=1}^n d\xi_i d\eta_i + \sum_{j=1}^{2n} t_j \int u_j \ e^{\frac{1}{2} \sum_{a,b} A_{ab}(\xi_a\eta_b - \eta_a\xi_b)} \prod_{i=1}^n d\xi_i d\eta_i$$ $$= \det(A) + \sum_{j=1}^{2n} t_j \langle u_j\rangle = \det(A)\tag{5}$$ since $$\langle u_j\rangle = 0$$. On the other hand, $$\xi_a \eta_b$$ behaves as a commuting variable (since it's a product of two Grassmann). So my guess is that I can put the variables $$u_i$$ above in the exponent, obtaining $$\mathbb E\left[e^{\langle t,u\rangle}\right] = \int e^{\sum_{j=1}^{2n} t_j u_j + \frac{1}{2} \sum_{a,b} A_{ab}(\xi_a\eta_b - \eta_a\xi_b)} \prod_{i=1}^n d\xi_i d\eta_i \ .\tag{6}$$ Using the last property here, we know this is equal to $$\det(A) \ e^{-t^T A^{-1} t}$$.

So, which one is the correct result for $$\mathbb E\left[e^{\langle t,u\rangle}\right]$$? $$\det(A)$$ or $$\det(A) \ e^{-t^T A^{-1} t}$$?

It is only true that $$\exp(a) = 1+a$$ if $$a$$ is a Grassman variable. A product of two Grassman variables is not Grassman.

For example, consider (with Grassman $$a,b,c,d$$)

$$\exp(ab+cd) = 1 + ab + cd + \frac{1}{2}(2abcd) + ...$$

In fact, this must be the case or else we would have

$$\int \exp\left(-\frac{1}{2}A_{ab}(\xi_a\eta_b - \eta_b\xi_a) \right)\prod d\xi d\eta = \int 1 - \frac{1}{2}A_{ab}(\xi_a\eta_b - \eta_b\xi_a) \prod d\xi d\eta$$

which is zero.

I believe the source of your issues is that the change of variables $$\eta \to \eta + t$$ only makes sense if $$t$$ is in the Grassman algebra - if it is an ordinary complex number, this operation is not defined. (see the Appendix A.3 of this paper).

However, the Grassmans are still a $$\mathbb{C}$$-vector space, so $$tu$$ is well defined for $$t\in\mathbb{C}$$, $$u$$ Grassman.

In that case, since the argument of the exponential is linear in all Grassman variables, $$\exp(\langle t,u \rangle) = 1 + \langle t,u \rangle$$ is correct, and $$\mathbb{E}(e^{tu}) = \det A$$.

• Funny enough, I ran across that paper yesterday. So basically the second derivation in my calculations wouldn't make sense for bosonic $t$. Does that mean that the first calculation (where I obtain the answer $\det(A)$) is the correct one? I know it sounds strange that I need bosonic $t$, but for my research I need to compute the characteristic function of a gaussian vector with fermionic variables, and compare it to the bosonic case. That's why I try and use boson $t$ in both cases. But maybe everything is ill-defined from the beginning. Commented Oct 20, 2022 at 10:40
• No, the second one is true (the Grassman analogy of the Gaussian integral). If the first one were true ($\det A$), then you would have shown that $\langle u_i u_j\rangle \propto \frac{\partial^2}{\partial t_{i} \partial t_j} \det A = 0$, etc. All correllators are exactly zero. Commented Oct 20, 2022 at 14:39
• That's for the case of fermionic $t$'s. But if they were bosonic? Can this calculation even be made? Wouldn't I obtain just $\det(A)$ in that case? Commented Oct 20, 2022 at 22:51