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 $\det(A) \ e^{-t^T A^{-1} t}$?


1 Answer 1


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 $\mathbb{E}(e^{tu}) = \det A$.

  • $\begingroup$ 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. $\endgroup$ Commented Oct 20, 2022 at 10:40
  • $\begingroup$ 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. $\endgroup$ Commented Oct 20, 2022 at 14:39
  • $\begingroup$ 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? $\endgroup$ Commented Oct 20, 2022 at 22:51

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