An Identity for a Gaussian Grassmann integral from Wikipedia

I found this identity on Wikipedia: $$\int\exp\left[\theta^T A\eta+\theta^T J+K^T\eta\right]d\theta d\eta =\det A\exp\left[-K^TA^{-1}J\right],$$

where the integration variables are Grassmann variable, and $A$ is an invertible matrix. Unfortunately the Wikipedia page gives little context, and I'm not clear if $K$ and $J$ are also supposed to be Grassmann variables.

Does anybody know how to prove this identity?

• Duplicate of this physics.stackexchange.com/q/20106 Commented Nov 14, 2017 at 13:55
• Possible duplicate of A different type of Gaussian Grassmann Integral Commented Nov 14, 2017 at 13:55
• $J$ and $K$ are $n$-dimensional sources, while $\theta$ and $\eta$ are spinor/fermionic fields. You might might want to consider deriving the simpler relation with one Grassmann field and source before proving this. e.g. $\int d\theta \exp\left[-\frac{1}{2}\theta^T Ax+ J^T\theta\right]$. Commented Nov 14, 2017 at 16:18
• ^$\int d\theta \exp\left[-\frac{1}{2}\theta^T A\theta+ J^T\theta\right]$ Commented Nov 14, 2017 at 16:26
• Commented Apr 11 at 17:11

Hint: Complete the square:

\begin{align} &\int\!d^n\theta ~d^n\eta\exp\left[\theta^T A\eta+\theta^T J+K^T\eta+K^TA^{-1}J\right] \cr ~=~& \int\!d^n\theta ~d^n\eta\exp\left[(\theta^T+K^TA^{-1})A(\eta+A^{-1}J)\right] \cr ~=~& \int\!d^n\theta ~d^n\eta\exp\left[\theta^TA\eta\right]\cr ~=~&\det A,\end{align} where the last/third equality is modulo sign conventions. In the second equality is used that Berezin integration is translation invariant.

$J$ and $K$ are $n$-dimensional sources, while $\theta$ and $\eta$ are spinor fields.

I can't provide a complete proof right now, but you might want to take a look here for an explicit calculation of the the analogous case for bosonic fields, which is much simpler.

I'll outline a heuristic derivation of the simpler case of only one spinor field with a single source $J$. In that case the relevant integral is

$\int d\theta \exp\left[-\frac{1}{2}\theta^T A\theta+ J^T\theta\right]$

where $\theta$ is understood to be a vector of $n$ Grassmann variables.

Since Grassmann integrals are invariant with respect to translation of the Grassmann integration variable, it would be nice if there was a transformation that would make the second term independent of $\theta$. Such a transformation is given by

$\theta \rightarrow (\theta - A^{-1}J)^T$,

where we assume that $A$ is anti-symmetric, $A^T = -A$.

Plugging this into the integrand and crunching through the algebra (mostly multiplication and recalling that we also have $J_i\theta_k = - \theta_kJ_i$) yields (up to sign errors)

$\int d\theta \exp\left[-\frac{1}{2}\theta^T A\theta+ J^T\theta\right] = \int d\theta \exp\left[-\frac{1}{2}\theta^T A\theta+ J^TA^{-1}J\right]$.

The second part of the exponent can come outside the integral, while the remaining part can be shown to be equal to the square root of the determinant of $A$.

If the above seems a bit flaky I would appreciate input from someone better acquainted with Grassmann integrals.

• Comment to the answer (v1): OP's identity holds even if $A$ is not anti-symmetric. Commented Nov 14, 2017 at 21:55
• Ah ok. That isn’t obvious to me. Could you outline why? My answer is presumably lacking then. @Qmechanic Commented Nov 15, 2017 at 8:09
• I wrote an answer. Commented Nov 17, 2017 at 14:52