A different type of Gaussian Grassmann Integral

Let $A$ be an antisymmetric matrix.

Usually, one proves that for a Grassmann integral of the type, $$\int d\psi d\theta \exp( \psi^T A \theta) = \det(A)$$ where $\psi$ and $\theta$ are vectors of Grassmann variables.

I have a different problem which I feel should trivially reduce to the same, but it can't figure it out:

$$\int d\psi d\theta \exp( 1/2 \psi^T A \psi + 1/2 \theta^T A \theta)$$

I think it should be possible to show that my expression is equivalent to the upper one, but the necessary variable substitution escapes me.

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The statement in the question(v1) would in general need the assumption that the matrix $A=-A^T$ is antisymmetric. – Qmechanic Jan 27 '12 at 22:03
Ah, okay. I know the version below doesn't require this assumption because the symmetric part of $A$ gives 0 contribution to the Grassmann integral anyway – Lagerbaer Jan 27 '12 at 22:05
In the first Grassmann integral of the question(v1) the symmetric part of $A$ contributes non-trivially. – Qmechanic Jan 27 '12 at 22:08
Okay. From this I take that I cannot transform the version 2 into version 1. But with the answer RE the Pfaffian that doesn't matter. – Lagerbaer Jan 27 '12 at 22:11

The second integral factors into the product of integrals over $\psi$ and $\theta$. In your notation, you want to show that the $\int d\psi \mbox{exp}(1/2 \psi^TA \psi) = \mbox{Pfaff}(A)$, the Pfaffian of $A$. Then use the fact that $\mbox{Pfaff}(A)^2 = \mbox{det}(A)$.

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Ah, I think I already did that, I just had no clue what a Pfaffian was, and got stuck at proving Pfaff(A)^2 = det(A). – Lagerbaer Jan 27 '12 at 22:01
...where the answer(v1) has implicitly used at one point that $A$ is antisymmetric, either in the definition of the Pfaffian, or in the very last equality... – Qmechanic Jan 28 '12 at 9:43

Define the Grassman variables $\sqrt{2}\eta=\psi+\phi$ and $\sqrt{2}\chi=\psi-\phi$.

Then $$\eta^{T} A \chi = {1\over 2} (\psi+\phi)^{T} A (\psi - \phi) = {1\over 2}(\psi^T A\psi - \phi^T A \phi)$$

Since the middle terms cancel by antisymmetry of A and the Grassmann property. The result proves the equality of the two actions, so that the determinant of A is equal to the product of Pfaffians.

This is not quite what you wanted--- there is a minus sign in the above. But the minus sign only gives an overall minus to the whole expression if the dimension of A is odd, and it is necessary.

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Thanks, that's a good idea. How about I define them as $\psi + i\phi$ and $\psi - i\phi$? Then, the middle terms should still cancel, but I'll get a $+$ sign. Also, for an antisymmetric matrix of odd dimension, the determinant is always 0, no? – Lagerbaer Jan 29 '12 at 18:06
@Lagerbaer: Yes, the "i" works in the action, but it changes the sign of the measure in the odd dimension case (because the $d\eta$ s get an "i" factor, so it is equivalent to what I did above. – Ron Maimon Jan 30 '12 at 2:21