# Gaussian Integral Identity with Grassmann Numbers

1. Feb 22, 2009

### Phileas.Fogg

Hi,
I read the chapter "Anticommuting Numbers" by Peskin & Schröder (page 299) about Grassmann Numbers and now I would like to prove

$$\int d \bar{\theta}_1 d \theta_1 ... d \bar{\theta}_N d \theta_N e^{-\bar{\theta} A \theta} = det A$$

$$\theta_i$$ are complex Grassmann Numbers.

$$\bar{\theta}_i$$ are the complex conjugates of $$\theta_i$$.

In Peskin & Schröder there is no derivation at all, so I tried to find it via google.

In lecture notes I found

But I don't understand, how the author expanded to get the first line. How he used the general ordering result and permutations to get the rest.

Could anybody explain that or refer to a website, with an explicit example, for maybe N=2 ?

Regards,
Mr. Fogg

2. Feb 22, 2009

### CompuChip

At first this calculation may be a bit hard to understand, but it exhibits a lot of features which make calculating with Grassmann numbers very straightforward (and tricks you might want to remember )

So first the author is expanding the exponent. As you know,
$$e^x = 1 + x + x^2 / 2 + \cdots + x^n / n! + \cdots$$
In this case,
$$e^{\theta_i^* A_{ij} \theta_j} = 1 + \theta_i A_{ij} \theta_j + (\theta_i^* A_{ij} \theta_j)^2 / 2 + \cdots + (\theta_i^* A_{ij} \theta_j)^n / n! + \cdots$$

Note that inside each pair of brackets, there is a summation going on. If I write out the second term:
$$(\theta_i^* A_{ij} \theta_j)^2 = \left( \sum_{i, j} \theta_i^* A_{ij} \theta_j \right)^2 = \left( \sum_{i_1, j_1} \theta_{i_1}^* A_{i_1 j_1} \theta_{j_1} \right) \left( \sum_{i_2, j_2} \theta_{i_2}^* A_{i_2j_2} \theta_{j_2} \right) = \sum_{i_1, i_2, j_1, j_2} \left[ \left( \theta_{i_1}^* A_{i_1j_1} \theta_{j_1} \right) \left( \theta_{i_2}^* A_{i_2 j_2} \theta_{j_2} \right) \right]$$

Now note, that when we integrate this term, it vanishes. We have N integrals over a theta, and N over a theta*, but the quadratic term which I wrote out only contains 2 of each. So even if the term doesn't vanish beforehand (because $i_1 = i_2$ for example, so there are two identical Grassmann numbers) we can start calculating the integrals and at some point we will integrate over a theta or theta* which is not in the expression.

On the other hand, we only have N theta's and theta*'s. So if we take the (N+m)th power (with m > 0) and expand the power, then we will get sums looking like (for m = 1)
$$\theta_{i_1}^* A_{i_1j_1} \theta_{j_1} \theta_{i_2}^* A_{i_2j_2} \theta_{j_2} \cdots \theta_{i_N}^* A_{i_Nj_N} \theta_{j_N} \theta_{i_{N+1}}^* A_{i_{N+1}j_{N+1}} \theta_{j_{N+1}}$$
in which there must necessarily be two equal theta's and two equal theta^*'s, so it vanishes.

On the second line, the author is just permuting stuff (pulling each theta through a theta or theta* gives a minus sign, a theta* through a theta or theta* gives a minus sign, and a theta or theta* through an A gives a plus sign - assuming that the matrix is real-valued and not Grassmann-valued) so from line 1 to line 2 you get at most a sign change (which is explained to be + in the text below).

3. Feb 22, 2009

### Phileas.Fogg

Thank You very much for the explanation.

To my better understanding, I try to do the proof for N=2

$$\int d \theta_1^* d \theta_1 d \theta_2^* d \theta_2 (-(1 + \theta_i^* A_{ij} \theta_j + (\theta_i^* A_{ij} \theta_j)^2 / 2 + \cdots + (\theta_i^* A_{ij} \theta_j)^n / n! + \cdots ))$$

Since the quadratic term and all higher powers vanish, I get

$$\int d \theta_1^* d \theta_1 d \theta_2^* d \theta_2 (-(1 + \sum_{i, j} \theta_i^* A_{ij} \theta_j))$$

The Integral over 1 is zero.

$$\int d \theta_1^* d \theta_1 d \theta_2^* d \theta_2 (- \sum_{i, j} \theta_i^* A_{ij} \theta_j) = \int d \theta_1^* d \theta_1 d \theta_2^* d \theta_2 (- \theta_1^* A_{11} \theta_1 - \theta_2^* A_{21} \theta_2 - \theta_1^* A_{12} \theta_2 - \theta_2^* A_{22} \theta_2)$$

Is that correct?

Regards,
Mr. Fogg

4. Feb 22, 2009

### CompuChip

The quadratic term does not vanish. You can explicitly work out the sum for n = 1:
$$\theta_i^* A_{ij} \theta_j = \theta_1^* A_{11} \theta_1 + \theta_1^* A_{12} \theta 2 + \theta_2^* A_{21} \theta_1 + \theta_2^* A_{22} \theta_2$$

Now square that, what do you get?

Also, what happens when you explicitly do your integration
$$\int d \theta_1^* d \theta_1 d \theta_2^* d \theta_2 (- \theta_1^* A_{11} \theta_1 - \theta_2^* A_{21} \theta_2 - \theta_1^* A_{12} \theta_2 - \theta_2^* A_{22} \theta_2)$$

Last edited: Feb 22, 2009
5. Feb 22, 2009

### Phileas.Fogg

Using $$(\theta_i)^2 = (\theta_i^*)^2 = 0$$

I get
$$(\theta_1^* A_{11} \theta_1) (\theta_2^* A_{22} \theta_2) + (\theta_1^* A_{12} \theta_2)(\theta_2^* A_{21} \theta_1) + (\theta_2^* A_{21} \theta_1)(\theta_1^* A_{12} \theta_2) + (\theta_2^* A_{22} \theta_2)(\theta_1^* A_{11} \theta_1)$$

But isn't this the case N = 2 , because i and j run till 2 ?

I will do this integration tomorrow morning, because I have to leave now.

Mr. Fogg

6. Feb 23, 2009

### CompuChip

Precisely.

Now observe how you have to be careful with sums. A simple expression like $\theta_1 \theta_2^*$ will vanish when squared, but sums of such terms do not necessarily (although stuff will cancel out).

It's really easy, you'll see.
Don't bring out the cigar yet, though... the square term is what you are interested in (see the general proof).

Welcome.

7. Feb 23, 2009

### Phileas.Fogg

The integration
$$\int d \theta_1^* d \theta_1 d \theta_2^* d \theta_2 (- \theta_1^* A_{11} \theta_1 - \theta_2^* A_{21} \theta_2 - \theta_1^* A_{12} \theta_2 - \theta_2^* A_{22} \theta_2)$$
gives
$$- \int d \theta_1^* d \theta_1 (\theta_1^* A_{11} \theta_1) - \int d \theta_2^* d \theta_2 (\theta_2^* A_{21} \theta_2) - \int d \theta_1^* d \theta_2 (\theta_1^* A_{12} \theta_2) - \int d \theta_2^* d \theta_2 (\theta_2^* A_{22} \theta_2)$$

$$A_{11} + A_{21} + A_{12} + A_{22}$$

There must be a mistake, because I don't see, how this helps me.
-----------------------------------------------------
If I work out the quadratic term explicitly, square it and do the integration (like above), I get

$$A_{11} A_{22} - A_{12} A_{21} + A_{21} A_{12} + A_{22} A_{11}$$

This looks a little more like a determinant, but there are some terms too much in this expression. Where is the mistake here?

Regards,
Mr. Fogg

8. Feb 23, 2009

### samwise

Remember that you are also integrating over the variables not in the expression. For example, the first term is actually
$$\int d \theta_1^* d \theta_1 d \theta_2^* d \theta_2 (\theta_1^* A_{11} \theta_1) = - A_{11} \int d \theta_1^* \theta_1^* d \theta_1 \theta_1 \int d \theta_2^* d \theta_2 = 0$$
since
$$\int d \theta_2 = 0 = \int d \theta_2^*$$
so you get zero for all those terms, and the only contributions to the integral come from the quadratic terms (where you can get terms that include all 4 variables and do not vanish). Higher than quadratic terms, in this case, contain squares and therefore vanish.

$$\theta_2^* A_{21} \theta_2$$ should read $$\theta_2^* A_{21} \theta_1$$

For the quadratic term, you should get
$$2(A_{11} A_{22} - A_{12} A_{21}) = 2 detA$$ where the $$2$$ cancels the factor $$\frac{1}{2!}$$ brought down with the expansion of the exponential. The sign difference you have on your $$A_{21} A_{12}$$ could be due to your summation where you have $$\theta_2^* A_{21} \theta_2$$ instead of $$\theta_2^* A_{21} \theta_1$$.

Regards,
samwise.

Last edited: Feb 23, 2009
9. Feb 23, 2009

### CompuChip

Note that in doing the integration, you need to re-order the terms. I don't readily see whether you took that into account, but for example:
$$\int d \theta_1^* d \theta_1 d \theta_2^* d \theta_2 \theta_1^* A_{12} \theta_2 \theta_2^* A_{21} \theta_1 = A_{12} A_{21} \int d \theta_1^* d \theta_1 d \theta_2^* d \theta_2 \theta_1^* \theta_2 \theta_2^* \theta_1 = (-1)^{2+1} A_{12} A_{21} \int d \theta_1^* d \theta_1 d \theta_2^* d \theta_2 \theta_1^* \theta_1\theta_2^*\theta_2 = - A_{12} A_{21}$$
where the factors of (-1) arise from ordering the theta's.
Which way you should order them depends on how Grassmann numbers are introduced, you can require that the theta's be in the same order as the integrals, so
$$\int d\theta_1 d\theta_2 \cdots d\theta_n \theta_1 \theta_2 \cdots \theta_n = 1$$
or that each theta has to be brought to its corresponding integration, so
$$\int d\theta_1 d\theta_2 \cdots d\theta_n \theta_n \theta_{n-1} \cdots \theta_2 \theta_1 = 1$$
(first do the inner integration, etc).

This will produce some minus signs. Then the result can be simplified, because the A's are just numbers, so
$$A_{11} A_{22} + A_{22} A_{11} = 2 A_{11} A_{22}$$.

10. Feb 24, 2009

### Phileas.Fogg

Hello,

In the quadratic term, I didn't make the same mistake with the indices.

To do the integral of the quadratic term, I used

$$\int d \theta^* d \theta \theta^* \theta = -1$$

Then I get

$$- \frac{1}{2} (A_{11}A_{22} - A_{12}A_{21} - A_{21}A_{12} + A_{22}A_{11})$$

The minus sign is left from the expansion.

Is $$A_{12}A_{21} = A_{21}A_{12} \; \text{and} \; A_{11}A_{22} = A_{22}A_{11}?$$

If that was true and I didn't have the minus sign, then it would be Samwise's solution.

What is right now?

Regards,
Mr. Fogg

11. Feb 25, 2009

### Avodyne

Yes, the elements of the A matrix are assumed to be commuting numbers.
The overall sign doesn't matter; it depends on the convention for ordering the differentials.