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I want to prove an identity from Peskin&Schroeder, namely that $$\left(\prod\limits_i^{} \int d \theta^*_i d\theta_i\right) \theta_m^* \theta_l \exp(\theta_j^* B_{jk} \theta_k)=\det(B) B^{-1}_{ml}$$ $B$ is a hermiteam $N\times N$ matrix and $N$ should be even. $\theta_j$ and $\theta_j^*$ are $N$ complex grassmann numbers.

I would like to do it similar to the case of complex and not grassmann gaussian integrals, where you artificially introduce a term in the exponential and then differentiate. I could prove, that $$\left(\prod\limits_i^{} \int d \theta^*_i d\theta_i\right)\exp(\theta_j^* B_{jk} \theta_k+\eta_j^* \theta_j+\theta_j^*\eta_j)=\det(B)\exp(-\eta_j^*B_{jk}^{-1}\eta_k)$$ where $\eta$ and $\theta$ are (complex) grassmann numbers.If they were just complex variables, the rest would be clear. I could get the first equation differentiating the second with respect to $\eta_m$ and $\eta_l^*$ and then let $\eta=\eta^*=0$. The problem is, that for a grassman number the exponential is $e^\theta=1+\theta$ since all higher order terms cancel. For a sum like we have here, there should also be terms of second order, but nevertheless the exponential is not reproduced by differentiation.

Can I nevertheless obtain the wished result in that way? It looks quite promising, I just don't see, how to come from the second equation (left side) to the first equation (left side). It should work with differentiating, but I don't see why.

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    $\begingroup$ While $e^{\theta} = 1+ \theta$, $e^{\theta+\phi} = 1+ \theta+\phi + \frac{1}{2}\theta\phi$, so be careful. $\endgroup$ – lionelbrits Nov 28 '13 at 1:33
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    $\begingroup$ The "fast" way of doing this calculation is to do a unitary transformation of variables that makes $B$ diagonal, (the integration measure is unchanged). All that remains is to do a product of one-dimensional integrals. $\endgroup$ – lionelbrits Nov 28 '13 at 16:42
  • $\begingroup$ @lionelbrits is it? the unitary transformation introduces the unitary matrix because of $\theta_m$ and $\theta_l$. how do you get rid of those? $\endgroup$ – Yossarian Apr 14 '15 at 10:39
  • $\begingroup$ Thanks, silv. My comment is wrong. I didn't think of the free indices. $\endgroup$ – lionelbrits Apr 15 '15 at 12:47
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EDIT

The method you want to use is ok, and gives a quick result. Here it is:

$$I=\int\prod d\theta^{*}d\theta\theta_{k}^{*}\theta_{l}exp(\theta^{*}B\theta+\eta^{*}\theta+\theta^{*}\eta)=\left(\frac{\partial}{\partial\eta_{k}^{*}}\right)\left(\frac{\partial}{\partial\eta_{l}}\right)\int\prod d\theta^{*}d\theta exp(\theta^{*}B\theta+\eta^{*}\theta+\theta^{*}\eta)$$

from where it follows that $I$ is equal to

$$\left.I=\mathrm{det}B\left(\frac{\partial}{\partial\eta_{k}^{*}}\right)\left(\frac{\partial}{\partial\eta_{l}}\right)\mathrm{exp}(\eta_{j}^{*}(B^{-1})_{ij}\eta_{i})\right\vert_{\eta^{*}=\eta=0}$$

$$I=\mathrm{det}B(B^{-1})_{kl}$$

Note: I use the shift $\theta_{i}\rightarrow\theta_{i} +(B^{-1})_{ij}\eta_{j}$ and $\theta_{i}^{*}\rightarrow\theta_{i}^{*} +\eta_{j}^{*}(B^{-1})_{ji}$. Here I used the fact that the second moments can be calculated as derivatives in the following way.

$$\left.\langle\eta_{l}\eta_{k}^{*}\rangle=\left(\frac{\partial}{\partial\eta_{k}^{*}}\right)\left(\frac{\partial}{\partial\eta_{l}}\right)\mathrm{exp}(\eta_{j}^{*}(B^{-1})_{ij}\eta_{i})\right\vert_{\eta^{*}=\eta=0}=(B^{-1})_{ij}$$

More or less, you can take this as a definition. But if you still want to prove this, do the following:

$$J=\left(\frac{\partial}{\partial\eta_{k}^{*}}\right)\left(\frac{\partial}{\partial\eta_{l}}\right)\mathrm{exp}(\eta_{j}^{*}(B^{-1})_{ij}\eta_{i})=\left(\frac{\partial}{\partial\eta_{k}^{*}}\right)\left(\frac{\partial}{\partial\eta_{l}}\right)\prod_{ij}(1-\eta_{j}^{*}(B^{-1})_{ij}\eta_{i})$$

After straight forward differentiation you get

$$J=\prod_{ij}\delta_{kj}\delta_{li}(B^{-1})_{ij}=(B^{-1})_{kl}$$

You can ignore whats below the line, that was my first answer. But I'll leave it because it may be instructive for others.



Since I cannot comment yet, i will sketch a proof for a simpler case

$$I=\int\prod_{i}d\theta_{i}^{*}d\theta_{i}exp(\theta_{i}^{*}B_{ij}\theta_{j})=det(B)$$

and hope that this will help you compute your integrals. After expanding the all the exponential we arrive at

$$I=\frac{1}{N!}\int d\theta_{1}^{*}d\theta_{1}\dots d\theta_{N}^{*}d\theta_{N}(\theta_{i_1}^{*}B_{i_{1}j_{1}}\theta_{j_{1}})(\theta_{i_2}^{*}B_{i_{2}j_{2}}\theta_{j_{2}})\dots (\theta_{i_N}^{*}B_{i_{N}j_{N}}\theta_{j_{N}})$$

At this point some explanations are in order. The factor $1/N!$ appears from the expansion of the exponentials. To see how the integrand was obtained, let's look at the case when $N=2$. We will have something like this

$$\int d\theta_{1}^{*}d\theta_{1}d\theta_{2}^{*}d\theta_{2}\left(1+\theta_{i}^{*}B_{ij}\theta_{j}+\frac{(\theta_{i}^{*}B_{ij}\theta_{j})^2}{2!}+\cdots\right)$$

It is obvious that only the quadratic term will contribute to the above integral, because only this term can saturate the number of grassmann variables in the integral measure.

$$\int d\theta_{1}^{*}d\theta_{1}d\theta_{2}^{*}d\theta_{2}\left(\frac{(\theta_{i}^{*}B_{ij}\theta_{j})^2}{2!}\right)=\frac{1}{2}\int d\theta_{1}^{*}d\theta_{1}d\theta_{2}^{*}d\theta_{2}\sum_{i_{1},i_{2},j_{1},j_{2}}^{2}(\theta_{i_1}^{*}B_{i_{1}j_{1}}\theta_{j_{1}})(\theta_{i_2}^{*}B_{i_{2}j_{2}}\theta_{j_{2}})$$

Expanding the sum and performing all four integrals, we get

$$\int d\theta_{1}^{*}d\theta_{1}d\theta_{2}^{*}d\theta_{2}\left(\frac{(\theta_{i}^{*}B_{ij}\theta_{j})^2}{2!}\right)=\frac{1}{2!}[2(B_{11}B_{22}-B_{12}B_{21})]=detB$$

Now, let us return to our original integral $I$. The next step before performing the integrals is to reorder the integrals and the grassmann numbers.

$$I=\frac{1}{N!}\int d\theta_{1}^{*}\dots d\theta_{N}^{*}\theta_{i_1}^{*}\dots\theta_{i_1}^{*}\int\theta_{1}\dots d\theta_{N}\theta_{j_1}\dots\theta_{j_1}B_{i_{1}j_1}\dots B_{i_{N}j_N}$$

From where we finally arrive at

$$I=\frac{1}{N!}\epsilon_{i_{1}\dots i_{N}}\epsilon_{j_{1}\dots j_{N}}B_{i_{1}j_1}\dots B_{i_{N}j_N}=\mathrm{det}B$$

(note: the ordering result $a_{1}b_{1}\dots a_{N}b_{N}=a_{1}\dots a_{N}b_{1}\dots b_{N}(-1)^{N(N-1)/2}$ has to be used twice for the integrals).

I found this method to be the most simple one of all when dealing with these sort of integrals. I hope this helps you prove those relations. The same method can be applied very easy in your case (just a strait forward extension). And with the method you described, it seams you are over complicating yourself. However, I'll work on it a bit and see what comes up.

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  • $\begingroup$ At what step are you referring to? $\endgroup$ – vnb Nov 28 '13 at 10:00
  • $\begingroup$ The problem is this step. The exponential function in the first integral is only a sum until degree 2 I guess and so I can not calculate with it like with a normal e-function. $$I=\mathrm{det}B\left(\frac{\partial}{\partial\eta_{k}^{*}}\right)\left(-\frac{\partial}{\partial\eta_{l}}\right)\mathrm{exp}(\eta^{*}B^{-1}\eta)$$ For example $$\partial_\theta \exp(\theta)=1 \neq \exp(\theta)$$ $\endgroup$ – TheoPhysicae Nov 28 '13 at 10:03
  • $\begingroup$ And of course also the step before that $$I=\int\prod d\theta^{*}d\theta\theta_{k}^{*}\theta_{l}exp(\theta^{*}B\theta+\eta^{*}\theta+ \theta^{*}\eta)=\left(\frac{\partial}{\partial \eta_{k}^{*}}\right)\left(-\frac{\partial}{\partial\eta_{l}}\right)\int\prod d\theta^{*}d\theta exp(\theta^{*}B\theta+\eta^{*}\theta+\theta^{*}\eta)$$ I'm not sure how to deal with the exponentials. To me it looks, like you just ignored that there are grassmann numbers. $\endgroup$ – TheoPhysicae Nov 28 '13 at 10:12
  • $\begingroup$ One step at a time. In one dimension you would be correct with $\partial_{\theta}\exp(\theta)=1$ but you deal with multiple dimensions. Maybe my notation was sloppy, but in the exponential, I should have wrote indices for $\eta$ and $\theta$. The trick was to take the derivatives with respect to indices $k,l$ not $i,j$ as found in the exponential. And in multiple dimension this holds $\frac{\partial\theta_j}{\partial\theta_i}=\delta_{ij}-\theta_{j} \frac{\partial}{\partial\theta_i}$. So you have to use this when making the derivative. $\endgroup$ – vnb Nov 28 '13 at 10:26
  • $\begingroup$ Yes, but in the integral we have a sum, where every index is contained. So I don't understand, why the exponential function is still there after differentiating. Could you please show how applying your just mentioned rule leads to the result? $\endgroup$ – TheoPhysicae Nov 28 '13 at 10:31

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