Integrals over Grassmann numbers

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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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TheoPhysicae
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Updated on November 15, 2020

Comments

  • TheoPhysicae
    TheoPhysicae almost 3 years

    I want to prove an identity from Peskin&Schroeder, namely that $$\left(\prod\limits_i^{} \int d \theta^*_i d\theta_i\right) \theta_m \theta_l^* \exp(\theta_j^* B_{jk} \theta_k)=\det(B) B^{-1}_{ml}\tag{9.70}$$ $B$ is a hermitean $N\times N$ 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 Grassmann 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.

    • lionelbrits
      lionelbrits almost 10 years
      While $e^{\theta} = 1+ \theta$, $e^{\theta+\phi} = 1+ \theta+\phi + \frac{1}{2}\theta\phi$, so be careful.
    • lionelbrits
      lionelbrits almost 10 years
      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.
    • Yossarian
      Yossarian over 8 years
      @lionelbrits is it? the unitary transformation introduces the unitary matrix because of $\theta_m$ and $\theta_l$. how do you get rid of those?
    • lionelbrits
      lionelbrits over 8 years
      Thanks, silv. My comment is wrong. I didn't think of the free indices.
    • MadMax
      MadMax almost 4 years
      @lionelbrits, "$e^{\theta+\phi} = 1+ \theta+\phi + \frac{1}{2}\theta\phi$", isn't it $e^{\theta+\phi} = 1+ \theta+\phi$? In your expression $e^{\theta+\phi} \neq e^{\phi+ \theta}$.
  • vnb
    vnb almost 10 years
    At what step are you referring to?
  • TheoPhysicae
    TheoPhysicae almost 10 years
    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 $$\partial_\theta \exp(\theta)=1 \neq \exp(\theta)$$
  • TheoPhysicae
    TheoPhysicae almost 10 years
    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.
  • vnb
    vnb almost 10 years
    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.
  • TheoPhysicae
    TheoPhysicae almost 10 years
    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?
  • vnb
    vnb almost 10 years
    Maybe this link will help you. Look at formula (104), but notice that they differentiate with respect to the same indices. Also that they take the derivatives at zero. Not as I did, taking the limit in the last result. workspace.imperial.ac.uk/theoreticalphysics/public/MSc/AQFT/‌​…
  • vnb
    vnb almost 10 years
    If you want, I can re edit my answer, according to the definition of the second moments.
  • TheoPhysicae
    TheoPhysicae almost 10 years
    That would be really nice.
  • vnb
    vnb almost 10 years
    @Wessi990 I've re edited my answer, I hope it's fine now.