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Would anyone be able to tell me how srednicki goes from step $(44.29)$ to $(44.30)$?

Here is the paragraph:

Now let us introduce the notion of complex Grassmann variables via $$\begin{align} \chi &= \frac{1}{\sqrt{2}}(\psi_1 + i \psi_2), \\ \bar\chi &= \frac{1}{\sqrt{2}}(\psi_1 - i \psi_2). \end{align}\tag{44.28}$$ We can invert this to get $$\begin{bmatrix} \psi_1 \\ \psi_2 \end{bmatrix}= \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1 \\ i & -i \\ \end{bmatrix}\begin{bmatrix} \bar\chi \\ \chi\end{bmatrix}.\tag{44.29}$$ The determinant of this transformation matrix is $-i$, and so $$d^2 \psi = d\psi_2 d\psi_1 = (-i)^{-1} d\chi d\bar\chi\tag{44.30}.$$ Also, $\psi_1 \psi_2 = -i \bar\chi\chi$.

When I compute the matrix given by $(44.29)$ I get that $$\psi_1 = \frac{1}{\sqrt{2}} (\bar\chi + \chi),$$ $$\psi_2 = \frac{1}{\sqrt{2}}i (\bar\chi - \chi).$$

Therefore I get that $$\psi_1 \psi_2 = \frac{1}{2} i ({\bar\chi}^2 - \chi^2) $$ and a similar form for the product of the derivatives... What am I missing, other than a fully functioning brain?

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    $\begingroup$ While I don't really know what Grassman numbers are, I do know that they anticommute. So be careful when expanding $(\bar{\chi} + \chi)(\bar{\chi}-\chi)$. $\endgroup$
    – Javier
    Nov 26 '14 at 12:53
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It's pretty much what Javier Badia said in the comments: Grassmann numbers anticommute.

$$\chi_1 \chi_2 = -\chi_2 \chi_1\tag{1}$$

or in this case, $\chi\bar\chi = -\bar\chi\chi$. Note that this implies the square of any Grassman number is zero, if you set $\chi_1 = \chi_2 = \chi$ in equation (1). Using these properties and some very careful algebra, you can show that

$$(\bar\chi + \chi)(\bar\chi - \chi) = -2\bar\chi\chi$$

When it comes to the differentials, you can still use $\mathrm{d}(\chi_1 + \chi_2) = \mathrm{d}\chi_1 + \mathrm{d}\chi_2$, because that doesn't rely on multiplication. Then the same careful algebra should get you to equation (44.30).

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What you are missing is that you are calculating the Jacobian, not simply multiplying $d\psi$ by $d\bar\psi$. The determinant also goes downstairs instead of upstairs, because that's how Grassmann numbers roll.

See http://en.m.wikipedia.org/wiki/Berezin_integral for details.

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  • $\begingroup$ I would add more but typing on my phone is murder. $\endgroup$ Nov 26 '14 at 13:14

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