Why doesn't this multiplication of Grassmann variables give the expected result?

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?

• While I don't really know what Grassman numbers are, I do know that they anticommute. So be careful when expanding $(\bar{\chi} + \chi)(\bar{\chi}-\chi)$. Nov 26 '14 at 12:53

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).

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.

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