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I have a question about a convention from Peskin & Schroeder, namely that $$\int d\theta^{*}\, d\theta \, (\theta \theta^*) = 1,$$ where $\theta$ and $\theta^*$ are independent Grassmann numbers. Always referring to that book, when performing a multiple integral over more than one Grassmann variable, the following convention is adopted

$$\int d\theta \int d\eta \, \eta \theta = 1. $$

Now we define $$\theta = \frac{\theta_1 +i\theta_2}{\sqrt{2}}, \;\;\;\;\;\; \theta^* = \frac{\theta_1 -i\theta_2}{\sqrt{2}}$$

At this point, perhaps naively, I would say that

$$\theta \theta^* = i \theta_2 \theta_1 $$ $$d\theta^*\, d\theta = i\, d\theta_1\, d\theta_2 $$ Then I will find that $$\int d\theta^*\, d\theta \, (\theta \theta^*) = - \int d\theta_1\, d\theta_2 \, \theta_2 \theta_1 =-1, $$ which differs by a sign from the adopted convention. So, what am I doing wrong?

It's probably a stupid question, but I can't understand the mistakes I'm making. I hope you understand, despite my bad English.

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2 Answers 2

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Integrating Grassman numbers does not work in a naive way. It is exactly opposite to the naive way. Consider $$ \int d\eta\, \eta =1 $$ Under the change of variables $\eta=2\eta_1$, the top expression would naively become $$ \int d(2\eta_1)\, (2\eta_1) =4\int d\eta_1 \eta_1=4\neq 1. $$ So, to make grassman integrals consistent under the change of variables $\eta=A\eta'$ (here $\eta=(\eta_1, \eta_2,...,\eta_n)$), one has to divide the integral by the Jacobian, instead of multiplying, unlike in the case of regular integration. In other words $$ \int d\eta_1 d\eta_2... d\eta_n f(\eta) =\frac{1}{\det{A}}\int d\eta_1'd\eta_2'...d\eta_n' f(A\eta'). $$ In your case it would mean $$ d\theta^*d\theta\rightarrow\frac{1}{i}d\theta_1d\theta_2. $$

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The answer already posted by Pavlo B. is not wrong, but I think it obscures the most important element of the explanation.

You have implicitly assumed that you can simply apply a linear operation to make a differential out of a complex Grassmann number. However, this is not correct. Consider, just $$1=\int d\theta\,\theta$$ for a complex-valued Grassmann variable $$\theta=\frac{\theta_{1}+i\theta_{2}}{\sqrt{2}}.$$ It is natural to expect that the integration measure $d\theta$ should be simply $$d\theta=\frac{d\theta_{1}+i\,d\theta_{2}}{\sqrt{2}},$$ but this is wrong!

That this cannot be correct can be seen simply by expanding out the expression that would result if this were the correct $d\theta$: $$\frac{1}{2}\int(d\theta_{1}+i\,d\theta_{2})(\theta_{1}+i\theta_{2})=\frac{1}{2}\left[\int d\theta_{1}\,\theta_{1}+i\int d\theta_{1}\,\theta_{2}+i\int d\theta_{2}\,\theta_{1}+i^{2}\int d\theta_{2}\,\theta_{2}\right].$$ The cross terms with $\int d\theta_{1}\,\theta_{2}$ and $\int d\theta_{2}\,\theta_{1}$ are zero, so what is left is $$\frac{1}{2}\int(d\theta_{1}+i\,d\theta_{2})(\theta_{1}+i\theta_{2})=\frac{1}{2}\left[\int d\theta_{1}\,\theta_{1}-\int d\theta_{2}\,\theta_{2}\right]=0.$$ Clearly, something has gone wrong.

In fact, in order to have $\int d\theta\,\theta=1$, it must instead be the case that $$d\theta=\frac{d\theta_{1}-i\,d\theta_{2}}{\sqrt{2}}.$$ There is an extra complex conjugate that appears in this expression. (Alternatively, the differential marker $d$ acts anti-linearly on complex-valued Grassmann numbers.) The extra minus sign in $d\theta$ flips the sign of the term with the $i^{2}$ in the preceding calculation, which changes it to a correct expression. Moreover, this is exactly what was needed to fix the sign problem you noticed. Combining the correct $d\theta$ with its complex conjugate, $$d\theta^{*}=\frac{d\theta_{1}+i\,d\theta_{2}}{\sqrt{2}}$$ (which obviously satisfies $\int d\theta^{*}\,\theta^{*}=1$), you get, instead of what you had in your question, $$d\theta^{*}\,d\theta=-i\,d\theta_{1}\,d\theta_{2},$$ and the extra negative sign fixes things.

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  • $\begingroup$ Interesting argument, but there is a problem. Your method of complex conjugation works only for unitary rotations among the grassman variables and does not work for a change of variables $\eta=2\eta'$. I personally was never able to make sense of Grassman integral as an integral with a measure. From my experience treating it as an integral over some measure was more misleading than helpful $\endgroup$
    – Pavlo. B.
    Commented Apr 29, 2021 at 8:54

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