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Consider the following transformation of the integration measure $dX d\psi_1 d\psi_2$: where $\psi_1$ $\psi_2$,$\varepsilon^1$, and $\varepsilon^2$ are Grassman variables.

$\delta_\varepsilon X= \varepsilon^1 \psi_1+ \varepsilon^2 \psi_2$

$\delta\psi_1= \varepsilon^2 \partial h$

and

$\delta\psi_2= -\varepsilon^1 \partial h$. The problem asked to show that the above integration measure is invariant under this transformation. My approach to the first one was to just use the fact that $\int \psi_1 \dots \psi_n d \psi_1 \dots d\psi_n=1$, and relate it to the bosonic "standard" variables which return $\int dX= \int 1 dX=X$. However, for the other two, I know that since $\int \psi d \psi=1$, it should follow easily. But what does the $h$ denote in the differential? and is this argument headed in the right direction?

Many thanks to all of you.

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Is this Homework?? I would expect a dh to be part of the integration measure, with $\delta h=0$.

To show that the integration measure $dX\,d\psi_1\,d\psi_2\,dh$ is invariant, you would need to write out the Jacobian super-determinant:

$$ J=\det\,\begin{pmatrix} \frac{\delta X'}{\delta X} & \frac{\delta X'}{\delta \psi_1} & \frac{\delta X'}{\delta \psi_2}&...\\ \frac{\delta \psi_1'}{\delta X} & \frac{\delta \psi_1'}{\delta \psi_1} & \frac{\delta \psi_1'}{\delta \psi_2}\\ \frac{\delta \psi_2'}{\delta X} & \frac{\delta \psi_2'}{\delta \psi_1} & \frac{\delta \psi_2'}{\delta \psi_2}\\ \vdots \end{pmatrix}\,. $$ Where primes indicate the transformed variables. Then, for small transformations, use $J=\det(1+\mathbb{A})=\exp\ln\det(1+\mathbb{A})\approx 1+\text{Tr}\mathbb{A}+\mathcal{O}(\mathbb{A})^2$. All you need to show is that trace of $\mathbb{A}$ vanishes to show that the measure remains invariant.

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    $\begingroup$ Comment to the answer(v1): The trace should be supertrace. $\endgroup$
    – Qmechanic
    Nov 25, 2012 at 17:46

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