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I am currently stuck with the following partition function

Let the action be $$S(X, \psi^1, \psi^2) = \frac{1}{2} (\partial h)^2 - \partial^2h\psi^1 \psi^2 ,\tag{9.29}$$ where $h$ is a real function of the bosonic variable $X$, $\partial h = h'$ and $ \psi^i$ are Grassmann odd variables. The action is invariant under $$ \delta X = \epsilon^1 \psi_1 + \epsilon^2 \psi_2$$ $$ \delta \psi_1 = \epsilon^2 \partial h,\tag{9.30}$$ $$ \delta \psi_2 = -\epsilon^1 \partial h,$$ with $\epsilon^i$ Grassmann odd variables.

Now Suppose $\partial h$ is nowhere $0$, i have to show that the corresponding partition function $$ Z = \int dX d\psi^1 d\psi^2 e^{-S}= 0.\tag{9.31}$$

After a change of variables $$ \hat{X} = X - \frac{\psi^1 \psi^2}{\partial h}$$ $$ \hat{\psi}^1= \alpha(X) \psi^1\tag{9.32}$$ $$ \hat{\psi}^2= \psi^1 + \psi^2,$$ such that $$S(X,\psi^1,\psi^2) = S(\hat{X}, 0,\hat{\psi}^2),\tag{9.33}$$ with $\alpha$ an arbitrary function of $X$, and some Grassmann integration, i get that

$$Z = -\int e^{-S(\hat{X}, 0, \hat{\psi}^2) } \frac{\partial^2h}{(\partial h)^2} \hat{\psi}^1 \hat{\psi}^2 d \hat{X} d \hat{\psi}^1 d \hat{\psi}^2\tag{9.35}$$

which should be zero. But i can not see why? I got the hint that the integrand is a total derivative in $ \hat{X}$ but that didn't help me.

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  • $\begingroup$ I do not understand what you mean by $\partial^2{h}{(\partial h)^2}$, this should be zero. $\endgroup$
    – lakehal
    Commented May 17, 2019 at 15:15
  • $\begingroup$ Possible duplicate: physics.stackexchange.com/q/376564/2451 , physics.stackexchange.com/q/324024/2451 $\endgroup$
    – Qmechanic
    Commented May 17, 2019 at 23:44
  • $\begingroup$ It is not quite a duplicate since i can not see why this last integral is a total derivative. $\endgroup$
    – Elskrt
    Commented May 18, 2019 at 7:44
  • $\begingroup$ Can we reopen this question please? I was just about to ask this question myself and I don't think the linked "answer" actually answers it. $\endgroup$
    – Gleeson
    Commented Oct 25, 2023 at 12:27
  • $\begingroup$ Is there something in particular that needs to be clarified? $\endgroup$
    – Qmechanic
    Commented Oct 26, 2023 at 7:10

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