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To do perturbation analysis of Supersymmetric Quantum Mechanical Hamiltonian, the superpotential is first scaled by a constant $\lambda >> 1$ and then expanded about it's critical point. Finally the Hamiltonian looks like (from equation 10.157 of Mirror Symmetry) as, $$ H = \lambda \{\dfrac{1}{2}\tilde{p}^2 + \dfrac{1}{2}h''(x_c)^2 (\tilde{x} - \tilde{x_c}^2) + \dfrac{1}{2} h''(x_c)[\overline{\psi}, \psi] \} + \lambda^{1/2} (...) + \lambda^{-1/2} (..)$$ where $\tilde{x} = \sqrt{\lambda} x$.
To the zeroth order, the Hamiltonian remains a Supersymmetric Harmonic oscillator, and hence the ground state can be easily written as $$ \psi_c = e^{-\frac{\lambda}{2}h''(x_c)(x-x_c)^2}|0\rangle \;\;\; ; h'' > 0$$

$$ \psi_c = e^{-\frac{\lambda}{2}h''(x_c)(x-x_c)^2} \overline{\psi}|0\rangle \;\;\; ; h'' < 0 $$.

The author claims that the $\psi_c$ remains a ground state for $\textit{all orders}$ in the perturbation series. On performing the first order correction, I came up with an integral, $$ \int e^{-h'' \eta^2} h'' h''' \eta^3 \;\;\; \text{and} \int e^{-h'' \eta^2} h''' \eta [\overline{\psi},\psi] $$ which on evaluation will be zero because they odd functions. But in second order correction, integrals $\propto \eta^2 $ and $\propto \eta^4 $ appears which are non-zero. So I am not able prove that the correction will remain zero for all orders. Is there any argument to say so ?

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Isn't it true just because the Hamiltonian only contains terms with at least one annihilation operator on the right side and it is normal-ordered? Note that the ground state is uniquely determined by its being annihilated by the appropriate annihilation operators. – Luboš Motl Apr 28 '13 at 5:47
@LubošMotl : Sir, do you mean if I expand $Q$ perturbatively in $\lambda$, and determine ground state upto some order 'n',and then take the Hamiltonian upto that order, and find that expectation value is zero. I understand that I can find uniquely the ground state by demanding it to be annihilated by Q, therefore the above result should hold true. I was just curious if I would be able to show it in various orders of perturbation. – Jaswin Apr 28 '13 at 7:02
As far as I know, it is only true that the ground energy is zero for all orders in $\lambda$. As you say, you can expand $Q$ and find the annihilated state, and this state will vary with the order. Well, I may be wrong, then try to do some integration by parts to cancel them. – Peter Kravchuk Apr 28 '13 at 8:17

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