Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 ?

share|cite|improve this question
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. – LuboŇ° Motl Apr 28 '13 at 5:47
@LuboŇ°Motl : 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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.