Source:Mirror Symmetry p.198

I have the Hamiltonian $$H = \lambda\bigg( \frac{1}{2} \tilde{p} + \frac{1}{2}h''(x_i)^2(\tilde{x}-\tilde{x_i})^2 + \frac{1}{2}h''(x_i)[\overline{\psi}, \psi] \bigg) + \lambda^{1/2} (\cdots) + (\cdots) + \mathcal{O}(\lambda^{-1/2})\tag{10.157}$$

where the leading term is

$$H_0 = \frac{1}{2}p^2 + \frac{\lambda^2}{2} h''(x_i)^2 (x-x_i)^2 + \frac{\lambda}{2}h''(x_i) [\overline{\psi}, \psi]\tag{10.158}$$

which is the Hamiltonian of the supersymmetric harmonic oscillator and therefore the ground states are:

$$\Psi_i = e^{-\frac{\lambda}{2}h''(x_i)(x-x_i)^2} \left\vert 0\right> + \cdots \ \text{if} \ h''(x_i) > 0\tag{10.159}$$ $$\Psi_i = e^{-\frac{\lambda}{2}\vert h''(x_i)\vert(x-x_i)^2} \overline{\psi} \left\vert 0\right> + \cdots \ \text{if} \ h''(x_i) < 0,\tag{10.160}$$ where the +... represent the subleading terms in $\lambda^{-1/2}$.

And now to my question: The author says that one can find the subleading terms so that the energy is strictly zero to all orders in $\lambda^{-1/2}$, by inserting the expansion of the potential $$h(x) = h(x_i) + \frac{1}{2 \lambda}h''(x_i) (x-x_i)^2 +...$$ into $e^{-\frac{\lambda}{2}h} \left\vert 0\right>$ or $e^{\frac{\lambda}{2}h} \overline{\psi} \left\vert 0\right>$ but i could not see it.


1 Answer 1


This is explained on p. 193-194. For $h^{\prime\prime}(x_i)\neq 0$ there is precisely one ground state (i.e. normalizable wave function with zero energy.) It is always a Gaussian function. The signature$^1$ of $h^{\prime\prime}(x_i)$ determines whether the ground state is proportional to $ \left\vert 0\right>$ or to $\overline{\psi} \left\vert 0\right>$.


$^1$ The square of $h^{\prime\prime}(x_i)$ in eqs. (10.157) & (10.158) are typos. Also there are implicitly written proportionality constants in eqs. (10.159) & (10.160).


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