# Precise zero energy bound for supersymmetry

Usually we can shift the energy $$E$$ by any amount $$\delta$$ to redefine the lowest energy as $$E + \delta.$$

However, in supersymmetry, there is a precise $$E=0$$ must be true, so that the supercharge $$Q$$ annihilates some state $$|\psi_{min}\rangle$$ to give the minimal energy $$Q |\psi_{min}\rangle =0$$ and also $$H|\psi_{min}\rangle =Q^2|\psi_{min}\rangle=0.$$

## Question

This raises the question that we have a precise zero energy bound for supersymmetry theory.

Does it mean that we cannot shift the energy $$E$$ to $$E + \delta$$ in supersymmetry theory? What is the deep reason behind it?

If the energy of the ground states of a supersymmetric theory were non exactly zero the underlying geometry of the background would be allowed to be curved in arbitrary ways. The latter is imposible in a supersymmetric theory because supersymmetry enforce at least an spin structure of the underlying background geometry or possibly a trivial canonical bundle or a special holonomy (as in the Calabi-Yau, $$G_{2}$$ or $$Spin(7)$$ cases) on the target manifold.
Supersymmetry in quantum mechanics: The hamiltonian in a $$(0+1)$$-supersymmetric theory can be schematically written (see Supersymmetry and Morse theory) as $$H=\frac{1}{2}(Q_{1}^{2}+Q_{2}^{2}).$$
If for some ground state $$\psi$$ $$H\psi \neq 0,$$ it would follow that $$\psi$$ wouldn't be annihilated by the (positive definite) squares of the supercharges; then $$\psi$$ wouldn't preserve some supersymmetry.
It is illustrative to notice that in the case of a Riemannian manifold $$\mathcal{M}$$ the hamiltonian coincides with the laplacian of $$\mathcal{M}$$ (see chapter 2 in Supersymmetry and Morse theory); if the laplacian was not zero, then you can make an analogy with electrodynamics, you can't be possibly describing the vaccum of the theory, sources must be present.