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 $$ H|\psi_{min}\rangle =Q^2|\psi_{min}\rangle=0. $$


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?


1 Answer 1


I will offer two arguments to try to exhibit what the problem with ground states with non zero energy in a supersymmetric theory is.

The case of gauged supersymmetry: Energy backreacts on the geometry.

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.


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