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  Precise zero energy bound for supersymmetry

+ 2 like - 0 dislike

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?

This post imported from StackExchange Physics at 2020-12-13 12:43 (UTC), posted by SE-user annie marie heart
asked Nov 4, 2020 in Theoretical Physics by annie marie heart (1,205 points) [ no revision ]
retagged Dec 13, 2020

1 Answer

+ 2 like - 0 dislike

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.

This post imported from StackExchange Physics at 2020-12-13 12:43 (UTC), posted by SE-user Ramiro Hum-Sah
answered Nov 8, 2020 by Ramiro Hum-Sah (80 points) [ no revision ]

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