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7

TL;DR: The supersymmetric partner potential to OP's potential is the constant potential, which is clearly reflectionless. Define for later convenience the constant $\kappa:=\hbar/\sqrt{2m}$. The constant potential and OP's potential are just the two first cases ($\ell=0$ and $\ell=1$) in an infinite sequence of reflectionless attractive$^1$ potentials ...


3

Your confusion comes from thinking that going to superstrings simply means adding fermions in the spectrum. The spectrum is instead different. For bosonic string (let's focus on NN boundary conditions and open strings) you have something like: $$\alpha' m^2=N-1$$ where N is the number operator of the transverse vibrational excitations of the bosonic ...


3

Let me first describe the basic idea without even mentioning supergravity. Consider some classical field theory of two fields $\phi$ and $\sigma$, with an action $S[\phi,\sigma]$. Suppose this theory has a continuous symmetry (with parameter $\epsilon$): $$S[\phi,\sigma]=S[\phi+\delta_\epsilon\phi,\sigma+\delta_\epsilon\sigma].$$ Say I find some particular ...


1

The explanation lies in the fact that you want to consider BPS states, or equivalently, short supersymmetric multiplets, because you are dealing with an extremal black hole. Now you could ask why we want to impose this restriction on the spectrum. The answer is that results based on the BPS spectrum can be generalized from weak coupling to strong coupling. ...


1

You cannot just substitute fermions for bosons to determine the SUSY equivalent of a given interaction. For example, the kinetic terms for the bosons and fermions look quite different--first order in derivatives for the fermions, versus second order for bosons. As another example, the SUSY equivalent of a four-boson interaction is a Yukawa interaction ( a ...



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