- #1

arivero

Gold Member

- 3,462

- 156

- TL;DR Summary
- Eigen(AA^+)=Eigen(A^+A)

As you know, a lot of SUSY examples, particularly from Witten's "SUSY QM", pivot on the factorisation trick: produce two hamiltonians $$H_0=AA^+, H_1=A^+A$$ and see they have the same eigenvalues except for ceros.

The proof usually goes by: let ##\Psi## be an eigenvector of ##H_0##, consider ##|A^+\Psi \rangle ##, then $$H_1 |A^+\Psi \rangle =A^+AA^+\Psi=A^+H_0\Psi=A^+\lambda\Psi=\lambda|A^+\Psi \rangle$$

Now this can be already seen if A a 2x3 matrix, and I think that I have sometimes this example as an starting point but just now I can not locate it, do any of you remember perhaps a blog entry or, better, any set of lecture notes doing this? With finite matrices, I mean.

I got the idea of searching simultaneously for "susy lectures" and "Cholesky factorisation" but no results.

The proof usually goes by: let ##\Psi## be an eigenvector of ##H_0##, consider ##|A^+\Psi \rangle ##, then $$H_1 |A^+\Psi \rangle =A^+AA^+\Psi=A^+H_0\Psi=A^+\lambda\Psi=\lambda|A^+\Psi \rangle$$

Now this can be already seen if A a 2x3 matrix, and I think that I have sometimes this example as an starting point but just now I can not locate it, do any of you remember perhaps a blog entry or, better, any set of lecture notes doing this? With finite matrices, I mean.

I got the idea of searching simultaneously for "susy lectures" and "Cholesky factorisation" but no results.

Last edited: