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Supersymmetric quantum mechanics

Pöschl-Teller potential

Let
$$
A=\frac{d}{dx}+\chi\tanh(\chi x),
\qquad
A^\dagger=-\frac{d}{dx}+\chi\tanh(\chi x).
$$
Then
$$
A^\dagger A=-\frac{d^2}{dx^2}+\chi^2-2\chi^2\operatorname{sech}^2(\chi x),
\qquad
AA^\dagger=-\frac{d^2}{dx^2}+\chi^2.
$$
Thus the one-soliton Schrodinger operator is paired with the free Schrodinger operator. Its normalizable state annihilated by $A$ is proportional to $\operatorname{sech}(\chi x)$.


Supersymmetric factorization of the one-soliton potential

Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 47 / 2 / Solution

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Supersymmetric factorization of the one-soliton potential

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Supersymmetric factorization of the one-soliton potential

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