Supersymmetric factorization of the one-soliton potential
= Supersymmetric factorization of the one-soliton 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 $\operatorname{sech}(\chi x)$.