Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 3 / 32D / b / Solution

LetOn $L^2(\mathbb{R})$, $A^{†}=-{\partial }_x+\chi \tanh (\chi x)$. The supersymmetric factorization of the one-soliton potential isIn particular, $A^{†}A$ is nonnegative, so every eigenvalue of $H$ satisfies

Equality is attained precisely when $A\psi =0$. Solving that first-order equation giveswhich is square-integrable. With unit normalization, $C=\sqrt{\chi /2}$, and

It remains to exclude another bound state. Since this Pöschl-Teller potential tends to zero at infinity, a bound-state energy must have $E<0$. If $H\psi =E\psi$, then applying $A$ togivesThe free one-dimensional Schrodinger operator has no nonzero square-integrable eigenfunction with $E<0$. Consequently $A\psi =0$, and the state must be proportional to ${\psi }_0$. Thus there is exactly one bound state, with

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