= Solution
Let
$$
H=-\partial_x^2-2\chi^2\operatorname{sech}^2(\chi x),
\qquad
A=\partial_x+\chi\tanh(\chi x).
$$
On $L^2(\mathbb R)$, $A^\dagger=-\partial_x+\chi\tanh(\chi x)$. The <supersymmetric factorization of the one-soliton potential> is
$$
A^\dagger A=H+\chi^2,
\qquad
AA^\dagger=-\partial_x^2+\chi^2.
$$
In particular, $A^\dagger A$ is nonnegative, so every eigenvalue of $H$ satisfies
$$
E\geq-\chi^2.
$$
Equality is attained precisely when $A\psi=0$. Solving that first-order equation gives
$$
\psi_0(x)=C\operatorname{sech}(\chi x),
$$
which is square-integrable. With unit normalization, $C=\sqrt{\chi/2}$, and
$$
H\psi_0=-\chi^2\psi_0.
$$
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 $A$ to
$$
A^\dagger A\psi=(E+\chi^2)\psi
$$
gives
$$
-\partial_x^2(A\psi)=E(A\psi).
$$
The 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
$$
\boxed{E_0=-\chi^2}.
$$
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