Aim: Show that $-{\hbar^2\over 2m} {d^2\over dx^2}\psi(x)-{\hbar^2\over m}\mathrm{sech}^2(x)\psi(x)=E\psi(x)$ is equivalent to $\hat{O}^\dagger\hat{O}\psi(x)=({2mE\over \hbar^2}+1)\psi(x)$ where $\hat{O}={d\over dx}+\tanh(x)$ and $\hat{O}^\dagger=-{d\over dx}+\tanh(x)$
What I did:
$\hat{O}^\dagger\hat{O}\psi(x)=({2mE\over \hbar^2}+1)\psi(x) \implies (-{d^2\over dx^2}+\tanh^2(x))\psi(x)=({2mE\over \hbar^2}+1)\psi(x)$
$\implies (-{d^2\over dx^2}-\mathrm{sech}^2(x))\psi(x)={2mE\over \hbar^2}\psi(x)$
$\implies -{\hbar^2\over 2m} {d^2\over dx^2}\psi(x)-{\hbar^2\over 2m}\mathrm{sech}^2(x)\psi(x)=E\psi(x)$
There is an extra "${1\over 2}$" in the 2nd term of LHS! Also, is there a better way of showing this. What about showing it in the other direction -- i.e. starting with $-{\hbar^2\over 2m} {d^2\over dx^2}\psi(x)-{\hbar^2\over m}\mathrm{sech}^2(x)\psi(x)=E\psi(x)$?
Thanks.