I was looking at how to solve the Schrodinger equation in momentum space and I came across this question and answer. If we have a potential $V(\hat{x})$ that can be expanded as a power series with finite order, then it is straightforward to make the transformation $\hat{x}\to i\hbar\frac{\partial}{\partial p}$ in Schrodinger's equation $\left(\frac{\hat{p^2}}{2m}+V(\hat{x})\right)|\psi\rangle=E|\psi\rangle$.
What if we have the Poschl-Teller potential, ${\displaystyle V(x)=-{\frac {\lambda (\lambda +1)}{2}}\mathrm {sech} ^{2}(x)}$? If we replace $x$ with $i\hbar\frac{\partial}{\partial p}$ in Schrodinger's equation, then we now have a derivative in the argument of $\mathrm{sech}^2(x)$. From Schrodinger's equation, we get
\begin{align} \frac{p^2}{2m}\psi(p)-\frac{\lambda(\lambda+1)}{8\left(e^{-i\hbar\frac{\partial}{\partial p}}+e^{i\hbar\frac{\partial}{\partial p}}\right)^2}\psi(p)=E\psi(p) \end{align}
Therefore,
\begin{align} \left(e^{-i\hbar\frac{\partial}{\partial p}}+e^{i\hbar\frac{\partial}{\partial p}}\right)^2\frac{p^2}{2m}\psi(p)-\frac{\lambda(\lambda+1)}{8}\psi(p)=E\left(e^{i\hbar\frac{\partial}{\partial p}}+e^{-i\hbar\frac{\partial}{\partial p}}\right)^2\psi(p) \end{align}
We see that we have translation operators acting on $\psi(p)$ and $p^2\psi(p)$, indicating that we have converted a differential equation into a difference equation, which could then be solved.
Is this process correct?
