# Analytical Solution to Schrödinger equation with Rosen-Morse potential

I have been trying for time to get analytical solution for Rosen-Morse potential, after I found the solution for Teller potential. Rosen-Morse potential is as follows: $$V(z) = V_0 \cdot \cosh^2\mu(\tanh(z) + \tanh(\mu))^2$$

First I used change of variable, $$u = \tanh(z)$$ which gives

$$\frac{\partial^2 \psi}{\partial z^2} = (1 - u^2) \frac{d}{du}\left((1 - u^2) \frac{d\psi}{du}\right)$$

Then I simply have the form of general hypergeometric function:

$$\frac{\partial^2 \psi}{\partial u^2} + \frac{\tau(u)}{\sigma(u)} \frac{\partial \psi}{\partial u} + {\frac{\varphi(u)}{\sigma^2(u)}}{\psi} = 0 \\ \\ \tau(u) = -2u \\ \sigma(u) = (1-u^2) \\ \varphi(u; \epsilon) = \epsilon - V_0 \cdot \cosh^2\mu(u - \tanh\mu)^2$$

I am stuck here. Should I use singular points $$(+1, -1)$$ and try to make asymptotic analysis at these points to try to get a $${_2F_1}$$ hypergeometric function. I am open other methods as well, but I want to solve it using hypergeometric functions if possible.

The full solution of this potential is rather tricky. Scattering states are dealt with in this paper, which cites a solution for bound states in the introduction of this one. They are rather long, so I will just cite the results.

Write the Rosen-Morse potential as

$$V(x) = -\alpha(\alpha+1)\textrm{sech}(x)+2\beta\tanh(x).$$

Bound states have a quantum number $$n$$, limited by $$0 \leq n<\alpha-\sqrt{\beta}$$. Their wave functions are

$$\psi_n(x) = A_n e^{-\beta x/(\alpha-n)}\textrm{sech}^{\alpha-n}(x)P_n^{\alpha-n+\beta/(\alpha-n),\alpha-n-\beta/(\alpha-n)}(\tanh x),$$ where $$P_n^{a,b}$$ denotes the $$n$$-th Jacobi polynomial with parameters $$a$$ and $$b$$. The normalization constant $$A_n$$ satisfies

$$|A_n|^{-2} = \frac{2^{2\alpha-2n}\Gamma(\alpha+\beta/(\alpha-n)+1)\Gamma(\alpha-\beta/(\alpha-n)+1)(\alpha-n)^3}{n!\Gamma(2\alpha-n+1)[(\alpha-n)^4-\beta^2]}.$$

For scattering states, define

$$D_\nu^{\mu,\eta}(x) = (1-x^2)^{-\frac{\eta}{2}}\left(\frac{1+x}{1-x}\right)^\frac{\mu}{2}F(-\nu-\eta,\nu+1-\eta;1-\mu-\eta;(1-x)/2),$$ where $$F$$ denotes the Gauss hypergeometric function.

Then, for a given wave number $$k$$, if $$k^2<4\beta$$, set

$$\mu = -\frac{\sqrt{4\beta-k^2}}{2}+\frac{ik}{2}, \qquad \eta = -\frac{\sqrt{4\beta-k^2}}{2}-\frac{ik}{2}.$$

If $$k^2 > 4\beta$$, set

$$\mu = \frac{i\textrm{sgn}(k)\sqrt{k^2-4\beta}}{2}+\frac{ik}{2}, \qquad \eta = -\frac{i\textrm{sgn}(k)\sqrt{k^2-4\beta}}{2}-\frac{ik}{2}.$$

Then scattering states are given by

$$\psi_k(x) = D^{\mu,\eta}_\nu(\tanh x).$$

When $$k > 4\beta$$, reflection and transmission coefficients are

$$R = \frac{\sin^2(\pi\alpha)+\sinh^2\left(\frac{\pi}{2}(k-\sqrt{k^2-4\beta})\right)}{\sin^2(\pi\alpha)+\sinh^2\left(\frac{\pi}{2}(k-\sqrt{k^2-4\beta})\right)},$$

$$T = \frac{\sinh(\pi k)\sinh(\pi\sqrt{k^2-4\beta})}{\sin^2(\pi\alpha)+\sinh^2\left(\frac{\pi}{2}(k-\sqrt{k^2-4\beta})\right)}.$$

If you write $$\bar{D}^{ik}_{\alpha,\beta}(\tanh x)$$ for the function with wave number $$k$$ and parameters $$\alpha$$ and $$\beta$$, their normalization is

$$\int_{-\infty}^\infty \bar{D}^{ip*}_{\alpha,\beta}(\tanh x)\bar{D}^{ik}_{\alpha,\beta}(\tanh x)dx = w_{\alpha,\beta}(k)\delta(k-p),$$

where

$$w_{\alpha,\beta}(k) = \begin{cases} 2\pi \frac{\sqrt{k^2-4\beta}}{|k|}\frac{\sin^2(\pi \alpha)+\sinh^2\left(\frac{\pi}{2}(\sqrt{k^2-4\beta}+k)\right)}{\sinh(\pi k)\sinh(\pi \sqrt{k^2-4\beta})}, & |k| > 2\sqrt{\beta} \\ \frac{2\pi^2 2^{-\sqrt{4\beta-k^2}}}{k\sinh(\pi k)}\frac{\Gamma\left(1+\sqrt{4\beta-k^2}\right)^2}{\left|\Gamma\left(1+\alpha+\frac{\sqrt{4\beta-k^2}}{2}+\frac{ik}{2}\right)\right|^2\left|\Gamma\left(-\alpha+\frac{\sqrt{4\beta-k^2}}{2}+\frac{ik}{2}\right)\right|^2}, & |k| \leq 2\sqrt{\beta} \end{cases}.$$

Even writing the solution took a lot of space. For details about the derivation, see the cited papers.