# 1D Quantum scattering from $V(x) = e^x$

Define $V(x) = e^{x}$, $x \in \mathbb{R}$ and consider the Hamiltonian $H = - \frac{d^2}{dx^2} + V(x)$. The eigenvalue problem is $$-\psi''(x) + e^{x} \psi(x) = E \psi(x)\,, \quad x \in \mathbb{R}\,.$$ A change of variables reduces this to a modified Bessel equation, so we can express the energy eigenfunctions in terms of well-known functions, see WolframAlpha.

Thus $$\psi(x) = c_1 I_{- \lambda} \left( 2 e^{ x/2} \right) + c_2 I_\lambda \left( 2 e^{x/2} \right)$$ with $\lambda = 2 \sqrt{E}i$ and where $I$ is the modified Bessel function of the first kind.

Using the asymptotic form for the modified Bessel function around the origin we see that in the limit $x \to - \infty$ $$\psi(x) \sim C_1 e^{-i \sqrt{E}x} + C_2 e^{i \sqrt{E}x}\,, \; x \to - \infty\,.$$ i.e. a solution to the free Schrödinger equation.

Now if $\psi_E$ is the eigenfunction of $H$ with energy $E$, I want to derive the asymptotic $$\psi_E(x) \sim e^{i \sqrt{E}x} + R(E) e^{- i \sqrt{E}x}\,, \; x \to - \infty\,,$$ where $R(E)$ is supposed to be the reflection coefficient, telling us how an incoming plane wave reflects from our potential. But to derive $R(E)$ I feel like I need some additional boundary condition? Maybe I need to define the behavior for $x \to \infty$? (Edit: indeed it is enough to assume that the solution vanishes at $+ \infty$).

• See page 13 of Liouville Field Theory. Jul 29, 2017 at 10:34
• I see you answered your own question, 0 at +infinity. Did it work?
– user196418
Dec 19, 2018 at 12:34

Indeed, there is a missing boundary condition: since the potential grows for $$x\rightarrow +\infty$$, the solution should vanish in this limit. Thus, even before solving the equation, we know that the wave is fully reflected, and the reflection coefficient has amplitude $$1$$.
Substitution $$z=2e^{x/2}$$ reduces the equation to the equation for the modified Bessel functions: $$z^2u''(z)+zu'(z)-[\lambda^2+z^2]u(z)=0$$ with $$\lambda = 2i\sqrt{E}$$. As the two linearly independent solutions of thsi equation one can take either $$I_{\pm\lambda}(z)$$ or $$I_\lambda(z),K_\lambda(z)$$. $$I_{\pm\lambda}(z)$$ diverges for $$x\rightarrow+\infty$$, but $$K_\lambda$$ (which is just a linear combination of $$I_{\pm\lambda}(z)$$) does not. We thus write the solution as $$\psi(x)=CK_\lambda(e^{x/2})=\tilde{C} \left[I_{-\lambda}(e^{x/2})- I_{\lambda}(e^{x/2})\right]$$ The asymptotic form of this solution for $$x\rightarrow-\infty$$ is: $$\psi(x)\rightarrow \tilde{C}\left[\frac{1}{\Gamma(1-2i\sqrt{E})}e^{-i\sqrt{E}x}-\frac{1}{\Gamma(1+2i\sqrt{E})}e^{i\sqrt{E}x}\right],$$ from which the reflection ciefficient is $$R=\frac{\Gamma(1+2i\sqrt{E})}{\Gamma(1-2i\sqrt{E})}.$$