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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 $+ \infty$).

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

1 Answer 1

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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 $$ R=\frac{\Gamma(1+2i\sqrt{E})}{\Gamma(1-2i\sqrt{E})}. $$

Remark: The last sentence in the OP indicates that they have already found the solution, whereas the math was done in the deleted answer by @2.1. However, I think it is worthwhile having it stated clearly.

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