A Solving Schrödinger's Equation with a Smooth Potential Wall: A Detailed Guide

[paul159753]

TL;DR Summary

I'm looking for help for an exercise of the chapter III. Schrödinger's equation, §25 The transmission coefficient (problem 3).

Hello everyone,

I'm looking for help for the problem 3 of the chapter III. Schrödinger's equation, §25 The transmission coefficient of the Volume 3 of the Landau-Lifshitz book (non-relativistic QM).

In this exercise Landau considers a smooth potential wall $$\frac{U_0}{1 + \exp{\left(-\alpha x \right)}}.$$ We search solutions for $E > U_0$.
So we have to solve $$\frac{d^2}{dx^2}\Psi + \frac{2m}{\hbar^2}\left(E - \frac{U_0}{1 + \exp{\left(-\alpha x \right)}}\right)\Psi = 0.$$ We do the change of variable $\xi = -\exp{\left(-\alpha x \right)}$. We write the derivative operator $$\frac{d^2}{dx^2}= \left(\frac{d\xi}{dx}\frac{d}{d\xi}\right)^2 = \alpha^2 \xi \frac{d}{d\xi} + \alpha^2 \xi ^2 \frac{d^2}{d\xi^2}.$$ Noting $k_2^2 = 2m(E-U_0)/\hbar^2,\ k_1^2 = 2mE/\hbar^2$ we can rewrite the equation $$\frac{d^2}{dx^2}\Psi + \left(\frac{k_2^2 - \xi k_1 ^2}{1 - \xi}\right)\Psi = 0.$$ We look for solutions of the form $\Psi = \xi ^{-ik_2/\alpha} \omega(\xi)$. Multiplying the equation by $1-\xi$ and dividing by $\alpha^2 \xi \xi ^{-ik_2/\alpha}$ we get the following equation for $\omega(\xi)$ : $$ \xi(1-\xi)\frac{d^2}{d\xi^2}\omega(\xi) + (1-\xi)\left(1 - 2ik_2 / \alpha\right)\frac{d}{d\xi}\omega(\xi) + \left((k_2^2 - k_1^2)/\alpha^2 - ik_2/\alpha \frac{(1-\xi)}{\xi} \right)\omega(\xi) = 0.$$
My problem is that Landau found $$\xi(1-\xi)\frac{d^2}{d\xi^2}\omega(\xi) + (1-\xi)\left(1 - 2ik_2 / \alpha\right)\frac{d}{d\xi}\omega(\xi) + \left((k_2^2 - k_1^2)/\alpha^2 \right)\omega(\xi) = 0, $$ so the same thing but without the $- ik_2/\alpha \frac{(1-\xi)}{\xi}\omega(\xi)$ term. So he gets an hypergeometric equation and I don't. I don't know where my mistakes are.

Any help would be much appreciated.

Thanks !

 

[TeethWhitener]

Multiplying the equation by $1-\xi$ and dividing by $\alpha^2 \xi \xi ^{-ik_2/\alpha}$ we get the following equation for $\omega(\xi)$ : $$ \xi(1-\xi)\frac{d^2}{d\xi^2}\omega(\xi) + (1-\xi)\left(1 - 2ik_2 / \alpha\right)\frac{d}{d\xi}\omega(\xi) + \left((k_2^2 - k_1^2)/\alpha^2 - ik_2/\alpha \frac{(1-\xi)}{\xi} \right)\omega(\xi) = 0.$$
My problem is that Landau found $$\xi(1-\xi)\frac{d^2}{d\xi^2}\omega(\xi) + (1-\xi)\left(1 - 2ik_2 / \alpha\right)\frac{d}{d\xi}\omega(\xi) + \left((k_2^2 - k_1^2)/\alpha^2 \right)\omega(\xi) = 0, $$ so the same thing but without the $- ik_2/\alpha \frac{(1-\xi)}{\xi}\omega(\xi)$ term. So he gets an hypergeometric equation and I don't. I don't know where my mistakes are.

Any help would be much appreciated.

Thanks !

When you expand the Schrodinger equation out, you should get cancellation of all the terms proportional to $i\omega(\xi)$. It looks like you just have a simple algebra error buried in your work somewhere. Try writing out explicitly your result for $$\frac{1}{\alpha^2}\frac{d^2}{dx^2}\Psi$$
The offending terms should cancel in that step.

 

[paul159753]

I'm really dumb, I see where my mistake was thanks !