Exploring Quantum Physics - Week 3 Bonus Question 1

Question:

Recall that in Lecture 5, Part II, we derived an energy quantization condition for the even parity bound states of a finite potential well of width a and depth $U_0$. What is the corresponding condition for the odd parity states in terms of the dimensionless variables $x=\frac{ka}{2}$ and $\xi^2 = \frac{ma^2U_0}{2\hbar^2}$?

Solution:

The derivation of the odd parity solution is almost exactly the same as the even version, so here it is:

$\begin{eqnarray*} \tilde{C}\sin(\frac{ka}{2}) &=& Ae^{-\gamma x} \\ \tilde{C}k\cos(\frac{ka}{2}) &=& -\gamma Ae^{-\gamma x} \\ k\cot(\frac{ka}{2}) &=& -\gamma \\ \cot(\frac{ka}{2}) &=& -\frac{\gamma}{k} \\ \cot(\frac{ka}{2}) &=& -\sqrt{\frac{\gamma^2}{k^2}} \\ &=& -\sqrt{\frac{\frac{2m}{\hbar^2}(U_0 - E)}{\frac{2m}{\hbar^2}E}} \\ &=& -\sqrt{\frac{U_0 - E}{E}} \\ &=& -\sqrt{\frac{U_0}{E} - 1} \\ &=& -\sqrt{(\frac{\xi}{x})^2 - 1} \\ \end{eqnarray*}$

So this is the answer!