# Finite Square Well

1. Jul 2, 2014

### Bashyboy

Hello everyone,

I am reading about the Finite Square Well in Griffiths Quantum Mechanics Text. Right now, I am reading about the case in which the particle can be in bound states, implying that it has an energy E < 0. After some derivations, the author comes across the equation

$\tan z = \sqrt{ \left(\frac{z_0}{z} \right)^2 - 1}$

where $z = la$ and $z_0 = \frac{a}{\hbar} \sqrt{2mV_0}$; additionally, $l= \frac{\sqrt{2m(E+V_0)}}{\hbar}$.

The author makes the claim that, "if z_0 is very large, the intersections occur just slightly below $z_n = n \frac{\pi}{2}$, with n odd; it follows that

$E_n + V_0 \approx \frac{n^2 \pi^2 \hbar^2}{2m(2a)^2}$" (1)

I don't quite see this. How does z_0 being large imply (1) is a true statement? I have tried to work out the details myself, but it appears that there are a lot of missing details which the author should have included; although, this is just my opinion.

2. Jul 2, 2014

### WannabeNewton

It seems very straightforward to me. $\frac{n\pi \hbar}{2a} \approx \sqrt{2m(E + V_0)}$ so $E + V_0 \approx \frac{n^2 \pi^2 \hbar^2}{2m(2a)^2}$. $z_0$ being very large implies that the solutions $z$ to $\tan z = \sqrt{(z_0/z)^2 - 1}$ are $z = \frac{n\pi}{2}$. Just make a plot of the two curves on mathematica for large $z_0$ and it should be obvious to you why the intersections are as they are for large $z_0$.