Finite square well bound states

Question

Let's suppose I have a finite potential well: $$ V(x)= \begin{cases} \infty,\quad x<0\\ 0,\quad 0<x<a\\ V_o,\quad x>a. \end{cases} $$

I solved the time-independent Schrodinger equation for each region and after applying the continuity conditions of $\Psi$ and its derivative I ended up with:

$$ \tan(k_1a)=-\frac{k_1}{k_2},$$ where $k_1=\sqrt{\frac{2mE}{\hbar^2}}$ and $k_2=\sqrt{\frac{2m(V_o-E)}{\hbar^2}}$.

I'm aware of the fact that solutions can only be calculated graphically, but what's the relation between the value of $V_o$ and the bound states? What if I want to find the acceptable values of $V_o$ for the bound states to be $1,2,3,\dots$ or none?

Answer 1

The simplest is to reorganize your equation so that $$ \tan\left(\sqrt{\frac{2m V_0 a^2}{\hbar^2}\frac{E}{V_0}}\right) =-\sqrt{\frac{E/V_0}{1-E/V_0}} $$ and define the dimensionless variables $$ z=\frac{E}{V_0}\, ,\qquad \xi= \frac{2m V_0 a^2}{\hbar^2} $$ The possible eigenvalues are then given by finding the roots of $$ \tan\sqrt{\xi z}=-\sqrt{\frac{z}{1-z}}\, . $$ In principle, you will know or fix the various variables in $\xi$ so that by changing $V_0$ you change $\xi$ and thus the values of $z$ for which there is intersection.

Obviously, the number of bound states will be determined by the number of times $\tan\sqrt{\xi z}$ intersects the other curves, which is always negative. Since $0\le z\le 1$, you will not get any bound state for $\xi<(\pi/2)^2$ since, in this region, $\tan\sqrt{\xi z}$ will never be negative. However, with (for instance) $\xi=1.1(\pi/2)^2$ you find a single intersection at $z=0.969$, meaning $E=0.969 V_0$, really close to the lip of your potential. The lowest value of $\xi$ for you will have 2 intersections is $\xi=(3\pi/2)^2$. For $\xi=1.1\times (3\pi/2)^2$ there are 2 solutions: $z_1=0.274$ and $z_2=0.973$ etc.