Finite square well: an alternate method to obtain k and b.

I showed this to some students today and they really liked it, so I thought I would share it here.
You can use: \(k^2+b^2=2mU/\hbar^2\), from the Schrodinger equation, and \(ktan(k)=b\) from the boundary conditions at x = 1nm.  Feed those to Wolfram alpha and it will give you values of k and b which "belong to" a particular energy eigenstate. This way is kind of appealing since you have a nice circle as part of your graph, with a radius proportional to U, and the values of b and k come together in pairs. (In general it is: \(ktan(kL/2)=b\). For our particular case, L/2=1 nm.)

Imagine rotating this by 90 degrees counter clockwise. Then it may be easier to see the function k*tan(k), which is an even function of k. How cool is that?!


The results will be totally the same as the method shown a few posts down in the Square Well Boundary Conditions post. All start from the same place, that is, the boundary conditions of continuity and smoothness at L/2, \(Acos(k)=B\) and \(-Aksin(k)=-Bb\);
and the Schrodinger eqn which leads to: \(\hbar^2k^2/(2m) = -\hbar^2 b^2/(2m) +U\), where U is the height of the wall.