Here is a big final post on this well.
To get the energies we needed to solve \( 1/kb = tan(kL/2) \). To do so I graphed the both sides as functions of E, so \( 1/kb = \sqrt(-\frac{E}{E-V})\) (red graph) and \(tan(kL/2) = tan(L/2 \cdot \frac{\sqrt(2m(E-V))}{\hbar}) \) (green graph).
But this only gives the even states. To get the odd states I chose the odd solution to the schrödinger equation inside the well, Asin(kx), and got
new equation for the energy \( 1/kb = - tan^{-1}(kL/2) \). The new left side is plotted as the pink lines.
Thank you Sophie. I would be interested to know: what are the b values for the 4 states? How far do they extend outside the well?
ReplyDeleteAlso, here is an interesting thing:
ReplyDeleteThe ground state is 3 eV below expectation (based on an inf sq well).
the 1st excited state is 12 eV below "expectation".
what about the next states? How far below "expectation" are they?