References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 8.12.
where is the left turning point, is the right turning point and Another example of this process is an application to the reflectionless potential that we considered earlier:
This gives a potential well centred at which approaches as . (It’s called ‘reflectionless’ since if , an incident particle passes straight through the potential without any reflection.)
We’ll consider the bound state and compare the WKB approximation to the exact answer, which we worked out as
To apply WKB, we need the turning points and where . Since is even, where
Since is an even function, we can write 1 as
Not surprisingly, Maple balks at this integral, so we need to help it along by trying a substitution. We can try
From 4, we get the limits in terms of . For , and for we have
Maple can do this integral provided we assume that (which is true from 10 since ) and ( just gives zero for the integral anyway). We get
From 1 we have
Since the smallest can be is zero, the largest can be is the greatest integer less than or equal to , so the only possible value of is . In that case
Comparing this to 3, we see that WKB gives a reasonable approximation.