Question

You do not believe that the ladder operator approach toward solving the Schrodinger equation for the Harmonic Oscillator potential V(x)=1/2mw^2x^2 gives you a reasonable answer. To verify the calculation, you can use the WKB approximation to determine the ground state energy. Do the results differ and if yes by how much? hint (integral a 0 sqrt(a^2-x^2)dx = pi*a^2/4.

Answer 1

The ladder operator approach gives the ground state energy as

The WKB approximation:

First find the classical turning points(where energy equals potential energy)

This gives the turning points as

and

note that

Then according to WKV approximation:

Therefore

since x2 = -x1

Since this is an even function and it is integrated from limit -x2 to x2

we can writw

or

where

Therefore using the given integral relation, we can write

or the ground state energy is

which matches with the ladder operator approach