00:01
In this following question, it is said that, demonstrate the wkb approximation yields the energy level of linear harmonic oscillator.
00:09
Then suppose this is the linear harmonic oscillator and according to wkb approximation 1 upon h cut minus a to plus a root under 2m, e minus v.
00:38
X, dx plus pi by 4 plus 5 by 4 is equal to n pi and n plus 1 pi where this n starts from 1 to 3 and this end starts from 0 1 2 and for vx equal half m omega square x square the above equation will be 1 upon h cut minus a to plus a root under 2m, root under p minus half m omega square x square d x and it is equal to n minus half pi, n from 1 to 3 and n plus half pi, where n is equal to 0 1 to 3.
01:46
So now minus a to a root under 2m, half m, omega square, a square minus half m, omega square, x square dx is equal to n minus half half 5 h cut.
02:13
And this is n plus half 5 h cut which will be equal to m omega minus a 2 plus a root under a square minus x square d x which is equal to n minus half pi h cut and n plus half pi h cut now we will assume x equal to a sine theta, which will give bx equal a phospheta, d theta.
02:51
So, m omega pi by 2 to 3 pi by 2, a phos theta into a hveta d theta is equal to n minus half pi h cut, n plus half, n plus half, pi h cut.
03:15
Now we will use this formula...