The WKB approximation (see Challenge Problem 40.64) can be...

The WKB approximation (see Challenge Problem 40.64) can be used to calculate the energy levels for a harmonic oscillator. In this approximation, the energy levels are the solutions to the equation $$\int ^b _a \sqrt{2m[E - U(x)]} dx = {nh \over 2} \space n = 1, 2, 3, . . .$$ Here $E$ is the energy, $U(x)$ is the potential-energy function, and $x = a$ and $x = b$ are the classical turning points (the points at which $E$ is equal to the potential energy, so the Newtonian kinetic energy would be zero). (a) Determine the classical turning points for a harmonic oscillator with energy $E$ and force constant $k'$. (b) Carry out the integral in the WKB approximation and show that the energy levels in this approximation are $E_n = \hslash\omega$, where $\omega = \sqrt{k'/m}$ and $n$ = 1, 2, 3, c. ($Hint$: Recall that $\hslash = h/2\pi$. A useful standard integral is $$\int \sqrt{A^2 - x^2} dx = {1\over2} [ x\sqrt{A^2 - x^2} + A^2 arcsin ({x \over \mid A\mid}) ]$$ where arcsin denotes the inverse sine function. Note that the integrand is even, so the integral from $-x$ to $x$ is equal to twice the integral from 0 to $x$.) (c) How do the approximate energy levels found in part (b) compare with the true energy levels given by Eq. (40.46)? Does the WKB approximation give an underestimate or an overestimate of the energy levels?

Transcript

00:01 All right, so problem number 65 talks about the wkb approximation.

00:06 So it can be used to calculate the energy levels for a harmonic oscillator.

00:10 So in this approximation, the energy levels are the solutions to the equation from a to b, the integral from a to b, square root of 2m times e minus u of x, dx, is equal to n, h, over 2, where n is equal to 1, 2, 3, or some other positive integer.

00:45 So e is the total energy of the oscillator, u is the potential energy.

00:50 So this term here is the kinetic energy, the e minus u.

00:54 And then a and b are the classical turning points of the harmonic oscillator, which the total energy is equal to the potential energy, so there's no kinetic energy.

01:02 This term becomes zero.

01:04 Part a asks you to determine the classical turning points for a harmonic oscillator with energy e and the force constant k -prime.

01:13 So for a classical harmonic oscillator, the potential energy is by one -half a prime x squared.

01:23 Now, for the classic turning points, e, the total energy is going to be just equal to this potential.

01:29 So we can say e is equal to one -half k -prime.

01:35 X squared.

01:38 All right, so now to determine where these classical turning points are, we just need to solve for x because that's going to be at the solutions, at which point the energy is this holds true.

01:50 So the solutions to that x is equal to plus or minus.

01:55 It's a really simple just 2e over k prime.

02:01 So these are your classical turning points.

02:03 Now for part b, it wants you to actually carry out the integral in the wkb approximation.

02:11 And then you want to show the energy levels in this approximation, e sub n is equal to h bar omega, where omega is equal to square root of k over m, and n is equal to some positive integer.

02:24 So our integral is this now, square root of 2m times e, minus 1 half k prime x squared d x because this is our potential energy here so what we need to find out is what our boundaries are so our boundaries are just our classical turning points so our integral actually becomes the from minus square root of 2e over k prime to plus square root of 2e over k prime to plus square root of 2e over k of 2m times e minus 1 half k prime x squared d x.

03:19 All right, so now this looks like a pretty daunting integral, but it does give us the standard integral here to help with the solution.

03:30 So that standard integral is given by this interval of a squared minus x squared d x is equal to 1 1�t times x times the square root of a squared minus x squared plus a squared arc sine of x over the amplitude of a.

04:03 All right.

04:04 So now our task becomes to put this integral into this form.

04:09 So i'm going to create a new page here.

04:12 And so let's, to do this, let's look at our intagrand.

04:18 This square root of 2m times e minus 1 half k prime x squared.

04:32 All right.

04:34 So we want to put this into the form of the square root of a squared minus x squared.

04:39 So first let's go ahead and expand this out a little bit to make this easier to work with for the purpose of simplifying.

04:50 So this is going to be, this can be written as two words.

04:53 Square root of 2m .e.

04:56 K prime over k prime minus m k prime x squared.

05:07 All right.

05:09 So now we can make the substitution of a squared is equal to 2e over k prime.

05:21 Now if we do that and plug that back in, we now get the integral of a squared mk prime minus mk prime x squared now it's starting to look a bit more like our what we want the standard integral so this can now be we can factor out this mk prime we get mk prime times the square root of a squared minus x squared so that's just what we want all right so now our integral become this here, so from a to b, times the square root of mk prime, times the square root of a squared minus x squared.

06:24 So one thing to note is that the symmetry of the integral, we can actually change the bounds, because we're making a substitution here.

06:40 This a squared is equal to e to k prime.

06:43 So we can change the boundaries of our integral to be from square root of m k prime from 2.

06:55 Sorry, i just got a little messy.

06:58 Got away from me a little bit...

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