HINTS for HW 11

1) Study Griffiths p. 267 to get the basic idea, especially Eq. 7.40! You will have a single (3-D) integral to do just like he did for "I". The phi part is trivial, and you should first integrate over theta. That integral will be familiar from a couple of weeks ago - it gives different answers when r<R and when r>R, so your radial integral has to get split up into two pieces, from zero to R and then R to infinity. (Those integrals are polynomials times exponentials, which can be done by parts, or MMA can do them for you) The final answer is Griffiths 7.47. (Note the factor of "a" out front in the definition of D)

2) Griffiths (P. 267-269) does most of the work, and in my notes (p. 175-9) I go into even slightly more detail. There's not too much you need to do except to figure out exactly which signs get changed in the formula for F(x). (That's the only difference, just a few sign flips!) You don't have to redo ANY integrals. When you get to comparing with Eq. 7.49, you should see two signs different from Griffiths. Don't forget to add the Vpp term, and divide through by -E1, and substitute in for what D, X, and I are (Griffiths gives you ALL those formulas, Eq. 7.42, 7.47, and 7.48. Remember that x=R/a, so be careful, some terms are x and some are 1/x...) The final result will be *very* similar to 7.51 (just some signs different here and there) although if you don't do the algebra simplifications precisely like Griffiths did, it may LOOK a little different. I'd strongly suggest plugging Griffith's F(x) into MMA and plotting it (to verify his figure 7.7), and then plugging YOUR new F(x) in to finish this problem...

3) You need the F(x) plot from the previous problem to solve this one. (This is F(x) from Eq 7.51, NOT the new F(x) you found in the previous problem!) 1st you'll need to find the minimum (numerically!) of F(x). I found the MMA expression

FindRoot[function'[x]==0,{x,1}]

to be very helpful! (Note the "prime" after the function name, that's the first derivative) Once you have the minimum you can trivially find the 2nd derivative, just use function''[xmin]. (two primes is the second derivative) Remember that the MMA syntax to define a function is function[x_] := -1+(2/x)*(stuff...)

Now that you have the VALUE of f''[xmin], you should think carefully about how the 2nd derivative of a potential curve is related to the allowed vibrational energies. Remember, for a harmonic oscillator with V=Vmin + (1/2) m omega^2 (x-xmin)^2,

the vibrational energies are Vmin+(n+1/2) hbar omega. (Do you see why?)

So, think about how the second derivative of V tells you what omega is. (In this problem, you know what "m" is, it's the REDUCED mass of two protons)

One important thing - the curve you've plotted, F(x), has had a factor of -E1=13.6 eV divided out, and ALSO it's a function of x=R/a, not R. So, you might want to restore the units in order to really get numbers out. (i.e. rewrite it as a true potential, V(R))

This problem is a little tricky, but it's really worth understanding - the idea of converting a potential into a "related" harmonic oscillator is one of the most common tricks in laboratory quantum mechanics.