1-ia) You can totally ignore the given potential - this is a pure review problem of basic Ch. 5 material (from last semester) Griffiths Eq 5.10 tells you how to construct a wavefunction for 2 identical particles (You have to use the + sign for bosons).

The 's here are just the usual wavefunctions in a box (Griffiths Eq. 2.24)

1-ib) If you get part a, part b is just a direct application of Eq. 6.9. The "bracket" involves two particle wave functions here, so when you write it as an integral, remember it's really a double integral over both "x1" and "x2", the locations of the two particles. But the Delta function in the potential collapses one integral immediately, so you're really just left with a regular old 1-D integral to do. (I used MMA)

1-ii) Fairly similar to the above, just remember that for fermions, you have to use the - sign in Eq. 5.10.

2i) Griffiths Eq. 6.14 is the master equation here. The integral you need in the bracket is easy, because of the Delta function. Stare at the "trig stuff" you're left with - if you did it right, those terms are all either 0 or 1, which helps simplify a little...

(Evaluating the sum is tricky but fun - try if you want, but it's not needed.)

ii) An application of Griffiths Eq. 6.13, with n=1 (the ground state). You might make more sense of your plot if you look back at Griffiths Ch. 2.5. (You can actually solve this problem exactly, without using perturbation theory, but there's really no need)

3) Griffith's only asks for the shift in the ground state, n=0, not a general state. That helps a little. The *idea* is the same as 2i, but the integrals look scarier. You have to use a trick -- go back to Ch. 2! Can you see from Eq. 2.41 that you can write

(You have to work out exactly what constant the ?? is there)

So, .

It's quite easy to apply a+ or a- to harmonic oscillator states - in fact, Griffiths Eq. 2.52 and 2.53 tell you exactly what they do, normalized and everything! So, it's not too hard to figure out <m|x^2|0>, exactly, without having to do any integrals at all.

Bottom line: You'll discover that although it looks like you need an infinite sum of terms , in fact only ONE term contributes, the rest all vanish.

4) Use the wave function table given in Griffiths Table 4.6 (p. 141).

Also, you will find life is much easier if you remember

for small r. (a is the Bohr radius: you should only *need* this approximation for r<1fm, which is 5 orders of magnitude smaller than a, hence this approximation is fine)