Here is the HOMEWORK for this week

1) The proof is similar to what Gas does in 3.28, only reversing the roles of p and x. (You may want to use 3.26 and/or the unnumbered equation two before 3.26) (I did not use Gas' hint in my solution)

2) Although you don't have to work it out, it may help you to first look at Gas prob. 3.4b. Note: .

(Taking derivatives of this with respect to a is a trick to give you other useful integrals!)

To get , use Gas. 3.26, and remember how we did the Gaussian Fourier transform in Ch. 2 (Eqs. 2.2 through 2.5) Just be careful about factors of 2 and pi and hbar now. (You really needn't redo any Gaussian integrals if you don't want to)

3) The main trick here is to figure out .

But you know (which Gas gives you), and (using Gas' suggested trick of taking derivatives), so can't you combine these?...

4) Gas' statement that is true, but not needed for this exercise! Study eqns. 3.37-38 (but work in p space instead of x space...)

5) This may be a somewhat challenging problem. It's a nice exercise with some useful math, though, so try to get it to work out!

Start with Gas 3.23 (except, with time dependent). Note that d/dt moves freely into an integral over x (but then becomes a partial). You must make use of the Schrodinger Eqn (Gas 4.1), and its conjugate. You will have to use partial integrations several times in this problem, . In this equation, the "uv" term is sometimes called a "surface term", because it's evaluated only at the limits of integration, and in this problem you should be able to argue that these surface terms vanish. (Why?!)

You will also want to take a look at the quantity

(I claim that, perhaps after some partial integrations, part of your integrand should look exactly like what you get for this derivative. And of course, when you integrate such a beast, it's an exact derivative, so you just get the surface term, which is again zero for well behaved wave functions)

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