Exercise [21.06] b

deant · 27 Aug 2012, 09:58

My solution to this problem is attached. I used a quite different approach to Robin.

Note that there are a couple of minor errors in the statement of this exercise in the book, which I've posted explicit corrections to in my response to Robin's solution, here. Specifically (reiterating them for reference), the exercise should state:

$Z = z + \tfrac{1}{2}t^2g$

(note the + sign), and

$\Psi = e^{\frac{i}{\hbar}\left(\frac{1}{6}mt^3g^2+mtzg\right)}\psi$

(note the additional factor of $\frac{1}{\hbar}$ in the exponent).

My solution to this problem builds upon my solution to the previous exercise, [21.05], which I didn't post because it's basically identical to Roberto's (posted as Exercise [21.05] b - although I've reiterated the relevant parts in my solution here where appropriate, so it should stand on its own).

Comparing my solution to [21.05] as used in this problem with Roberto's solution, the main difference is that Roberto uses r to represent the coordinate vector, whereas I use x for that. I use r in place of Roberto's u and k, the precise relationship being (my r; all other variables Roberto's):

$\boldsymbol{r} = \sqrt{2m}k\boldsymbol{u}$

and hence

$k^2 = \tfrac{1}{2m}\|\boldsymbol{r}\|^2$

My r is in fact just the momentum at t=0

.