# A problem from Sakurai

1. Oct 26, 2008

### gulsen

1. The problem statement, all variables and given/known data
(Sakurai 1.27)
[...] evaluate
$$\langle \mathbf{p''} | F(r) | \mathbf{p'} \rangle$$
Simplify your expression as far as you can. Note that $r = \sqrt{x^2 + y^2 + z^2}$, where x, y and z are operators.

2. Relevant equations
$$\langle \mathbf{x'} | \mathbf{p'} \rangle = \frac{1}{ {(2 \pi \hbar)}^{3/2} }exp(i \mathbf{p'} \cdot \mathbf{x'} / \hbar)$$,
$$F(r) | \mathbf{x'} \rangle = F(|\mathbf{x'}|) | \mathbf{x'} \rangle$$
and
$$\langle \mathbf{x''} | \mathbf{x'} \rangle = \delta(\mathbf{x''} - \mathbf{x'})$$

3. The attempt at a solution
Using the resolution of identity, I wrote
$$\langle \mathbf{p''} | F(r) | \mathbf{p'} \rangle = \int \int \langle \mathbf{p''} | \mathbf{x''} \rangle \langle \mathbf{x''} |F(r) | \mathbf{x'} \rangle \langle \mathbf{x'} | \mathbf{p'} \rangle d \mathbf{x''} d \mathbf{x'}$$
Using the above "relevant equations", I get
$$\frac{1}{(2 \pi \hbar)^3}\int exp(i (\mathbf{p'} - \mathbf{p''}) \cdot \mathbf{x'} / \hbar) F(|\mathbf{x'}|) d \mathbf{x'}$$

Is this the correct answer? (wish that answers were avaiable at the backside of the book...)

Last edited: Oct 26, 2008
2. Oct 29, 2008

### borgwal

So far this looks good: but you can simplify even further by using spherical coordinates, with the z-axis chosen in a smart way: you will end up with an integral only over r, no more angles.