Solve Sakurai 1.27: Evaluate $\langle \mathbf{p''} | F(r) | \mathbf{p'} \rangle$

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Homework Statement


(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.

Homework 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'})

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...)
 
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gulsen said:

Homework Statement


(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.

Homework 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'})

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...)

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.
 
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