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

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The discussion focuses on solving Sakurai problem 1.27, which requires evaluating the expression $\langle \mathbf{p''} | F(r) | \mathbf{p'} \rangle$. The user applies the resolution of identity and relevant equations to derive the integral $\frac{1}{(2 \pi \hbar)^3}\int exp(i (\mathbf{p'} - \mathbf{p''}) \cdot \mathbf{x'} / \hbar) F(|\mathbf{x'}|) d \mathbf{x'}$. Further simplification is suggested using spherical coordinates to eliminate angular dependencies, resulting in an integral solely over the radial variable r.

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  • Quantum mechanics principles, specifically wave functions and inner products.
  • Understanding of the resolution of identity in quantum mechanics.
  • Familiarity with spherical coordinates and their application in integrals.
  • Knowledge of Fourier transforms in quantum mechanics, particularly involving momentum and position representations.
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Students and professionals in quantum mechanics, particularly those tackling advanced problems involving operator functions and integrals in momentum space. This discussion is beneficial for anyone looking to deepen their understanding of quantum state evaluations and simplifications.

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Homework Statement


(Sakurai 1.27)
[...] evaluate
[tex]\langle \mathbf{p''} | F(r) | \mathbf{p'} \rangle[/tex]
Simplify your expression as far as you can. Note that [itex]r = \sqrt{x^2 + y^2 + z^2}[/itex], where x, y and z are operators.

Homework Equations


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

The Attempt at a Solution


Using the resolution of identity, I wrote
[tex]\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'}[/tex]
Using the above "relevant equations", I get
[tex]\frac{1}{(2 \pi \hbar)^3}\int exp(i (\mathbf{p'} - \mathbf{p''}) \cdot \mathbf{x'} / \hbar) F(|\mathbf{x'}|) d \mathbf{x'}[/tex]

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
[tex]\langle \mathbf{p''} | F(r) | \mathbf{p'} \rangle[/tex]
Simplify your expression as far as you can. Note that [itex]r = \sqrt{x^2 + y^2 + z^2}[/itex], where x, y and z are operators.

Homework Equations


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

The Attempt at a Solution


Using the resolution of identity, I wrote
[tex]\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'}[/tex]
Using the above "relevant equations", I get
[tex]\frac{1}{(2 \pi \hbar)^3}\int exp(i (\mathbf{p'} - \mathbf{p''}) \cdot \mathbf{x'} / \hbar) F(|\mathbf{x'}|) d \mathbf{x'}[/tex]

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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