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

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In summary, the expression for evaluating\langle \mathbf{p''} | F(r) | \mathbf{p'} \rangleis simplified using the resolution of identity and relevant equations to\frac{1}{(2 \pi \hbar)^3}\int exp(i (\mathbf{p'} - \mathbf{p''}) \cdot \mathbf{x'} / \hbar) F(|\mathbf{x'}|) d \mathbf{x'}and can be further simplified using spherical coordinates to only an integral over r.
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gulsen
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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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  • #2
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.
 

1. What does the notation $\langle \mathbf{p''} | F(r) | \mathbf{p'} \rangle$ mean?

This notation represents the expectation value of the operator $F(r)$ between the states $|\mathbf{p''}\rangle$ and $|\mathbf{p'}\rangle$. Essentially, it is a way of calculating the average value of the operator when operating on the two given states.

2. How does one evaluate $\langle \mathbf{p''} | F(r) | \mathbf{p'} \rangle$?

To evaluate this expression, you would first need to determine the form of the operator $F(r)$. Then, you can use the appropriate mathematical techniques (such as integration or differentiation) to calculate the expectation value between the two states $|\mathbf{p''}\rangle$ and $|\mathbf{p'}\rangle$.

3. What is the significance of the states $|\mathbf{p''}\rangle$ and $|\mathbf{p'}\rangle$ in this expression?

The states $|\mathbf{p''}\rangle$ and $|\mathbf{p'}\rangle$ represent the momentum states of a particle. This notation is commonly used in quantum mechanics to describe the properties of a particle, such as its position, momentum, or energy.

4. Can $\langle \mathbf{p''} | F(r) | \mathbf{p'} \rangle$ be used to solve any physical problems?

Yes, this expression can be used to solve various physical problems, such as calculating the energy levels of a quantum system or determining the probability of a particle being in a certain state. However, the specific applications of this expression will depend on the form of the operator $F(r)$ and the states being used.

5. Are there any limitations to using $\langle \mathbf{p''} | F(r) | \mathbf{p'} \rangle$ in solving problems?

While this expression can be a useful tool in solving physical problems, it does have some limitations. For example, it may not be applicable in cases where the system is not in a well-defined state or when the particles involved have interactions with each other. Additionally, the calculation of this expression can become more complex for systems with higher dimensions or more particles.

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