Theoretical quantum mechanics problem

[EliotHijano]

Homework Statement

Demonstrate that <$$\alpha_{1},j_{1},m_{1}|r·p+p·r|\alpha_{2},j_{2},m_{2}>$$ is 0 when $$m_{1}\neq m_{2}$$

Homework Equations

The Attempt at a Solution

I have demonstrated that r·p+p·r is equal to $$\frac{i·m}{\hbar}[H,r^{2}]$$

I would appreciate your help. Thank you

 

[Jhero]

for your post.

To prove that <\alpha_{1},j_{1},m_{1}|r·p+p·r|\alpha_{2},j_{2},m_{2}> is 0 when m_{1}\neq m_{2}, we can use the fact that the commutator of two operators is equal to 0 when the operators commute. In this case, r and p commute because they are both position and momentum operators, respectively. Therefore, we can rewrite the expression as:

<r·p+p·r> = <r·p> + <p·r>

Next, we can use the fact that the expectation value of the product of two operators is equal to the product of the expectation values of each individual operator. This means we can rewrite the expression as:

<r·p+p·r> = <r><p> + <p><r>

Since r and p commute, we can interchange them without changing the overall expression. Therefore, we can rewrite the expression as:

<r·p+p·r> = <p><r> + <r><p>

Now, we know that the expectation value of a position operator is equal to 0 when the state has a definite momentum, and vice versa. In other words, <r> = 0 when the state has a definite momentum, and <p> = 0 when the state has a definite position. Since we are dealing with two different states, we can assume that either <r> or <p> will be equal to 0, depending on the states <\alpha_{1},j_{1},m_{1}> and <\alpha_{2},j_{2},m_{2}>.

Therefore, in either case, one of the terms <p><r> or <r><p> will be equal to 0, which means that the overall expression <r·p+p·r> will also be equal to 0. This proves that <\alpha_{1},j_{1},m_{1}|r·p+p·r|\alpha_{2},j_{2},m_{2}> is 0 when m_{1}\neq m_{2}.