Solving Uncoupled Commutator Homework

[Activeuser]

Homework Statement

In some cases mj1 and mj2 may be specified at the
same time as j because although [j2,j1z] is non-zero, the
effect of [j2,j1z] on the state with mj1=j1, mj2=j2 is zero.
Confirm that [j2,j1z]| j1j1; j2j2=0 and [j2,j1z] | j1,- j1; j2,- j2=0.

Homework Equations

what is the procedure to solve this problem

The Attempt at a Solution

I evaluated the coummutator of [j2, j1z] = 2ih (L1y L2x- L1x L2y), how to plug in m1, m2.. mj= + or - J.
Thank you.

 

[mark726]

Hello,

Thank you for your post. In order to solve this problem, we need to first understand the meaning of the symbols and equations given. The notation [j2, j1z] represents the commutator of the operators j2 and j1z, which is defined as [j2, j1z] = j2j1z - j1zj2. This is a mathematical operation that helps us understand the relationship between these two operators.

Next, we need to understand the notation for the state |j1j1; j2j2. This represents a state with quantum numbers j1 and j2 for the operators j1z and j2z, respectively. This state can also be written as |j1, mj1; j2, mj2>, where mj1 and mj2 are the quantum numbers for the operators j1z and j2z.

Now, we can use these definitions to solve the problem. We need to show that the commutator [j2, j1z] applied to the state |j1j1; j2j2> is equal to 0, and the same for the state |j1, -mj1; j2, -mj2>. This can be done by plugging in the appropriate values for mj1 and mj2 into the commutator and showing that the result is 0.

I hope this helps. Please let me know if you need any further clarification or assistance. Good luck with your problem!