# Homework Help: Angular momentum commutation relations

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1. Jun 1, 2017

### McLaren Rulez

1. The problem statement, all variables and given/known data
Show that $|l, m\rangle$ for $l=1$ vanishes for the commutator $[l_i^2, l_j^2]$.

2. Relevant equations
$L^2 = l_1^2 + l_2^2 + l_3^2$ and $[l_i^2,L^2]=0$

3. The attempt at a solution
I managed to so far prove that $[l_1^2, l_2^2] = [l_2^2, l_3^2] = [l_3^2, l_1^2]$. I know that for $l=1$, I have $m=-1,0,1$ but I'm not really sure how to proceed from here though. Any tips?

2. Jun 1, 2017

### BvU

What is the complete problem statement ? Or: what do you mean when you write $|l, m\rangle$ vanishes ??

3. Jun 1, 2017

### McLaren Rulez

That is the full question. I believe it means that the commutator acting on the state $|l,m\rangle$ i.e. $\langle m, l| [l_i^2, l_j^2] |l, m \rangle = 0$. Of course $l$ and $m$ are the usual quantum numbers corresponding to $L^2$ and $l_3$

4. Jun 2, 2017

### BvU

Ah, clear. Sorry to have pulled you off the unanswred threads list

So the thing to do is work out $\left ( l_i^2 l_j^2 - l_j^2 l_i^2 \right ) \left | 1,m \right \rangle$ and conclude that it's zero |
But you knew that already.

I tried working them out but had a hard time keeping it from getting longer and longer. Still looking for a shorter path through.

5. Jun 2, 2017

### McLaren Rulez

That's okay, thank you for helping. I guess the idea is to somehow write $l_1$ and $l_2$ in terms of the ladder operators but some basic manipulations haven't really led me anywhere.

6. Jun 2, 2017

### BvU

Same idea here. I'm looking at \begin{align*} L_- L_+ & = L_x^2 + L_y^2 +i \left ( L_x L_y - L_y L_x \right ) \\ & = L^2 -L_z^2 -\hbar L_z \end{align*}which is from Ballentine (7.9) but haven't worked it towards our commutator yet.

7. Jun 2, 2017

### Staff: Mentor

You can prove that $\langle l, m| [l_i^2, l_j^2] |l, m \rangle = 0$ when i or j is equal to 3 in about two lines by calculating explicitly the commutator and using the fact that $|l, m \rangle$ is an eigenfunction of $l_3$.

Combining that with
should complete the proof.

8. Jun 2, 2017

### McLaren Rulez

I feel silly for not trying the obvious. Thank you Dr. Claude.

9. Jun 3, 2017

### BvU

For the admiring spectators (including me ): can you show it ?

10. Jun 3, 2017

### McLaren Rulez

Of course. Let's start with the $[l_2, l_3]$ commutator

\begin{align*} \langle m, l | [l_2^2, l_3^2]| l,m \rangle &= \langle m, l | l_2^2l_3^2 - l_3^2l_2^2| l,m \rangle \\ &= \langle m, l | l_2^2l_3^2| l,m \rangle - \langle m, l | l_3^2l_2^2| l,m \rangle \\ \end{align*}

Utilize the fact that $l_i$ are Hermitian and the eigenvalue equation $l_3| l,m \rangle = m| l,m \rangle$, we get
\begin{align*} \langle m, l | l_2^2l_3^2| l,m \rangle - \langle m, l | l_3^2l_2^2| l,m \rangle &= m^2(\langle m, l | l_2^2| l,m \rangle - \langle m, l | l_2^2| l,m \rangle) \\ & = 0 \end{align*}

As for the commutator result, we have
\begin{align*} [l_1^2, l_2^2] &= [l_1^2, L^2 - l_1^2 - l_3^2] \\ &= -[l_1^2, l_3^2] \\ & = [l_3^2, l_1^2], \end{align*}
where we use that any component of angular momentum commutes with the total angular momentum. A similar proof can be done for the $[l_2^2, l_3^2]$ commutator.

Also, correct me if I am wrong but it appears that this is true for all $|l,m\rangle$, not just the specific case of $l=1$.

Last edited: Jun 3, 2017
11. Jun 7, 2017

### Staff: Mentor

You indeed never need to make use of that. I wonder if this is a red herring or if the the person who wrote the question had another proof in mind, involving ladder operators (where square ladder operators only need to involve two states, $|l, l\rangle$ and $|l, -l\rangle$).

12. Jun 7, 2017

### TSny

Maybe this means to show that $[l_i^2, l_j^2] \, |l, m\rangle = 0$ for $l = 1$.

You might consider using the matrix representation of the angular momentum operators for $l = 1$.
http://quantummechanics.ucsd.edu/ph130a/130_notes/node247.html