# Quantum Mechanics: Angular Momentum Operators

1. Feb 28, 2015

### Robben

1. The problem statement, all variables and given/known data

Use the spin$-1$ states $|1,1\rangle, \ |1,0\rangle, \ |1, -1\rangle$ as a basis to form the matrix representations of the angular momentum operators.

2. Relevant equations

$\mathbb{\hat{S}}_+|s,m\rangle = \sqrt{s(s+1)-m(m+1)}\hbar|s,m+1\rangle$
$\mathbb{\hat{S}}_-|s,m\rangle = \sqrt{s(s+1)-m(m-1)}\hbar|s,m-1\rangle$

3. The attempt at a solution

I am wonder how exactly do I compute the equations listed above? So I have $\langle1,1|\mathbb{\hat{S}}_+|1,1\rangle$ but why would this equal $0$.

Also, $\langle1,1|\mathbb{\hat{S}}_+|1,0\rangle = \sqrt{2}\hbar$ and $\langle1,0|\mathbb{\hat{S}}_+|1,0\rangle = 0$, why is that? How do I use the equations, given above, to substitute the given states?

2. Feb 28, 2015

### MisterX

To find the matrix elements we can use these formulas (with $s = 1$)
$\mathbb{\hat{S}}_+|s,m\rangle = \sqrt{s(s+1)-m(m+1)}\hbar|s,m+1\rangle$
$\mathbb{\hat{S}}_-|s,m\rangle = \sqrt{s(s+1)-m(m-1)}\hbar|s,m-1\rangle$
then we "bra" them from the left by $\langle 1, m' \mid$ and use the orthogonality property
$$\langle 1, m' \mid 1, m \rangle = \begin{cases} 1 & m = m' \\ 0 & m \neq m'\end{cases}$$

So, for example
$\langle 1, 1|\mathbb{\hat{S}}_+|1,0\rangle = \sqrt{1(1+1)-0(0+1)}\hbar\langle 1, 1|1,1\rangle = \sqrt{2} \hbar$

3. Feb 28, 2015

### Robben

Can you elaborate please? So no matter the state, i.e. it could be a state $\frac{3}{2}$, if $m\ne m'$ it will always be $0$?

4. Feb 28, 2015

### MisterX

Yes it will always be zero if they are not equal, regardless of $s$.

5. Feb 28, 2015

### Robben

I see, thank you very much!

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