Quantum Mechanics: Angular Momentum Operators

[Robben]

Homework Statement

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.

Homework Equations

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

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?

 

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

 

[Robben]

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

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 $0$?

 

[MisterX]

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

 

[Robben]

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

I see, thank you very much!