# Homework Help: Angular momentum operators on matrix form

1. Jun 12, 2013

### Denver Dang

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

I'm given a 3-dimensional subspace $H$ that is made up of the states $|1,-1\rangle$, $|1,0\rangle$ and $|1,1\rangle$ with the states defined as $|l,m\rangle$ and $l=1$ as you can see.

The usual operator relations for $L_{z}$ and $L^{2}$ applies, and also:
$$L_{+} = L_{x}+iL_{y}$$
$$L_{-} = L_{x}-iL_{y}$$

Then I'm told to express the operators $L_{+}$ and $L_{-}$ in $H$.

The answer for $L_{+}$ is supposed to be:
$${{L}_{+}}=\sqrt{2}\hbar \left[ \begin{matrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ \end{matrix} \right]$$
And the transposed for $L_{-}$

But I'm really not sure how that is found.

2. Relevant equations

3. The attempt at a solution
The $\sqrt{2}\hbar$ probably comes from the fact that:
$${{L}_{\pm }}\left| l,m \right\rangle =\sqrt{l\left( l+1 \right)-m\left( m\pm 1 \right)}\hbar \left| l,m \right\rangle$$

But I can't figure out why the entries in the matrix is like that.
My first thought was that it should be diagonal, as it was made up of the 3 bases, but as you can see, it is not diagonal.

So I was hoping someone could explain what I'm missing out ?

2. Jun 13, 2013

### MisterX

I don't think that is right; it should change the value of m, and also m may not be greater than $\ell$.

This acts on column vector that has the first row for m = -1, the second row for m =0, and the third row for m= 1.

The first row of the matrix is all zero because no matter what, when $L_{+}$ acts on a state, there is never anything remaining in the m = -1 position. The second row takes whatever was in the m = -1 position and moves it to the m=0 position (multiplied by a constant). The third row of the the matrix takes whatever is in the m=0 and moves it to m =1 position, multiplied by a constant.

This was a conceptual question and not a homework exercise, right?