Angular momentum operators on matrix form

[Denver Dang]

Homework Statement

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}<br /> 0 &amp; 0 &amp; 0 \\<br /> 1 &amp; 0 &amp; 0 \\<br /> 0 &amp; 1 &amp; 0 \\<br /> \end{matrix} \right]<br />
And the transposed for L_{-}

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

Homework Equations

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 ?Thanks in advance.

 

[MisterX]

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

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

The answer for L_{+} is supposed to be:
{{L}_{+}}=\sqrt{2}\hbar \left[ \begin{matrix}<br /> 0 &amp; 0 &amp; 0 \\<br /> 1 &amp; 0 &amp; 0 \\<br /> 0 &amp; 1 &amp; 0 \\<br /> \end{matrix} \right]

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?