- #1
Denver Dang
- 148
- 1
Homework Statement
Hi.
I'm given a 3-dimensional subspace [itex]H[/itex] that is made up of the states [itex]|1,-1\rangle[/itex], [itex]|1,0\rangle[/itex] and [itex]|1,1\rangle[/itex] with the states defined as [itex]|l,m\rangle[/itex] and [itex]l=1[/itex] as you can see.
The usual operator relations for [itex]L_{z}[/itex] and [itex]L^{2}[/itex] applies, and also:
[tex]L_{+} = L_{x}+iL_{y}[/tex]
[tex]L_{-} = L_{x}-iL_{y}[/tex]
Then I'm told to express the operators [itex]L_{+}[/itex] and [itex]L_{-}[/itex] in [itex]H[/itex].
The answer for [itex]L_{+}[/itex] is supposed to be:
[tex]{{L}_{+}}=\sqrt{2}\hbar \left[ \begin{matrix}
0 & 0 & 0 \\
1 & 0 & 0 \\
0 & 1 & 0 \\
\end{matrix} \right]
[/tex]
And the transposed for [itex]L_{-}[/itex]
But I'm really not sure how that is found.
Homework Equations
The Attempt at a Solution
The [itex]\sqrt{2}\hbar[/itex] probably comes from the fact that:
[tex]{{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[/tex]
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.