Spherical harmonics, angular momentum, quantum

[Chronos000]

Homework Statement

I have to construct 3, 3X3 matrices for L_z, L_x, L_y for the spherical harmonics Y(l,m) given l=1 and m = 1,0,-1

So I can determine the relevant harmonics for these values of l and m.

I act with L_z on Y to get

L Y(1,0) = 0

L Y(1,1) = hbar Y(1,1)

L Y(1,-1) = -hbar Y(1,-1)

I'm not sure what to do with this however

 

[fzero]

If you choose the basis

$$Y(1,1) = \begin{pmatrix} 1 \\ 0 \\ 0\end{pmatrix}, ~ Y(1,0) = \begin{pmatrix} 0 \\ 1 \\ 0\end{pmatrix},~ Y(1,-1) = \begin{pmatrix} 0 \\ 0 \\ 1\end{pmatrix},$$

then you should be able to confirm that you've shown that

$$L_z = \begin{pmatrix} \hbar & 0 & 0 \\ 0 & 0 & 0 \\ 0& 0 & -\hbar \end{pmatrix}.$$

You need to work out the action of $$L_x$$ and $$L_y$$ and rewrite them in terms of the basis vectors above. This will give you the matrix representations of the other generators.

 

[Chronos000]

so can I choose whatever basis I want? Y(1,1) could be (0,1,0)?

Do these vectors make up the vertical components of the matrix?

 

[fzero]

so can I choose whatever basis I want? Y(1,1) could be (0,1,0)?

You could choose another basis. The one I suggested is particularly nice if you ever work with the ladder operators

$$L_\pm = L_x \pm i L_y$$

Do these vectors make up the vertical components of the matrix?

The way to see the matrix is to write, say

$$L_z Y(1,1) = m_{11} Y(1,1) + m_{12} Y(1,0) + m_{13} Y(1,-1)$$

$$L_z Y(1,0) = m_{21} Y(1,1) + m_{22} Y(1,0) + m_{23} Y(1,-1)$$

$$L_z Y(1,-1) = m_{31} Y(1,1) + m_{32} Y(1,0) + m_{33} Y(1,-1)$$

Then the coefficients $$m_{ij}$$ are the entries of the corresponding matrix. You should be able to do this for the other components.