the angular momentum operator L has follow properties:

L^2 \left| l, m \right> = l(l+1) \hbar \left|l,m\right>

L_z \left |l,m \right> = m \hbar \left|l, m \right>

for a spin 1 operator, we know that it is 3 dimensional. thus:

S^2\left|m\right>=2\left|m\right>

S_z\left |m\right>=m\left|m\right>

Thus, we have:

L^2(l=1) = \frac{\hbar}{2} S^2

where the size of L is narrowed to l =1 .

with a position bra act from the left, the L becomes:

L^2 Y_l^m = l(l+1) \hbar Y_l^m

L_z Y_l^m = m \hbar Y_l^m

Y_l^m = \sqrt{\frac{3}{4 \pi}} \left< \hat{r} | l,m \right>

on the counter part of spin 1 operator, the ket can be replaced by a real vector. and for any vector v, we can expand it by the 3 eigenvectors:

\vec{v} = \left| v \right> = \sum \left|m \right> \left< m| v \right> = \sum v_m \left|m\right>

thus, we suitable constant, a unit vector can be written in:

\hat{r} =\sqrt{\frac{4\pi}{3}}\sum\left|m\right>Y_1^{m *}

this is how the spherical harmonic enter the vector and then, later in matrix, and become operator.

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