for general angular momentum state, it is a “ugly”combination of the orbital state and the spin state:

\left|l,m_l,s,m_s \right> = \left|l,m_l\right>\bigotimes \left|s,m_s \right>

the above combination only possible when:

[L_i,S_j]=0

OK, we start with basic result.

L = ( L_x,L_y,L_z)

L^2 \left|l,m_l \right> = l(l+1) \left |s,m_s \right>

L_z \left|l,m_l \right> = m_l \left |s,m_s \right>

[L_x,L_y]= i L_z

L_{\pm} \left|l,m_l\right> = \sqrt{ l(l+1) - m(m\pm1)} \left|l,m_l \right>

the above are basic properties shared by spin angular momentum operator.

thus, we have follow result:

[L_z, L_{\pm}]=\pm L_{\pm}

[L^2,L_{\pm}]=0

and the total angular momentum operator is

J= L\bigotimes 1 + 1 \bigotimes S

and the state is

\left| j, m_j \right> = \left| l, m_l, s, m_s \right>

J_z \left| j, m_j \right> = m_j \left| j, m_j \right>

by J_z= L_z\bigotimes 1 + 1 \bigotimes S_z

m_j =m_l+m_s

by same on x component and y component:

J_{\pm} = L_{\pm} + S_{\pm}

and by [L_i,S_j]=0

[J_x,J_y ] = i J_z

J^2 = L^2 + S^2 + 2 L \cdot S

To find the possible j , we can use the highest state withe J_{\pm} and J_z . Since the highest state are always symmetric. we have to find the others orthogonal states, which are anti-symmetric.

the result is

\left| l-s \right| \leq j \leq \left| l + s \right|

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