Spectroscopic Notation

$²$P\(_{1/2}\) state. l = 1, s = 1/2, j = 1/2. The total angular momentum is up along z-axis. \(m_j = +\frac{1}{2}\) What is the probability to find electron with spin-down. So, \(j_1=\ell=1\), \(j_2 = s = \frac{1}{2}\), \(j=\ell - s\). The electron spin down relates to \(m_2 = m_s = -\frac{1}{2}\).

\((|jm_j\rangle=)|1/2,1/2\rangle = |3/2-1,3/2-1\rangle = \sqrt{\frac{1}{3/2}}|l,s-1\rangle\rangle - \sqrt{\frac{1/2}{3/2}}|l-1,s\rangle = \sqrt{\frac{2}{3}}|1,-1/2\rangle - \sqrt{\frac{1}{3}}|0, 1/2\rangle\). So, the probability is 2/3.

We could relate this to spinors by using spherical harmonics with \(\ell\) and \(m_\ell\). \((jm_j\rangle=)|1/2,1/2\rangle = \sqrt{\frac{2}{3}}Y_1^1(\Omega)\otimes|-\rangle - \sqrt{\frac{1}{3}}Y_1^0(\Omega)\otimes|+\rangle\doteq \frac{1}{\sqrt{3}}\begin{bmatrix}-Y_1^0\\\sqrt{2}Y_1^1\end{bmatrix}\).

For \(j_1=j_2=1\) \(|2 1\rangle = |2(2-1)\rangle = \frac{1}{\sqrt{2}}|01\rangle + \frac{1}{\sqrt{2}}|10\rangle\)

More than 2 angular momenta. \(\vec{J}=\vec{J}_1 + \vec{J}_2 + \vec{J}_3=(\vec{J}_1 + \vec{J}_2) + \vec{J}_3=\vec{J}' + \vec{J}_3\)

Assume we have 3 spins. \(\vec{S}_1,\vec{S}_2,\vec{S}_3\). Assume 1,2 and 2,3 are neighbors.

\(H = -2J(\vec{S}_1\cdot\vec{S}_2 + \vec{S}_2\cdot\vec{S}_3)\). \(s_1=s_2=s_3=3/2\). Let \(\vec{S}_T = \vec{S}_1 + \vec{S}_2 + \vec{S}_3\), \(\vec{S}_{13} = \vec{S}_1 + \vec{S}_3\). Then, \(\{H,\vec{S}_1^2,\vec{S}_2^2,\vec{S}_3^2,\vec{S}_{13}^2,\vec{S}_T^2,\vec{S}_{Tz}\}\) So, \(H=-2J\frac{\vec{S}_T^2-\vec{S}_{13}^2-\vec{S}_2^2}{2}\). Thus, \(H|s_1s_2s_3s_{13}s_Tm_T\rangle = -\hbar^2J(s_T(s_T+1) - s_{13}(s_{13}+1) - s_2(s_2+1))|s_1s_2s_3s_{13}s_Tm_T\rangle = J\hbar^2(s_{13}(s_{13}+1) + 15/4 - s_T(s_T+1))|s_1s_2s_3s_{13}s_Tm_T\rangle\)

\(s_{13} = s_1+s_3, |s_1-s_3| = 3,2,1,0\)

$s_T = $ see worksheet.