Addition of Angular Momenta

For a coupled state: \(|s_1s_2sm_s\rangle\). \(\vec{S}_{1,2}^2|sm\rangle = \hbar^2 s_{1,2}(s_{1,2}+1)|sm\rangle = \hbar^2\frac{3}{4}|sm\rangle\). \(\vec{S}^2|sm\rangle = \hbar^2 s(s+1)|sm\rangle\) and \(S_z|sm\rangle = \hbar m|sm\rangle\). What are the possible \(s\) and \(m_s\)? \(s=|s_1-s_2|,|s_1-s_2+1|,\cdots,|s_1+s_2|\), \(|m_s|\leq s\).

\(\vec{S}^2 = (\vec{S}_1+\vec{S}_2)^2 = \vec{S}_1^2+\vec{S}_2^2 + 2S_{1z}S_{2z}+S_{1+}S_{2-}+S_{1-}S_{2+}\). So, \(\vec{S}^2|++\rangle = 2\hbar^2\frac{3}{4}|++\rangle + 2\hbar^2 \frac{1}{4}|++\rangle + 0|?-\rangle + |-?\rangle = 2\hbar^2|++\rangle\). Similarly, \(\vec{S}^2|--\rangle = 2\hbar^2|--\rangle\). For \(\vec{S}^2|+-\rangle = \frac{3}{2}\hbar^2|+-\rangle - \frac{1}{2}\hbar^2 + \hbar^2|-+\rangle = \hbar^2|+-\rangle + \hbar^2|-+\rangle\). Thus, the matrix representation is,

\begin{align*} \vec{S}^2 \doteq \hbar^2\begin{pmatrix} 2 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 2 \end{pmatrix}. \end{align*}

Diagonalizing this matrix, we see that we get eigenvalues \(\lambda=2\hbar^2,0\) with \(2\hbar^2\) triply degenerate.

We then get \(|11\rangle = |++\rangle, |1-1\rangle = |--\rangle, |10\rangle = \frac{1}{\sqrt{2}}(|+-\rangle+|-+\rangle), |00\rangle = \frac{1}{\sqrt{2}}(|+-\rangle - |-+\rangle)\). These give the triplet and singlet states.

\(\frac{\vec{S}^2-\vec{S}_1^2-\vec{S}_2^2}{2} = \vec{S}_1\cdot\vec{S}_2\) and \(S_{1z}+S_{2z} = S_z\). \(H=A(S_{1z}+S_{2z}) + B\vec{S}_1\cdot\vec{S}_2 = AS_z + B\frac{\vec{S}^2-\vec{S}_1^2-S_2^2}{2}\). The energy levels are then, \(\hbar Am_s + B\frac{\hbar^2 s(s+1)-\hbar^2 s_1(s_1+1) -\hbar^2 s_2(s_2+1)}{2} = \hbar Am_s + \frac{B\hbar^2}{2}\left[s(s+1)-\frac{3}{2}\right]\).

Total allowed energies: \(s=0,s=1:m_s=0,\pm 1\), then, \(E_{singlet} = -\frac{3}{2}\hbar^2 B,E_{triplet,-1} = -A\hbar + \frac{B\hbar^2}{4},E_{triplet,0} = \frac{B\hbar^2}{4},E_{triplet,1} = A\hbar+\frac{B\hbar^2}{4}\).

\(\vec{S} \doteq \frac{\hbar}{2}\vec{\sigma}\).

\(H=-A(\sigma_x^{(1)}\sigma_x^{(2)}+\sigma_y^{(1)}\sigma_y^{(2)}) = -A(\vec{\sigma^{(1)}}\cdot\vec{\sigma}^{(2)}-\sigma_z^{(1)}\sigma_z^{(2)}) = -A\left(\frac{(\vec{\sigma}^{(1)}+\vec{\sigma}^{(2)}) - \vec{\sigma}^{(1)}-\vec{\sigma}^{(2)}}{2} - \sigma_z^{(1)}\sigma_z^{(2)}\right) = -A\left(\frac{(\vec{\sigma}^{(1)}+\vec{\sigma}^{(2)}) - 3\mathbb{I} - 3\mathbb{I}}{2} - \sigma_z^{(1)}\sigma_z^{(2)}\right) = -A\left(\frac{(\vec{\sigma}^{(1)}+\vec{\sigma}^{(2)}) - 6\mathbb{I}}{2} - \frac{(\sigma_z^{(1)}+\sigma_z^{(2)})^2-\sigma_z^{(1)2}-\sigma_z^{(2)2}}{2}\right) = -A\left(\frac{(\vec{\sigma}^{(1)}+\vec{\sigma}^{(2)}) - 6\mathbb{I}}{2} - \frac{(\sigma_z^{(1)}+\sigma_z^{(2)})^2-2\mathbb{I}}{2}\right) = -\frac{A}{2}\left((\vec{\sigma}^{(1)}+\vec{\sigma}^{(2)}) - 6\mathbb{I} - (\sigma_z^{(1)}+\sigma_z^{(2)})^2+2\mathbb{I}\right) = -\frac{A}{2}\left((\vec{\sigma}^{(1)}+\vec{\sigma}^{(2)}) - 4\mathbb{I} - (\sigma_z^{(1)}+\sigma_z^{(2)})^2\right) = -A(\frac{2}{\hbar^2}\vec{S}^2 - \frac{2}{\hbar^2}S_z^2-2\mathbb{I})\).

\(H=-A\left(\frac{2}{\hbar^2}(S^2-S_z^2)-2\mathbb{I}\right)\). The energy levels are then \(-A\left(\frac{2}{\hbar^2}(s(s+1)\hbar^2 + m_s^2\hbar^2) - 2\right) = -2A[s(s+1) - m_s^2 - 1]\). Due to the \(m_s^2\) we have some degeneracies.

\(E_{singlet} = 2A\), \(E_{triplet,\pm 1} = 0\), \(E_{triplet, 0} = -2A\).

We may apply a magnetic field to lift the degeneracy.

Spin-1/2 particle \(\psi(\vec{r}) = \begin{pmatrix} \Psi_+(\vec{r}) \\ \Psi_-(\vec{r}) \end{pmatrix}\) in \(S_z\) basis. With \(\Psi_+(\vec{r}) = R(r)[Y_0^0+Y_1^0/\sqrt{3}], \Psi_-(\vec{r}) = R(r)/\sqrt{3}[Y_1^1-Y_1^0]\). Alternatively, \(\psi(\vec{r}) = \psi_+(\vec{r})|+\rangle + \psi_-(\vec{r})|-\rangle\)

Normalizing: \(2\int_0^\infty |rR(r)|^2 dr\). Note that the two came from the spherical harmonics coefficients adding to 2.

\(\mathcal{P}(S_z=+\hbar/2) = \int_0^\infty |\psi_+|^2 dV = \frac{4}{6} = \frac{2}{3}\).

\(\langle S_z\rangle = \frac{\hbar}{2}(2/3 - 1/3) = \frac{\hbar}{6}\).

Consider \(S_x = \frac{\hbar}{2}\sigma_x=\frac{\hbar}{2}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\). Diagonalizing, we get \(\lambda = \pm \frac{\hbar}{2}\) with vectors \(\frac{1}{\sqrt{2}}\cdot\{(|+\rangle + |-\rangle),(|+\rangle - |-\rangle)\}\).

Then, \(\psi(\vec{r}) = \psi_+|+\rangle + \psi_-|-\rangle = \frac{\psi_+ + \psi_-}{2}(|+\rangle + |-\rangle) - \frac{\psi_+ - \psi_-}{2}(|+\rangle - |-\rangle)\).

Recall for s=1/2, \(|00\rangle = \frac{1}{\sqrt{2}}(|+-\rangle - |-+\rangle)\), \(|10\rangle = \frac{1}{\sqrt{2}}(|+-\rangle + |-+\rangle)\), \(|1\pm1\rangle = |\pm\pm\rangle\)

1. \(\langle m_1m_2|jm\rangle \neq 0\) only if \(m_j=m_1+m_2,|j_1-j_2|\leq j\leq j_1+j_2\)
2. The coefficients are real
3. \(\langle m_1m_2|jm_j\rangle=\langle j_1 (j-j_1)|jj\rangle>0\)
4. \(\langle m_1m_2|jm_j\rangle = (-1)^{j_1+j_2-j}\langle (-m_1)(-m_2)|j(-m_j)\rangle\)
5. Recursion: \(J_{\pm}\sum_{m_1m_2}\langle m_1m_2|jm_j\rangle|j_1j_2m_1m_2 = c_\pm^{(j)}|j(m_j\pm1)\rangle (J_{1\pm}+J_{2\pm})\sum = \sum C(m_1m_2;jm_j)(c_{\pm}^{(j_1)}|(m_1\pm 1)m_2\rangle + c_{\pm}^{(j_2)}|m_1(m_2\pm1)\rangle)\) With \(c_\pm^{(j)} = \hbar\sqrt{j(j+1)-m_j(m_j\pm 1)}\)
6. Convention: \((|jm_j\rangle=)|(j_1+j_2)(j_1+j_2)\rangle = |j_1j_2\rangle(=|m_1m_2\rangle)\) I.e. \(|11\rangle = |++\rangle\) We can then act the lowering operator on this to derive the other states.