# Addition of Angular Momenta

$$\vec{J} = \vec{L} + \vec{S}$$ or $$\vec{J}_1 + \vec{J}_2$$. $$\vec{L} \in\mathcal{E}_{\vec{r}},\vec{S}\in\mathcal{E}_{S},\vec{J}\in\mathcal{E}_J =\mathcal{E}_{\vec{r}}\otimes\mathcal{E}_S\Rightarrow \vec{L}\otimes\mathbb{I}_S + \mathbb{I}_{\vec{r}}\otimes\vec{S}$$.

$$\{H,\vec{L}^2,L_z,\vec{S}^2,\vec{S}_z\}\Rightarrow |n\ell m_\ell;sm_s\rangle$$. $$\{H,\vec{J}^2,\vec{J}_z,\vec{L}^2,\vec{S}^2\}\Rightarrow |njm_j;\ell s\rangle$$.

Choosing the CSCO depends on the Hamiltonian. For the coulomb interaction the first is good. But for more complicated interactions the second set may be a better representation.

Example: Spin-orbit: $$H_{so} = A\vec{L}\cdot\vec{S}$$. $$H=H_{coulomb} + H_{so}$$. Now, $$[L_z,H_{so}] = A[L_z,L_xS_x, L_yS_y+L_zS_z] = Ai\hbar(L_yS_x - L_xS_y)\neq 0$$. So the Hamiltonian no longer commutes with $$L_z$$. Similarly with $$S_z$$, $$[S_z,H_{so}] = Ai\hbar(L_xS_y-L_yS_x) = -[L_z,H_{so}]\neq 0$$. But, $$[L_z+S_z,H_{so}] = [J_z,H_{so}] = 0$$. Similarly, $$[\vec{J}^2,H_{so}] = 0$$.

So, $$\{H,\vec{J}^2,J_z,\vec{L}^2,\vec{S}^2\}$$ provides a complete set of commuting observables giving eigenstates $$|njm_j;\ell s\rangle$$.

## Two Particle Systems

$$\{\vec{J}_1^2,\vec{J}_2^2,J_{1z},J_{2z}\}$$, $$\vec{J} = \vec{J}_1+\vec{J}_2$$. We could also use $$\{\vec{J}_1^2,\vec{J}_2^2,\vec{J}^2,J_z\}$$. The first is the coupled system $$|j_1j_2m_1m_2\rangle$$ and the second is the uncoupled system $$|j_1j_2jm\rangle$$. When we have a dot product, $$H=A\vec{L}\cdot\vec{S}$$ then we have the coupled system, $$[\vec{L}\cdot\vec{S},L_z+S_z]=[\vec{L}\cdot\vec{S},J_z]=0$$.

### Addition of Two Spin 1/2 Particles

$$\ell_1,\ell_2=0, s_1=s_2=\frac{1}{2}$$.

$$\mathcal{E}_s = \mathcal{E}_{s_1} \otimes \mathcal{E}_{s_2}$$ gives a 4 dimensional space.

If we have an uncoupled state: $$|s_1s_2m_1m_2\rangle\equiv |m_1m_2\rangle$$, we can abbreviate since we dictated $$s_i$$ at the start. Then, we have the 4 possible states: $$|\pm\pm\rangle$$. If we have $$\vec{S}_{1,2}^2|m_1m_2\rangle = \hbar^2s_{1,2}(s_{1,2}+1)|m_1m_2\rangle = \hbar^2\frac{3}{4}|m_1m_2\rangle$$. Similarly, $$S_{1z,2z}|m_1m_2\rangle = \hbar m_{1,2}|m_1m_2\rangle$$.

#### Coupled Basis

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.

### Examples

#### Example 1

$$\frac{\vec{S}^2-\vec{S}_1^2-\vec{S}_2^2}{2} = \vec{S}_1\cdot\vec{S}_2$$ 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, $$E_{singlet} = -\frac{3}{2}\hbar^2 B,E_{triplet,-1} = -A\hbar + \frac{B\hbar^2}{4},E_{triplet,0} = \frac{B\hbar^2}{4},E_{triplet,1} = A\hbar+\frac{B\hbar^2}{4}$$.

#### Example 2

$$\vec{S} \doteq \frac{\hbar}{2}\vec{\sigma}$$.

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

$$H=-A\left(\frac{2}{\hbar^2}(S^2-S_z^2)-2\mathbb{I}\right)$$. 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.

#### Example 3 - Spinors

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

### Bounds for Angular Momenta Addition

$$J=|\vec{J}_{max}| = |\vec{J}_1| + |\vec{J}_2|$$, $$j=|\vec{J}_{min}| = ||\vec{J}_1|-|\vec{J}_2||$$.

Then, the allowed are $$j,j+1,\cdots,J-1,J$$. $$m_j = m_1+m_2$$.

If we want to switch bases from the coupled to the uncoupled,

$$|j m_j\rangle = \sum_{m_1,m_2}|m_1m_2\rangle\langle m_1 m_2|jm_j\rangle$$. Thus, the coefficients are $$\langle m_1m_2|jm_j\rangle$$ called $$C(j_1j_2j;m_1m_2m_j)$$ Clebsch-Gordan Coefficients.

The dimensionality of the added system is $$N_{j_1j_2} = (2j_1+1)(2j_2+1) = \sum_{j=|j_1-j_2|}^{|j_1+j_2|}(2j+1)$$.

#### Properties of CG-Coefficients

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.
• Examples

$$\langle j_1j_2m_1m_2|jm_j\rangle=\langle 1/2,1,1/2,0|3/2,1/2\rangle$$. $$|3/2,1/2\rangle = \sqrt{\frac{1/2}{1/2+1}}|1/2 1-1/21\rangle + \sqrt{\frac{1}{1/2+1}}|1/211/20\rangle$$. Thus, our coefficient is attached to the second one, $$=\sqrt{\frac{2}{3}}$$. $$|(j_1+j_2)(j_1+j_2-1)\rangle=\sqrt{\frac{j_1}{j_1+j_2}}|(j_1-1)j_2\rangle + \sqrt{\frac{j_2}{j_1+j_2}}|j_1(j_2-1)\rangle$$

HW: $$(|jm_j\rangle=)|(j_1+j_2-1)(j_1+j_2-1) = \sqrt{\frac{j_1}{j_1+j_2}}|j_1(j_2-1)\rangle - \sqrt{\frac{j_2}{j_1+j_2}}|(j_1-1)j_2\rangle(=|m_1m_2\rangle)$$

Created: 2023-06-25 Sun 02:29

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