Addition of Angular Momenta

$\vec{J} = \vec{L} + \vec{S}$ or ${\vec{J}}_{1} + {\vec{J}}_{2}$ . $\vec{L} \in E_{\vec{r}}, \vec{S} \in E_{S}, \vec{J} \in E_{J} = E_{\vec{r}} \otimes E_{S} \Rightarrow \vec{L} \otimes I_{S} + I_{\vec{r}} \otimes \vec{S}$ .

${H, {\vec{L}}^{2}, L_{z}, {\vec{S}}^{2}, {\vec{S}}_{z}} \Rightarrow | n ℓ m_{ℓ}; s m_{s} ⟩$ . ${H, {\vec{J}}^{2}, {\vec{J}}_{z}, {\vec{L}}^{2}, {\vec{S}}^{2}} \Rightarrow | n j m_{j}; ℓ s ⟩$ .

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: $H_{s o} = A \vec{L} \cdot \vec{S}$ . $H = H_{c o u l o m b} + H_{s o}$ . Now, $[L_{z}, H_{s o}] = A [L_{z}, L_{x} S_{x}, L_{y} S_{y} + L_{z} S_{z}] = A i ℏ (L_{y} S_{x} - L_{x} S_{y}) \neq 0$ . So the Hamiltonian no longer commutes with $L_{z}$ . Similarly with $S_{z}$ , $[S_{z}, H_{s o}] = A i ℏ (L_{x} S_{y} - L_{y} S_{x}) = - [L_{z}, H_{s o}] \neq 0$ . But, $[L_{z} + S_{z}, H_{s o}] = [J_{z}, H_{s o}] = 0$ . Similarly, $[{\vec{J}}^{2}, H_{s o}] = 0$ .

So, ${H, {\vec{J}}^{2}, J_{z}, {\vec{L}}^{2}, {\vec{S}}^{2}}$ provides a complete set of commuting observables giving eigenstates $| n j m_{j}; ℓ s ⟩$ .

Two Particle Systems

${{\vec{J}}_{1}^{2}, {\vec{J}}_{2}^{2}, J_{1 z}, J_{2 z}}$ , $\vec{J} = {\vec{J}}_{1} + {\vec{J}}_{2}$ . We could also use ${{\vec{J}}_{1}^{2}, {\vec{J}}_{2}^{2}, {\vec{J}}^{2}, J_{z}}$ . The first is the coupled system $| j_{1} j_{2} m_{1} m_{2} ⟩$ and the second is the uncoupled system $| j_{1} j_{2} j m ⟩$ . 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

$ℓ_{1}, ℓ_{2} = 0, s_{1} = s_{2} = \frac{1}{2}$ .

$E_{s} = E_{s_{1}} \otimes E_{s_{2}}$ gives a 4 dimensional space.

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

Coupled Basis

For a coupled state: $| s_{1} s_{2} s m_{s} ⟩$ . ${\vec{S}}_{1, 2}^{2} | s m ⟩ = ℏ^{2} s_{1, 2} (s_{1, 2} + 1) | s m ⟩ = ℏ^{2} \frac{3}{4} | s m ⟩$ . ${\vec{S}}^{2} | s m ⟩ = ℏ^{2} s (s + 1) | s m ⟩$ and $S_{z} | s m ⟩ = ℏ m | s m ⟩$ . What are the possible $s$ and $m_{s}$ ? $s = | s_{1} - s_{2} |, | s_{1} - s_{2} + 1 |, \dots, | s_{1} + s_{2} |$ , $| m_{s} | \leq s$ .

${\vec{S}}^{2} = ({\vec{S}}_{1} + {\vec{S}}_{2})^{2} = {\vec{S}}_{1}^{2} + {\vec{S}}_{2}^{2} + 2 S_{1 z} S_{2 z} + S_{1 +} S_{2 -} + S_{1 -} S_{2 +}$ . So, ${\vec{S}}^{2} | + + ⟩ = 2 ℏ^{2} \frac{3}{4} | + + ⟩ + 2 ℏ^{2} \frac{1}{4} | + + ⟩ + 0 | ? - ⟩ + | - ? ⟩ = 2 ℏ^{2} | + + ⟩$ . Similarly, ${\vec{S}}^{2} | - - ⟩ = 2 ℏ^{2} | - - ⟩$ . For ${\vec{S}}^{2} | + - ⟩ = \frac{3}{2} ℏ^{2} | + - ⟩ - \frac{1}{2} ℏ^{2} + ℏ^{2} | - + ⟩ = ℏ^{2} | + - ⟩ + ℏ^{2} | - + ⟩$ . Thus, the matrix representation is,

\begin{array}{r} {\vec{S}}^{2} ≐ ℏ^{2} (\begin{array}{c} 2 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 2 \end{array}) . \end{array}

Diagonalizing this matrix, we see that we get eigenvalues $λ = 2 ℏ^{2}, 0$ with $2 ℏ^{2}$ triply degenerate.

We then get $| 11 ⟩ = | + + ⟩, | 1 - 1 ⟩ = | - - ⟩, | 10 ⟩ = \frac{1}{\sqrt{2}} (| + - ⟩ + | - + ⟩), | 00 ⟩ = \frac{1}{\sqrt{2}} (| + - ⟩ - | - + ⟩)$ . 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 $S_{1 z} + S_{2 z} = S_{z}$ . $H = A (S_{1 z} + S_{2 z}) + B {\vec{S}}_{1} \cdot {\vec{S}}_{2} = A S_{z} + B \frac{{\vec{S}}^{2} - {\vec{S}}_{1}^{2} - S_{2}^{2}}{2}$ . The energy levels are then, $ℏ A m_{s} + B \frac{ℏ^{2} s (s + 1) - ℏ^{2} s_{1} (s_{1} + 1) - ℏ^{2} s_{2} (s_{2} + 1)}{2} = ℏ A m_{s} + \frac{B ℏ^{2}}{2} [s (s + 1) - \frac{3}{2}]$ .

Total allowed energies: $s = 0, s = 1 : m_{s} = 0, \pm 1$ , then, $E_{s i n g l e t} = - \frac{3}{2} ℏ^{2} B, E_{t r i p l e t, - 1} = - A ℏ + \frac{B ℏ^{2}}{4}, E_{t r i p l e t, 0} = \frac{B ℏ^{2}}{4}, E_{t r i p l e t, 1} = A ℏ + \frac{B ℏ^{2}}{4}$ .

Example 2

$\vec{S} ≐ \frac{ℏ}{2} \vec{σ}$ .

$H = - A (σ_{x}^{(1)} σ_{x}^{(2)} + σ_{y}^{(1)} σ_{y}^{(2)}) = - A (\vec{σ^{(1)}} \cdot {\vec{σ}}^{(2)} - σ_{z}^{(1)} σ_{z}^{(2)}) = - A (\frac{({\vec{σ}}^{(1)} + {\vec{σ}}^{(2)}) - {\vec{σ}}^{(1)} - {\vec{σ}}^{(2)}}{2} - σ_{z}^{(1)} σ_{z}^{(2)}) = - A (\frac{({\vec{σ}}^{(1)} + {\vec{σ}}^{(2)}) - 3 I - 3 I}{2} - σ_{z}^{(1)} σ_{z}^{(2)}) = - A (\frac{({\vec{σ}}^{(1)} + {\vec{σ}}^{(2)}) - 6 I}{2} - \frac{(σ_{z}^{(1)} + σ_{z}^{(2)})^{2} - σ_{z}^{(1) 2} - σ_{z}^{(2) 2}}{2}) = - A (\frac{({\vec{σ}}^{(1)} + {\vec{σ}}^{(2)}) - 6 I}{2} - \frac{(σ_{z}^{(1)} + σ_{z}^{(2)})^{2} - 2 I}{2}) = - \frac{A}{2} (({\vec{σ}}^{(1)} + {\vec{σ}}^{(2)}) - 6 I - (σ_{z}^{(1)} + σ_{z}^{(2)})^{2} + 2 I) = - \frac{A}{2} (({\vec{σ}}^{(1)} + {\vec{σ}}^{(2)}) - 4 I - (σ_{z}^{(1)} + σ_{z}^{(2)})^{2}) = - A (\frac{2}{ℏ^{2}} {\vec{S}}^{2} - \frac{2}{ℏ^{2}} S_{z}^{2} - 2 I)$ .

$H = - A (\frac{2}{ℏ^{2}} (S^{2} - S_{z}^{2}) - 2 I)$ . The energy levels are then $- A (\frac{2}{ℏ^{2}} (s (s + 1) ℏ^{2} + m_{s}^{2} ℏ^{2}) - 2) = - 2 A [s (s + 1) - m_{s}^{2} - 1]$ . Due to the $m_{s}^{2}$ we have some degeneracies.

$E_{s i n g l e t} = 2 A$ , $E_{t r i p l e t, \pm 1} = 0$ , $E_{t r i p l e t, 0} = - 2 A$ .

We may apply a magnetic field to lift the degeneracy.

Example 3 - Spinors

Spin-1/2 particle $ψ (\vec{r}) = (\begin{matrix} Ψ_{+} (\vec{r}) \\ Ψ_{-} (\vec{r}) \end{matrix})$ in $S_{z}$ basis. With $Ψ_{+} (\vec{r}) = R (r) [Y_{0}^{0} + Y_{1}^{0} / \sqrt{3}], Ψ_{-} (\vec{r}) = R (r) / \sqrt{3} [Y_{1}^{1} - Y_{1}^{0}]$ . Alternatively, $ψ (\vec{r}) = ψ_{+} (\vec{r}) | + ⟩ + ψ_{-} (\vec{r}) | - ⟩$

Normalizing: $2 \int_{0}^{\infty} | r R (r) |^{2} d r$ . Note that the two came from the spherical harmonics coefficients adding to 2.

$P (S_{z} = + ℏ / 2) = \int_{0}^{\infty} | ψ_{+} |^{2} d V = \frac{4}{6} = \frac{2}{3}$ .

$⟨ S_{z} ⟩ = \frac{ℏ}{2} (2 / 3 - 1 / 3) = \frac{ℏ}{6}$ .

Consider $S_{x} = \frac{ℏ}{2} σ_{x} = \frac{ℏ}{2} (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix})$ . Diagonalizing, we get $λ = \pm \frac{ℏ}{2}$ with vectors $\frac{1}{\sqrt{2}} \cdot {(| + ⟩ + | - ⟩), (| + ⟩ - | - ⟩)}$ .

Then, $ψ (\vec{r}) = ψ_{+} | + ⟩ + ψ_{-} | - ⟩ = \frac{ψ_{+} + ψ_{-}}{2} (| + ⟩ + | - ⟩) - \frac{ψ_{+} - ψ_{-}}{2} (| + ⟩ - | - ⟩)$ .

Bounds for Angular Momenta Addition

$J = | {\vec{J}}_{m a x} | = | {\vec{J}}_{1} | + | {\vec{J}}_{2} |$ , $j = | {\vec{J}}_{m i n} | = | | {\vec{J}}_{1} | - | {\vec{J}}_{2} | |$ .

Then, the allowed are $j, j + 1, \dots, J - 1, J$ . $m_{j} = m_{1} + m_{2}$ .

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

$| j m_{j} ⟩ = \sum_{m_{1}, m_{2}} | m_{1} m_{2} ⟩ ⟨ m_{1} m_{2} | j m_{j} ⟩$ . Thus, the coefficients are $⟨ m_{1} m_{2} | j m_{j} ⟩$ called $C (j_{1} j_{2} j; m_{1} m_{2} m_{j})$ Clebsch-Gordan Coefficients.

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

Properties of CG-Coefficients

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

$⟨ m_{1} m_{2} | j m ⟩ \neq 0$ only if $m_{j} = m_{1} + m_{2}, | j_{1} - j_{2} | \leq j \leq j_{1} + j_{2}$
The coefficients are real
$⟨ m_{1} m_{2} | j m_{j} ⟩ = ⟨ j_{1} (j - j_{1}) | j j ⟩ > 0$
$⟨ m_{1} m_{2} | j m_{j} ⟩ = (- 1)^{j_{1} + j_{2} - j} ⟨ (- m_{1}) (- m_{2}) | j (- m_{j}) ⟩$
Recursion: $J_{\pm} \sum_{m_{1} m_{2}} ⟨ m_{1} m_{2} | j m_{j} ⟩ | j_{1} j_{2} m_{1} m_{2} = c_{\pm}^{(j)} | j (m_{j} \pm 1) ⟩ (J_{1 \pm} + J_{2 \pm}) \sum = \sum C (m_{1} m_{2}; j m_{j}) (c_{\pm}^{(j_{1})} | (m_{1} \pm 1) m_{2} ⟩ + c_{\pm}^{(j_{2})} | m_{1} (m_{2} \pm 1) ⟩)$ With $c_{\pm}^{(j)} = ℏ \sqrt{j (j + 1) - m_{j} (m_{j} \pm 1)}$
Convention: $(| j m_{j} ⟩ =) | (j_{1} + j_{2}) (j_{1} + j_{2}) ⟩ = | j_{1} j_{2} ⟩ (= | m_{1} m_{2} ⟩)$ I.e. $| 11 ⟩ = | + + ⟩$ We can then act the lowering operator on this to derive the other states.

Examples

$⟨ j_{1} j_{2} m_{1} m_{2} | j m_{j} ⟩ = ⟨ 1 / 2, 1, 1 / 2, 0 | 3 / 2, 1 / 2 ⟩$ . $| 3 / 2, 1 / 2 ⟩ = \sqrt{\frac{1 / 2}{1 / 2 + 1}} | 1 / 21 - 1 / 21 ⟩ + \sqrt{\frac{1}{1 / 2 + 1}} | 1 / 211 / 20 ⟩$ . Thus, our coefficient is attached to the second one, $= \sqrt{\frac{2}{3}}$ . $| (j_{1} + j_{2}) (j_{1} + j_{2} - 1) ⟩ = \sqrt{\frac{j_{1}}{j_{1} + j_{2}}} | (j_{1} - 1) j_{2} ⟩ + \sqrt{\frac{j_{2}}{j_{1} + j_{2}}} | j_{1} (j_{2} - 1) ⟩$

HW: $(| j m_{j} ⟩ =) | (j_{1} + j_{2} - 1) (j_{1} + j_{2} - 1) = \sqrt{\frac{j_{1}}{j_{1} + j_{2}}} | j_{1} (j_{2} - 1) ⟩ - \sqrt{\frac{j_{2}}{j_{1} + j_{2}}} | (j_{1} - 1) j_{2} ⟩ (= | m_{1} m_{2} ⟩)$