Part 4 · Spin and angular momentum

4.4 Addition of angular momenta

Source: tensor-product states, two spins ½, Clebsch–Gordan coefficients and orbital plus spin ½ from the 2024/25 board #17; the exercise from board #18; this year’s course has not reached it yet

Objectives

Understand why two angular momenta can be labelled two ways, product or coupled, and why both sets of labels cannot be sharp at once.
Recipe: build coupled states from the top one by lowering; expand one basis in the other with Clebsch–Gordan coefficients.
Picture: the grid of product states, with the diagonals of equal total $m$ that each coupled state lives on.
Theorem: two spins ½ make a triplet and a singlet; $ℓ$ and $s = \frac{1}{2}$ make $J = ℓ \pm \frac{1}{2}$ .
Vocabulary: tensor product, product and coupled basis, triplet, singlet, Clebsch–Gordan coefficient, spin–orbit coupling.

States of two angular momenta

An electron has a position and a spin. Its state needs both, the tensor product $∣ \overset{r}{ˉ} ⟩ \otimes ∣ \pm ⟩$ of a position state and a spin state, “this orbital, with this spin”; in practice a product $ψ (\overset{r}{ˉ}) ∣ \pm ⟩$ . Each operator acts on its own factor, so the total angular momentum is $\hat{\overset{ˉ}{J}} = \hat{\overset{ˉ}{L}} \otimes \hat{I} + \hat{I} \otimes \hat{\overset{ˉ}{S}}, written \hat{\overset{ˉ}{J}} = \hat{\overset{ˉ}{L}} + \hat{\overset{ˉ}{S}} .$ Components of different factors commute, so $\hat{\overset{ˉ}{J}}$ satisfies the angular momentum relations, and everything in §4.2 Angular momentum applies to it (beyond the boards).

For an electron in an atom, the orbital part is $∣ n, ℓ, m_{ℓ} ⟩$ , with the energy, angular momentum and magnetic quantum numbers, and the spin part $∣ s, m_{s} ⟩$ . The angular part has two labellings.

Definition 4.4.1Product and coupled bases2024/25 board #17

Product (uncoupled): $∣ ℓ, m_{ℓ} ⟩ \otimes ∣ s, m_{s} ⟩$ , with $L_{z}$ and $S_{z}$ sharp separately.
Coupled: $∣ J, m, ℓ, s ⟩$ , with the total $\hat{J}^{2} = ℏ^{2} J (J + 1)$ and $\hat{J}_{z} = m ℏ$ sharp.

==They are different bases of the same space==: $\hat{J}^{2}$ does not commute with $\hat{L}_{z}$ or $\hat{S}_{z}$ , so the two sets of labels cannot all be sharp at once. Which one is useful depends on the Hamiltonian; when the energy depends on $\hat{\overset{ˉ}{L}} \cdot \hat{\overset{ˉ}{S}}$ , it is the coupled one (Exercise 4.4.4).

Two spins ½

Two particles of spin ½, total $\hat{\overset{ˉ}{S}} = \hat{\overset{ˉ}{S}}_{1} + \hat{\overset{ˉ}{S}}_{2}$ . In the product basis, each up or down, four states; $∣ +, - ⟩$ means particle 1 up, particle 2 down.

	$∣ +, + ⟩$	$∣ +, - ⟩$	$∣ -, + ⟩$	$∣ -, - ⟩$
$S_{z}$	$ℏ$	$0$	$0$	$- ℏ$
$S^{2}$	$2 ℏ^{2}$	?	?	$2 ℏ^{2}$
$∣ s, m_{s} ⟩$	$∣ 1, 1 ⟩$			$∣ 1, - 1 ⟩$

$∣ +, + ⟩$ has $m_{s} = 1$ , so $s \geq 1$ , and 1 is the most two spins ½ can give: $∣ +, + ⟩ = ∣1, 1 ⟩$ . The middle two have $S_{z} = 0$ sharp but $S^{2}$ not, the board’s question marks: $[\hat{S}^{2}, \hat{S}_{1 z}] \neq = 0$ , because $\hat{S}^{2} = \hat{S}_{1}^{2} + \hat{S}_{2}^{2} + 2 \hat{\overset{ˉ}{S}}_{1} \cdot \hat{\overset{ˉ}{S}}_{2}$ contains $\hat{S}_{1 x}$ and $\hat{S}_{1 y}$ (the reason is beyond the boards). $s = 1$ needs a third state, $m_{s} = 0$ , and one state is left over.

Recipe 4.4.2Building coupled states by lowering2024/25 board #17

The top state is a single product state: $∣ J_{max}, J_{max} ⟩$ has every part at its top.
Apply $\hat{J}_{-} = \hat{J}_{1 -} + \hat{J}_{2 -}$ : on the left with the ladder coefficient of $J$ , on the right each operator on its own particle.
Divide through; repeat for each $m$ .
The state of the next lower $J$ with the same $m$ is the combination orthogonal to the ones found.

$\hat{S}_{-} ∣1, 1 ⟩ = ℏ 2 ∣1, 0 ⟩$ , and $(\hat{S}_{1 -} + \hat{S}_{2 -}) ∣ +, + ⟩ = ℏ∣ -, + ⟩ + ℏ∣ +, - ⟩$ : one up and one down, in equal parts, with a plus sign. The leftover is the same pair with a minus.

Theorem 4.4.3Triplet and singlet2024/25 board #17

Two spins ½ combine into the triplet, $s = 1$ , $∣1, 1 ⟩ = ∣ +, + ⟩, ∣1, 0 ⟩ = \frac{1}{2} (∣ +, - ⟩ + ∣ -, + ⟩), ∣1, - 1 ⟩ = ∣ -, - ⟩,$ and the singlet, $s = 0$ , $∣0, 0 ⟩ = \frac{1}{2} (∣ +, - ⟩ - ∣ -, + ⟩) .$ $2 \times 2 = 4 = 3 + 1$ .

The singlet has $s = 0$ because $\hat{S}_{\pm}$ give zero on it: $\hat{S}_{+} ∣0, 0 ⟩ = \frac{ℏ}{2} (∣ +, + ⟩ - ∣ +, + ⟩) = 0$ . A state that can be neither raised nor lowered from $m = 0$ is the whole ladder (beyond the boards).

A check with the total spinbeyond the boards

$\hat{\overset{ˉ}{S}}_{1} \cdot \hat{\overset{ˉ}{S}}_{2} = \hat{S}_{1 z} \hat{S}_{2 z} + \frac{1}{2} (\hat{S}_{1 +} \hat{S}_{2 -} + \hat{S}_{1 -} \hat{S}_{2 +})$ sends $∣ +, - ⟩$ to $- \frac{ℏ ^{2}}{4} ∣ +, - ⟩ + \frac{ℏ ^{2}}{2} ∣ -, + ⟩$ . On the sum it gives $+ \frac{ℏ ^{2}}{4}$ , so $\hat{S}^{2} = (\frac{3}{4} + \frac{3}{4} + \frac{1}{2}) ℏ^{2} = 2 ℏ^{2}$ ; on the difference $- \frac{3}{4} ℏ^{2}$ , so $\hat{S}^{2} = 0$ .

==Swapping the particles leaves the triplet unchanged and flips the sign of the singlet== (beyond the boards). This decides which spin state two electrons can have in a given orbital (§5.1 Identical particles).

Clebsch–Gordan coefficients

Definition 4.4.4Clebsch–Gordan coefficients2024/25 board #17

A coupled state expands on the product states, $∣ J, m, J_{1}, J_{2} ⟩ = m_{1}, m_{2} \sum ⟨ J_{1}, J_{2}, m_{1}, m_{2} ∣ J, m, J_{1}, J_{2} ⟩ ∣ J_{1}, J_{2}, m_{1}, m_{2} ⟩ .$ The components are the Clebsch–Gordan coefficients, as $b_{n} = ⟨ u_{n} ∣ ψ ⟩$ in Recipe 1.4.5.

For two spins ½ (the table is beyond the boards):

	$∣ +, + ⟩$	$∣ +, - ⟩$	$∣ -, + ⟩$	$∣ -, - ⟩$
$∣ 1, 1 ⟩$	1	0	0	0
$∣ 1, 0 ⟩$	0	$\frac{1}{2}$	$\frac{1}{2}$	0
$∣ 1, - 1 ⟩$	0	0	0	1
$∣ 0, 0 ⟩$	0	$\frac{1}{2}$	$- \frac{1}{2}$	0

Key observation 4.4.5Rules of additionbeyond the boards

A coefficient is zero unless $m = m_{1} + m_{2}$ : $J_{z}$ adds term by term.
$J$ runs from $∣ J_{1} - J_{2} ∣$ to $J_{1} + J_{2}$ in steps of one, and the states add up: $(2 J_{1} + 1) (2 J_{2} + 1) = \sum_{J} (2 J + 1)$ .
The table is an orthogonal matrix, so it reads backwards too: $∣ +, - ⟩ = \frac{1}{2} (∣1, 0 ⟩ + ∣0, 0 ⟩)$ .

Loading the demo…

Figure 4.4.1 Each dot is a product state

∣ m_{1} ⟩ ∣ m_{2} ⟩

; here

ℓ = 1

across and

s = \frac{1}{2}

up. The diagonals join states with the same total

m

. Pick a coupled state

∣ J, M ⟩

: the dots on its diagonal grow with

∣ C ∣^{2}

filled

for

C > 0

and hollow for

C < 0

. Off the diagonal every coefficient is zero. Switch to two spins ½ for the triplet and singlet, or to larger pairs.

Orbital and spin ½

Theorem 4.4.6Coupling ℓ with s = ½2024/25 board #17

An electron with $ℓ \geq 1$ and $s = \frac{1}{2}$ has $J = ℓ \pm \frac{1}{2}$ , and $∣ ℓ \pm \frac{1}{2}, m ⟩ = C_{1}^{\pm} ∣ ℓ, m - \frac{1}{2} ⟩ ∣ + ⟩ + C_{2}^{\pm} ∣ ℓ, m + \frac{1}{2} ⟩ ∣ - ⟩,$ $C_{1}^{\pm} = \pm \frac{ℓ \pm m + \frac{1}{2}}{2 ℓ + 1}, C_{2}^{\pm} = \frac{ℓ \mp m + \frac{1}{2}}{2 ℓ + 1},$ upper signs for $J = ℓ + \frac{1}{2}$ , lower for $J = ℓ - \frac{1}{2}$ . With $ℓ = 0$ there is only $J = \frac{1}{2}$ .

Each coupled state with total $m$ mixes the two product states with $m_{ℓ} + m_{s} = m$ . The board gives the formula without derivation; it comes from lowering the top state. The coefficients are normalised, $(C_{1}^{\pm})^{2} + (C_{2}^{\pm})^{2} = 1$ , and the two $J$ are orthogonal: $C_{1}^{+} C_{1}^{-} + C_{2}^{+} C_{2}^{-} = 0$ , which is what the minus in $C_{1}^{-}$ is for (beyond the boards). At the ends only one product state has the extreme $m$ : $∣ ℓ + \frac{1}{2}, ℓ + \frac{1}{2} ⟩ = ∣ ℓ, ℓ ⟩ ∣ + ⟩$ .

Example 4.4.7ℓ = 1, s = ½2024/25 board #17

$3 \times 2 = 6$ product states become $J = \frac{3}{2}$ (4 states) and $J = \frac{1}{2}$ (2 states): $∣ \frac{3}{2}, \frac{1}{2} ⟩ = \frac{2}{3} ∣1, 0 ⟩ ∣ + ⟩ + \frac{1}{3} ∣1, 1 ⟩ ∣ - ⟩, ∣ \frac{3}{2}, - \frac{1}{2} ⟩ = \frac{1}{3} ∣1, - 1 ⟩ ∣ + ⟩ + \frac{2}{3} ∣1, 0 ⟩ ∣ - ⟩,$ $∣ \frac{1}{2}, \frac{1}{2} ⟩ = - \frac{1}{3} ∣1, 0 ⟩ ∣ + ⟩ + \frac{2}{3} ∣1, 1 ⟩ ∣ - ⟩, ∣ \frac{1}{2}, - \frac{1}{2} ⟩ = - \frac{2}{3} ∣1, - 1 ⟩ ∣ + ⟩ + \frac{1}{3} ∣1, 0 ⟩ ∣ - ⟩ .$ By lowering, as a check (beyond the boards): $\hat{J}_{-} ∣ \frac{3}{2}, \frac{3}{2} ⟩ = ℏ 3 ∣ \frac{3}{2}, \frac{1}{2} ⟩$ , and $(\hat{L}_{-} + \hat{S}_{-}) ∣1, 1 ⟩ ∣ + ⟩ = ℏ 2 ∣1, 0 ⟩ ∣ + ⟩ + ℏ∣1, 1 ⟩ ∣ - ⟩$ .

==Each $m$ has a $2 \times 2$ block==: the two product states with that $m$ turn into one $J = \frac{3}{2}$ and one $J = \frac{1}{2}$ state. In an atom these are the $p_{3/2}$ and $p_{1/2}$ levels, split slightly by spin–orbit coupling (beyond the boards).

Exercises

Exercise 4.4.1Zettili ex. 6.13, board 182024/25 board #18

An electron is in $ψ = R_{21} (r) [\frac{1}{3} Y_{10} ∣ \frac{1}{2}, \frac{1}{2} ⟩ + \frac{2}{3} Y_{11} ∣ \frac{1}{2}, - \frac{1}{2} ⟩]$ . (a) Is $ψ$ an eigenfunction of $\hat{J}_{z}$ , and with what eigenvalue? (b) $S_{z}$ is measured: values and probabilities? © $\hat{J}^{2}$ is measured: values and probabilities?

Solution

(a) $\hat{J}_{z} = \hat{L}_{z} + \hat{S}_{z}$ . $Y_{10} ∣ + ⟩$ has $0 + \frac{1}{2}$ , $Y_{11} ∣ - ⟩$ has $1 - \frac{1}{2}$ : both $\frac{1}{2}$ , so yes, $\hat{J}_{z} ψ = \frac{ℏ}{2} ψ$ . (The board drops the $∣ - ⟩$ of one term.) (b) $+ \frac{ℏ}{2}$ with $\frac{1}{3}$ , $- \frac{ℏ}{2}$ with $\frac{2}{3}$ . Not an eigenfunction of $\hat{S}_{z}$ or $\hat{L}_{z}$ ; only their sum is sharp. © $J = \frac{3}{2}$ , $\frac{15}{4} ℏ^{2}$ , or $J = \frac{1}{2}$ , $\frac{3}{4} ℏ^{2}$ . Project on the $m = \frac{1}{2}$ coupled states of the example: $c_{1} = \frac{2}{3} \frac{1}{3} + \frac{1}{3} \frac{2}{3} = \frac{2 2}{3}$ , $c_{2} = - \frac{1}{3} + \frac{2}{3} = \frac{1}{3}$ . So $\frac{15}{4} ℏ^{2}$ with $\frac{8}{9}$ , $\frac{3}{4} ℏ^{2}$ with $\frac{1}{9}$ .

Exercise 4.4.2beyond the boards

Two spins ½ are in $∣ +, - ⟩$ . The total $S^{2}$ is measured: values and probabilities? What is $S_{z}$ ?

Solution

$∣ +, - ⟩ = \frac{1}{2} (∣1, 0 ⟩ + ∣0, 0 ⟩)$ : $2 ℏ^{2}$ or $0$ , each $\frac{1}{2}$ . $S_{z} = 0$ with certainty.

Exercise 4.4.3beyond the boards

An electron in $∣1, 0 ⟩ ∣ + ⟩$ . Possible $J^{2}$ and probabilities?

Solution

Invert the $m = \frac{1}{2}$ block: $∣1, 0 ⟩ ∣ + ⟩ = \frac{2}{3} ∣ \frac{3}{2}, \frac{1}{2} ⟩ - \frac{1}{3} ∣ \frac{1}{2}, \frac{1}{2} ⟩$ . $\frac{15}{4} ℏ^{2}$ with $\frac{2}{3}$ , $\frac{3}{4} ℏ^{2}$ with $\frac{1}{3}$ .

Exercise 4.4.4Spin–orbit couplingbeyond the boards

Show $\hat{\overset{ˉ}{L}} \cdot \hat{\overset{ˉ}{S}} = \frac{1}{2} (\hat{J}^{2} - \hat{L}^{2} - \hat{S}^{2})$ , and find its eigenvalues for $ℓ = 1$ , $s = \frac{1}{2}$ .

Solution

$\hat{J}^{2} = \hat{L}^{2} + \hat{S}^{2} + 2 \hat{\overset{ˉ}{L}} \cdot \hat{\overset{ˉ}{S}}$ , since $\hat{\overset{ˉ}{L}}$ and $\hat{\overset{ˉ}{S}}$ act on different factors and commute. On $∣ J, m ⟩$ : $\frac{ℏ ^{2}}{2} [J (J + 1) - 2 - \frac{3}{4}]$ , which is $\frac{ℏ ^{2}}{2}$ for $J = \frac{3}{2}$ and $- ℏ^{2}$ for $J = \frac{1}{2}$ . Diagonal in the coupled basis, not in the product one: it splits $p_{3/2}$ from $p_{1/2}$ . The weighted sum $4 \cdot \frac{1}{2} - 2 \cdot 1 = 0$ : the centre of the level does not move.

Exercise 4.4.5beyond the boards

How many states, and which $J$ , for $ℓ = 2$ with $s = \frac{1}{2}$ ? Write $∣ \frac{5}{2}, \frac{5}{2} ⟩$ and $∣ \frac{3}{2}, \frac{3}{2} ⟩$ .

Solution

$5 \times 2 = 10 = 6 + 4$ , $J = \frac{5}{2}, \frac{3}{2}$ . $∣ \frac{5}{2}, \frac{5}{2} ⟩ = ∣2, 2 ⟩ ∣ + ⟩$ . For $∣ \frac{3}{2}, \frac{3}{2} ⟩$ , $m = \frac{3}{2}$ : $C_{1}^{-} = - \frac{1}{5}$ on $∣2, 1 ⟩ ∣ + ⟩$ and $C_{2}^{-} = \frac{2}{5}$ on $∣2, 2 ⟩ ∣ - ⟩$ .