University Quantum Mechanics II · Rotations and the Addition of Angular Momentum · 5.5

The allowed values of j, from |j₁ − j₂| to j₁ + j₂

This is where you stop memorising |j₁ − j₂| ≤ j ≤ j₁ + j₂ and start deriving it. A table of how many product states share each M settles which total j exist, how many states each holds and whether any has gone missing, all before a single Clebsch-Gordan coefficient is written down.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Two angular momenta on a product space can be described by the product states |j₁ m₁⟩|j₂ m₂⟩ or by states of definite total j, and the question here is which j occur and how often. The answer is bookkeeping, not algebra. Because J_z = J₁z + J₂z, every product state already has a definite M = m₁ + m₂, so M is a label both descriptions share, and the number n(M) of product states at each M follows by counting pairs: one at the top, M = j₁ + j₂, one more at each step down, until the count levels off at 2 min(j₁, j₂) + 1 once |M| reaches |j₁ − j₂|.

A multiplet of total j owns exactly one state at every M from −j to j, so a column can gain a state only where a new multiplet begins. The staircase therefore starts one multiplet per step and none on the plateau: every j from |j₁ − j₂| to j₁ + j₂ in integer steps, each once, and Σ(2j + 1) = (2j₁ + 1)(2j₂ + 1) confirms nothing is lost. The count is exact and contains no interaction and no power of α.

It fixes how many levels a rotation-invariant coupling can make and how many states each holds, but not their order, their spacing or the superpositions that realise them. And the one-multiplet-per-j rule is special to two angular momenta: three spin-halves give one quartet and two separate doublets, so j and M alone no longer name a state.

Simple definition: Coupling angular momenta j₁ and j₂ produces exactly one multiplet of total angular momentum j for each j = |j₁ − j₂|, |j₁ − j₂| + 1, …, j₁ + j₂, and the 2j + 1 states of those multiplets add up to the (2j₁ + 1)(2j₂ + 1) product states.
Example: An l = 1 electron with spin ½ can only have j = ½ or 3/2, holding 2 + 4 = 6 = 3 × 2 states; sodium's 3p level splits exactly this way, into the upper levels of the D lines at 589.6 nm and 589.0 nm (measured).

Product states at each Mn(M) = min(j₁ + j₂ + 1 − |M|, 2 min(j₁, j₂) + 1), for |M| ≤ j₁ + j₂

Derived by counting pairs (m₁, m₂) with m₁ + m₂ = M: one state at the top, one more per step down, then a flat plateau for |M| ≤ |j₁ − j₂|.

j₁, j₂ and M are pure numbers; J_z = Mħ, ħ = 1.055 × 10⁻³⁴ J s; M = m₁ + m₂ because J_z = J₁z + J₂z

Multiplets of each j, by peelingN(j) = n(j) − n(j + 1)

Derived. Each multiplet owns one state at every M from −j to j (quoted from QM I), so n(M) = Σ N(j) over j ≥ |M| and neighbouring columns differ by one term.

N(j) counts independent multiplets of total j; n = 0 above j₁ + j₂. Pure numbers

Allowed total angular momentumj = |j₁ − j₂|, |j₁ − j₂| + 1, …, j₁ + j₂, each exactly once

Derived for two angular momenta: the triangle rule of vector addition made discrete. With three or more, the same j can occur several times.

2 min(j₁, j₂) + 1 values, all integer or all half-integer; |J|² = j(j + 1)ħ², in J² s²

Dimension checkΣ (2j + 1), summed from j = |j₁ − j₂| to j₁ + j₂, equals (2j₁ + 1)(2j₂ + 1)

Derived as an arithmetic series: 2 min(j₁, j₂) + 1 terms averaging 2 max(j₁, j₂) + 1. A list of j that fails it has lost or doubled a multiplet.

left side counts coupled states |j, M⟩, right side product states |j₁ m₁⟩|j₂ m₂⟩; both are pure counts

Two spin-ones1 ⊗ 1 = 0 ⊕ 1 ⊕ 2, with 9 = 1 + 3 + 5

Derived by peeling: N(2) = 1 − 0, N(1) = 2 − 1, N(0) = 3 − 2. The middle value j = 1 is not optional; the column count forces it.

columns n(M) = 1, 2, 3, 2, 1 from M = 2 down to −2; seen as carbon's ³P levels J = 0, 1, 2 (tables)

Three spin-halves½ ⊗ ½ ⊗ ½ = 3/2 ⊕ ½ ⊕ ½, with 8 = 4 + 2 + 2

Derived in two stages, (0 ⊕ 1) ⊗ ½, and confirmed by N(½) = 3 − 1 = 2. Two orthogonal doublets share j and M, so a third label is needed.

columns 1, 3, 3, 1 from M = 3/2 down to −3/2; the doublets carry S₁₂ = 0 or 1 in one coupling order

Count the product states by M, the label both bases share

Start from what is already established. J = J₁ + J₂ obeys the angular-momentum algebra, so its eigenstates come in multiplets |j, M⟩ with M = −j, …, j; and J_z = J₁z + J₂z, so the product state |j₁ m₁⟩|j₂ m₂⟩ already has J_z = (m₁ + m₂)ħ. M is therefore a label the coupled and uncoupled bases share, and nothing that commutes with J_z mixes different M. Count the product states in each M column. For j₁ ≥ j₂ the pairs with m₁ + m₂ = M have m₁ running from max(−j₁, M − j₂) to min(j₁, M + j₂), which gives n(M) = j₁ + j₂ + 1 − |M| when |M| ≥ j₁ − j₂, and n(M) = 2j₂ + 1 when |M| ≤ j₁ − j₂. For j₁ = 2 and j₂ = 1 the columns from M = 3 down to −3 hold 1, 2, 3, 3, 3, 2, 1 states: a staircase that climbs by one per step and then flattens into a plateau. The fifteen states are 5 × 3, as they must be.

Peel off one multiplet per step, and none on the plateau

Now read the columns from the top. A multiplet of total j owns exactly one state at every M from −j to j, the ladder-operator result quoted from Quantum Mechanics I, so n(M) = Σ N(j) over all j ≥ |M|, where N(j) counts the independent multiplets with total j. Subtracting neighbouring columns leaves one term: N(j) = n(j) − n(j + 1). The top column, M = j₁ + j₂, holds one state, and only a multiplet with j = j₁ + j₂ reaches it. One step down there are two states, provided neither angular momentum is zero; that multiplet claims one, so exactly one new multiplet, j = j₁ + j₂ − 1, must begin. Each further step adds one state and one multiplet until M = |j₁ − j₂|. Below that the columns stop growing, every state is already claimed, and nothing new begins. Derived here without a single Clebsch-Gordan coefficient: j = |j₁ − j₂|, |j₁ − j₂| + 1, …, j₁ + j₂, each once, 2 min(j₁, j₂) + 1 values in all, integers when j₁ + j₂ is an integer and half-integers otherwise.

Check the dimension, then work 1 ⊗ 1 = 0 ⊕ 1 ⊕ 2

A list of j is complete only if its multiplets use up every product state. Summing 2j + 1 from |j₁ − j₂| to j₁ + j₂ is an arithmetic series of 2 min(j₁, j₂) + 1 terms whose average is 2 max(j₁, j₂) + 1, so Σ(2j + 1) = (2j₁ + 1)(2j₂ + 1) exactly. For two spin-ones the columns from M = 2 down to −2 hold 1, 2, 3, 2, 1 states. Peeling gives N(2) = 1 − 0, N(1) = 2 − 1 and N(0) = 3 − 2, one of each, and 5 + 3 + 1 = 9 = 3 × 3. Nature shows the count wherever an orbital L = 1 meets a total spin S = 1. Carbon's ground term ³P has exactly three fine-structure levels, J = 0, 1 and 2, at 0, 16.4 and 43.4 cm⁻¹ in the spectroscopic tables, and submillimetre astronomers observe the transitions between them at 609 μm and 370 μm. There is no fourth level, and a magnetic field resolves the three into 1, 3 and 5 sublevels, nine in all.

Limiting cases: a spinless partner, a spin-half and large j

Three limits test the rule. With j₂ = 0 every column holds one state and only j = j₁ survives: adding nothing changes nothing. With j₂ = ½ it gives j = j₁ ± ½, the two levels of every l ≥ 1 electron that fine structure will split, and j = ½ alone when j₁ = 0. For large quantum numbers it becomes the triangle rule for adding vectors of lengths j₁ħ and j₂ħ, whose sum can be anything from |j₁ − j₂|ħ to (j₁ + j₂)ħ; quantum mechanics keeps the range and allows only integer steps. The discreteness has a measurable edge. Even the top multiplet has |J| = √((j₁ + j₂)(j₁ + j₂ + 1)) ħ, shorter than √(j₁(j₁ + 1)) ħ + √(j₂(j₂ + 1)) ħ: for two spin-halves 1.414ħ against 1.732ħ, so the spins are never exactly parallel, while for j₁ = j₂ = 50 the ratio is 0.995 and the classical picture has returned. The units never change: every |J|² eigenvalue is j(j + 1)ħ², in J² s².

Add a third spin-half: one quartet and two distinct doublets

Three spin-halves have 2³ = 8 states. Couple the first pair, ½ ⊗ ½ = 0 ⊕ 1 from the previous lesson, then add the third to each result: 0 ⊗ ½ = ½ and 1 ⊗ ½ = ½ ⊕ 3/2. The total is one j = 3/2 quartet and two j = ½ doublets, 4 + 2 + 2 = 8. The M count agrees without choosing any order: the columns at M = 3/2, ½, −½ and −3/2 hold 1, 3, 3, 1 states, so N(3/2) = 1 and N(½) = 3 − 1 = 2. The two doublets are orthogonal multiplets with the same j, so j and M no longer name a state. Coupling in this order supplies the extra label, S₁₂ = 0 or 1, the spin of the first pair. Couple spins 2 and 3 first and you get a different pair of doublets spanning the same four-dimensional j = ½ subspace, while the quartet, the only j = 3/2 multiplet, is the same in every order. Nitrogen's ground term, ⁴S₃/₂ in the tables, is that quartet: three 2p electron spins in the largest total spin they can form.

What the count fixes, and what it leaves to the Hamiltonian

The count is kinematics. It uses only the commutation relations, so it is exact, involves no interaction and carries no power of the fine-structure constant α. For two angular momenta it fixes the levels any rotation-invariant coupling such as A J₁⋅J₂ can make: at most one per j, each (2j + 1)-fold degenerate. The energies are dynamics, and arrive at very different orders. Sodium's 3p level splits into j = ½ and 3/2 by a measured 17.2 cm⁻¹, or 2.13 meV, about 10⁻³ of the 2.10 eV D-line photon: fine structure, of relative order α² ≈ 5 × 10⁻⁵ in hydrogen and enlarged in sodium by the nuclear charge the 3p electron feels near the core. The 3s ground level splits into F = 1 and 2 by a measured 1771.6 MHz, 7.33 μeV, a hyperfine effect held down by the nucleus's small magnetic moment. Neither interaction changes the list the count produced, j for the fine structure and F for the hyperfine structure, or the number of states in each level. What counting cannot give is the ordering, the spacing, and the superpositions of product states that form each |j, M⟩; the next lesson builds those by lowering and orthogonality.

Change one variable at a time

Make the relationship visible.

Interactive model

First angular momentum j₁2.0

Second angular momentum j₂1.0

Shaded column, steps below top1

Set j₂ = 0 and one row is left; set j₂ = 0.5 and every j₁ ≥ 0.5 gives two rows, j₁ ± ½. At j₁ = j₂ = 1 read the columns 1, 2, 3, 2, 1 as rows j = 2, 1, 0, then step the shaded column down: a new row starts only where a column grows.

SHADED COLUMN M2.0

PRODUCT STATES THERE n(M)2

MULTIPLETS WITH j = M1

TOTAL STATES15

Live interpretationSHADED COLUMN M: 2.0. PRODUCT STATES THERE n(M): 2. MULTIPLETS WITH j = M: 1. TOTAL STATES: 15

Catch the common trap

Explain before calculating.

Misconception check

The commonest slip is to generalise the two cases already met, ½ ⊗ ½ = 0 ⊕ 1 and l ⊗ ½ giving l ± ½, into the rule that j is either j₁ + j₂ or |j₁ − j₂|. It is right whenever one partner is 0 or a spin-half, because then nothing lies between the extremes, and wrong for every other pair. For two spin-ones it predicts j = 0 and 2 only, 1 + 5 = 6 states out of 9: the j = 1 multiplet that the column count N(1) = 2 − 1 forces into existence has vanished, and carbon's measured J = 1 level at 16.4 cm⁻¹ has nowhere to come from. The opposite slip, running j down to 0 or ½ whatever j₁ and j₂ are, overfills the space: for j₁ = 2 and j₂ = 1 it adds j = 0 and claims 16 states out of 15. Both are caught in one line by Σ(2j + 1) = (2j₁ + 1)(2j₂ + 1).

A ³F term has orbital angular momentum L = 3 and total spin S = 1. Counting the product states at each M = M_L + M_S, which fine-structure levels J does the term contain, and how many states does it hold in all?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyIn ground-state deuterium the electron's j = ½ couples to the deuteron's nuclear spin I = 1 to give a total F. Find the allowed F by counting M-values, check the dimension, and find the wavelength and photon energy of the transition between the two levels, whose measured frequency is 327.384 MHz.

Take j₁ = I = 1 and j₂ = ½: 3 × 2 = 6 product states. By M = m₁ + m₂ the columns are M = 3/2 with one state (m₁ = 1, m₂ = ½), M = ½ with two, M = −½ with two and M = −3/2 with one.
Peel from the top: N(3/2) = n(3/2) − n(5/2) = 1 − 0 = 1, and N(½) = n(½) − n(3/2) = 2 − 1 = 1. So F = 3/2 and F = ½, once each, which is j₁ ± ½ as the spin-half limit requires.
Dimension check: (2 × 3/2 + 1) + (2 × ½ + 1) = 4 + 2 = 6 = 3 × 2 ✓.
The frequency is quoted from measurement. Wavelength: λ = c/ν = (2.998 × 10⁸ m s⁻¹)/(3.27384 × 10⁸ s⁻¹) = 0.9157 m.
Photon energy: hν = (6.626 × 10⁻³⁴ J s)(3.27384 × 10⁸ s⁻¹) = 2.169 × 10⁻²⁵ J = 1.354 × 10⁻⁶ eV.

AnswerF = ½ and F = 3/2, holding 2 + 4 = 6 states; the photon has λ = 91.6 cm and energy 2.17 × 10⁻²⁵ J (1.35 μeV).

MediumSodium's 3p electron has l = 1 and s = ½, and the ²³Na nucleus has spin I = 3/2. Couple l and s to j, then j and I to F. List every hyperfine level of the 3p configuration with its degeneracy, check the total, and say which F values occur more than once.

Product space: (2l + 1)(2s + 1)(2I + 1) = 3 × 2 × 4 = 24 states.
First stage, 1 ⊗ ½: j = ½ and j = 3/2, holding 2 + 4 = 6 = 3 × 2 states.
Second stage for j = ½ with I = 3/2: F = 1 and 2, holding 3 + 5 = 8 = 2 × 4. For j = 3/2 with I = 3/2 the columns from M = 3 down to 0 hold 1, 2, 3, 4 states, so N(3) = N(2) = N(1) = N(0) = 1: F = 0, 1, 2, 3, holding 1 + 3 + 5 + 7 = 16 = 4 × 4.
Total 8 + 16 = 24 ✓, in six hyperfine levels. F = 1 and F = 2 each occur twice, once under each j, so F and M alone do not name a state; the intermediate j does.
Independent check over all 24 states: the columns M = 3, 2, 1, 0 hold 1, 3, 5, 6 product states, so N(3) = 1, N(2) = 3 − 1 = 2, N(1) = 5 − 3 = 2 and N(0) = 6 − 5 = 1, matching the list. Measured spectra resolve the four j = 3/2 hyperfine levels, F = 0, 1, 2, 3, spaced by 15.8, 34.3 and 58.3 MHz (quoted from tables).

AnswerSix levels: F = 1 (3 states) and F = 2 (5) under j = ½; F = 0 (1), 1 (3), 2 (5) and 3 (7) under j = 3/2. That is 24 states, with F = 1 and F = 2 each appearing twice.

HardThe four hydrogen nuclei of methane each have spin ½. Find how many multiplets of each total nuclear spin I the 16 spin states form, check the result both by dimension and by the trace of I², and predict the population ratio of the spin isomers when all 16 spin states are equally populated.

Count by M = m₁ + m₂ + m₃ + m₄. Choosing which k of the four spins point down gives C(4, k) states at M = 2 − k, so the columns from M = 2 down to −2 hold 1, 4, 6, 4, 1.
Peel: N(2) = 1, N(1) = 4 − 1 = 3, N(0) = 6 − 4 = 2. Dimension: 1 × 5 + 3 × 3 + 2 × 1 = 16 = 2⁴ ✓. Coupling in pairs agrees: (0 ⊕ 1) ⊗ (0 ⊕ 1) = 0 ⊕ 1 ⊕ 1 ⊕ (0 ⊕ 1 ⊕ 2).
Trace in the product basis: I² is the sum of the four Iₖ² plus twice the six pair products Iₖ⋅Iₗ. Each Iₖ² equals (3/4)ħ² on all 16 states, a trace of 12ħ² apiece and 48ħ² for four. Each pair product is traceless, because every Pauli matrix is, so Tr I² = 48ħ².
Trace in the coupled basis: Σ N(I)(2I + 1)I(I + 1)ħ² = (1 × 5 × 6 + 3 × 3 × 2 + 2 × 1 × 0)ħ² = (30 + 18 + 0)ħ² = 48ħ² = 5.34 × 10⁻⁶⁷ J² s² ✓, with no coefficient needed.
Equal population of all 16 states puts 5, 9 and 2 of them in I = 2, 1 and 0. The isomers are called meta (I = 2), ortho (I = 1) and para (I = 0), and 5 : 9 : 2 is the ratio quoted for warm methane.

AnswerI = 2 once, I = 1 three times and I = 0 twice (5 + 9 + 2 = 16); Tr I² = 48ħ² in both bases; meta : ortho : para = 5 : 9 : 2, or 31.3%, 56.3% and 12.5%.