University Quantum Mechanics II · Rotations and the Addition of Angular Momentum · 5.4

Two spin-halves: the triplet and the singlet

The smallest case of adding angular momentum, and the one you will reuse most: the two electrons of helium, the electron and proton behind the 21 cm line, an exchange-coupled pair in a magnet. Build the four coupled states by hand, then read S², S_z and exchange symmetry off them without writing a matrix.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Two spin-halves span four states, and each uncoupled state |m₁ m₂⟩ is already an eigenstate of S_z = S₁z + S₂z: one state at M = +1, two at M = 0, one at M = −1. The coupled basis asks a different question, what is S², and its answer is built rather than guessed. |↑↑⟩ is the only M = +1 state, so it tops an s = 1 multiplet; S₋ = S₁₋ + S₂₋ lowers it to (|↑↓⟩ + |↓↑⟩)/√2 and then to |↓↓⟩, and the one M = 0 direction left over, orthogonal to |1, 0⟩, is the s = 0 singlet (|↑↓⟩ − |↓↑⟩)/√2. Because S² = (3/2)ħ² + 2S₁⋅S₂, the operator S₁⋅S₂ is fixed by s alone, +ħ²/4 on the triplet and −3ħ²/4 on the singlet, so any spin Hamiltonian built from it is diagonal in the coupled basis before a matrix is written.

Exchange symmetry comes along free: the triplet is symmetric and the singlet antisymmetric, which makes (1 + σ₁⋅σ₂)/2 the swap operator. The model stops where its assumptions do. A term that treats the two spins differently, such as unequal g-factors in a magnetic field, mixes |1, 0⟩ with |0, 0⟩ and hands the problem back to the uncoupled basis.

With three or more spins the same s appears more than once (½ ⊗ ½ ⊗ ½ gives s = 3/2 once and s = ½ twice), so total spin no longer fixes the exchange symmetry, and the two-spin lock between them is lost.

Simple definition: The triplet and the singlet are the four simultaneous eigenstates of S² and S_z for two spin-½ particles: three states with s = 1, symmetric under exchange, and one state with s = 0, antisymmetric.
Example: |1, 0⟩ = (|↑↓⟩ + |↓↑⟩)/√2 and |0, 0⟩ = (|↑↓⟩ − |↓↑⟩)/√2 both have M = 0, yet S² is 2ħ² = 2.22 × 10⁻⁶⁸ J² s² on the first and exactly 0 on the second: one relative sign separates them.

One lowering step from the top|1, 0⟩ = S₋|↑↑⟩/(√2 ħ) = (|↑↓⟩ + |↓↑⟩)/√2

Derived here: each Sᵢ₋ lowers only its own spin. One more step gives |1, −1⟩ = |↓↓⟩, and a third gives zero, as m = −s requires.

S₋ = S₁₋ + S₂₋ in J s, with Sᵢ₋|↑⟩ = ħ|↓⟩; √2 ħ is the ladder factor ħ√(s(s + 1) − m(m − 1)) at s = m = 1

The singlet, by orthogonality|0, 0⟩ = (|↑↓⟩ − |↓↑⟩)/√2

Derived: S₊ and S₋ both annihilate it, so s = 0. The relative minus sign is forced by orthogonality; only the overall sign is a convention.

The only M = 0 state orthogonal to |1, 0⟩; Condon–Shortley puts the plus sign on the m₁ = +½ term; kets are dimensionless

Spin–spin product fixed by the total spinS₁⋅S₂ = ½[S² − (3/2)ħ²], eigenvalue ½[s(s + 1) − 3/2]ħ²

Derived from S² = S₁² + S₂² + 2S₁⋅S₂ with Sᵢ² = (3/4)ħ². Any Hamiltonian in S₁⋅S₂ is diagonal in the coupled basis, no matrix needed.

S₁⋅S₂ in J² s², with ħ = 1.055 × 10⁻³⁴ J s; the value is +ħ²/4 for s = 1 and −3ħ²/4 for s = 0

The exchange operatorP₁₂ = (1 + σ₁⋅σ₂)/2

Derived: it is +1 on the three symmetric triplet states and −1 on the antisymmetric singlet, so it equals the swap on a complete basis.

σ₁⋅σ₂ = 4S₁⋅S₂/ħ² is dimensionless, +1 on the triplet and −3 on the singlet; P₁₂|a⟩|b⟩ = |b⟩|a⟩

Singlet weight of any two-spin stateP(s = 0) = ⟨χ|(1 − σ₁⋅σ₂)/4|χ⟩

Derived: (1 − σ₁⋅σ₂)/4 is 0 on the triplet and 1 on the singlet. It peaks at ½ for antiparallel spins: no product state is more than half singlet.

Dimensionless; for a product state |n₁⟩|n₂⟩ with unit Bloch vectors n₁ and n₂ it is (1 − n₁⋅n₂)/4

Splitting by a spin–spin couplingE(s = 1) − E(s = 0) = A, for H = (A/ħ²)S₁⋅S₂

Triplet at +A/4, singlet at −3A/4, weighted centroid unmoved: 3 × A/4 − 3A/4 = 0. Unit 9 supplies the coupling constant; this lesson quotes it and does not derive it.

A in J or eV; hydrogen 1s measures A = h × 1420.405751768 MHz = 5.874 μeV, the 21 cm line

Count the states at each M before building any

Two spin-halves span a four-dimensional product space: |↑↑⟩, |↑↓⟩, |↓↑⟩ and |↓↓⟩, with particle 1's arrow written first. Each is already an eigenstate of S_z = S₁z + S₂z with M = m₁ + m₂, so the count is one state at M = +1, two at M = 0 and one at M = −1. Only |↑↑⟩ reaches M = +1, so it must top the largest multiplet. Test it with S² = S₋S₊ + S_z² + ħS_z, quoted from Quantum Mechanics I: S₊ = S₁₊ + S₂₊ annihilates |↑↑⟩, so S²|↑↑⟩ = (0 + ħ² + ħ²)|↑↑⟩ = 2ħ²|↑↑⟩, which is s(s + 1)ħ² with s = 1. An s = 1 multiplet needs one state at each of M = +1, 0 and −1. That uses three of the four; the M = 0 state left over must form a multiplet on its own, with s = 0. The dimensions agree, 2 × 2 = 3 + 1, and the next topic turns this count into the rule for any j₁ and j₂.

Lower from the top, and let the ladder check you

Apply S₋ = S₁₋ + S₂₋ to both sides of |1, 1⟩ = |↑↑⟩. On the coupled side the ladder rule S₋|s, m⟩ = ħ√(s(s + 1) − m(m − 1))|s, m − 1⟩, quoted from Quantum Mechanics I, gives √2 ħ|1, 0⟩. On the uncoupled side each Sᵢ₋ lowers only its own spin, with Sᵢ₋|↑⟩ = ħ|↓⟩, so S₋|↑↑⟩ = ħ(|↓↑⟩ + |↑↓⟩). Equating the two gives |1, 0⟩ = (|↑↓⟩ + |↓↑⟩)/√2, and it arrives normalised: the first check. Lower again: S₋|1, 0⟩ = ħ(|↓↓⟩ + |↓↓⟩)/√2 = √2 ħ|↓↓⟩, while the ladder rule predicts √(2 − 0) ħ|1, −1⟩, so |1, −1⟩ = |↓↓⟩ with coefficient exactly 1, the only M = −1 state there is: the second check. A third lowering gives zero, as it must at m = −s. Every coefficient of the triplet is derived here; only the ladder rule is quoted.

Orthogonality leaves one state; a convention fixes its sign

The M = 0 subspace is two-dimensional and |1, 0⟩ already occupies one direction in it. Whatever is left must be a|↑↓⟩ + b|↓↑⟩ with (a + b)/√2 = 0, so up to an overall phase the singlet is |0, 0⟩ = (|↑↓⟩ − |↓↑⟩)/√2. Confirm that it really has s = 0: S₊|↑↓⟩ = ħ|↑↑⟩ and S₊|↓↑⟩ = ħ|↑↑⟩, so S₊ annihilates the difference, and S₋ does the same through |↓↓⟩. A state that neither raises nor lowers is a one-member multiplet. The relative minus sign is physics, forced by orthogonality; only the overall sign is a choice. The Condon–Shortley convention makes the coefficient with m₁ = +j₁ real and positive, which puts the plus sign on |↑↓⟩. The convention depends on which particle is called 1: write particle 2 first and the same rule gives minus the state, a first sign that the singlet is odd under exchange.

S₁⋅S₂ from S², with no matrix

S₁ and S₂ act on different factors, so they commute and S² = S₁² + S₂² + 2S₁⋅S₂ with no ordering to worry about. Each Sᵢ² is (3/4)ħ² on every state, so S₁⋅S₂ = ½[S² − (3/2)ħ²]: +ħ²/4 on all three triplet states and −3ħ²/4 on the singlet. Two checks. The trace: 3 × ¼ − ¾ = 0, as it must be, because every Pauli matrix is traceless. The vector model: |S₁||S₂| = (3/4)ħ², so the angle between the spins has cos θ = ⅓ in the triplet, 70.5°, and cos θ = −1 in the singlet. The triplet's spins are never parallel; the singlet's are as antiparallel as quantum mechanics allows. In the uncoupled basis the same operator reads S₁zS₂z + ½(S₁₊S₂₋ + S₁₋S₂₊), and its flip-flop term turns |↑↓⟩ into |↓↑⟩. That is why neither uncoupled M = 0 state is an eigenstate, and why S₁⋅S₂ is diagonal only in the coupled basis.

Exchange symmetry comes free: the swap operator

Let P₁₂ swap the two spins, P₁₂|a⟩|b⟩ = |b⟩|a⟩. All three triplet states are unchanged by the swap, eigenvalue +1, and the singlet changes sign, eigenvalue −1. For two spin-halves the labels are locked together: s = 1 is symmetric and s = 0 antisymmetric. Now σ₁⋅σ₂ = 4S₁⋅S₂/ħ² is +1 on the triplet and −3 on the singlet, so (1 + σ₁⋅σ₂)/2 is +1 on the triplet and −1 on the singlet. It agrees with P₁₂ on a complete basis, so the two are the same operator: P₁₂ = (1 + σ₁⋅σ₂)/2, Dirac's exchange identity. Test it on |↑↓⟩: σ₁zσ₂z gives −|↑↓⟩ and σ₁xσ₂x + σ₁yσ₂y gives 2|↓↑⟩, so (1 + σ₁⋅σ₂)/2 sends |↑↓⟩ to |↓↑⟩. The projectors follow the same way, (3 + σ₁⋅σ₂)/4 onto the triplet and (1 − σ₁⋅σ₂)/4 onto the singlet. Unit 7 leans on this lock: for identical fermions the spin symmetry fixes the spatial one.

Zero spin is blind to rotations, and hydrogen measures the gap

Every component of S annihilates the singlet, so exp(−i n⋅S φ/ħ)|0, 0⟩ = |0, 0⟩ for every axis n and angle φ: rotate both spins together and it returns unchanged. Along any common axis it therefore keeps the form (|+n, −n⟩ − |−n, +n⟩)/√2 up to an overall phase, which is why Unit 4 found it alike along every axis; that it is entangled is quoted from UQMII-04.02 and not developed here. A spin Hamiltonian built from S₁⋅S₂ is rotation invariant too, and a measured number tests its eigenvalues. Hydrogen's 1s contact hyperfine term is (A/ħ²)Sₑ⋅Sₚ, with A quoted from Unit 9: the triplet sits at +A/4 and the singlet at −3A/4, so the gap equals the coupling constant itself, and the 21 cm line measures it: h × 1420.405751768 MHz = 5.874 μeV. The quoted leading-order A = (4/3)gₚ(mₑ/mₚ)α⁴mₑc² gives 5.877 μeV: order α⁴, suppressed by mₑ/mₚ, and within 0.1% of the measurement.

Change one variable at a time

Make the relationship visible.

Interactive model

Spin 1 direction θ₁ from +z30 °

Angle from spin 1 to spin 2120 °

Set the angle between the spins to 0° and sweep θ₁: the singlet bar stays at zero while the triplet weight slides from |1,1⟩ through |1,0⟩ to |1,−1⟩. Then open the angle to 180°: the singlet bar stops at the dashed ½ line, since no product state is more than half singlet.

SINGLET WEIGHT P(s = 0)0.375

⟨S₁⋅S₂⟩ / ħ²-0.125

⟨S²⟩ / ħ²1.250

SWAP ⟨P₁₂⟩0.250

Live interpretationSINGLET WEIGHT P(s = 0): 0.375. ⟨S₁⋅S₂⟩ / ħ²: −0.125. ⟨S²⟩ / ħ²: 1.250. SWAP ⟨P₁₂⟩: 0.250

Catch the common trap

Explain before calculating.

Misconception check

Reading the triplet as spins parallel and the singlet as spins antiparallel, and so taking |↑↓⟩, one up and one down, to be the singlet or at least a state of zero spin. M = 0 says only that the z-components cancel; s = 0 needs Sₓ and S_y to annihilate the state as well, and on |↑↓⟩ they do not: Sₓ|↑↓⟩ = (ħ/2)(|↑↑⟩ + |↓↓⟩). The flip-flop part of 2S₁⋅S₂ gives S²|↑↓⟩ = ħ²(|↑↓⟩ + |↓↑⟩), so |↑↓⟩ = (|1, 0⟩ + |0, 0⟩)/√2, half triplet and half singlet, with ⟨S²⟩ = ħ². Meanwhile |1, 0⟩, whose z-components also cancel, has the same S₁⋅S₂ = +ħ²/4 as |↑↑⟩, and the state with both spins along +x, which anyone would call parallel, is ½|1, 1⟩ + (1/√2)|1, 0⟩ + ½|1, −1⟩: pure triplet with half its weight at M = 0. What separates triplet from singlet is the relative sign between |↑↓⟩ and |↓↑⟩, not how many arrows point up.

Two spin-½ particles are prepared in |χ⟩ = 0.8|↑↓⟩ + 0.6|↓↑⟩, and the square of their total spin, S², is measured. Which results can the measurement return, and with what probabilities?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyPositronium's ground level splits into ortho (s = 1) and para (s = 0) states, with a measured gap of 203.39 GHz. Treat its spin dependence as (A/ħ²)S₁⋅S₂. Find A in meV, the shift of each level from the centroid, and compare A with the leading-order QED value (7/12)α⁴mₑc², which is quoted here, not derived.

Any rotation-invariant spin operator on two spin-halves takes one value on the triplet and one on the singlet, so it has the form a + bS₁⋅S₂; the virtual-annihilation term, which acts only on s = 1, is included.
The gap is A[¼ − (−¾)] = A, so A = hf = (4.1357 × 10⁻¹⁵ eV s)(2.0339 × 10¹¹ s⁻¹) = 8.412 × 10⁻⁴ eV = 0.8412 meV.
Ortho, three states: +A/4 = +0.2103 meV. Para, one state: −3A/4 = −0.6309 meV. Weighted centroid: 3(0.2103) − 0.6309 = 0.0000 meV, as a traceless S₁⋅S₂ requires. The coupling is positive, so para is the ground state.
Order in α: α⁴mₑc² = (7.29735 × 10⁻³)⁴ × 510.999 keV = 1.4490 meV, and 7/12 of it is 0.8453 meV, or 204.39 GHz. The measurement is 0.49% lower, the size of the next, order-α⁵ correction quoted from QED.
Hydrogen's gap is 5.874 μeV, 143 times smaller: its proton moment carries a factor mₑ/mₚ, while the positron's moment equals the electron's.

AnswerA = 0.8412 meV; ortho +0.2103 meV (three states), para −0.6309 meV (ground). Leading-order QED gives 0.8453 meV, 0.49% high, an α⁵ effect.

MediumSpin 1 is prepared along +x and spin 2 along the direction 150° from +z in the xz-plane, in the product state |n₁⟩|n₂⟩. Find the probability of each coupled state |s, m⟩, then ⟨S₁⋅S₂⟩ and ⟨S²⟩, and check ⟨S₁⋅S₂⟩ against the 60° angle between the spins.

With |n⟩ = cos(θ/2)|↑⟩ + sin(θ/2)|↓⟩, quoted from Quantum Mechanics I: |n₁⟩ = 0.7071(|↑⟩ + |↓⟩) and |n₂⟩ = 0.2588|↑⟩ + 0.9659|↓⟩. The product has amplitudes 0.1830 on |↑↑⟩, 0.6830 on |↑↓⟩, 0.1830 on |↓↑⟩ and 0.6830 on |↓↓⟩.
Project: ⟨1, 1|χ⟩ = 0.1830, ⟨1, −1|χ⟩ = 0.6830, ⟨1, 0|χ⟩ = (0.6830 + 0.1830)/√2 = 0.6124 and ⟨0, 0|χ⟩ = (0.6830 − 0.1830)/√2 = 0.3536.
Square: P = 0.0335, 0.4665, 0.3750 and 0.1250, which sum to 1.0000.
⟨S₁⋅S₂⟩ = 0.8750(ħ²/4) + 0.1250(−3ħ²/4) = 0.1250ħ² = 1.39 × 10⁻⁶⁹ J² s². For a product state it must equal ⟨S₁⟩⋅⟨S₂⟩ = (ħ/2)² cos 60° = 0.1250ħ², and it does. Then ⟨S²⟩ = 2ħ² × 0.8750 = 1.75ħ².
P(s = 0) = (1 − cos 60°)/4 = 0.125 depends only on the angle between the spins: rotate both together and it stays fixed while the three triplet weights redistribute.

AnswerP(|1, 1⟩) = 0.033, P(|1, 0⟩) = 0.375, P(|1, −1⟩) = 0.467, P(|0, 0⟩) = 0.125; ⟨S₁⋅S₂⟩ = 0.125ħ² = 1.39 × 10⁻⁶⁹ J² s² and ⟨S²⟩ = 1.75ħ².

HardAn exchange-coupled pair of electron spins has H = (J/ħ²)S₁⋅S₂ + (gμ_B B/ħ)(S₁z + S₂z), with J = 1.20 meV, both g = 2.00, and B along z. Choose the basis that diagonalises H, find the four levels at B = 6.0 T, and find the field at which the ground state changes character.

S₁⋅S₂ = ½[S² − (3/2)ħ²] and S₁z + S₂z = S_z are both diagonal in |s, m⟩, so the coupled basis diagonalises H exactly. In the uncoupled basis the flip-flop part of S₁⋅S₂ would mix |↑↓⟩ and |↓↑⟩.
Levels: E(1, m) = J/4 + m gμ_B B and E(0, 0) = −3J/4. The plus sign holds because the electron's moment is −gμ_B S/ħ.
At 6.0 T, gμ_B B = 2.00 × (5.788 × 10⁻⁵ eV T⁻¹) × 6.0 T = 0.695 meV, so |1, 1⟩ is at +0.995 meV, |1, 0⟩ at +0.300 meV, |1, −1⟩ at −0.395 meV and |0, 0⟩ at −0.900 meV. The singlet is still the ground state, 0.505 meV below |1, −1⟩.
Crossing: J/4 − gμ_B B_c = −3J/4 gives B_c = J/(gμ_B) = 1.20 × 10⁻³ eV ÷ 1.158 × 10⁻⁴ eV T⁻¹ = 10.4 T, and eV ÷ eV T⁻¹ is T. Above it |↓↓⟩ is lowest and the pair's moment jumps from 0 to 2μ_B. H conserves S² and S_z, so the levels cross rather than repel.
Limits: B → 0 recovers the zero-field gap J. If g₁ ≠ g₂, H gains ½(g₁ − g₂)μ_B B(S₁z − S₂z)/ħ, and (S₁z − S₂z)|1, 0⟩ = ħ|0, 0⟩, so the coupled basis is exact only for equal g.

AnswerCoupled basis |s, m⟩. At 6.0 T: +0.995, +0.300, −0.395 and −0.900 meV, singlet lowest. The ground state switches to |1, −1⟩ = |↓↓⟩ at B_c = 10.4 T.