University Physics V · Multi-Electron Atoms · 12.2
Electron Spin as a Two-State Space
One number from the Stern-Gerlach magnet — two beams, not one and not three — buys a whole new Hilbert-space factor of dimension two. This lesson teaches you to work in it: Pauli matrices as the operators, half-angles as the geometry, and a tensor product with L²(R³) that usually refuses to factorise.
Build the model
Connect the measurement to the mechanism.
A silver atom's ground state has one valence electron in a 5s orbital, so l = 0 and a field gradient should not deflect it. It splits into two beams. Two is the whole experimental input, and it is decisive: an observable with a two-point spectrum and no leftover degeneracy acts on a two-dimensional complex space, and 2l + 1 = 2 has no integer solution, so that space cannot be built from functions of the angles.
The model is then forced. Attach a factor C² to every electron, giving L²(R³) ⊗ C², and put on it the smallest faithful representation of the angular-momentum algebra, S = (ħ/2)σ, so that [Sᵢ, Sⱼ] = iħ εᵢⱼₖ Sₖ exactly as for orbital angular momentum while S² = (3/4)ħ² I is a multiple of the identity — s = ½ is a permanent label of the particle, not a quantum number any state can change. The cost is the classical picture.
Nothing rotates: reproducing ħ/2 from a charged sphere of the classical electron radius needs an equatorial speed near 170c, and the electron is pointlike far below that radius. The factor gₛ = 2 is not derivable from this algebra either; it arrives with the Dirac equation. And because C² is a tensor factor rather than a label, a general one-electron state is a spinor field whose two components need not be proportional — spin and position can entangle, which is exactly how the magnet reads the spin out.
- Simple definition
- Electron spin is a two-dimensional complex vector space that every electron carries alongside its wavefunction, acted on by S = (ħ/2)σ, whose measurement along any axis returns only +ħ/2 or −ħ/2.
- Example
- An electron prepared spin-up along z meets an analyser turned 90° from z and takes each channel with probability cos²(45°) = 0.500; turn that analyser to 180° and the +ħ/2 channel empties completely, because antiparallel — not perpendicular — is what orthogonal means in C².
Every spin question becomes 2×2 linear algebra — eigenvectors of a Hermitian matrix, with no differential equation and no boundary condition to impose.
σ is dimensionless, S carries J s. Basis |+z⟩ = (1,0)ᵀ and |−z⟩ = (0,1)ᵀ, the one in which σz is diagonal.
σᵢ² = I confines every component's eigenvalues to ±1 and tracelessness then forces one of each, while the commutator half returns [Sᵢ, Sⱼ] = iħ εᵢⱼₖ Sₖ, the orbital algebra.
Symmetric part (σᵢ, σⱼ) = 2δᵢⱼ I; antisymmetric part [σᵢ, σⱼ] = 2i εᵢⱼₖ σₖ. All three matrices are dimensionless, Hermitian and traceless.
S² is a multiple of the identity, so s = ½ labels the particle the way charge does — no state of an electron has any other value.
|S| = (√3/2)ħ = 0.866ħ while the largest Sz is 0.500ħ, so S never lies along the axis it is sharp on.
Axes 90° apart give 0.500 each way; only antiparallel axes, Δ = 180°, give orthogonal states. Every pure spin state is up along some n.
θ, φ are the polar angles of n. For axes Δ apart, |⟨+n|+m⟩|² = cos²(Δ/2) — the angle is halved.
Two values of μz in one gradient give two beams. The 2 in gₛ comes from the Dirac equation, the trailing 0.00232 from QED.
μB = eħ/2mₑ = 9.274×10⁻²⁴ J T⁻¹; gₛ = 2.00232, so μz = ∓(gₛ/2)μB = ∓1.00116 μB, and ∂Bz/∂z is in T m⁻¹.
The exchange operator is already diagonal in the coupled basis: symmetric triplet, antisymmetric singlet — the spin half of a Slater determinant.
S² = (3/2)ħ² I + (ħ²/2)σ₁⋅σ₂ returns 2ħ² for the triplet and 0 for the singlet.
Two beams are a statement about dimension
Stern and Gerlach sent a beam of silver atoms through an inhomogeneous field in 1922. Silver is [Kr]4d¹⁰5s¹: the closed shells contribute nothing, and the valence electron sits in an s orbital with l = 0, so the orbital magnetic moment vanishes and there should be no deflection at all. The beam split into two. Count what that requires. A magnetic moment along the field axis takes exactly two values, so the operator measuring it has a two-point spectrum; if nothing else is degenerate, the space it acts on has dimension two. Orbital angular momentum cannot supply it: an l multiplet has 2l + 1 states, and 2l + 1 = 2 needs l = ½, which single-valuedness of the spherical harmonics under a 2π rotation forbids. So the two-fold degeneracy is not hiding anywhere in L²(R³); it is a new tensor factor C², carried by the electron everywhere it goes. Read the history carefully, because it is a trap. Bohr-Sommerfeld space quantisation assigned the atom a k = 1 orbit and admitted only two orientations of it — the orientation with the orbit plane containing the field, m = 0 in modern labels, was ruled out — so two beams were exactly what the old theory predicted, and the 1922 doublet was hailed as its confirmation. Modern quantum mechanics gives l = 1 three states rather than two, and silver's valence electron is not in an l = 1 state at all; the right reading waited for Uhlenbeck and Goudsmit in 1925.
The smallest faithful representation of the angular-momentum algebra
Build the operators from the algebra rather than importing them. Demand [Sᵢ, Sⱼ] = iħ εᵢⱼₖ Sₖ on a two-dimensional space and choose the basis in which Sz is diagonal with eigenvalues ±ħ/2. Up to a choice of phase for |−z⟩, that fixes S = (ħ/2)σ, with σₓ = [[0,1],[1,0]], σᵧ = [[0,−i],[i,0]] and σz = [[1,0],[0,−1]]. Check one product by hand: σₓσᵧ = [[i,0],[0,−i]] = iσz while σᵧσₓ = −iσz, so [σₓ, σᵧ] = 2iσz and [Sₓ, Sᵧ] = iħSz. The ladder operators come free: S± = Sₓ ± iSᵧ gives S₊ = ħ[[0,1],[0,0]], so S₊|−⟩ = ħ|+⟩, S₊|+⟩ = 0 and S₊² = 0 — the ladder has exactly two rungs, which is the algebraic form of the statement that the space is two-dimensional. Each σᵢ is Hermitian and traceless with σᵢ² = I: the square confines its eigenvalues to ±1 and the vanishing trace forces one of each, so every component of S has spectrum ±ħ/2 — along z, along x, along any axis at all. There is no direction in which a spin measurement returns a third answer, or zero. Note also that spin-½ is the smallest faithful representation, not the smallest one: the trivial s = 0 representation is one-dimensional but sends every generator to zero.
Every direction is a basis, and every angle is halved
A general normalised state is |χ⟩ = a|+z⟩ + b|−z⟩ with |a|² + |b|² = 1. Four real parameters, minus one for normalisation and one for the unobservable global phase, leave two — the coordinates of a point on a sphere. Writing |+n⟩ = cos(θ/2)|+z⟩ + e(iφ)sin(θ/2)|−z⟩ makes the identification explicit: this is the eigenvector of n⋅σ with eigenvalue +1, for n with polar angles (θ, φ). Two consequences matter. First, every pure state of an isolated spin is spin-up along some direction; no state fails to point somewhere. Second, the angle is halved: the overlap of two states whose axes are Δ apart is |⟨+n|+m⟩|² = cos²(Δ/2), so axes 90° apart give 0.500 each way, axes 60° apart give 0.750 and 0.250, and zero overlap arrives only at Δ = 180°, where the axes are antiparallel. The same half-angle runs the rotation operator: U(θ, n) = exp(−iθ n⋅S/ħ) = cos(θ/2) I − i sin(θ/2) n⋅σ, so U(2π) = −I. A spinor turned once through a full circle comes back with a minus sign and needs 4π to return, which neutron interferometry has measured.
The state space is L²(R³) ⊗ C², and it rarely factorises
The C² factor is a tensor factor, not a relabelling. One electron's state space is L²(R³) ⊗ C², and a general state is a two-component spinor field Ψ(r) = (ψ₊(r), ψ₋(r))ᵀ, normalised by ∫(|ψ₊|² + |ψ₋|²) d³r = 1. The Born rule reads it jointly: |ψ₊(r)|² d³r is the probability of finding the electron in d³r and spin-up along z, not the probability of position times the probability of spin. Only when the spinor factorises, Ψ(r) = ψ(r)χ, is there such a thing as the spin state, and factorisation is the exception. A Stern-Gerlach magnet is a machine for destroying it: the term −μ⋅B with ∂Bz/∂z ≠ 0 pushes the two components in opposite directions, so an atom entering as ψ₀(r)(|+⟩ + |−⟩)/√2 leaves as (ψᵤₚ(r)|+⟩ + ψdown(r)|−⟩)/√2 with the two packets separated in space. Position and spin are now entangled, which is exactly what makes the spin readable — you look at where the atom landed. This is entanglement between two degrees of freedom of a single particle; no second particle is needed.
Two electrons: four dimensions splitting three and one
Two electrons carry C² ⊗ C², four dimensions, and the total spin S = S₁ + S₂ splits them as 3 ⊕ 1. Start at the top with |↑↑⟩ and lower: S₋|↑↑⟩ = ħ(|↓↑⟩ + |↑↓⟩), which normalises to (|↑↓⟩ + |↓↑⟩)/√2, and lowering again gives |↓↓⟩. Those three are symmetric under exchange and form the S = 1 triplet; the remaining orthogonal state, (|↑↓⟩ − |↓↑⟩)/√2, is antisymmetric and is the S = 0 singlet. Verify with the Casimir: S² = S₁² + S₂² + 2S₁⋅S₂ = (3/2)ħ² I + (ħ²/2)σ₁⋅σ₂, and σ₁⋅σ₂ = +1 on the triplet, −3 on the singlet, giving 2ħ² = 1(1+1)ħ² and 0. The exchange operator on the spin factor is P₁₂ = (I + σ₁⋅σ₂)/2, with eigenvalue +1 on the triplet and −1 on the singlet — already diagonal, which is why the coupled basis is the working basis for identical particles. Antisymmetry of the total state then forces the pairing: a triplet spin function demands an antisymmetric spatial function, which vanishes at r₁ = r₂ and digs the Fermi hole that puts helium's 1s2s triplet 0.796 eV below its singlet.
Nothing is spinning, and gₛ = 2 is not free
Take the classical picture seriously for one calculation and watch it fail. A uniform sphere of mass m and radius r has I = (2/5)mr², so an angular momentum of ħ/2 requires an equatorial speed v = 5ħ/(4mr). With the classical electron radius r = 2.82 × 10⁻¹⁵ m and mₑ = 9.11 × 10⁻³¹ kg that gives v ≈ 5.1 × 10¹⁰ m s⁻¹, about 170c. Scattering experiments bound any electron substructure below 10⁻¹⁸ m, which makes the required speed worse by three further orders of magnitude. Nothing is rotating. The magnetic moment says the same thing from another direction: a circulating charge gives μ = −(e/2m)L, a g-factor of 1, but the spin moment is μ = −gₛ(e/2m)S with gₛ = 2.00232 — twice as magnetic per unit angular momentum as any charge circulation can manage. The algebra of this lesson cannot supply that 2. It falls out of the Dirac equation, where the four-component spinor and minimal coupling together generate the term −(eħ/2m)σ⋅B. The residual 0.00232 is QED, whose leading piece is Schwinger's α/2π = 0.00116 in (g − 2)/2.
Change one variable at a time
Make the relationship visible.
Set the state to 0° and drag the analyser. At 90° the bars are equal; the + bar only vanishes at 180°, where the axes are antiparallel rather than perpendicular. Orthogonal in C² means antipodal on the sphere, because every angle in the probability is halved.
RELATIVE ANGLE Δ120 °
P(+ħ/2) = cos²(Δ/2)0.250
P(−ħ/2) = sin²(Δ/2)0.750
⟨S⋅n⟩ IN STATE m-0.250 ħ
Live interpretationRELATIVE ANGLE Δ: 120 °. P(+ħ/2) = cos²(Δ/2): 0.250. P(−ħ/2) = sin²(Δ/2): 0.750. ⟨S⋅n⟩ IN STATE m: −0.250 ħ
Catch the common trap
Explain before calculating.
An electron is prepared spin-up along z. It then passes through a Stern-Gerlach analyser whose axis n lies at 60° from z. What fraction of the electrons leaves through the +ħ/2 channel of that analyser?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyAn electron is in the state |χ⟩ = ½|+z⟩ + (√3/2)|−z⟩. Find the probability of each Sz outcome, the expectation value ⟨Sz⟩, and the direction n in the x–z plane along which this state is spin-up.
- Normalisation first: |½|² + |√3/2|² = ¼ + ¾ = 1, so the state is already normalised and the Born rule applies directly to the amplitudes.
- Probabilities are squared moduli: P(+ħ/2) = ¼ = 0.250 and P(−ħ/2) = ¾ = 0.750.
- ⟨Sz⟩ = (+ħ/2)(0.250) + (−ħ/2)(0.750) = −ħ/4 = −0.250ħ. No single measurement ever returns −ħ/4; that is the ensemble mean of outcomes each of magnitude ħ/2.
- Match to |+n⟩ = cos(θ/2)|+z⟩ + e(iφ)sin(θ/2)|−z⟩. Both amplitudes are real and positive, so φ = 0, and cos(θ/2) = ½ gives θ/2 = 60°, hence θ = 120°.
- Check against the general result ⟨Sz⟩ = (ħ/2)cos θ: (ħ/2)cos 120° = (ħ/2)(−0.500) = −ħ/4 ✓, with n = (sin 120°, 0, cos 120°) = (0.866, 0, −0.500).
AnswerP(+ħ/2) = 0.250 and P(−ħ/2) = 0.750; ⟨Sz⟩ = −ħ/4 = −0.250ħ; the state is spin-up along n at θ = 120° from +z, that is n = (0.866, 0, −0.500).
MediumThree analysers in a row: the first selects |+z⟩, the second sits at angle θ from z in the x–z plane and its +ħ/2 beam alone is kept, and the third measures Sz. Taking the beam leaving the first analyser as the unit, find the fraction that emerges in the third analyser's −ħ/2 channel as a function of θ, evaluate it at θ = 90° and θ = 60°, and say what happens if the middle analyser is removed.
- Stage one to two: the amplitude is ⟨+n|+z⟩ = cos(θ/2), so the transmitted fraction is cos²(θ/2). The surviving beam is now in |+n⟩, not in |+z⟩ — the analyser prepared it, it did not merely filter it.
- Stage two to three: ⟨−z|+n⟩ = sin(θ/2), so a further fraction sin²(θ/2) leaves through the −ħ/2 channel.
- Multiply the independent stages: f(θ) = cos²(θ/2) sin²(θ/2) = ¼ sin²θ, which peaks at θ = 90°.
- At θ = 90°: f = ¼(1)² = 0.250. At θ = 60°: f = ¼(sin 60°)² = ¼(0.750) = 0.1875.
- Remove the middle analyser and the beam stays in |+z⟩, so the −ħ/2 fraction is |⟨−z|+z⟩|² = 0. Inserting an extra filter raised the output from nothing to ¼ sin²θ — a quarter at θ = 90°, where the middle analyser measures Sₓ, and 0.1875 at θ = 60°. The reason is algebraic, not mechanical: for n = (sin θ, 0, cos θ), [S⋅n, Sz] = sin θ [Sₓ, Sz] = −iħ sin θ Sᵧ, which vanishes only at sin θ = 0, so for every other axis the middle analyser's eigenstates are not Sz eigenstates.
Answerf(θ) = ¼ sin²θ, so f(90°) = 0.250 and f(60°) = 0.1875, against exactly 0 when the middle analyser is removed.
HardTwo electrons are described by the effective spin Hamiltonian H = (J/ħ²) S₁⋅S₂ with J = 0.60 eV. (a) Express S₁⋅S₂ through S² = (S₁ + S₂)². (b) Give the triplet and singlet energies and the splitting. (c) Confirm the same numbers from the exchange operator P₁₂ = (I + σ₁⋅σ₂)/2. (d) State which spatial symmetry each spin sector forces, and which sign of J helium's 1s2s levels correspond to.
- (a) S² = S₁² + S₂² + 2S₁⋅S₂, and each electron has Sᵢ² = (3/4)ħ² I, so S₁⋅S₂ = ½S² − (3/4)ħ².
- (b) Triplet: S² = 1(1+1)ħ² = 2ħ², so S₁⋅S₂ = ħ² − 0.75ħ² = +0.25ħ² and ET = +0.25J = +0.150 eV. Singlet: S² = 0, so S₁⋅S₂ = −0.75ħ² and ES = −0.75J = −0.450 eV. The splitting is ET − ES = J = 0.600 eV, with the singlet lower whenever J > 0.
- (c) S₁⋅S₂ = (ħ²/4)σ₁⋅σ₂ and σ₁⋅σ₂ = 2P₁₂ − I. The triplet has P₁₂ = +1, so σ₁⋅σ₂ = +1 and S₁⋅S₂ = +0.25ħ² ✓; the singlet has P₁₂ = −1, so σ₁⋅σ₂ = −3 and S₁⋅S₂ = −0.75ħ² ✓. The Hamiltonian is the exchange operator in disguise: H = (J/4)(2P₁₂ − I).
- (d) The full two-electron state must be antisymmetric under exchange. A symmetric triplet spin function therefore demands an antisymmetric spatial function, which vanishes at r₁ = r₂; the antisymmetric singlet spin function demands a symmetric spatial function, which does not.
- Helium's 1s2s triplet lies 0.796 eV below its singlet, so there the triplet is the lower sector and the fitted constant is negative: J = ET − ES = −0.796 eV. The +0.60 eV used above is the opposite sign, the antiferromagnetic case a bonding electron pair shows.
AnswerET = +0.150 eV, ES = −0.450 eV, splitting J = 0.600 eV with the singlet lower; triplet ⇒ antisymmetric spatial, singlet ⇒ symmetric spatial. Helium's 1s2s corresponds to J = −0.796 eV.