University Physics V · Quantum Spin · 10.2

Spin-Half States in C²

Before any Pauli matrix appears, get the arena right. A spin-half state is two complex numbers, and this lesson is about what they are not: not probabilities, not four free parameters, and not always the whole state. Count what survives normalisation and the global phase, and you know exactly what a spin-half can hold.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Spin-half is the smallest quantum system that is not trivial, and its state space is C²: a two-dimensional inner-product space over the complex numbers. The dimension is not a choice — the su(2) ladder argument gives 2s + 1 = 2 states at s = ½ — but the basis is. Diagonalise the Hermitian operator S_z, name its eigenvectors |↑⟩ and |↓⟩, and every state becomes a column (α, β)ᵀ, with |↑⟩⟨↑| + |↓⟩⟨↓| = I guaranteeing nothing is missing. Two complex components look like four real numbers, but the model spends two of them at once.

Normalisation, |α|² + |β|² = 1, is the Born rule insisting the two S_z outcomes exhaust the possibilities; and multiplying the whole vector by e^(iγ) changes no probability and no expectation value, so a physical state is a ray, not a vector. Two real parameters survive, and they are exactly the polar and azimuthal angles of a direction: every pure spinor is spin-up along some axis. The Casimir S² = (3/4)ħ² I is a multiple of the identity here, so it labels the space rather than sorting states within it. The price of the whole picture is a factorisation assumption.

The electron really lives in L²(R³) ⊗ C², and a lone spinor describes its spin only while the full state is a product of a spatial part and a spin part. Couple spin to position, as a Stern-Gerlach gradient does, and no spinor exists — only a reduced density matrix.

Simple definition: A spin-half state is a normalised vector |ψ⟩ = α|↑⟩ + β|↓⟩ in the two-dimensional complex space C², where |↑⟩ and |↓⟩ are the S_z eigenvectors, |α|² + |β|² = 1, and vectors differing only by an overall phase e^(iγ) are the same physical state.
Example: |ψ⟩ = (√3/2)|↑⟩ + (i/2)|↓⟩ is normalised, since 3/4 + 1/4 = 1, and gives P(S_z = +ħ/2) = 0.75. Multiplying both components by e^(iπ/4) changes nothing measurable; multiplying only the second by −1 flips ⟨S_y⟩ from +0.433ħ to −0.433ħ.

State vector in the S_z eigenbasis|ψ⟩ = α|↑⟩ + β|↓⟩ ↔ (α, β)ᵀ, α, β ∈ ℂ

Fixes the coordinates. Once S_z is diagonalised, every spin state is a two-component complex column and every operator a 2 × 2 matrix with entries ⟨i|A|j⟩.

|↑⟩, |↓⟩: eigenvectors of S_z with eigenvalues +ħ/2 and −ħ/2, ħ = 1.055 × 10⁻³⁴ J s; α, β dimensionless

Orthonormality and completeness⟨↑|↓⟩ = 0, ⟨↑|↑⟩ = ⟨↓|↓⟩ = 1, |↑⟩⟨↑| + |↓⟩⟨↓| = I

Lets you insert I anywhere: ⟨φ|ψ⟩ = ⟨φ|↑⟩⟨↑|ψ⟩ + ⟨φ|↓⟩⟨↓|ψ⟩ = φ↑*α + φ↓*β, which turns every inner product into component arithmetic.

I is the 2 × 2 identity; each |i⟩⟨i| is a dimensionless rank-one projector

Normalisation and the S_z Born rule⟨ψ|ψ⟩ = |α|² + |β|² = 1, P(+ħ/2) = |α|², P(−ħ/2) = |β|²

One real constraint on four real numbers, and the reason the components are amplitudes: |α|² is a probability, α is not.

probabilities dimensionless; they refer to a measurement of S_z only, not of S along any other axis

Global phase and the parameter count|ψ⟩ ≃ e^(iγ)|ψ⟩ ⇒ 4 − 1 (norm) − 1 (phase) = 2 real parameters

Tells you a physical state is a ray, and that the space of spin-half states is a two-parameter surface, not a four-dimensional blob.

γ ∈ [0, 2π) is invisible in every |⟨a|ψ⟩|² and ⟨ψ|A|ψ⟩; the relative phase arg β − arg α is observable

Bloch-angle form of a spinor|ψ⟩ = cos(θ/2)|↑⟩ + e^(iφ) sin(θ/2)|↓⟩

Turns the two surviving parameters into a direction n̂: every pure spin-half state is spin-up along some axis. The half-angle keeps θ = π pointing down, not back up.

θ ∈ [0, π] polar, φ ∈ [0, 2π) azimuthal; ⟨S⟩ = (ħ/2)(sin θ cos φ, sin θ sin φ, cos θ)

S² is a multiple of the identityS² = Sₓ² + S_y² + S_z² = s(s + 1)ħ² I = (3/4)ħ² I

A measurement of S² cannot tell two spin-half states apart. It labels the representation and commutes with everything on C² by construction.

units J² s²; on C² every vector is an eigenvector with the same eigenvalue

Fix the basis by diagonalising S_z

The ladder argument of the previous lesson gives 2s + 1 = 2 states at s = ½, so the spin factor of the state space is C². What it does not give is a basis. S_z is Hermitian, so its eigenvectors span C² and can be chosen orthonormal; call them |↑⟩ = |½, +½⟩ and |↓⟩ = |½, −½⟩, with S_z|↑⟩ = (ħ/2)|↑⟩ and S_z|↓⟩ = −(ħ/2)|↓⟩. That is a choice of coordinates, not a fact about spin — diagonalising Sₓ instead gives an equally good pair. The completeness relation |↑⟩⟨↑| + |↓⟩⟨↓| = I is the promise that two vectors are enough. Once the basis is fixed, every ket is a column (α, β)ᵀ = (⟨↑|ψ⟩, ⟨↓|ψ⟩)ᵀ, every bra the conjugate row (α*, β*), and every operator the matrix Aᵢⱼ = ⟨i|A|j⟩, so S_z = (ħ/2) diag(1, −1). Inner products become arithmetic: for |φ⟩ = (1, 1)/√2 and |ψ⟩ = (1, i)/√2, ⟨φ|ψ⟩ = (1⋅1 + 1⋅i)/2 = (1 + i)/2, and |⟨φ|ψ⟩|² = ½. Conjugate the bra, not the ket: ⟨ψ|φ⟩ = (1 − i)/2 is the complex conjugate, and forgetting that flips the sign of every relative phase you compute.

Normalise, then read the moduli as S_z probabilities

The Born rule says P(S_z = +ħ/2) = |⟨↑|ψ⟩|² = |α|² and P(S_z = −ħ/2) = |β|², so ⟨ψ|ψ⟩ = |α|² + |β|² = 1 is not decoration: it is the two outcomes exhausting the possibilities. Take the un-normalised column (2, 1 + i). Its norm squared is |2|² + |1 + i|² = 4 + 2 = 6 — the modulus of 1 + i is √2, not 2 — so the normalised state is (2, 1 + i)/√6. Then P(↑) = 4/6 = 0.667, P(↓) = 2/6 = 0.333, and ⟨S_z⟩ = (ħ/2)(|α|² − |β|²) = (ħ/2)(1/3) = ħ/6 ≈ 1.76 × 10⁻³⁵ J s. Two cautions follow. First, the moduli speak about S_z only: the same state has ⟨Sₓ⟩ = ħ Re(α*β) = ħ Re[2(1 + i)]/6 = ħ/3, hence P(Sₓ = +ħ/2) = ½(1 + 2⟨Sₓ⟩/ħ) = ½(1 + 2/3) = 5/6, a number no amount of staring at |α|² would reveal. Second, normalisation is a convention on the representative: (2, 1 + i) and (2, 1 + i)/√6 are the same physical state, and the second is merely the version whose components can be read as amplitudes without a further division.

Count the parameters and throw away the global phase

α and β are four real numbers. Normalisation spends one. The global phase spends another: replace |ψ⟩ by e^(iγ)|ψ⟩ and every probability |⟨a|ψ⟩|² picks up |e^(iγ)|² = 1, every expectation ⟨ψ|A|ψ⟩ picks up e^(−iγ)e^(iγ) = 1, so no measurement on the spin alone can see γ. A physical state is therefore a ray — a vector up to complex scale — and two real parameters remain. Spend the phase freedom making α real and non-negative; normalisation then lets you write α = cos(θ/2), β = e^(iφ) sin(θ/2) with θ ∈ [0, π] and φ ∈ [0, 2π). The two survivors are the polar and azimuthal angles of a direction, and the space of rays in C² is the Bloch sphere. The relative phase φ is not a global phase and is fully physical: (1, 1)/√2 and (1, −1)/√2 are orthogonal, with ⟨Sₓ⟩ = +ħ/2 for one and −ħ/2 for the other. Worked: (√3/2, i/2) has cos(θ/2) = √3/2, so θ/2 = 30° and θ = 60°, and arg β − arg α = 90° = φ. Its Bloch vector (sin 60° cos 90°, sin 60° sin 90°, cos 60°) = (0, 0.866, 0.5) gives ⟨S⟩ = (ħ/2)(0, 0.866, 0.5): tilted 60° from z toward +y.

See why S² is trivial on C²

With Sᵢ = (ħ/2)σᵢ and σᵢ² = I, S² = Sₓ² + S_y² + S_z² = 3(ħ/2)² I = (3/4)ħ² I, which is s(s + 1)ħ² at s = ½. Every vector in C² is an eigenvector with the same eigenvalue, so S² has expectation 3ħ²/4 and zero variance in every state, commutes with every 2 × 2 matrix, and a measurement of it returns only what the dimension already told you. Contrast orbital angular momentum, where L² takes different values l(l + 1)ħ² on different subspaces of L²(S²) and genuinely sorts states. The identity is still worth one look. A pure spinor has |⟨S⟩| = ħ/2, so |⟨S⟩|² = ħ²/4 while ⟨S²⟩ = 3ħ²/4: the missing ħ²/2 is quantum spread. For |↑⟩ the pieces are ⟨S_z²⟩ = ħ²/4 and ⟨Sₓ²⟩ = ⟨S_y²⟩ = ħ²/4 each, even though ⟨Sₓ⟩ = ⟨S_y⟩ = 0 — the transverse components are never zero, only zero on average. So ⟨S⟩ never reaches the full length √3ħ/2: the ratio (ħ/2)/(√3ħ/2) = 1/√3 is cos 54.7°, the cone half-angle in every vector-model picture.

Know when two components are the whole state

An electron's state lives in L²(R³) ⊗ C², so in full it is a two-component wavefunction Ψ(r) = ψ↑(r)|↑⟩ + ψ↓(r)|↓⟩ normalised by ∫(|ψ↑|² + |ψ↓|²) d³r = 1. A lone spinor (α, β) describes the spin exactly only when Ψ factorises, Ψ = φ(r) ⊗ (α|↑⟩ + β|↓⟩), which means ψ↑ and ψ↓ are the same spatial function up to the constants α and β. Hamiltonians of the form H_space ⊗ I + I ⊗ Hₛₚᵢₙ keep products as products: a uniform field precesses the spinor and leaves φ alone. A gradient does not. H = −γₑ S_z B′z, with γₑ the gyromagnetic ratio, couples the two factors, and the Stern-Gerlach state α|↑⟩⊗|χ₊⟩ + β|↓⟩⊗|χ₋⟩ with ⟨χ₊|χ₋⟩ → 0 has no spinor at all. What survives is the reduced density matrix ρᵢⱼ = ∫ψᵢ ψⱼ* d³r, a 2 × 2 Hermitian matrix with unit trace; for that state it is diag(|α|², |β|²), whose purity Tr ρ² = |α|⁴ + |β|⁴ falls to ½ at α = β = 1/√2. Every result in this unit that quotes a single (α, β) is silently assuming the product form.

Check it all in NumPy

Represent kets as complex arrays and let the machine police the conjugates. With up = np.array([1, 0], complex) and dn = np.array([0, 1], complex), the completeness check is np.allclose(np.outer(up, up.conj()) + np.outer(dn, dn.conj()), np.eye(2)). For ψ = np.array([2, 1 + 1j]) / np.√(6), the inner product is np.vdot(φ, ψ) — vdot conjugates its first argument, np.dot does not, and that one letter is the difference between ⟨φ|ψ⟩ and a meaningless bilinear form. The S_z probabilities are np.abs(ψ)**2, which returns [0.667, 0.333]. Build S2 = sum((0.5 * s) @ (0.5 * s) for s in (sx, sy, sz)) in units of ħ and assert np.allclose(S2, 0.75 * np.eye(2)). The Bloch vector is [np.vdot(ψ, s @ ψ).real for s in (sx, sy, sz)], which gives (0.667, 0.667, 0.333) with norm 1.000 — unit length is the numerical signature of a pure state — and multiplying psi by np.exp(0.7j) first leaves all three components unchanged to machine precision, which is the global phase dropping out in code.

Change one variable at a time

Make the relationship visible.

Interactive model

Polar angle θ60 °

Relative phase φ45 °

Global phase γ0 °

Drag γ through a full turn: both phasors on the left rotate together while the stacked bar and the Bloch arrow never move — that is the global phase dropping out. Then drag φ and watch only the Bloch arrow swing about z, and θ to trade population between ↑ and ↓.

|α|² = P(S_z = +ħ/2)0.750

|β|² = P(S_z = −ħ/2)0.250

⟨Sₓ⟩/ħ = Re(α*β)0.306

⟨S_z⟩/ħ0.250

Live interpretation|α|² = P(S_z = +ħ/2): 0.750. |β|² = P(S_z = −ħ/2): 0.250. ⟨Sₓ⟩/ħ = Re(α*β): 0.306. ⟨S_z⟩/ħ: 0.250

Catch the common trap

Explain before calculating.

Misconception check

The components are not a recipe. Students read α|↑⟩ + β|↓⟩ as "partly up and partly down" — a fuzzy blend in which the spin has no definite direction and the relative phase is a technicality. It is the reverse. (1, 1)/√2, (1, i)/√2 and (1, −1)/√2 all have |α|² = |β|² = ½, yet they are three different states, pointing along +x, +y and −x, each an exact eigenvector of the spin component along its own axis with zero uncertainty there. The state that genuinely is half up and half down with no preferred direction is the mixture ρ = I/2, the centre of the Bloch sphere, which returns ½ for every axis; no pure spinor does that. |α|² and |β|² are the S_z probabilities and nothing more: they say what a z-oriented magnet would see, and the phase between α and β says what every other magnet would see.

A spin-half is prepared in |ψ⟩ = (|↑⟩ + i|↓⟩)/√2. Which statement about this state is correct?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA spin-half is described by the un-normalised column (3, 4i) in the S_z basis. Normalise it, find the probability of each S_z outcome, and compute ⟨S_z⟩ in J s.

Norm squared: |3|² + |4i|² = 9 + 16 = 25, so the normalising factor is 1/√25 = 1/5 and |ψ⟩ = (3/5)|↑⟩ + (4i/5)|↓⟩. Check: 9/25 + 16/25 = 1.
Born rule for S_z: P(+ħ/2) = |α|² = 9/25 = 0.36 and P(−ħ/2) = |β|² = 16/25 = 0.64. The factor i in β has modulus 1 and touches neither probability.
⟨S_z⟩ = (ħ/2)(|α|² − |β|²) = (ħ/2)(0.36 − 0.64) = −0.14ħ.
With ħ = 1.055 × 10⁻³⁴ J s: ⟨S_z⟩ = −0.14 × 1.055 × 10⁻³⁴ = −1.48 × 10⁻³⁵ J s. A negative mean, but any single measurement still returns +ħ/2 or −ħ/2, never −0.14ħ.

Answer|ψ⟩ = (3/5)|↑⟩ + (4i/5)|↓⟩; P(+ħ/2) = 0.36, P(−ħ/2) = 0.64; ⟨S_z⟩ = −0.14ħ = −1.48 × 10⁻³⁵ J s.

MediumFind the Bloch angles θ and φ and the expectation vector ⟨S⟩ for |ψ⟩ = ((1 + i)/2)|↑⟩ + (1/√2)|↓⟩, then confirm ⟨Sₓ⟩ directly from ⟨Sₓ⟩ = ħ Re(α*β).

Normalisation: |(1 + i)/2|² = 2/4 = ½ and |1/√2|² = ½, summing to 1, so no rescaling is needed.
Polar form of α: |α| = 1/√2 and arg α = arctan(1/1) = π/4, so α = (1/√2)e^(iπ/4). β = 1/√2 has arg 0.
Strip the global phase by multiplying the whole state by e^(−iπ/4), which changes nothing physical: |ψ⟩ ≃ (1/√2)|↑⟩ + (1/√2)e^(−iπ/4)|↓⟩. Now α is real and non-negative, as the Bloch form requires.
Read off: cos(θ/2) = 1/√2 gives θ/2 = 45°, θ = 90°; φ = arg β − arg α = 0 − π/4 = −π/4, i.e. 315°.
Bloch vector (sin θ cos φ, sin θ sin φ, cos θ) = (cos 45°, −sin 45°, 0) = (0.707, −0.707, 0), so ⟨S⟩ = (ħ/2)(0.707, −0.707, 0): on the equator, midway between +x and −y.
Direct check with the original components: α*β = ((1 − i)/2)(1/√2) = (1 − i)/(2√2), so ⟨Sₓ⟩ = ħ Re(α*β) = ħ/(2√2) = 0.354ħ = (ħ/2)(0.707), and ⟨S_y⟩ = ħ Im(α*β) = −0.354ħ. The global phase we removed never entered.

Answerθ = 90°, φ = −45° (equivalently 315°); ⟨S⟩ = (ħ/2)(0.707, −0.707, 0), i.e. ⟨Sₓ⟩ = +0.354ħ, ⟨S_y⟩ = −0.354ħ, ⟨S_z⟩ = 0.

HardAn electron's full state is Ψ(r) = √0.6 φₐ(r)|↑⟩ + √0.4 [cos η φₐ(r) + sin η φ_b(r)]|↓⟩, with φₐ and φ_b orthonormal. For which η is the spin described by a single spinor? Find the reduced spin density matrix, its purity Tr ρ² at η = 0 and η = 90°, and the Bloch-vector length at η = 60°.

Normalisation holds for every η: ∫(|ψ↑|² + |ψ↓|²) d³r = 0.6 + 0.4(cos²η + sin²η) = 1, using the orthonormality of φₐ and φ_b.
A single spinor needs the product form Ψ = φ(r) ⊗ χ, i.e. ψ↓ ∝ ψ↑ as functions of r. That requires sin η = 0: η = 0 or 180°, where Ψ = φₐ ⊗ (√0.6|↑⟩ ± √0.4|↓⟩). Any other η entangles spin with position.
Trace out position: ρᵢⱼ = ∫ψᵢ ψⱼ* d³r. ρ↑↑ = 0.6, ρ↓↓ = 0.4, and ρ↑↓ = √0.6 · √0.4 · ⟨φₐ|cos η φₐ + sin η φ_b⟩ = √0.24 cos η = 0.490 cos η = ρ↓↑.
Purity: Tr ρ² = ρ↑↑² + ρ↓↓² + 2|ρ↑↓|² = 0.36 + 0.16 + 0.48 cos²η = 0.52 + 0.48 cos²η. At η = 0 it is 1.00, a pure spinor; at η = 90° it is 0.52, a mixture — the spatial factor has recorded which spin branch, exactly as a Stern-Gerlach does.
Bloch vector from ρ = (I + a⋅σ)/2: a_z = ρ↑↑ − ρ↓↓ = 0.20, aₓ = 2 Re ρ↑↓ = 0.980 cos η, a_y = 0. At η = 60°: aₓ = 0.490, so |a| = √(0.490² + 0.20²) = √(0.240 + 0.040) = √0.280 = 0.529. Inside the sphere: no spinor has this ⟨S⟩.

AnswerOnly η = 0 or 180° gives a product state. Tr ρ² = 0.52 + 0.48 cos²η: 1.00 at η = 0, 0.52 at η = 90°. At η = 60° the Bloch vector has length 0.529 < 1, so the spin is mixed and no single spinor describes it.

Fix the basis by diagonalising Sz

Normalise, then read the moduli as Sz probabilities