University Physics V · Quantum Spin · 10.3
Pauli Matrices & the Spin Operator
Spin-half needs no wavefunction. Diagonalise Sz and every observable on the space becomes a 2×2 matrix built from three fixed arrays. This lesson is about the one product rule that generates all of them — the commutators, the anticommutators, and every rotation you will ever apply to a qubit — and about where that rule stops.
Build the model
Connect the measurement to the mechanism.
Unit 10 began with an algebra, [Sᵢ, Sⱼ] = iħ εᵢⱼₖ Sₖ, acting on a two-dimensional space. Diagonalising Sz and fixing the one leftover relative phase turns that abstract statement into three concrete arrays — the Pauli matrices — and the single definition S = (ħ/2)σ. What you buy is that one line of arithmetic, σᵢσⱼ = δᵢⱼ I + i εᵢⱼₖ σₖ, now generates everything: its antisymmetric half returns the su(2) commutators, and its symmetric half, (σᵢ, σⱼ) = 2δᵢⱼ I, is a Clifford relation the commutators never implied.
From σᵢ² = I follows (n̂⋅σ)² = I for every unit vector, which is why a spin component along any axis has eigenvalues ±ħ/2, and why exp(−iθ n̂⋅σ/2) collapses from an infinite series to two terms. And because I together with the three σᵢ span the Hermitian 2×2 matrices, every qubit observable, Hamiltonian and density matrix is a₀I + a⋅σ — four real numbers, with eigenvalues a₀ ± |a|. The cost is paid twice.
The arrays are a representation, not the physics: another basis or phase convention hands you different arrays obeying the same algebra, so only the trace, determinant, eigenvalues and commutators are claims about spin. And the anticommutator is a two-dimensional accident. Spin-1 obeys the same commutators with 3×3 matrices for which Sz² is not a multiple of the identity, and every shortcut here that leaned on σᵢ² = I dies there.
- Simple definition
- The Pauli matrices are the three Hermitian, traceless 2×2 matrices that square to the identity and represent spin-half in the Sz eigenbasis, the spin operator itself being S = (ħ/2)σ.
- Example
- σz = diag(1, −1) gives Sz = (ħ/2)diag(1, −1) with eigenvalues ±ħ/2 = ±5.27×10⁻³⁵ J s, while σₓ² + σy² + σz² = 3I makes S² = (3/4)ħ²I, the same on every spinor.
Parks every unit in one prefactor, so the algebra below is pure numbers.
Sᵢ in J s; each σᵢ dimensionless, Hermitian, traceless, with σᵢ² = I.
Only σz is diagonal, so only Sz has these basis kets as eigenstates.
Rows and columns ordered up, down. The semicolon separates the two rows.
Nine matrix products stored in one line — every relation below is read off it.
i, j, k run over x, y, z; εᵢⱼₖ is +1 on cyclic order; sum over repeated k.
The first holds at every spin; the second only in two dimensions.
Equivalently [Sᵢ, Sⱼ] = iħ εᵢⱼₖ Sₖ and {Sᵢ, Sⱼ} = (ħ²/2) δᵢⱼ I.
Truncates the exponential to exp(−iα n̂⋅σ) = cos α I − i sin α (n̂⋅σ).
a, b any real 3-vectors in whatever unit they carry. For a = b = n̂: (n̂⋅σ)² = I.
Four numbers fix any qubit Hamiltonian, and the eigenvalues are a₀ ± |a|.
a₀ and a real, in the unit of A; uses Tr σᵢ = 0 and Tr(σᵢ σⱼ) = 2 δᵢⱼ.
Diagonalise Sz first, and the arrays are forced
Nothing here is postulated. The previous topic fixed a two-dimensional space and the one before it fixed the algebra; choose the basis that diagonalises Sz, label its eigenvectors |↑⟩ and |↓⟩ with eigenvalues ±ħ/2, and the rest follows. Sz = (ħ/2)diag(1, −1) by construction. For the other two use the ladder operators S_± = Sₓ ± iSy, whose only non-zero element is ⟨s, m±1|S_±|s, m⟩ = ħ√(s(s+1) − m(m±1)); at s = ½ and m = −½ this is ħ√(3/4 + 1/4) = ħ, so S+ = ħ|↑⟩⟨↓| and S_− = ħ|↓⟩⟨↑| are the strictly upper and lower triangular arrays. Then Sₓ = (S+ + S_−)/2 and Sy = (S+ − S_−)/2i give the familiar σₓ and σy. The only freedom left is the phase of |↓⟩ relative to |↑⟩. The Condon–Shortley convention takes that ladder element real and positive, and that single choice is what puts −i in the top right of σy rather than +i.
One product rule, and the whole algebra is done
σᵢσⱼ = δᵢⱼ I + i εᵢⱼₖ σₖ is nine matrix products compressed into one line, and it is worth checking once by hand: σₓσy = (0 1; 1 0)(0 −i; i 0) = (i 0; 0 −i) = iσz. Now split it by symmetry in i ↔ j. The antisymmetric half is [σᵢ, σⱼ] = 2i εᵢⱼₖ σₖ, which with S = (ħ/2)σ becomes [Sᵢ, Sⱼ] = (ħ²/4)(2i)εᵢⱼₖ(2/ħ)Sₖ = iħ εᵢⱼₖ Sₖ — the su(2) relation you started from, recovered, which is the consistency check that these arrays really are a representation of it. The symmetric half is (σᵢ, σⱼ) = 2δᵢⱼ I, and this is genuinely new information: nothing in the commutators implies it. Setting i = j gives σᵢ² = I, hence Sᵢ² = (ħ²/4)I for each component separately, and summing the three gives S² = (3/4)ħ²I. A spin-half therefore returns ⟨S²⟩ = 3ħ²/4 in every state, pure or mixed — S² is a multiple of the identity and carries no information at all.
Squaring to the identity is the workhorse
Multiply two Pauli vectors out: (a⋅σ)(b⋅σ) = Σᵢⱼ aᵢ bⱼ σᵢσⱼ = (a⋅b)I + i(a×b)⋅σ. Put a = b = n̂ and the cross product vanishes, leaving (n̂⋅σ)² = I for every unit vector, not merely for the three axes. Two consequences do most of the work in this unit. First, an operator that squares to I and has zero trace can only have eigenvalues +1 and −1, one of each, so S⋅n̂ = (ħ/2) n̂⋅σ has eigenvalues ±ħ/2 along every axis and no measurement of a spin component ever returns anything in between. Second, powers collapse: n̂⋅σ raised to an even power is I, to an odd power is n̂⋅σ itself. Feed that into the series for exp(−iα n̂⋅σ) and it splits cleanly into the cosine and sine series, giving exp(−iα n̂⋅σ) = cos α I − i sin α (n̂⋅σ) exactly, in closed form and with no approximation. Every single-qubit gate you will meet is this identity with α = θ/2.
Four real numbers describe any qubit operator
Tr σᵢ = 0 and Tr(σᵢσⱼ) = 2δᵢⱼ make (I, σₓ, σy, σz) an orthogonal basis for the 2×2 matrices under the Hilbert–Schmidt product Tr(A†B). So any Hermitian A is A = a₀I + a⋅σ with a₀ = ½Tr A and aₖ = ½Tr(σₖ A), all four real. In components A = (a₀+az, aₓ−iay ; aₓ+iay, a₀−az), so the top-right entry hands you aₓ and ay by inspection. Because (n̂⋅σ)² = I, the eigenvalues are a₀ ± |a|: the identity part shifts both levels alike and the vector part opens a gap 2|a| about it. Two payoffs follow at once. A Zeeman Hamiltonian H = −γB⋅S = −(γħ/2)B⋅σ has a₀ = 0, so its splitting is ħ|γ||B|; for an electron at B = 0.10 T, with |γ| = 1.761×10¹¹ s⁻¹ T⁻¹, that is 2.80 GHz in frequency, or 1.86×10⁻²⁴ J. And any density matrix, Hermitian with Tr ρ = 1, is forced into ρ = (I + a⋅σ)/2 with a = ⟨σ⟩ — the Bloch vector.
The arrays are a representation; the algebra is the physics
Change basis with a unitary U and every operator becomes UσU†. That preserves Hermiticity, the trace, the determinant, the eigenvalues and every commutator, while scrambling the arrays themselves. Diagonalise Sₓ instead of Sz and, in that basis, the array diag(1, −1) is what you would call σₓ, while σz has moved off-diagonal. So the claim that σz is diagonal records a choice you made, not a property of the z axis, and the column (1, 0)ᵀ means spin-up only relative to a stated basis. Two things do pin the set down. The triple (σₓ, σy, σz) transforms as a vector under rotations, which is exactly what makes n̂⋅σ invariant when n̂ and the axes turn together. And handedness is fixed by σₓσyσz = iI: flipping the sign of σy alone would leave every anticommutator intact while reversing the commutators, giving a left-handed set that is not the same physics.
Where the two-dimensional shortcuts stop
Every step above that leaned on σᵢ² = I is spin-half only. For spin-1 the same su(2) commutators are carried by 3×3 matrices with Sz = ħ diag(1, 0, −1), so Sz² = ħ² diag(1, 0, 1), which is not a multiple of the identity. The Clifford relation fails, and with it the two-valued spectrum — spin-1 components have eigenvalues +ħ, 0 and −ħ — and the two-term rotation formula. The exponential still truncates for spin-1, but at three terms, because (n̂⋅S/ħ)³ = n̂⋅S/ħ there. Completeness fails too: a Hermitian 3×3 needs nine real numbers, and I with the three Sᵢ supply only four; the missing five are the rank-2 quadrupole operators, which is why a spin-1 nucleus can carry an electric quadrupole moment and a spin-half nucleus cannot. What survives at every spin is the commutator algebra and S² = ħ²s(s+1)I. Keep the order of dependence straight: su(2) first, the Clifford relation second, and only the first generalises.
Change one variable at a time
Make the relationship visible.
Turn a without changing its length — try (aₓ, ay, az) = (2, 2, 1), then (1, 2, 2), then (2, 1, 2): the printed matrix changes completely and the two levels never move, because only the length of a sets the gap. Then push one slider outward and watch the gap open as twice that length.
LENGTH OF a3.00 meV
LEVEL SPLITTING 2|a|6.00 meV
det A-9.00 meV²
POLAR ANGLE OF n48.2 deg
Live interpretationLENGTH OF a: 3.00 meV. LEVEL SPLITTING 2|a|: 6.00 meV. det A: −9.00 meV². POLAR ANGLE OF n: 48.2 deg
Catch the common trap
Explain before calculating.
The identity σᵢσⱼ = δᵢⱼ I + i εᵢⱼₖ σₖ carries two independent pieces of information. Which statement separates them correctly?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyMultiply σₓ σy and σy σₓ out by hand in the Sz basis, then use the two products to obtain [σₓ, σy] and (σₓ, σy). Check both against σᵢσⱼ = δᵢⱼ I + i εᵢⱼₖ σₖ, and convert the commutator into spin operators.
- σₓσy = (0 1; 1 0)(0 −i; i 0). Row 1: (0⋅0 + 1⋅i, 0⋅(−i) + 1⋅0) = (i, 0). Row 2: (1⋅0 + 0⋅i, 1⋅(−i) + 0⋅0) = (0, −i). So σₓσy = (i 0; 0 −i) = i σz.
- σyσₓ = (0 −i; i 0)(0 1; 1 0). Row 1: (−i, 0). Row 2: (0, i). So σyσₓ = −i σz. Order matters: swapping the factors flipped the sign.
- Commutator: [σₓ, σy] = iσz − (−iσz) = 2i σz, matching 2i εxyz σz with εxyz = +1.
- Anticommutator: (σₓ, σy) = iσz + (−iσz) = 0, matching 2δxy I with δxy = 0.
- In spin operators: [Sₓ, Sy] = (ħ/2)²[σₓ, σy] = (ħ²/4)(2iσz) = (iħ²/2)σz, and σz = 2Sz/ħ, so [Sₓ, Sy] = iħ Sz as required.
Answerσₓσy = iσz and σyσₓ = −iσz, so [σₓ, σy] = 2iσz while (σₓ, σy) = 0 — that is [Sₓ, Sy] = iħ Sz, with Sₓ and Sy also anticommuting.
MediumA two-level defect has Hamiltonian H = (3, 1−2i ; 1+2i, −1) meV in the up/down basis, the semicolon separating its rows. Expand it as H = a₀I + a⋅σ, find both energy levels and the splitting, and check the result against the trace and determinant of H.
- H is Hermitian: the diagonal is real and the off-diagonal entries are conjugates, so the expansion coefficients are real.
- a₀ = ½Tr H = ½(3 + (−1)) = 1 meV, and az = ½(H₁₁ − H₂₂) = ½(3 − (−1)) = 2 meV.
- The top-right entry of a₀I + a⋅σ is aₓ − i ay, so 1 − 2i gives aₓ = 1 meV and ay = 2 meV. Hence a = (1, 2, 2) meV and |a| = √(1 + 4 + 4) = 3 meV.
- Since (n̂⋅σ)² = I with n̂ = a/|a| = (1, 2, 2)/3, the eigenvalues of a⋅σ are ±|a|, so E± = a₀ ± |a| = 1 ± 3, giving +4 meV and −2 meV.
- Check: E₊ + E₋ = 2 meV = Tr H, and E₊E₋ = −8 meV² = det H = (3)(−1) − (1−2i)(1+2i) = −3 − 5. Both agree.
- The identity part only lifts both levels by 1 meV; the observable splitting is 2|a| = 6 meV, which is resonant at f = 6 meV / h = 1.45 THz.
AnswerH = (1 meV)I + (1, 2, 2) meV · σ, with E₊ = +4 meV and E₋ = −2 meV, a 6 meV splitting about the 1 meV offset. Its eigenvectors point along n̂ = (1, 2, 2)/3, at 48.19° from the z axis.
HardStarting from σᵢσⱼ = δᵢⱼ I + i εᵢⱼₖ σₖ, prove (a⋅σ)(b⋅σ) = (a⋅b)I + i(a×b)⋅σ, deduce the closed form of exp(−iα n̂⋅σ), and evaluate it for n̂ = (1, 2, 2)/3 and α = π/3. Verify the result is unitary with unit determinant.
- Expand: (a⋅σ)(b⋅σ) = Σᵢⱼ aᵢ bⱼ σᵢσⱼ = Σᵢⱼ aᵢ bⱼ (δᵢⱼ I + i εᵢⱼₖ σₖ) = (a⋅b)I + i(a×b)⋅σ, since Σᵢⱼ aᵢ bⱼ εᵢⱼₖ = (a×b)ₖ.
- Set a = b = n̂ with |n̂| = 1. Then n̂⋅n̂ = 1 and n̂×n̂ = 0, so (n̂⋅σ)² = I. Every even power of n̂⋅σ is therefore I and every odd power is n̂⋅σ.
- In exp(−iα n̂⋅σ) = Σ (−iα)ⁿ(n̂⋅σ)ⁿ/n!, the even terms sum to cos α times I and the odd terms to −i sin α times n̂⋅σ, giving exp(−iα n̂⋅σ) = cos α I − i sin α (n̂⋅σ) exactly.
- Numbers: cos(π/3) = 0.5000, sin(π/3) = 0.8660, and n̂⋅σ = (nz, nₓ−iny ; nₓ+iny, −nz) = (0.6667, 0.3333−0.6667i ; 0.3333+0.6667i, −0.6667).
- So U = 0.5000 I − 0.8660i (n̂⋅σ) = (0.500 − 0.577i, −0.577 − 0.289i ; 0.577 − 0.289i, 0.500 + 0.577i).
- Checks: the first column has squared norm (0.250 + 0.333) + (0.333 + 0.083) = 1.000, and det U = 0.583 − (−0.417) = 1.000. Writing α = θ/2, this U rotates ⟨S⟩ by θ = 120° about n̂.
Answerexp(−iα n̂⋅σ) = cos α I − i sin α (n̂⋅σ). For n̂ = (1,2,2)/3 and α = π/3 it is (0.500 − 0.577i, −0.577 − 0.289i ; 0.577 − 0.289i, 0.500 + 0.577i), unitary with det U = 1 — a 120° rotation about n̂.