Pauli matrices and the spin operator

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 S_z, label its eigenvectors |↑⟩ and |↓⟩ with eigenvalues ±ħ/2, and S_z = (ħ/2)diag(1, −1) by construction. The other two arrays follow from the ladder operators, S₊ = ħ|↑⟩⟨↓| and S₋ = ħ|↓⟩⟨↑|, through Sₓ = (S₊ + S₋)/2 and S_y = (S₊ − S₋)/2i; UPV-09.08 derives those ladder elements, so take them as given here. What no derivation can fix is the phase of |↓⟩ relative to |↑⟩, and it is worth seeing what that phase controls. Replace |↓⟩ by e^(iχ)|↓⟩: σ_z is untouched, but the off-diagonal elements pick up e^(±iχ), so in terms of the standard arrays σₓ is now represented by cos χ σₓ − sin χ σ_y and σ_y by sin χ σₓ + cos χ σ_y. The phase convention rotates the x and y arrays into each other about z. The Condon–Shortley convention takes ⟨↑|S₊|↓⟩ real and positive, which sets χ = 0 and puts −i in the top right of σ_y; choose χ = π/2 instead and S_y is represented by the real array (ħ/2)(0 1; 1 0) while Sₓ takes an imaginary one.

σᵢσⱼ = δᵢⱼ I + i εᵢⱼₖ σₖ is nine matrix products compressed into one line, and it is worth checking once by hand: σₓσ_y = (0 1; 1 0)(0 −i; i 0) = (i 0; 0 −i) = iσ_z. 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.

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. So any function given by a power series needs only two terms, f(n̂⋅σ) = ½[f(1) + f(−1)] I + ½[f(1) − f(−1)] n̂⋅σ. For f(x) = exp(−iαx) that is exp(−iα n̂⋅σ) = cos α I − i sin α (n̂⋅σ) exactly; UPV-09.08 reads it, with α = θ/2, as the SU(2) rotation through θ, and every single-qubit gate is one of these up to a global phase.

Tr σᵢ = 0 and Tr(σᵢσⱼ) = 2δᵢⱼ make (I, σₓ, σ_y, σ_z) 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+a_z, aₓ−ia_y ; aₓ+ia_y, a₀−a_z), so the top-right entry hands you aₓ and a_y 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.

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 S_z and, in that basis, the array diag(1, −1) is what you would call σₓ, while σ_z has moved off-diagonal. So the claim that σ_z 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, σ_y, σ_z) 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 σₓσ_yσ_z = iI: flipping the sign of σ_y alone would leave every anticommutator intact while reversing the commutators, giving a left-handed set that is not the same physics.

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 S_z = ħ diag(1, 0, −1), so S_z² = 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.