Wigner's theorem: every symmetry is unitary or antiunitary

Units are SI with ħ written out; nothing here is relativistic, so no metric enters, and when four-vectors arrive the course's signature is (+, −, −, −). A pure state is a ray, the set e^(iα)|a⟩ of unit vectors differing by a phase, because every prediction is built from transition probabilities |⟨a|b⟩|², and those cannot see α. So a symmetry is a one-to-one map T of rays onto rays with |⟨Ta|Tb⟩|² = |⟨a|b⟩|² for every pair. Nothing says T is linear; it is not even defined on vectors. A spin half shows what that permits. Its rays are the points n of the Bloch sphere, and |⟨a|b⟩|² = (1 + nₐ ⋅ n_b)/2, so a symmetry is any map of the sphere that keeps every angle: an orthogonal 3 × 3 matrix. Rotations, with determinant +1, are the unitaries exp(−iθ n̂ ⋅ σ/2) of Quantum Mechanics II. Reflections, with determinant −1, keep every angle just as well, and no unitary produces them, as the invariant B below will show. Wigner's theorem says what does: an antiunitary operator, and no third kind is ever needed.

Take an orthonormal basis |n⟩. Because |⟨n|m⟩|² = δₙₘ is kept, the image rays are mutually orthogonal; unit vectors |n′⟩ on them form a basis because T is onto, but their phases are arbitrary. Fix them with the sums (|1⟩ + |n⟩)/√2. The image ray has probability ½ with T(1) and with T(n) and none with the rest, so it holds e^(iα)|1′⟩ + e^(iβ)|n′⟩, and redefining |n′⟩ → e^(i(β−α))|n′⟩ makes it the ray of |1′⟩ + |n′⟩. One overall phase is left. Now take |ψ⟩ = Σ cₙ|n⟩ with c₁ real and positive, and write its image as Σ c′ₙ|n′⟩ with c′₁ real and positive, so c′₁ = c₁. Keeping |⟨n|ψ⟩|² gives |c′ₙ| = |cₙ|; keeping the probability with (|1⟩ + |n⟩)/√2 gives |c₁ + c′ₙ| = |c₁ + cₙ|, so Re c′ₙ = Re cₙ. Each coefficient therefore goes to cₙ or to cₙ*. That this is one choice for every n and every vector, including vectors with c₁ = 0 and infinite dimension, is the step Bargmann's 1964 proof secures. It is quoted; the Medium example shows why a mixed choice fails.

Pairwise probabilities cannot separate a unitary from an antiunitary, since |z*| = |z|. Three states can. The product B = ⟨a|b⟩⟨b|c⟩⟨c|a⟩ is untouched when any vector is re-phased, because each appears once as a bra and once as a ket, so B belongs to the rays. A unitary keeps every factor and so keeps B; an antiunitary conjugates every factor and sends B to B*. For a qubit, writing each projector as (1 + n ⋅ σ)/2 and tracing their product gives B = [1 + nₐ ⋅ n_b + n_b ⋅ n_c + n_c ⋅ nₐ + i nₐ ⋅ (n_b × n_c)]/4, derived, and a spherical-geometry identity, quoted, makes arg B half the solid angle of the triangle abc. So Im B carries the orientation of the triangle: a unitary, being a rotation, keeps it, and an antiunitary is a mirror that reverses it. Two limits check this. For a single state B = 1 and the kinds agree. For three Bloch vectors coplanar with the origin, all real in one basis, Im B = 0, and no test on those states can tell the kinds apart; it takes a state off that plane.

Kinds compose like signs. If A₁ and A₂ are antiunitary, A₁A₂(c|a⟩) = A₁(c*A₂|a⟩) = c A₁A₂|a⟩, so the product is linear, and ⟨A₁A₂a|A₁A₂b⟩ = ⟨A₂a|A₂b⟩* = ⟨a|b⟩, so it is unitary; a unitary times an antiunitary is antiunitary. Now take a one-parameter group, T(θ₁)T(θ₂) = T(θ₁ + θ₂): rotations by θ about a fixed axis, translations by a along x, evolution through a time t. Its operators obey V(θ) = ωV(θ/2)² with |ω| = 1, and whichever kind V(θ/2) is, its square is unitary, so every V(θ) is unitary, derived from the group law alone. Stone's theorem, quoted, then writes V(θ) = exp(−iθG/ħ) with G Hermitian: J, p or H. Dimensions check each exponent: θJ/ħ is rad × J s ÷ J s, ap/ħ is m × kg m s⁻¹ ÷ J s and tH/ħ is s × J ÷ J s, all pure numbers, and θ → 0 returns V → 1. So the linearity of time evolution is not an extra postulate: it follows once evolution keeps probabilities and composes as a group.

The square-root argument needs a square root inside the group. Parity has none: P² = 1, so its group holds only 1 and P, and neither squares to P. Time reversal, charge conjugation and the point reflections of a crystal are in the same position, so Wigner's theorem lets each be either kind and the choice is physics rather than mathematics. Parity will be unitary in topic 1.03 and time reversal antiunitary in topic 1.04, for reasons this topic does not supply; "only discrete symmetries can be antiunitary" is a permission, not a prediction. The rule of signs fixes the bookkeeping. Labelling unitary +1 and antiunitary −1 is a homomorphism of the symmetry group onto the two signs, or onto +1 alone, so the antiunitary elements, if there are any, form exactly one coset of the unitary subgroup: an antiunitary symmetry times any rotation is again antiunitary, and everything continuously connected to the identity is unitary. The qubit makes this concrete. O(3) splits into rotations and reflections, and det(O²) = (det O)² = +1 is the square-root argument in miniature.

V is fixed only up to a phase, so the operators of a symmetry group compose as V(g₁)V(g₂) = ω(g₁, g₂)V(g₁g₂) with |ω| = 1, a projective representation, and the phases are not decoration. A rotation by 2π sends every ray to itself, so its operator need only be a number times 1, and exp(−i2πJ_z/ħ) on |j, m⟩ gives e^(−i2πm) = (−1)^(2j), the same for every m because j − m is an integer: −1 for half-integer j. The sign Quantum Mechanics II met at 2π is therefore no paradox. R(2π) is the identity symmetry, carried out by the operator −1. For rotations Bargmann (1954), quoted, showed that the phases can be chosen so that the operators form a true representation of SU(2), the double cover of SO(3), which leaves ω = ±1. No measurement on the rotated system alone can see the sign, since every |⟨a|b⟩|² is untouched; it appears only when a rotated wave interferes with an unrotated one, as in the neutron interferometer analysed in the Hard example.