Time reversal with spin: Θ = e^{(−iπJ_y/ħ)}K, Θ² = (−1)^(2j)

Units are SI with ħ = 1.055 × 10⁻³⁴ J s kept explicit; nothing here is relativistic, so no metric signature enters. Topics 1.02 and 1.04 supply the frame: Θ is antiunitary, so Θ = UK for some unitary U once a basis for K is chosen, and it must keep x and reverse p. Let K conjugate components in the position and S_z basis. Then KxK = x and, because p = −iħ∇, KpK = −p, which is why topic 1.05 could stop at Θ = K. Spin spoils that. With Condon–Shortley phases Sₓ and S_z are real matrices and S_y is imaginary, so KSₓK = Sₓ, KS_yK = −S_y and KS_zK = S_z: conjugation reverses one component in three. Time reversal must reverse all of them. If ΘSΘ⁻¹ = εS with one sign ε = ±1 for all three components, as a rotation-invariant Θ must treat a vector, applying Θ to [Sₓ, S_y] = iħS_z gives ε²iħS_z on the left and −iħεS_z on the right, because Θ conjugates the i. Only ε = −1 survives. K's own pattern, two components kept and one reversed, also satisfies the commutators; what it fails is rotation invariance.

Write Θ = UK. K already treats x and p correctly, so U must commute with both and act on spin alone, with USₓU⁻¹ = −Sₓ, US_yU⁻¹ = +S_y and US_zU⁻¹ = −S_z. A rotation by π about y does exactly that: it sends x̂ → −x̂ and ẑ → −ẑ, leaves ŷ alone, and vector operators follow their axes. So U = e^{(−iπS_y/ħ)} works, and it is the only choice up to a phase: if U′ also works, U⁻¹U′ commutes with Sₓ, S_y and S_z, and on an irreducible spin space Schur's lemma, quoted from linear algebra, makes it a number. Hence Θ = η e^{(−iπS_y/ħ)}K, derived. For spin half, S_y = (ħ/2)σ_y and σ_y² = 1 give e^{(−iπσ_y/2)} = cos(π/2) − iσ_y sin(π/2) = −iσ_y, the real matrix with rows (0, −1) and (1, 0). Check one rule: Θσ_zΘ⁻¹ = (−iσ_y)σ_z(iσ_y) = σ_yσ_zσ_y = −σ_z. In the spinless limit S = 0 the rotation is the identity and Θ = ηK, topic 1.05 again.

The action on one multiplet follows without any matrices. Θ leaves J² alone and reverses J_z, so J_zΘ|j, m⟩ = −ΘJ_z|j, m⟩ = −mħΘ|j, m⟩, and therefore Θ|j, m⟩ = cₘ|j, −m⟩ for some numbers cₘ. Next, Θ flips Jₓ and J_y and also conjugates the i in J₊ = Jₓ + iJ_y, so ΘJ₊Θ⁻¹ = −Jₓ + iJ_y = −J₋. Apply ΘJ₊ = −J₋Θ to |j, m⟩: J₊ on |j, m⟩ and J₋ on |j, −m⟩ carry the same real coefficient ħ√[(j − m)(j + m + 1)], so cₘ₊₁ = −cₘ. One sign change per rung gives cₘ = cⱼ(−1)^(j−m), derived. Within the multiplet Θ therefore acts as cⱼ e^{(−iπJ_y/ħ)}K, with K now conjugating in the |j, m⟩ basis, because QM-II's rotation matrices give ⟨j, −m|e^{(−iπJ_y/ħ)}|j, m⟩ = (−1)^(j−m) and zero elsewhere. For an orbital multiplet with Θ = K in position space, Yₗᵐ* = (−1)^m Yₗ⁻ᵐ = (−1)^l (−1)^(l−m) Yₗ⁻ᵐ, so η′ = (−1)^l; for l = 1 the signs on m = 1, 0, −1 are −1, +1, −1. Topic 1.05's rule is this formula written in another basis.

In the |j, m⟩ basis −iJ_y is a real matrix, so U = e^{(−iπJ_y/ħ)} is real and K passes through it: Θ² = ηUKηUK = ηη*UU*K² = |η|²U² = e^{(−2πiJ_y/ħ)}. A rotation by 2π has eigenvalues e^(−2πim) = (−1)^(2j) on every m, so Θ² = (−1)^(2j), derived. The stretched state shows it directly: |j, j⟩ is 2j aligned spin-halves, each with amplitude cos(β/2) to survive a rotation β about y, so ⟨j, j|e^{(−iβJ_y/ħ)}|j, j⟩ = cos(β/2)^(2j), real at every β and equal to (−1)^(2j) at β = 2π. The ladder form agrees: Θ²|j, m⟩ = cₘ*c₋ₘ|j, m⟩ = |cⱼ|²(−1)^(j−m)(−1)^(j+m)|j, m⟩. The phase enters as ηη* because Θ conjugates the η standing to its right; a linear ηU would square to η²(−1)^(2j), and η = i would turn −1 into +1. Antilinearity is what makes the sign physical. Checks: 2πJ_y/ħ is dimensionless, and j = 0 gives K² = +1, topic 1.05. Measured: since Θ² = R(2π), a neutron whose spin precesses through 2π in one arm of an interferometer returns with sign −1, and Rauch and co-workers (1978) found the interference period to be 716.8 ± 3.8° of precession, 0.84 standard errors from 720°.

The sign does not depend on the construction. Apply the rules twice: Θ²xΘ⁻² = x, Θ²pΘ⁻² = (−1)²p = p and Θ²SΘ⁻² = S. So Θ² commutes with x, p and S, which act irreducibly on the states of one particle of spin s, and Schur's lemma makes Θ² = c, a number. A product of two antiunitary operators is linear and unitary, so |c| = 1. Now use associativity, Θ(Θ²) = (Θ²)Θ. On the left Θc = c*Θ, because Θ conjugates numbers; on the right the product is cΘ. So c* = c, and a real number of modulus one is +1 or −1, derived. Nothing here chose a basis for K or a phase η. Any two operators obeying the rules differ by a phase, Θ′ = λΘ, by the same Schur argument, and Θ′² = λΘλΘ = λλ*Θ² = Θ²: c belongs to the rules, not to a convention. The explicit construction only says which sign occurs: (−1)^(2j) with j = s.

For N particles the rules hold particle by particle, so Θ = Θ₁ ⊗ Θ₂ ⊗ … ⊗ Θ_N with K taken in the product basis, and Θ² is the product of the single-particle squares: (−1)^N when N of them have half-integer spin. Angular-momentum addition agrees, since N spin-halves couple to integer J when N is even and to half-integer J when N is odd, so (−1)^(2J) = (−1)^N. Hydrogen, an electron and a proton, has Θ² = +1, and its ground level splits into F = 0 and F = 1, the 1420.406 MHz line. Deuterium adds a neutron, so Θ² = −1, and its ground level splits into F = 1/2 and F = 3/2, 327.384 MHz apart. Θ² is a property of the state space, settled before a Hamiltonian is written: whatever H contains, a crystal field that keeps the symmetry or a magnetic field that breaks it, it decides only whether H commutes with Θ, never what Θ² is. Topic 1.07 needs both facts, Θ² = −1 and [H, Θ] = 0, to pair levels.