Spinless time reversal: Θ = K and real wavefunctions

Topic 1.04 showed that Θ must be antiunitary with ΘxΘ⁻¹ = x and ΘpΘ⁻¹ = −p; quote that and ask which operator it is. Let K conjugate a state's components in the position basis, Kψ(x) = ψ*(x). Position multiplies by a real number, so KxK⁻¹ = x. Momentum acts as −iħ∇, and conjugating −iħ∇ψ* gives +iħ∇ψ, so KpK⁻¹ = −p. Any rival can be written UK with U unitary (topic 1.02), and it must give UxU⁻¹ = x and UpU⁻¹ = p; for a spinless particle x and p act irreducibly (quoted), so U is a constant phase, invisible in every ΘAΘ⁻¹ and every probability. Hence Θ = K, derived. Add spin and U is free to act on it, which is where topic 1.06 starts. The statement is tied to the basis: Θ|p⟩ = |−p⟩, so a momentum-space wavefunction goes to φ*(−p), not φ*(p). Units are SI throughout, and nothing here is relativistic, so the course metric diag(+1, −1, −1, −1) is not needed.

Let iħ∂ψ/∂t = Hψ with H = −(ħ²/2M)∇² + V(x), M the particle's mass, and define ψ_T(x, t) = ψ*(x, −t). Conjugating the Schrödinger equation and relabelling t as −t gives iħ∂ψ_T/∂t = [−(ħ²/2M)∇² + V*]ψ_T, so ψ_T solves the original equation exactly when V* = V: when H is a real operator in the position basis (derived). Drop the conjugation and ψ(x, −t) obeys iħ∂ψ/∂t = −Hψ, the negative-energy trouble of topic 1.04. In the classical limit Ehrenfest's theorem retraces ⟨x⟩ with ⟨p⟩ reversed: the film run backwards. Two failures matter later. With a vector potential, K(−iħ∇ − qA)K⁻¹ = −(−iħ∇ + qA), so ψ_T solves the problem with A reversed: an applied magnetic field is odd under T, as the force qv × B already says. An optical potential V − iW with W = 1.0 meV makes |ψ|² decay at 2W/ħ = 3.0 × 10¹² s⁻¹; its conjugate grows at that rate, so absorption is not T-invariant.

Since L = x × p, ΘLΘ⁻¹ = x × (−p) = −L; equivalently L = −iħ x × ∇ is i times a real operator, and K flips it (derived). L² is even, and K turns an L_z eigenfunction of eigenvalue mħ into one of −mħ, so (Yₗᵐ)* is Yₗ⁻ᵐ up to a phase. The ladder operators fix the phase. Conjugating L± = Lₓ ± iL_y gives KL₊K⁻¹ = −L₋. Define Zₘ = (−1)ᵐ(Yₗ⁻ᵐ)*; that rule gives L₊Zₘ = ħ√((l − m)(l + m + 1)) Zₘ₊₁, the Condon–Shortley recursion with its real, positive coefficients, and the L₋ recursion follows the same way. Both families also start together: Z₀ = (Yₗ⁰)* = Yₗ⁰, because Yₗ⁰ ∝ Pₗ(cos θ) is real. Same start, same recursions, so Zₘ = Yₗᵐ for every m: (Yₗᵐ)* = (−1)ᵐYₗ⁻ᵐ, derived. Check at l = 1: Y₁¹ = −√(3/(8π)) sin θ e^(iφ) conjugates to −√(3/(8π)) sin θ e^(−iφ) = −Y₁⁻¹. So Θ|l, m⟩ = (−1)ᵐ|l, −m⟩.

Let ΘH = HΘ and Hψₙ = Eₙψₙ. Then H(Θψₙ) = Θ(Hψₙ) = Eₙ(Θψₙ), because Eₙ is real and survives the conjugation: Θψₙ has the same energy. If that energy is non-degenerate, Θψₙ can only be ψₙ times a number of modulus one, ψₙ* = e^(iα)ψₙ, and χ = e^(iα/2)ψₙ then satisfies χ* = χ: the state can be chosen real (derived). Two consequences follow. The current j = (ħ/M) Im(ψ*∇ψ), in m⁻² s⁻¹, vanishes everywhere, because a real function times its own gradient is real and a global phase cancels. And ⟨L⟩ = −iħ∫χ(x × ∇χ) d³x is −i times a real number yet must be real, so ⟨L⟩ = 0; topic 1.02's matrix-element rule gives the same for any Hermitian T-odd observable. Every one-dimensional bound state, every oscillator Hermite function and hydrogen's 1s are real for this reason. The free particle is the limit the theorem does not cover: ±k are degenerate, and e^(ikx) duly carries current.

Θ maps an eigenstate to one of the same energy, but nothing makes it the same state when the level is degenerate. A particle of mass M on a ring of radius R has Eₘ = m²ħ²/(2MR²) with partners e^(±imφ)/√(2π), and K turns one into the other. Each circulates: ⟨L_z⟩ = ±mħ, and the current formula gives probability passing a fixed point at mħ/(2πMR²) per second (derived). For an electron with m = 1 on a ring of R = 1.0 nm that is 1.84 × 10¹³ s⁻¹, an electric current of 2.95 μA around the area πR², and a moment of 9.27 × 10⁻²⁴ J T⁻¹: exactly one μ_B, whatever R is. The real combinations cos mφ and sin mφ are equally good eigenstates and carry nothing. The perturbation decides which basis nature uses: a real bump on the ring is T-even and splits the pair into standing waves, while a magnetic flux, T-odd, splits it into the circulating ones. The figure sets the two against each other.

A 3d ion in a crystal feels its neighbours' electrostatic field, a real potential. When spin-orbit coupling is weaker than that field, the orbital factor of the state obeys a real, spin-free Hamiltonian, and the spinless theorem applies to it with spin as a spectator. If the orbital ground level is non-degenerate, ⟨L⟩ = 0 and the first-order orbital Zeeman shift vanishes; what survives is of order (μ_B B)²/Δ, about 1.5 × 10⁻⁹ eV at 1 T for a splitting Δ = 2.2 eV, against 5.8 × 10⁻⁵ eV for a free orbital moment. Cr³⁺ (3d³, free-ion term ⁴F with J = 3/2) sits in an orbital singlet at octahedral sites. Quantum Mechanics II's formulas (quoted) give a free-ion Landé value of 0.77 μ_B and a spin-only value 2√(S(S + 1)) μ_B = 3.87 μ_B; its salts measure about 3.8 μ_B, so the orbital moment is gone. Co²⁺, whose octahedral ground level is orbitally threefold degenerate, measures about 4.8 μ_B against a spin-only 3.87 μ_B, keeping part of its orbital moment as the theorem permits.