Why time reversal must be antiunitary

Units here are SI, with ħ = 1.055 × 10⁻³⁴ J s, and nothing is relativistic, so no metric signature enters. Film a ball in flight and play the film backwards: every frame shows the same position with the velocity reversed, and the reversed flight is a perfectly good motion under the same gravity. Quantum mechanically the reversal is an operator Θ on states, defined by Θ U(t) = U(−t) Θ with U(t) = exp(−iHt/ħ): evolving for t and then reversing gives the state you get by reversing first and evolving backward for t. Multiply on the left by U(t) and it reads U(t)ΘU(t) = Θ, the echo form: run forward, reverse, run forward again for the same time, and you hold the reversed initial state. Wigner's theorem, quoted from the first topic of this unit, leaves two options, unitary or antiunitary, and only a discrete operation may take the second. Parity, in the previous topic, commutes with U(t) whenever it is a symmetry. Time reversal cannot simply commute, because it has to turn t into −t.

Try the linear option. For small t, U(t) = 1 − iHt/ħ + O(t²), and a linear Θ carries the −i through unchanged, so the first-order terms of ΘU(t)Θ⁻¹ = U(−t) read −iΘHΘ⁻¹t/ħ = +iHt/ħ, that is ΘHΘ⁻¹ = −H (derived). Both sides are energies, in J. Act on an eigenstate: HΘ|E⟩ = −ΘH|E⟩ = −EΘ|E⟩, so every level E would need a partner at −E. A free particle's energies p²/2m fill [0, ∞), so its partners would fill (−∞, 0]: a spectrum with no bottom, into which any atom could radiate without end. The hydrogen 1s level at −13.6 eV is stable, and no free electron has ever been measured with negative kinetic energy. Impose the classical rule p → −p as well and the contradiction needs no spectrum at all: a linear Θ would leave p²/2m unchanged while the requirement makes it change sign, so H = −H and H = 0.

An antilinear Θ obeys Θ(c|α⟩) = c*Θ|α⟩ for every complex c, and an antiunitary one also keeps |⟨α|β⟩|, both quoted from the topic on antiunitary operators. Carrying the i through is the whole point: Θ(−iHt/ħ)Θ⁻¹ = +iΘHΘ⁻¹t/ħ, because t and ħ are real. The first-order terms of ΘU(t)Θ⁻¹ = U(−t) now give +iΘHΘ⁻¹ = +iH, so ΘHΘ⁻¹ = H, or [Θ, H] = 0 (derived). Since the exponential series has real coefficients 1/n!, ΘU(t)Θ⁻¹ = exp(+iΘHΘ⁻¹t/ħ) exactly, and commuting is also sufficient at every order. The eigenvalue argument now runs the other way: HΘ|E⟩ = ΘH|E⟩ = Θ(E|E⟩) = EΘ|E⟩, because E is real, so Θ|E⟩ has the same energy. The free spectrum maps onto itself and the 1s level stays the bottom. What has not been shown is that any particular H commutes with Θ; that is a property of the H, tested below.

What Θ does to observables comes from classical mechanics by correspondence: a reversed trajectory passes through the same positions with the velocities flipped, so we impose ΘxΘ⁻¹ = x and ΘpΘ⁻¹ = −p. Apply Θ to [x, p] = iħ. The left side becomes [x, −p] = −iħ. The right side becomes Θ(iħ)Θ⁻¹, which is +iħ for a linear Θ and −iħ for an antilinear one, so only the antilinear choice is consistent. The argument never mentions H, so it holds for every system (derived). Angular momentum repeats it: L = x × p gives ΘLΘ⁻¹ = −L, and [Lₓ, L_y] = iħL_z survives because the left side is unchanged while the right becomes (−iħ)(−L_z); a linear Θ would give (iħ)(−L_z), the wrong sign. Spin has no x or p, but it must reverse with L so that J = L + S does, and it obeys the same algebra, so its reversal S → −S is forced to be antilinear for the same reason; the construction waits for a later topic. As ħ → 0 the commutator vanishes and only p → −p is left, which is classical reversal.

Let ψ(t) solve iħ∂ψ/∂t = Hψ and define ψ_R(t) = Θψ(−t). With ψ′ the derivative with respect to the argument, iħ∂ψ_R/∂t = −iħΘψ′(−t) = Θ[iħψ′(−t)], the minus sign from the chain rule cancelled by the conjugated i, and Θ[iħψ′(−t)] = ΘHψ(−t), which equals Hψ_R(t) exactly when [Θ, H] = 0 (derived). The clock still runs forward; the motion is what turns round. A free electron at 1.00 eV has ψ = exp[i(kx − ωt)] with k = 5.12 × 10⁹ m⁻¹ and ω = 1.52 × 10¹⁵ s⁻¹. Running the clock backwards gives ψ(x, −t) = exp[i(kx + ωt)], whose e^(+iωt) is the time dependence of a −1.00 eV state, a solution of iħ∂ψ/∂t = −Hψ instead. For a spinless particle Θ acts as complex conjugation, constructed in the next topic, giving ψ*(x, −t) = exp[i(−kx − ωt)]: momentum −ħk, energy still +1.00 eV. Expectation values confirm it, ⟨x⟩_R(t) = ⟨x⟩(−t) and ⟨p⟩_R(t) = −⟨p⟩(−t), and |ψ_R(x, t)|² = |ψ(x, −t)|² is the film played backwards.

Whether Θ is a symmetry is a property of H, and an applied magnetic field is the standard way to lose it. With minimal coupling, quoted from Quantum Mechanics II, H = (p − qA)²/2m + qφ, where A and φ are real functions of position that pass through Θ unchanged while p flips: ΘHΘ⁻¹ = (p + qA)²/2m + qφ = H(−A) (derived). Reversed motion is a solution only if B = ∇ × A is reversed too, as reversing the source currents would do. Classically, an electron in 1.00 T circles at eB/2πmₑ = 28.0 GHz, the measured cyclotron resonance, in one sense only; the reversed orbit would need the field flipped. On a ring of radius R enclosing flux Φ, H = (L_z − qΦ/2π)²/2mR², and L_z = ℓħ gives E(ℓ) = (ħ²/2mR²)(ℓ − qΦ/h)² (derived). Θ sends ℓ to −ℓ, whose level sits 4ℓ(qΦ/h)ħ²/2mR² away until Φ is reversed, as the figure shows. The spin Zeeman term −γS⋅B is odd for the same reason, S → −S. At B = 0 the invariance returns.