Energy conservation, Q values and thresholds

The Schrödinger equation iħ ∂|ψ⟩/∂t = H|ψ⟩ is solved by |ψ(t)⟩ = U(t)|ψ(0)⟩ with U(t) = exp(−iHt/ħ), so H is the Hermitian generator of the unitary family that shifts the system in time. Take the Heisenberg equation for any observable, d⟨A⟩/dt = (i/ħ)⟨[H, A]⟩ + ⟨∂A/∂t⟩, and set A = H. The commutator [H, H] vanishes identically, leaving d⟨H⟩/dt = ⟨∂H/∂t⟩. Energy is conserved precisely when the Hamiltonian does not depend explicitly on the clock — when the laws are the same on Tuesday as on Monday. An atom driven by a laser has H(t) = H₀ − 𝐝⋅𝐄₀ cos ωt and its energy is not conserved; the atom plus field has a time-independent H and its energy is. In quantum field theory the same symmetry appears vertex by vertex: the ∫dt at each interaction point produces δ(ΣE), so energy balances exactly at every vertex of a Feynman diagram, not only between the initial and final states.

Q = (Σmᵢₙᵢₜᵢₐₗ − Σm_final)c² is the rest energy that turns into kinetic energy, and it is a Lorentz scalar, so a negative Q cannot be rescued by a change of frame. Use consistent masses. With nuclear masses the bookkeeping is literal; with the atomic masses tables actually list, the electrons cancel for α decay, β⁻ decay and most reactions, but β⁺ decay needs 2mₑ c² = 1.022 MeV subtracted. Rutherford's transmutation ¹⁴N + α → ¹⁷O + p: atomic masses 14.003074 + 4.002602 = 18.005676 u in, 16.999132 + 1.007825 = 18.006957 u out, so Q = −0.001281 u × 931.494 MeV/u = −1.193 MeV. Endothermic: the α particle must bring energy, and it will turn out to need more than 1.193 MeV of it. Neutron decay: 939.565 − 938.272 − 0.511 = 0.782 MeV, exothermic, and that 0.782 MeV is the total kinetic energy the proton, electron and antineutrino share.

Add the four-momenta of everything on one side and square: s = (Σp)² = (ΣE)² − |Σ𝐩|² with c = 1. It is the same number in every inertial frame. In the centre-of-momentum frame Σ𝐩 = 0, so √s is the total CM energy, and the least it can be while still making the final state is when every product sits at rest there: √s = Σm_f. That is the whole threshold condition. For beam a on target b at rest, s = mₐ² + m_b² + 2Eₐ m_b, which you evaluate in the laboratory and compare with (Σm_f)². Antiproton production p + p → p + p + p + p̄ needs √s = 4mₚ = 3.753 GeV. A fixed target requires Eₐ = (16mₚ² − 2mₚ²)/(2mₚ) = 7mₚ, a kinetic energy of 6mₚ = 5.63 GeV — the Bevatron was built for 6.2 GeV precisely to clear it. A proton–proton collider reaches the same √s with 2mₚ per beam: 0.94 GeV of kinetic energy each.

Expanding the fixed-target result gives Tₜₕᵣ = [(Σm_f)² − (mₐ + m_b)²]/(2m_b) = −Q(Σm_f + mₐ + m_b)/(2m_b), exact in relativity, and in the non-relativistic limit Tₜₕᵣ ≈ −Q(1 + mₐ/m_b). The excess over |Q| is what momentum conservation charges: the beam's momentum must be carried away by the products, so they cannot all be at rest in the laboratory and part of the kinetic energy you supplied is locked into their centre-of-mass motion. For ¹⁴N(α, p)¹⁷O the factor is (18.006957 + 18.005676)/(2 × 14.003074) = 1.286, so Tₜₕᵣ = 1.193 × 1.286 = 1.53 MeV rather than 1.19 MeV. For p + p → p + p + π⁰ the factor is (2011.5 + 1876.5)/1876.5 = 2.07, so 135 MeV of pion costs 280 MeV of beam. The factor tends to 1 only as m_b → ∞, a target too heavy to recoil — which is why the tax is smallest on a heavy nucleus and largest on a light one.

Because four-momentum is conserved exactly at each vertex, an internal line's momentum is fixed by the external legs: in electron–muon scattering the photon carries q = p₁ − p₃, and q² = (p₁ − p₃)² is negative, spacelike, nothing like the m² = 0 of a real photon. That line is off shell. The amplitude contains the propagator 1/(q² − m²c² + iε), and the mass-shell condition q² = m²c² is simply where that denominator has its pole — where an external, detectable particle would sit. Nothing is borrowed and nothing is paid back: the virtual particle is a term in a perturbation series, and its 'wrong' mass is a statement about a denominator. The same bookkeeping makes threshold a hard wall. Below √s = Σm_f there is no final state with the right energy, so the transition amplitude is zero however short the interaction: a reaction 0.1 MeV short of threshold does not happen briefly, it does not happen.

The Robertson inequality σ_A σ_B ≥ |⟨[A, B]⟩|/2 needs two Hermitian operators, and time is not one. If some T satisfied [T, H] = iħ, then exp(−iaT/ħ) would shift energy by any real a, so the spectrum of H would be the whole real line and no ground state could exist — Pauli's argument. What survives is different in kind. A state whose survival amplitude is ⟨ψ(0)|ψ(t)⟩ = e^{(−iE₀t/ħ − t/2τ)}, so that |⟨ψ(0)|ψ(t)⟩|² = e^(−t/τ), has an energy distribution whose Fourier transform is that amplitude: a Lorentzian of full width Γ = ħ/τ, since transforming the probability e^(−t/τ) instead would give twice the width. The Z boson has Γ = 2.495 GeV, so τ = 6.582 × 10⁻²⁵ GeV s ÷ 2.495 GeV = 2.64 × 10⁻²⁵ s; the neutron's 878 s life gives Γ = 7.5 × 10⁻¹⁹ eV, unmeasurably narrow. The Mandelstam–Tamm form σ_A / |d⟨A⟩/dt| ≥ ħ/(2σ_H) is legitimate too, but its 'Δt' is a derived time scale, not the spread of an operator. Neither statement licenses energy non-conservation.