Nucleons, isospin and the two-nucleon system

Heisenberg's observation was arithmetic: mₚ c² = 938.272 MeV and mₙ c² = 939.565 MeV differ by 1.293 MeV, one part in 726. Treat that near-degeneracy as a symmetry and give the nucleon an internal two-dimensional space, |p⟩ = (1, 0)ᵀ and |n⟩ = (0, 1)ᵀ, with operators Tᵢ = τᵢ/2 built from the same Pauli matrices spin uses. The algebra is identical: [Tᵢ, Tⱼ] = i εᵢⱼₖ Tₖ, T² = ¾ on the doublet, and T_± = T₁ ± iT₂ move you along it, T₊|n⟩ = |p⟩. Charge depends on the third component alone, Q = e(T₃ + B/2) with B the baryon number, which fixes the proton at T₃ = +½ in the convention used here. Nothing has yet been assumed about the force. This is a relabelling, and its entire content is the claim that the two labels can be rotated into each other.

Two distinct claims hide behind "the strong force ignores charge". Charge symmetry is invariance under the π rotation exp(−iπT₂), which sends |p⟩ → |n⟩ and |n⟩ → −|p⟩ and therefore equates only pp with nn. Charge independence is [Hₛ, Tᵢ] = 0 for all three i: full SU(2) invariance, so Hₛ may depend on a pair's total T but never on T₃, and it brings np in a T = 1 state into the same equality. Different data test the two. Coulomb-corrected ¹S₀ scattering lengths give aₚₚ = −17.3 fm and aₙₙ = −18.9 fm, a 1.6 fm gap measuring charge-symmetry breaking; aₙₚ = −23.74 fm sits 5.6 fm from their mean, measuring the larger charge-independence breaking. Both look enormous as lengths and are tiny as forces: fitted as square wells of radius 2.1 fm they are 21.12, 21.28 and 21.65 MeV deep, a spread of 2.5%.

Once proton and neutron are one particle, two nucleons are identical fermions, and what must be antisymmetric is the product of space, spin and isospin. Exchange multiplies the spatial factor by (−1)^L; the spin triplet is symmetric and the singlet antisymmetric, giving (−1)^(S+1); isospin does the same, giving (−1)^(T+1). The product is (−1)^(L+S+T), so L + S + T must be odd. Apply it. At L = 0 exactly two channels survive, ¹S₀ with T = 1 and ³S₁ with T = 0. At L = 1, S + T must be even, so ¹P₁ carries T = 0 while ³P₀, ³P₁ and ³P₂ carry T = 1. Now use the projection: a pp or nn pair has |T₃| = 1, hence T = 1, and is barred from ³S₁ and ¹P₁ outright. The deuteron is J^π = 1⁺, so L is even and S = 1, making L + S odd and forcing T = 0. A T = 0 state has T₃ = 0 only. It can have no charged partners, and none is observed.

Put numbers on it with the crudest model that works. With μ = mₚ mₙ/(mₚ + mₙ), ħ²/2μ = (ħc)²/2μc² = 38938/938.9 = 41.47 MeV fm². For a square well of radius R the first S-wave bound state appears when a quarter wavelength just fits, k₀R = π/2, so V_crit = π²(41.47)/4R² = 102.3/R², or 23.20 MeV at R = 2.1 fm. Fit the triplet: matching u = sin kr inside to exp(−κr) outside gives k cot kR = −κ, and with κ = √(B/41.47) = 0.2316 fm⁻¹ the solution is kR = 1.8305, V₀ = 33.73 MeV. That one well then returns a zero-energy scattering length of 5.41 fm against the measured 5.424 fm, so a single parameter fits two independent numbers. Fit the singlet to a = −23.74 fm instead and V₀ = 21.65 MeV, 6.7% under threshold. The deuteron survives on a few MeV of margin in a 34 MeV well, and it shows: 1/κ = 4.32 fm, so most of the pair sits outside the 1.41 fm range of the force holding it.

A pure ³S₁ state is spherical and has a strictly zero quadrupole moment, yet Q_d = +0.2859 fm². Something must mix in L = 2, and the only static operator that can is the tensor S₁₂ = 3(σ₁⋅r̂)(σ₂⋅r̂) − σ₁⋅σ₂, which commutes with J², S² and parity but not with L², and whose matrix elements in the (|³S₁⟩, |³D₁⟩) basis are 0, 2√2 and −2. The radial problem is therefore two coupled equations for u(r) and w(r), with a 6ħ²/2μr² centrifugal term in the D channel, boundary conditions u(0) = w(0) = 0 and both decaying as exp(−κr); in Python it is a 2N × 2N Hermitian matrix passed to numpy.linalg.eigh, or a two-channel shooting problem. Q = (1/20)∫r²w(√8u − w)dr is linear in w to leading order, which is why a 4 to 6% probability — a 20 to 25% amplitude — yields a measurable Q. The magnetic moment gives a rough independent estimate: μ_d = 0.857438 μ_N against μₚ + μₙ = 0.879804 μ_N gives P_D ≈ 3.9% in the impulse approximation, which ignores meson-exchange currents and relativistic terms; P_D is model-dependent, not an observable.

Two culprits are known. Electromagnetism is the obvious one, since it couples to T₃ rather than to T. The subtler one is that the up and down quarks have different masses, m_d − mᵤ ≈ 2.5 MeV, which is why the neutron is heavier at all; lattice calculations split mₙ − mₚ = 1.293 MeV into roughly +2.3 MeV of quark-mass effect against −1.0 MeV of QED. Together they leave isospin good at the percent level, which is enough for T to stay a working label: isobaric analogue states, the Fermi operator Στ_±(i) of beta decay, and ΔT selection rules all survive. But do not read scattering lengths as a measure of the breaking. Near a pole a is hypersensitive, and the 27% spread from −17.3 to −23.74 fm comes from a 2.5% spread in well depth. A quantity that magnifies its input is a superb detector of small differences and a terrible ruler for them.