University Physics V · Nuclear Physics · 14.1

Nucleons, Isospin & the Two-Nucleon System

Nature made exactly one bound state out of two nucleons, and it is a strange one: barely held, mostly outside its own potential, and not spherical. This topic hands you the bookkeeping that makes all three follow from a single antisymmetry rule, rather than from a table of measured numbers you have to trust.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Heisenberg's move was to stop treating the proton and the neutron as two particles and start treating them as two states of one, separated by a label the strong interaction cannot read. Give the nucleon an internal two-dimensional space, |p⟩ = (1, 0)ᵀ and |n⟩ = (0, 1)ᵀ, act on it with Tᵢ = τᵢ/2 built from the very Pauli matrices spin already uses, and the whole SU(2) apparatus transfers untouched: ladder operators T_± that turn a neutron into a proton, Clebsch-Gordan addition, total-T multiplets. Charge independence then becomes a statement about operators, [Hₛ, Tᵢ] = 0 for i = 1, 2, 3, so the strong Hamiltonian reads a pair's total isospin and never its projection; charge symmetry, invariance under the π rotation exp(−iπT₂) alone, is the weaker claim that equates only pp with nn.

The evidence is that the three ¹S₀ scattering lengths agree to a few fermi out of twenty; the cost is that they agree only to a few fermi, because Coulomb and the up-down quark mass difference make this a percent-level symmetry rather than an exact one. What it buys is a counting rule. Two nucleons are now identical fermions, so space, spin and isospin must be antisymmetric together, (−1)^(L+S+T) = −1.

At L = 0 that leaves exactly two channels — ³S₁ with T = 0 and ¹S₀ with T = 1 — and since pp and nn are stuck at T = 1, they are locked out of the only one that binds.

Simple definition: Isospin is a two-valued internal label that treats the proton and the neutron as one particle, acted on by operators Tᵢ = τᵢ/2 obeying the same SU(2) algebra as spin, and conserved by the strong interaction alone.
Example: The two nucleons of a deuteron sit in the isospin singlet |0,0⟩ = (|pn⟩ − |np⟩)/√2, so T = 0; its T = 1 partner (|pn⟩ + |np⟩)/√2 is the ¹S₀ state, unbound by about 66 keV, and its charged relatives pp and nn are unbound too.

Isospin doublet and its algebraTᵢ = τᵢ/2, [Tᵢ, Tⱼ] = i εᵢⱼₖ Tₖ, T₃|p⟩ = +½|p⟩, T₃|n⟩ = −½|n⟩

Every spin identity you own — ladders, Clebsch-Gordan, multiplets — transfers to charge with no new algebra to learn.

τᵢ the Pauli matrices; T is dimensionless; Q = e(T₃ + B/2) for a nucleon, B = 1

Charge independence against charge symmetry[Hₛ, Tᵢ] = 0, i = 1, 2, 3 (CI)[Hₛ, exp(−iπT₂)] = 0 (CS only)

CS equates only pp with nn; CI adds np in the same T = 1 state, so the two are tested by different experiments.

exp(−iπT₂) is a π rotation about the 2-axis: |p⟩ → |n⟩, |n⟩ → −|p⟩

Generalised Pauli principle(−1)^(L+S+T) = −1, so L + S + T is odd

One line decides which channels exist: ³S₁ demands T = 0, ¹S₀ demands T = 1, and the ³P states are T = 1 only.

L, S, T the pair's orbital, spin and isospin quantum numbers, all integers

Deuteron binding from the mass tableB = [m(¹H) + mₙ − m(²H)]c² = 0.002388 u × 931.494 MeV/u = 2.2244 MeV

Gives the only energy the two-nucleon system has, and κ = √(B/41.47 MeV fm²) = 0.2316 fm⁻¹ with it.

Neutral atomic masses in u; one electron on each side, so they cancel exactly

Effective-range expansionk cot δ₀ = −1/a + ½ r₀k²aₜ = +5.424 fm, aₛ = −23.74 fm

The sign of a is the whole answer: positive means a bound state below threshold, negative means a virtual one above it.

k in fm⁻¹, δ₀ the S-wave phase shift, a and r₀ in fm; rₜ = 1.759 fm

Tensor operator and the S-D mixingS₁₂ = 3(σ₁⋅r̂)(σ₂⋅r̂) − σ₁⋅σ₂⟨³S₁|S₁₂|³D₁⟩ = 2√2, ⟨³D₁|S₁₂|³D₁⟩ = −2

The only term that can put D-wave into a J = 1 state, so it alone accounts for Q_d ≠ 0.

S₁₂ is dimensionless, acts only on S = 1, and commutes with J² and parity but not L²

Two states of one particle, and the operators that say so

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.

Charge independence is the stronger of two symmetries

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%.

Antisymmetry now runs over three labels at once

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.

One channel binds, and only just

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.

The tensor force is what makes the deuteron non-spherical

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% probability — a 20% amplitude — yields a measurable Q. The magnetic moment agrees independently: μ_d = 0.857438 μ_N against μₚ + μₙ = 0.879804 μ_N gives P_D = 3.9%.

What breaks isospin, and what survives the breaking

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.

Change one variable at a time

Make the relationship visible.

Interactive model

Well depth V₀21.50 MeV

Well radius R2.10 fm

Raise the depth through threshold and watch the exterior line flatten: a runs out to minus infinity and returns from plus infinity, and only then does the node at r = a walk in from the right. At R = 2.10 fm, 21.5 MeV is the unbound ¹S₀ channel and 33.75 MeV the bound ³S₁ one.

INTERIOR PHASE k₀R1.512 rad

THRESHOLD DEPTH23.20 MeV

SCATTERING LENGTH a-21.5 fm

DEPTH OVER THRESHOLD-1.70 MeV

Live interpretationINTERIOR PHASE k₀R: 1.512 rad. THRESHOLD DEPTH: 23.20 MeV. SCATTERING LENGTH a: −21.5 fm. DEPTH OVER THRESHOLD: −1.70 MeV

Catch the common trap

Explain before calculating.

Misconception check

The usual story is that the deuteron binds because the neutron-proton force is stronger than the neutron-neutron force. It is not. Charge independence holds the T = 1 interaction equal in all three pairs to a percent or two, and the test is direct: the np ¹S₀ state, built from exactly the pair that does bind in ³S₁, is itself unbound by about 66 keV, with a = −23.74 fm against aₙₙ = −18.9 fm. The same two particles fail in one spin state and succeed in the other; what separates them is channel access, not strength. L + S + T odd means an L = 0 pair with T = 1 must have S = 0, and identical nucleons are locked to T = 1; only np can reach S = 1 with T = 0, where the tensor force operates. The second half of the error is reading a = −23.74 fm as strong attraction because it is large. A large |a| says only that a pole sits close to threshold; the minus sign says which side of it.

The np system has a bound ³S₁ state (a = +5.42 fm) and an unbound ¹S₀ state (a = −23.74 fm), built from the very same proton and neutron. A neutron pair has no bound state at all. Which account covers both facts?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyNeutral atomic masses are m(¹H) = 1.007825 u, mₙ = 1.008665 u and m(²H) = 2.014102 u, with 1 u = 931.494 MeV/c². Find the deuteron's binding energy and B/A, then convert B into the asymptotic decay length 1/κ of its wavefunction using ħ²/2μ = 41.47 MeV fm². Compare that length with the 1.41 fm range of one-pion exchange.

Electron bookkeeping first. The left side carries one atomic electron (in ¹H) and the right side one (in ²H), so neutral atomic masses may be subtracted directly; the electron binding that does not cancel is a few eV, far below the precision here.
Mass defect: Δm = 1.007825 + 1.008665 − 2.014102 = 0.002388 u.
Binding energy: B = 0.002388 u × 931.494 MeV/u = 2.2244 MeV, so B/A = 1.112 MeV per nucleon against the 8.79 MeV plateau at the Ni-62 peak. The deuteron is the most weakly bound nucleus there is.
Outside the range of the force the radial function goes as exp(−κr) with κ = √(B ÷ 41.47 MeV fm²) = √0.05364 = 0.2316 fm⁻¹, so 1/κ = 4.32 fm.
Compare: 4.32 fm against ħ/m_πc = 197.327/139.57 = 1.41 fm. The decay length is three times the range, so most of the probability lies where the potential is already negligible.

AnswerB = 2.2244 MeV and B/A = 1.112 MeV; 1/κ = 4.32 fm, about three times the 1.41 fm range, so the deuteron is mostly outside the force that binds it.

MediumTwo nucleons are in a relative S or P state. (a) Write the isospin triplet and singlet and confirm their exchange symmetry with P^τ = (1 + τ₁⋅τ₂)/2. (b) Use L + S + T odd to list every allowed L = 0 and L = 1 channel. (c) Say which are open to a neutron pair, and deduce the deuteron's isospin from J^π = 1⁺.

Triplet: |1,+1⟩ = |pp⟩, |1,0⟩ = (|pn⟩ + |np⟩)/√2, |1,−1⟩ = |nn⟩. Singlet: |0,0⟩ = (|pn⟩ − |np⟩)/√2.
Since T = T₁ + T₂, τ₁⋅τ₂ = 4T₁⋅T₂ = 2[T(T+1) − 3/2], which is +1 for T = 1 and −3 for T = 0. So P^τ = (1 + τ₁⋅τ₂)/2 gives +1 on the triplet and −1 on the singlet: symmetric and antisymmetric, as required.
Antisymmetry over all three labels: (−1)^L (−1)^(S+1) (−1)^(T+1) = (−1)^(L+S+T) = −1, so L + S + T is odd. L = 0 then allows (S, T) = (0,1), the ¹S₀, and (1,0), the ³S₁. L = 1 needs S + T even, allowing (0,0), the ¹P₁, and (1,1), the ³P₀, ³P₁ and ³P₂.
A neutron pair has T₃ = −1, so T = 1 is forced. It may occupy ¹S₀ and the three ³P states, and is excluded from ³S₁ and ¹P₁ — the two T = 0 channels.
The deuteron has J^π = 1⁺. Positive parity forces even L, and J = 1 with S ≤ 1 then means S = 1 and L = 0 or 2. Either way L + S is odd, so T must be even: T = 0. A T = 0 state has only T₃ = 0, so no pp or nn analogue can exist — and none is seen.

AnswerAllowed channels: ¹S₀ (T = 1), ³S₁ (T = 0), ¹P₁ (T = 0), ³P₀,₁,₂ (T = 1). A neutron pair is confined to T = 1 and so locked out of ³S₁; the deuteron is T = 0, which is why it has no charged partners.

HardModel the S-wave np interaction as a square well of radius R = 2.1 fm, with ħ²/2μ = 41.47 MeV fm². (a) Find the depth at which the first S-wave bound state appears. (b) Find the depth reproducing B = 2.2246 MeV, and test it against the measured triplet scattering length +5.424 fm. (c) Find the depth reproducing the singlet a = −23.74 fm, and say by what margin that channel misses binding.

Threshold. A bound state first appears when the interior quarter wavelength just fits the well, k₀R = π/2 with k₀ = √(V₀ ÷ 41.47 MeV fm²). So V_crit = π²(41.47)/(4R²) = 102.3/R² = 102.3/4.41 = 23.20 MeV.
Triplet, bound state. Match u = sin kr inside to exp(−κr) outside: k cot kR = −κ, with k = √((V₀−B)/41.47) and κ = √(2.2246/41.47) = 0.2316 fm⁻¹, so κR = 0.4864. Solving y cot y = −0.4864 gives y = kR = 1.8305.
Hence k = 1.8305/2.1 = 0.8717 fm⁻¹, V₀ − B = 41.47 × 0.7598 = 31.51 MeV, and V₀ = 33.73 MeV — 45% above threshold, yet binding only 2.22 MeV. Binding energy climbs very slowly out of threshold.
Test that well at zero energy: k₀ = √(33.73/41.47) = 0.9019 fm⁻¹, k₀R = 1.8940, tan(k₀R) = −2.985, and a = R − tan(k₀R)/k₀ = 2.1 + 3.310 = 5.41 fm against the measured 5.424 fm. One fitted parameter reproduces a second, independent number.
Singlet, from a. At zero energy a = R[1 − tan y/y], so tan y/y = 1 − a/R = 1 + 23.74/2.1 = 12.305, whose root is y = 1.5173 — just below π/2, so no bound state. Then k₀ = 0.7225 fm⁻¹ and V₀ = 41.47 × 0.5220 = 21.65 MeV.
Compare: 21.65 MeV against the 23.20 MeV threshold is a shortfall of 1.55 MeV, or 6.7%. Strengthen the ¹S₀ interaction by one part in fifteen and the dineutron would bind, and the chart of nuclides would look nothing like it does.

AnswerV_crit = 23.20 MeV; the triplet well is 33.73 MeV and returns aₜ = 5.41 fm against the measured 5.424 fm; the singlet well is 21.65 MeV, 6.7% short of threshold.