University Quantum Mechanics II · Hydrogen in External Fields and Hyperfine Structure · 9.7

The hyperfine interaction of electron and proton spins

Fine structure leaves the hydrogen ground state as four degenerate spin states. The proton is a magnet too, and this lesson builds the coupling that splits them: the delta function hiding in every dipole field, why an s electron feels nothing else, and the 5.88 μeV scale of the result.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Fine structure moves all four spin states of 1s together, so the level stays degenerate until something couples the electron's spin to the proton's. The proton carries a moment μₚ = gₚ(e/2mₚ)Sₚ with gₚ = 5.5857 quoted from experiment; a structureless Dirac particle would have 2, and the excess records a quark structure this course does not compute. Because mₚ = 1836 mₑ, the moment is 657 times weaker than the electron's.

Its field is the familiar 1/r³ dipole field plus a term easy to drop that turns out to carry everything, (2μ₀/3)μₚδ³(r). That term is fixed by a rule of magnetostatics, read off the uniformly magnetised sphere: the field summed over any sphere enclosing the currents is (2μ₀/3)μₚ, however small the sphere. The electron's spin moment, with gₑ taken as 2, sits in this field.

In an s state the electron's density is spherical, the 1/r³ part cancels on every shell, and first-order degenerate perturbation theory on the four spin states leaves (A/ħ²)Sₑ ⋅ Sₚ, derived here with A = μ₀gₚe²ħ²/(3πmₚmₑa³) = (4/3)gₚ(mₑ/mₚ)α⁴mₑc² = 5.88 μeV: order α⁴ like fine structure, suppressed by the mass ratio. The model stops where its idealisations do. gₑ is really 2.00232, the proton has a measured radius near 0.84 fm, the true orbit carries the reduced mass, and for l ≥ 1 the wavefunction vanishes at the proton, so the contact term goes and the dipolar and orbital terms take over.

Simple definition: The hyperfine interaction is the magnetic coupling of the electron, through its spin and its orbital motion, to the nucleus's moment; in an s state only the spin's contact part survives, an energy proportional to Sₑ ⋅ Sₚ and to the electron density at the nucleus.
Example: In hydrogen 1s the density at the proton is |ψ₁₀₀(0)|² = 1/(πa³) = 2.15 × 10³⁰ m⁻³, which makes the coupling (A/ħ²)Sₑ ⋅ Sₚ with A = 5.88 μeV, or A/h = 1421 MHz; with both spins up the energy is A/4 = 1.47 μeV.

Proton spin momentμₚ = gₚ (e/2mₚ) Sₚ

657 times weaker than μ_B because mₚ = 1836 mₑ, which is why hyperfine sits below fine structure. A Dirac point proton would have gₚ = 2; the rest comes from its quarks.

gₚ = 5.5857, measured; e/2mₚ = 4.789 × 10⁷ C kg⁻¹; μₚ in J T⁻¹, Sₚ in J s; spin up gives 1.4106 × 10⁻²⁶ J T⁻¹

Point-dipole field, contact term includedB(r) = (μ₀/4πr³)[3(μ ⋅ r̂)r̂ − μ] + (2μ₀/3) μ δ³(r)

The first term integrates to zero over any sphere centred on the dipole. The delta term restores what the source holds inside, and in an s state it is the only term that counts.

B in T, μ in J T⁻¹, μ₀ = 1.2566 × 10⁻⁶ N A⁻², δ³(r) in m⁻³; the first term holds only outside the source

Inside a uniformly magnetised sphereBᵢₙ = (2/3) μ₀ M = μ₀ μ / (2π R³)

Times the volume (4/3)πR³ it gives (2μ₀/3)μ for every R: the weight of the delta term. A dipole of magnetic charges would give −(1/3)μ₀M inside instead.

Quoted from magnetostatics: M in A m⁻¹, R in m, μ = (4/3)πR³M; the interior field is uniform

Contact Hamiltonian, gₑ = 2H_c = (μ₀gₚe² / 3mₚmₑ) Sₚ ⋅ Sₑ δ³(r)

The coefficient is positive: parallel spins cost energy, because the electron's moment points against its spin while the proton's points along its own.

From −μₑ ⋅ (2μ₀/3)μₚδ³(r) with μₑ = −(e/mₑ)Sₑ; spins in J s, δ³(r) in m⁻³, H_c in J

The dipolar term dies in an s state∫ (3cos²θ − 1) dΩ = 2π ∫ (3x² − 1) dx = 0, with x = cos θ from −1 to 1

Any spherical density, so every s state, gets nothing from the 1/r³ term, shell by shell. Take the angles first: done radially first, the integral diverges at r = 0.

θ from the z axis, dΩ = sin θ dθ dφ in sr; mixed parts such as sin θ cos θ cos φ vanish in the φ integral

The 1s hyperfine operatorH_hf(1s) = (A/ħ²) Sₑ ⋅ Sₚ, with A = μ₀gₚe²ħ² / (3πmₚmₑa³)

In α: A = (4/3)gₚ(mₑ/mₚ)α⁴mₑc², order α⁴ like fine structure but smaller by gₚmₑ/mₚ ≈ 0.003. It lies 0.053% above the measured 1420.406 MHz.

A = 9.417 × 10⁻²⁵ J = 5.877 μeV, A/h = 1421.2 MHz, from |ψ₁₀₀(0)|² = 1/(πa³), a = 5.2918 × 10⁻¹¹ m

A magnet 657 times weaker than the electron's

After fine structure the 1s level is still four states, two electron spins times two proton spins, all at one energy, because nothing so far has involved the proton's spin. The proton does carry a magnetic moment, μₚ = gₚ(e/2mₚ)Sₚ, with gₚ = 5.5857 quoted from experiment. The Dirac equation of relativistic quantum mechanics gives g = 2 for a structureless spin-½ particle; the proton's excess comes from its quarks and gluons, which nothing at this level computes. Because a moment scales as charge over mass, μₚ is small: its spin-up z-component is 1.4106 × 10⁻²⁶ J T⁻¹, against μ_B = 9.274 × 10⁻²⁴ J T⁻¹, a ratio of 1/657. A rough estimate sets the scale. The proton's dipole field at r = a is μ₀μₚ/(4πa³) = 9.5 mT, and μ_B times that is 0.55 μeV, or (gₚ/8)(mₑ/mₚ)α⁴mₑc². That is the scaling to expect: order α⁴ like fine structure, reduced by the mass ratio. The coefficient, and even the sign, need the field where this estimate never looked, inside the proton.

The field inside the source is a delta function

The familiar dipole field, (μ₀/4πr³)[3(μ ⋅ r̂)r̂ − μ], holds outside the currents. Integrate it over a sphere centred on the dipole and it gives zero, because 3(μ ⋅ r̂)r̂ averages to μ over directions. Yet inside a current loop the field points strongly along μ, so the formula has lost something at the origin. The uniformly magnetised sphere of magnetostatics, quoted here, shows what. Its bound surface currents produce a uniform interior field Bᵢₙ = (2/3)μ₀M, and with μ = (4/3)πR³M the interior adds Bᵢₙ × (4/3)πR³ = (2μ₀/3)μ to the volume integral, whatever R is. Shrink R at fixed μ and Bᵢₙ grows as 1/R³; a sphere of proton size, R = 0.84 fm, would hold 4.8 × 10¹² T, yet its integral never changes. That limit is a delta function, so the complete field of a point dipole is the 1/r³ term plus (2μ₀/3)μδ³(r). The same integral holds for any current distribution inside the sphere, so nothing hinges on the proton really being a uniform ball.

Coupling the electron's spin moment to that field

The electron's spin moment is μₑ = −(gₑe/2mₑ)Sₑ with gₑ = 2, the Dirac value; the measured 2.00232 comes from quantum electrodynamics and is set aside here. Its energy in the proton's field is −μₑ ⋅ B, and inserting μₚ = gₚ(e/2mₚ)Sₚ gives two spin-spin terms, both derived here: a dipole-dipole term (μ₀gₚe²/8πmₚmₑr³)[3(Sₚ ⋅ r̂)(Sₑ ⋅ r̂) − Sₚ ⋅ Sₑ] and a contact term (μ₀gₚe²/3mₚmₑ)Sₚ ⋅ Sₑ δ³(r). A third term couples μₚ to the electron's orbital motion through μₚ ⋅ L/r³, and it vanishes on any l = 0 state because L annihilates it. Then the bookkeeping. The four 1s spin states are degenerate, so this is degenerate perturbation theory, and all four share the spatial function ψ₁₀₀. Every matrix element therefore factorises into one spatial integral times a spin matrix element, and the whole problem reduces to a number multiplying a spin operator on the four-dimensional space.

An s electron feels only the contact term

The dipole-dipole term needs ∫|ψ₁₀₀|² [3(Sₚ ⋅ r̂)(Sₑ ⋅ r̂) − Sₚ ⋅ Sₑ]/r³ d³r, with d³r = r² dr dΩ. The density depends on r alone, so on each spherical shell the angles integrate by themselves. With θ from the z axis, the zz part gives ∫(3cos²θ − 1)dΩ = 2π[x³ − x] between x = −1 and 1, which is 0, and mixed parts such as sin θ cos θ cos φ vanish in the φ integral: every shell contributes zero. Order matters. Done radially first, ∫|R₁₀|² r² dr/r³ diverges logarithmically at r = 0 and the answer looks ill-defined; angles first, on shells outside a vanishing sphere, is exactly the prescription under which the interior was given to the delta term, so nothing is counted twice or lost. What remains is the contact term, and ⟨δ³(r)⟩ = |ψ₁₀₀(0)|² = 1/(πa³) = 2.148 × 10³⁰ m⁻³, from Unit 2. Only s states reach this far: for l ≥ 1 the wavefunction vanishes as rˡ at the proton.

The size of the coupling, and four checks

Assembled, the 1s operator is μ₀gₚe²/(3πmₚmₑa³) times Sₚ ⋅ Sₑ, written (A/ħ²)Sₑ ⋅ Sₚ and derived here, with A = 9.417 × 10⁻²⁵ J = 5.877 μeV, or A/h = 1421.2 MHz. Dimensions: μ₀e² is kg m and ħ²/(mₚmₑa³) is m s⁻², so A is kg m² s⁻², a joule. Order in α: with μ₀e² = 4παħ/c and a = ħ/(αmₑc), A = (4/3)gₚ(mₑ/mₚ)α⁴mₑc², order α⁴ like fine structure but smaller by gₚmₑ/mₚ = 0.0030. Limits: A vanishes as mₚ → ∞ or gₚ → 0, and it scales as 1/a³, so any tighter s orbit raises it. One state needs no diagonalising: with both spins up the flip-flop parts of Sₑ ⋅ Sₚ annihilate it, the eigenvalue is ħ²/4, and its energy A/4 = 1.47 μeV is μ_B times an effective contact field (2μ₀/3)μₚ|ψ(0)|² = 25.4 mT. Measurement: the hydrogen maser runs at 1420.406 MHz, which the next lesson identifies as A/h, and this coefficient sits 0.053% above it.

What the point-dipole model leaves out

Each simplification in the derivation is a named approximation, and each touches only the coefficient. The electron's g-factor is 2.00232 rather than 2, quoted from quantum electrodynamics, which raises A by 0.116%. The Bohr radius assumed an infinitely heavy proton; the orbit really carries the reduced mass, and since A ∝ 1/a³ that lowers A by about 3mₑ/mₚ = 0.163%. The proton is not a point: the electron density falls across it by a fraction near 2r/a, a few parts in 10⁵ for r ≈ 1 fm, which is where hyperfine measurements begin to see proton structure. The next lesson weighs these against 1420.406 MHz. What changes the form of the answer is angular momentum. For l ≥ 1, ψ(0) = 0, the contact term is gone, and the dipolar and orbital terms, which no longer average away, set a smaller splitting: quoted values are 59.2 MHz in 2P₁/₂ against 177.56 MHz in 2S₁/₂.

Change one variable at a time

Make the relationship visible.

Interactive model

Model proton radius R0.40 a

Electron state ns, n1

Slide R down to 0.05 a: the solid plateau climbs toward the point-dipole line while the curve beyond R stays flat, so no energy comes from outside the source. The real proton is 1.6 × 10⁻⁵ a. Then switch to 2s: A/h falls eightfold, yet the size correction barely moves.

A/h, POINT PROTON1421.2 MHz

A/h, SPHERE OF RADIUS R789.8 MHz

SIZE CORRECTION44.4 %

A/h, MAGNETIC CHARGES-394.9 MHz

Live interpretationA/h, POINT PROTON: 1421.2 MHz. A/h, SPHERE OF RADIUS R: 789.8 MHz. SIZE CORRECTION: 44.4 %. A/h, MAGNETIC CHARGES: −394.9 MHz

Catch the common trap

Explain before calculating.

Misconception check

It is tempting to picture the 1s electron circling at about a, sitting in the proton's 1/r³ dipole field, and to estimate the hyperfine energy as μ_B × μ₀μₚ/(4πa³). That gives 0.55 μeV with the right dependence on gₚ, mₑ/mₚ and α⁴, which is why the picture survives, but its physics is wrong. In an s state the outside field averages to exactly zero on every spherical shell, so it contributes nothing at any distance. The entire energy comes from the part of the electron's density that overlaps the proton, where the field is the interior field, and |ψ(0)|² measures that overlap. The effective field is 25.4 mT, not 9.5 mT, and its direction is set by what the proton is made of: a source built of magnetic charges would have the same outside field and an opposite, halved, interior one.

Suppose the proton's moment μₚ came from a pair of magnetic charges instead of circulating currents, with the same μₚ and the same field everywhere outside a tiny sphere around the proton. What would happen to the 1s coupling A in (A/ħ²)Sₑ ⋅ Sₚ?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyThe contact term acts on the 1s electron like an effective field B_c = (2μ₀/3)μₚ|ψ₁₀₀(0)|², where μₚ is the proton's moment for spin up. Taking gₚ = 5.5857, evaluate μₚ, |ψ₁₀₀(0)|² and B_c, and find the energy of the 1s state in which the electron and proton spins both point up.

Proton moment for spin up, with gₚ quoted: μₚ = gₚ(e/2mₚ)(ħ/2) = gₚμ_N/2 = 5.5857 × 5.0508 × 10⁻²⁷ J T⁻¹ ÷ 2 = 1.4106 × 10⁻²⁶ J T⁻¹, 657 times smaller than μ_B.
Density at the proton, from Unit 2: |ψ₁₀₀(0)|² = 1/(πa³) = 1/(π × 1.4818 × 10⁻³¹ m³) = 2.148 × 10³⁰ m⁻³.
Effective field: B_c = (2 × 1.2566 × 10⁻⁶ N A⁻² ÷ 3) × 1.4106 × 10⁻²⁶ J T⁻¹ × 2.148 × 10³⁰ m⁻³ = 2.538 × 10⁻² T. Units: J T⁻¹ is A m², and N A⁻² × A m² × m⁻³ = N A⁻¹ m⁻¹, which is a tesla.
With gₑ = 2 the spin-up electron's moment is −μ_B along z, so E = −μₑ ⋅ B_c = +μ_B B_c = 9.274 × 10⁻²⁴ J T⁻¹ × 0.02538 T = 2.354 × 10⁻²⁵ J = 1.469 μeV. Check: Sₑ ⋅ Sₚ gives ħ²/4 on this state, so E = A/4 with A = 5.877 μeV, the same number.

Answerμₚ = 1.411 × 10⁻²⁶ J T⁻¹, |ψ₁₀₀(0)|² = 2.15 × 10³⁰ m⁻³, B_c = 25.4 mT, and the up-up state is raised by 1.47 μeV (h × 355 MHz), because the electron's moment points against its spin.

MediumThe hydrogen 2s function is ψ₂₀₀ = (8πa³)^(−1/2)(1 − r/2a)e^(−r/2a). Find the 2s coupling A(2s) as a multiple of A(1s), evaluate it in MHz, and compare with the measured 2s value, 177.5568 MHz. Then use the measured 1s value, 1420.4058 MHz, to test the scaling, and say why the ratio is a sharper test than either number alone.

The derivation used the radial shape only at r = 0. The 2s density is spherical, so the dipolar term cancels shell by shell exactly as in 1s, radial node included, and the contact term samples |ψ₂₀₀(0)|².
At r = 0 the bracket and the exponential are both 1, so |ψ₂₀₀(0)|² = 1/(8πa³), one eighth of 1/(πa³). The spin operator is unchanged, so A(2s) = A(1s)/8.
A(2s)/h = 1421.16 MHz ÷ 8 = 177.645 MHz, which is 0.050% above the measured 177.5568 MHz. The 1s coefficient sits 0.053% above its measured value: the same fractional excess.
Measured ratio: 1420.4058 ÷ 177.5568 = 7.9997, equal to 8 within 3.4 parts in 10⁵. The corrections the model drops, gₑ − 2, the reduced mass and the leading proton-size term, multiply every s state by the same factor, so they cancel in the ratio although two of them, at 0.116% and 0.163%, are 34 and 48 times larger than the leftover, which is relativistic and radiative n-dependence.

AnswerA(2s) = A(1s)/8, so A(2s)/h = 177.6 MHz against 177.557 MHz measured; the measured 1s to 2s ratio, 7.9997, confirms the 1/(8πa³) contact density to 3.4 parts in 10⁵.

HardThe ion ³He⁺ has one electron bound to a spin-½ helion (Z = 2) whose measured moment is −2.1276 μ_N: it points against the helion's spin. Adapt the hydrogen contact result to find the 1s coupling A of ³He⁺ in MHz with its sign and the energy of the state with both spins up, compare |A| with the measured 8665.65 MHz, and explain why the raw result errs in the opposite direction to hydrogen's.

The nucleus enters only through its moment and the density at its position: A = 4μ_B μ_I (2μ₀/3)|ψ(0)|², with μ_I the nuclear moment along z for spin up. For hydrogen, μ_I = μₚ = 2.7928 μ_N reproduces 9.417 × 10⁻²⁵ J.
A charge Ze pulls the orbit in to a/Z, so ψ₁₀₀ = (Z³/πa³)¹⁄²e^(−Zr/a) and |ψ(0)|² = Z³/(πa³), eight times hydrogen's for Z = 2, quoted from Unit 2.
Ratio: A(³He⁺)/A(H) = Z³ × μ_I/μₚ = 8 × (−2.1276 ÷ 2.7928) = 8 × (−0.761816) = −6.0945, so A/h = −6.0945 × 1421.16 MHz = −8661.3 MHz.
Sign: the helion's moment, and so its contact field, points against its spin. The up-up state therefore moves by A/4 = h × (−2165 MHz), down, where hydrogen's moves up by h × 355 MHz: the level order the next lesson finds for hydrogen is inverted in ³He⁺.
Compare: 8661.3 MHz is 0.050% below the measured 8665.65 MHz, while hydrogen's raw value was 0.053% above. gₑ − 2 raises A by 0.116% in both; the reduced mass lowers it by about 3mₑ/M, 0.163% for the proton but 0.055% for the helion, three times heavier. Applying both factors gives 1420.49 MHz and 8666.6 MHz, 0.006% and 0.011% above measurement.

AnswerA/h = −8661 MHz for ³He⁺: 6.09 times hydrogen's in size and opposite in sign, so the up-up state shifts down by 2165 MHz. |A| is 0.05% below the measured 8665.65 MHz, because the helion's smaller reduced-mass correction lets gₑ − 2 win.