University Quantum Mechanics II · Hydrogen in External Fields and Hyperfine Structure · 9.8
The 21 cm line: F = 1 and F = 0
This is where the contact term becomes a line in the sky. Couple the two spins into a total spin F, read Sₑ⋅Sₚ off it, and hydrogen's ground state splits by 5.88 μeV: a 21.1 cm photon within 0.053% of the line radio telescopes tune to. Then the harder part: account for that 0.053%, and say which corrections are derived and which quoted.
Build the model
Connect the measurement to the mechanism.
The previous lesson reduced the proton's field, as a 1s electron feels it, to one operator, Hhf = (A/ħ²)Sₑ⋅Sₚ, with A fixed by |ψ₁₀₀(0)|² = 1/(πa³) and the proton's measured gₚ. Nothing else in hydrogen competes with it: with l = 0 only j = ½ exists, so fine structure and the Lamb shift move all four 1s spin states together. Degenerate perturbation theory therefore needs only the basis in which Sₑ⋅Sₚ is diagonal, and that is the total spin F = Sₑ + Sₚ.
Squaring F gives Sₑ⋅Sₚ = +ħ²/4 on the three F = 1 states and −3ħ²/4 on the singlet, so about its old energy the three F = 1 states rise by A/4 and the singlet falls by 3A/4, a splitting ΔE = A = (4/3)gₚ(mₑ/mₚ)α⁴mₑc² = 5.88 μeV: the α⁴ order of fine structure, cut by mₑ/mₚ because a heavy particle carries a small moment. The triplet lies on top because the electron's moment points against its spin. The model is non-relativistic, holds the proton fixed as a point dipole and takes gₑ = 2, and it lands 0.053% above the measured 1420.406 MHz.
Restoring the reduced mass and gₑ = 2.0023 closes that to 56 ppm; the rest belongs to relativistic, radiative and proton-structure corrections this course names but does not derive. Pushed to positronium, with the positron's moment for the proton's and the reduced mass restored, the formula gives 117 GHz, about 4/7 of the measured 203 GHz, because there the pair can also annihilate virtually.
- Simple definition
- The 21 cm line is the photon hydrogen emits when its ground state drops from F = 1, where the electron and proton spins combine into a triplet, to the F = 0 singlet 5.87 μeV below it.
- Example
- The line is measured at f = 1420.406 MHz, so F = 1 lies hf = 5.874 μeV above F = 0, and the photon's wavelength is c/f = 21.106 cm, about 4 × 10⁹ Bohr radii: an atomic transition in the radio band.
Derived by squaring F, since Sₑ² = Sₚ² = 3ħ²/4 on every state. Sₑ⋅Sₚ is diagonal in |F, mF⟩, so no matrix needs diagonalising.
F = 1: +ħ²/4 on three states; F = 0: −3ħ²/4 on one. ħ = 1.055 × 10⁻³⁴ J s, so Sₑ⋅Sₚ is in J² s²
Derived. Weighted by degeneracy, 3(+A/4) + 1(−3A/4) = 0: a traceless operator splits the level without moving its centroid.
A = 5.877 μeV, the previous lesson's coefficient of Sₑ⋅Sₚ/ħ²; F = 1 at +A/4 = +1.469 μeV, F = 0 at −3A/4 = −4.408 μeV
Derived: the two eigenvalues of Sₑ⋅Sₚ differ by exactly ħ². A positive A puts F = 1 on top, because the electron's moment opposes its spin and the singlet's moments are parallel.
ħ in J s, masses in kg, c in m s⁻¹, a in m: J⁴ s⁴/(kg³ m⁶ s⁻²) = J. Value 9.417 × 10⁻²⁵ J = 5.877 μeV
Derived with a = ħ/(αmₑc): fine-structure order α⁴, times (4/3)gₚmₑ/mₚ = 4.06 × 10⁻³, so ΔE is about 1/31 of the 1s fine-structure shift, α⁴mₑc²/8 = 0.181 meV.
α = 1/137.036, mₑ/mₚ = 1/1836.15 and gₚ = 5.5857 (quoted) are pure numbers; α⁴mₑc² = 1.449 meV
The derived line is 531 ppm, 0.053%, above the measured one. That gap measures what the model leaves out; it is not a rounding error.
hc = 1.2398 × 10⁻⁶ eV m. Derived: 1421.16 MHz and 21.095 cm. Measured: 1420.405 751 768 MHz and 21.106 cm
Leaves +56 ppm for the omitted corrections, which together come to about −56 ppm: Breit's relativistic +80 ppm and QED's −96 ppm, both quoted, then higher-order QED and proton structure, named but not derived.
f₀ = 1421.16 MHz. gₑ = 2.002 32, quoted, raises it 0.116%; the reduced mass, derived, lowers it 0.163%: f₁ = 1420.49 MHz
Nothing competes, so F is the good quantum number
Before the proton's moment is counted, hydrogen's 1s level holds four states: two electron spin states times two proton spin states. Nothing so far splits them. With l = 0 there is only j = ½, so Unit 8's relativistic, Darwin and Lamb terms move all four together, and the nearest other level is 10.2 eV away, 1.7 million times A. So, unlike the Zeeman and Stark problems of this unit, no second perturbation has to be weighed against the hyperfine term, and degenerate perturbation theory asks one question: which basis makes (A/ħ²)Sₑ⋅Sₚ diagonal inside the four-state space? The uncoupled states fail. Writing Sₑ⋅Sₚ = Sez Spz + ½(Sₑ₊Sₚ₋ + Sₑ₋Sₚ₊) exposes a flip-flop term that turns |↑↓⟩, electron up and proton down, into |↓↑⟩. What commutes with Sₑ⋅Sₚ is F² and Fz for F = Sₑ + Sₚ, so the good states are Unit 5's |F, mF⟩: the triplet |1, 1⟩ = |↑↑⟩, |1, 0⟩ = (|↑↓⟩ + |↓↑⟩)/√2 and |1, −1⟩ = |↓↓⟩, and the singlet |0, 0⟩ = (|↑↓⟩ − |↓↑⟩)/√2.
Reading Sₑ⋅Sₚ off the total spin
Square F = Sₑ + Sₚ. The two spins belong to different particles and commute, so F² = Sₑ² + Sₚ² + 2Sₑ⋅Sₚ. Each spin-half has S² = ½(½ + 1)ħ² = 3ħ²/4 on every state, so Sₑ⋅Sₚ = ½[F(F + 1) − 3/2]ħ². On the triplet F(F + 1) = 2 gives +ħ²/4; on the singlet 0 gives −3ħ²/4. The first-order shifts are +A/4 on each of three states and −3A/4 on one, and the gap is exactly A. Two checks come free. The weighted sum 3(+¼) + 1(−¾) = 0: Sₑ⋅Sₚ has zero trace, so the mean of the four states stays where 1s was. And |1, 0⟩ and |0, 0⟩ share the diagonal value Sez Spz = −ħ²/4; only the relative sign of |↑↓⟩ and |↓↑⟩ separates them. The flip-flop term, whose element between |↑↓⟩ and |↓↑⟩ is ħ²/2, raises the symmetric combination by ħ²/2 and lowers the antisymmetric one by ħ²/2, which is the whole difference between the F = 1 and F = 0 values.
Where 5.88 μeV sits among hydrogen's energy scales
Written with a = ħ/(αmₑc), μ₀ = 1/(ε₀c²) and e²/(4πε₀) = αħc, the previous lesson's A becomes ΔE = (4/3)gₚ(mₑ/mₚ)α⁴mₑc², and each factor places the splitting on hydrogen's ladder of scales. The Bohr energy is ½α²mₑc² = 13.6 eV. Fine structure is one factor of α² lower: relativistic and Darwin terms together lower 1s by α⁴mₑc²/8 = 0.181 meV. The Lamb shift, quoted from QED, is of order α⁵mₑc² times a logarithm: 1057.8 MHz, or 4.37 μeV, between 2S₁/₂ and 2P₁/₂. Hyperfine keeps the α⁴ of fine structure but pays (4/3)gₚmₑ/mₚ = 4.06 × 10⁻³ for a moment that falls as 1/mₚ, and lands at 5.88 μeV, beside the Lamb shift in size though not in origin. In a spectroscopist's other units 5.877 μeV is 1421.16 MHz, 21.09 cm, 0.0682 K, and μB B at B = 0.10 T, which is why fields of tens of millitesla already compete with it, the subject of the next lesson. Against the 13.6 eV binding energy the ratio is 4.3 × 10⁻⁷: an atom whose first optical line is at 10.2 eV also has a radio line.
Why the triplet is on top, and three checks
A is positive because of two signs. The electron's moment, −gₑμB Sₑ/ħ, points against its spin; the proton's, +gₚμN Sₚ/ħ with gₚ = +5.586, points along its spin. The contact energy −(2μ₀/3)μₑ⋅μₚ δ³(r) favours parallel moments, because inside a magnetised sphere the field points along the magnetisation. In the singlet the spins are opposed, so the moments are parallel and the energy is lowest: F = 0 sits 3A/4 below the old 1s energy and F = 1 sits A/4 above it. The order is not a law of two spins; a nucleus whose moment points against its spin, such as the helion of ³He⁺, inverts it. Three checks. Dimensions: ħ⁴/(mₚmₑ²c²a⁴) is J⁴ s⁴/(kg³ m⁶ s⁻²) = J. Limits: let mₚ → ∞ at fixed gₚ and the nuclear magneton eħ/2mₚ, and with it ΔE, goes to zero, which is why fine structure could treat the proton as a bare point charge; set gₚ = 0 and the level stays four-fold degenerate. Order: α⁴mₑc² times a mass ratio, one step below fine structure.
Accounting for the missing 0.053%
The derived 1421.16 MHz exceeds the measured 1420.405 751 768 MHz by 531 ppm. The two largest omissions pull opposite ways. The reduced mass, derived here: the electron circles the centre of mass, so a becomes a(1 + mₑ/mₚ) and |ψ₁₀₀(0)|² = 1/(πa³) falls by (1 + mₑ/mₚ)⁻³, 0.163% down, while the moments keep the particles' own masses. The electron's g-factor, quoted: gₑ = 2.002 32 from QED and experiment, whose first correction α/2π is Schwinger's, scales everything by gₑ/2, 0.116% up. Together they give 1420.49 MHz, still 56 ppm high. What remains is quoted, not derived: Breit's relativistic correction from the Dirac equation, +(3/2)α² = +80 ppm; the leading QED binding correction, α²(ln 2 − 5/2) = −96 ppm; smaller higher-order QED terms; and the proton's structure, led by its finite magnetic size at about −40 ppm and partly offset by recoil and polarisability terms of a few ppm. Their net closes the gap to within a few ppm, and there theory stops: the proton's magnetic size is known far less well than the line, which is measured to better than 10⁻¹².
A line the whole galaxy emits
At 21 cm the line passes through the dust that hides visible light, and neutral hydrogen is the commonest gas in the Milky Way. In a cloud at 80 K, kT = 6.9 meV is over a thousand times ΔE, so collisions keep F = 1 and F = 0 filled almost in the 3 : 1 ratio of their degeneracies. Each atom radiates very rarely: the spontaneous rate, quoted here and computed in Unit 13, is about 2.9 × 10⁻¹⁵ s⁻¹, a lifetime near eleven million years. Columns of 10²⁰ atoms or more per square centimetre still make the line bright. Because the rest frequency is known to thirteen figures, every shift reads as a velocity: 1 km s⁻¹ moves the line by 4.74 kHz. Radio astronomers turn those shifts into the rotation of the galactic disk and a map of its spiral arms, and the same line, redshifted, traces hydrogen in the young universe.
Change one variable at a time
Make the relationship visible.
Switch on the reduced mass alone and the dot falls from +531 to −1102 ppm; gₑ alone sends it to +1691 ppm; both together leave +56 ppm, the part Breit, QED and proton structure explain. Drag the moment to zero and the four states merge; below zero F = 0 rises above F = 1, as for ³He⁺.
ΔE = E(F=1) − E(F=0)5.877 μeV
LINE FREQUENCY |ΔE|/h1421.16 MHz
HYDROGEN λ = c/f21.095 cm
H: PREDICTED − MEASURED531 ppm
Live interpretationΔE = E(F=1) − E(F=0): 5.877 μeV. LINE FREQUENCY |ΔE|/h: 1421.16 MHz. HYDROGEN λ = c/f: 21.095 cm. H: PREDICTED − MEASURED: 531 ppm
Catch the common trap
Explain before calculating.
Hydrogen's 1s level, with the proton's moment ignored, is four-fold degenerate. The contact term is then switched on; its coupling constant, the coefficient of Sₑ⋅Sₚ/ħ², is 5.88 μeV. Where do the four states end up, measured from the energy 1s had before?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyThe measured ground-state splitting of hydrogen is ΔE = 5.874 μeV. Find the shift of each F level from the energy 1s would have without the contact term, the temperature at which kT equals ΔE, and the fraction of atoms in F = 1 if the spins were in equilibrium with the cosmic microwave background at 2.725 K.
- The shifts are (ΔE/ħ²)⟨Sₑ⋅Sₚ⟩: +ΔE/4 for F = 1 and −3ΔE/4 for F = 0. So F = 1 rises by 5.874/4 = 1.4685 μeV and F = 0 falls by 3 × 5.874/4 = 4.4055 μeV.
- Centroid check: three states at +1.4685 μeV and one at −4.4055 μeV give 3(1.4685) − 4.4055 = 0, as the traceless Sₑ⋅Sₚ requires.
- Temperature scale: T = ΔE/k = 5.874 × 10⁻⁶ eV ÷ 8.617 33 × 10⁻⁵ eV K⁻¹ = 0.06816 K. Far above this temperature the Boltzmann factor is close to 1, and populations follow degeneracy alone.
- At 2.725 K, kT = 234.8 μeV and ΔE/kT = 0.02501. Boltzmann with degeneracies 3 and 1 gives N₁/N₀ = 3e(−ΔE/kT) = 3 × 0.97530 = 2.9259.
- Fraction in F = 1: 2.9259/(1 + 2.9259) = 0.7453. Limits: as T → ∞ it tends to 3/4, set by counting states; as T → 0 it tends to 0, with every atom in the singlet.
AnswerF = 1 lies 1.469 μeV above and F = 0 lies 4.406 μeV below the unshifted 1s energy; ΔE/k = 0.0682 K; at 2.725 K a fraction 0.745 of the atoms is in F = 1, close to the 3/4 set by degeneracy.
MediumSuppose one distant gas cloud absorbs both the 21 cm line and Lyman α (1s–2p, frequency fL) of hydrogen, and the ratio of their frequencies is measured. Using this lesson's splitting and the Bohr value (3/8)α²mₑc² for Lyman α, find the combination of constants f₂₁/fL measures, evaluate it and compare with the laboratory ratio of 1420.406 MHz to 2466.07 THz. Then find the change in the ratio if α in the cloud were larger by 2.0 parts per million.
- Ratio: f₂₁/fL = [(4/3)gₚ(mₑ/mₚ)α⁴mₑc²]/[(3/8)α²mₑc²] = (32/9)gₚ(mₑ/mₚ)α². The mₑc² cancels and two powers of α survive, the hyperfine line's extra α² over the optical one.
- Evaluate: (32/9) × 5.5857 = 19.860; times mₑ/mₚ = 1/1836.15 gives 1.0816 × 10⁻²; times α² = 5.3251 × 10⁻⁵ gives 5.7598 × 10⁻⁷.
- Laboratory: 1.420406 × 10⁹ Hz ÷ 2.46607 × 10¹⁵ Hz = 5.7598 × 10⁻⁷, a match within 2 ppm. That is partly luck: the derived 21 cm value runs 531 ppm above its measured line and the Bohr Lyman α, 2467.38 THz, runs 532 ppm above its own, for different reasons, and the two almost cancel in the ratio.
- Sensitivity: the ratio goes as α², so δ(ratio)/ratio = 2δα/α = 2 × 2.0 × 10⁻⁶ = 4.0 × 10⁻⁶. Against the optical line, that moves the 21 cm line by 4.0 × 10⁻⁶ × 1420.406 MHz = 5.7 kHz.
- A common Doppler shift scales both lines by the same factor and leaves the ratio alone, so only a velocity difference between the gas absorbing each line can fake the effect: Δv = c × 4.0 × 10⁻⁶ = 1.2 km s⁻¹. And the ratio constrains gₚ(mₑ/mₚ)α², not α by itself.
Answerf₂₁/fL = (32/9)gₚ(mₑ/mₚ)α² = 5.760 × 10⁻⁷, within 2 ppm of the laboratory ratio; a 2.0 ppm rise in α raises it by 4.0 ppm, which a 1.2 km s⁻¹ velocity difference between the two absorbing gases would mimic.
HardAt t = 0 a ground-state hydrogen atom in zero field is prepared in |↑↓⟩: electron spin up and proton spin down along z. Using the measured splitting, A/h = 1420.406 MHz, find the probability of finding |↓↑⟩ at time t and the time for a complete swap, the first time at which the two spins are maximally entangled, and how ⟨Sez⟩ and ⟨Hhf⟩ evolve.
- Expand in the good states. From |1, 0⟩ = (|↑↓⟩ + |↓↑⟩)/√2 and |0, 0⟩ = (|↑↓⟩ − |↓↑⟩)/√2, the initial state is |↑↓⟩ = (|1, 0⟩ + |0, 0⟩)/√2: equal parts F = 1 and F = 0, so it is not stationary.
- Each part gains its own phase e(−iEt/ħ), with E = +A/4 and −3A/4. Taking out the common phase e(iAt/4ħ) leaves |ψ(t)⟩ ∝ cos(At/2ħ)|↑↓⟩ − i sin(At/2ħ)|↓↑⟩.
- So P(↓↑) = sin²(At/2ħ) = sin²(πft), with A = hf. The swap is complete when πft = π/2: t = 1/(2f) = 1/(2 × 1.420406 × 10⁹ s⁻¹) = 0.3520 ns, and the spins swap back and forth with period 1/f = 0.7040 ns.
- Maximal entanglement comes first at πft = π/4, t = 1/(4f) = 0.1760 ns, where |ψ⟩ ∝ (|↑↓⟩ − i|↓↑⟩)/√2. Tracing out the proton leaves the electron with ρₑ = I/2, I the 2 × 2 identity: each spin alone is unpolarised, though the pair is in a pure state.
- Expectation values: ⟨Sez⟩ = (ħ/2)[cos²(πft) − sin²(πft)] = (ħ/2)cos(2πft), and ⟨Spz⟩ = −⟨Sez⟩, so Fz = 0 is conserved. ⟨Hhf⟩ = ½(A/4) + ½(−3A/4) = −A/4 = −1.469 μeV at all times, equal to ⟨↑↓|Hhf|↑↓⟩ = (A/ħ²)(−ħ²/4).
- Limit check: keep only Sez Spz and |↑↓⟩ is an eigenstate, so P(↓↑) = 0 forever. The whole oscillation is the flip-flop term, and it runs at the 21 cm frequency itself.
AnswerP(↓↑) = sin²(πft): a complete swap at 0.3520 ns, repeating every 0.7040 ns; maximal entanglement first at 0.1760 ns, as (|↑↓⟩ − i|↓↑⟩)/√2; ⟨Sez⟩ = (ħ/2)cos(2πft), while ⟨Hhf⟩ = −A/4 = −1.469 μeV throughout.