University Quantum Mechanics II · Hydrogen in External Fields and Hyperfine Structure · 9.9
Hyperfine levels in a magnetic field, and the hydrogen maser
This is where the hyperfine story leaves the blackboard for the clock room. Add a field to the two-spin problem and it solves exactly, block by block. It tells you when F stops being a good label, why every Zeeman line drifts with B while one frequency barely notices, and how a magnet sorts atoms so a maser can oscillate on that line.
Build the model
Connect the measurement to the mechanism.
Put the hydrogen ground state in a uniform field B along z and only two spins are left to act, so the problem is a 4 × 4 matrix: the contact term A Sₑ⋅Sₚ/ħ² of the previous topic plus one Zeeman term for each moment. Neither F² nor mₛ is conserved any more, but Fz = Sₑz + Sₚz commutes with every term, and that cuts the four states into blocks of one, two and one. The stretched states mF = ±1 have nothing to mix with, so their energies are exactly linear in B.
The two mF = 0 states form a two-level system with splitting A and coupling Ax/2, where x = (gₑμB + gₚμN)B/A; its exact eigenvalues, −A/4 ± (A/2)√(1 + x²), are the Breit–Rabi result for hydrogen, derived here. Because A is only 5.874 μeV, of order α⁴mₑc²(mₑ/mₚ), x reaches 1 at a modest Bc = 50.6 mT: well below it F is good, well above it mₛ and mI are good and the lines pair off by electron spin. The frequency between the two mF = 0 levels has zero slope at B = 0 and rises as 2.77 × 10¹¹ Hz T⁻² × B², which is why the hydrogen maser runs on it inside magnetic shields.
The model is exact for two spins in a static, uniform field with the 1s orbital frozen. It stops being enough at the order-α² binding correction to gₑ, at the wall and cavity shifts of a real maser, and as B nears ħ/(ea₀²) = 2.35 × 10⁵ T, where the field distorts the orbital itself.
- Simple definition
- The Breit–Rabi levels of hydrogen are the exact energies of the 1s electron and proton spins in a uniform field: two levels linear in B for mF = ±1, and a repelling pair −A/4 ± (A/2)√(1 + x²) for mF = 0.
- Example
- In a 2.0 mT field, x = 0.0395: the F = 1, mF = +1 level has risen by 27.98 MHz, but the F = 1, mF = 0 → F = 0 frequency by only 1.109 MHz, because the pair's gap has no term linear in B.
Fz = Sₑz + Sₚz commutes with all three terms and F² does not, so mF labels blocks of one, two and one state, and only the middle block needs diagonalising.
A/h = 1420.406 MHz measured; μB = 9.274 × 10⁻²⁴ J T⁻¹, μN = 5.051 × 10⁻²⁷ J T⁻¹; gₑ = 2.00232, gₚ = 5.58569 quoted
Derived here. A one-state block has nothing to mix with, so this is exact at every field: the lines never curve, from 1 nT to 10 T.
E in J, B in T; the slope over h is ±13.991 GHz T⁻¹, which is ±1.399 MHz per 0.1 mT
Derived here: the exact eigenvalues of the mF = 0 block, whose diagonal is A/4 and −3A/4 in the |F, 0⟩ basis and whose off-diagonal element is Ax/2. It is the two-state problem of QM I.
E in J; the upper sign is F = 1 at low field and e↑p↓ at high field, the lower is F = 0, becoming e↓p↑
x ≪ 1: F and mF are good. x ≫ 1: mₛ and mI are good. Compare the field with 50.6 mT before computing anything.
x is dimensionless, (J T⁻¹)(T)/J; (gₑμB + gₚμN)/h = 28.068 GHz T⁻¹; the rough 2μB B = A gives 50.7 mT
Derived here. No linear term, because the Zeeman operator has no diagonal element in either mF = 0 state: it only couples them, and they repel at second order.
f₀ = 1420.405 751 768 MHz; K = 2.77 × 10¹¹ Hz T⁻² = 277 kHz mT⁻²; the quadratic form needs x ≪ 1
mₛ = +½ atoms are pushed toward weak field and focused by a hexapole; mₛ = −½ atoms are thrown out, which leaves the upper maser level populated and the lower one empty.
U in J, ∇|B| in T m⁻¹, force in N; the strong-field limit, valid where x ≫ 1, with mₛ = ±½
Two spins, one field, and the number that survives
From the previous topic, the 1s level carries two spins coupled by A Sₑ⋅Sₚ/ħ², with A/h = 1420.405 751 768 MHz quoted as measured. A field B along z adds the Zeeman energy of each moment. The electron's moment points against its spin, so its term is +(gₑμB B/ħ)Sₑz; the proton's points along its spin, giving −(gₚμN B/ħ)Sₚz, 658 times weaker. The orbital Zeeman term vanishes in 1s, and the diamagnetic term e²B²(x² + y²)/8mₑ shifts all four spin states equally, by 1.2 × 10⁻¹⁰ eV at 1 T, so it cancels from every line. Now ask what is conserved. Sₑ⋅Sₚ = (F² − Sₑ² − Sₚ²)/2 commutes with every component of F, but the Zeeman term weights the two spins differently, so it commutes with Fz = Sₑz + Sₚz and not with F². So mF = mₛ + mI is good at every field, and the four product states sort into mF = +1 (↑↑), mF = 0 (↑↓ and ↓↑) and mF = −1 (↓↓), with the electron's arrow written first: blocks of one, two and one.
The stretched states cannot bend
A one-state block is already diagonal. For |↑↑⟩, Sₑ⋅Sₚ/ħ² = +¼ and the two Zeeman terms give +gₑμB B/2 and −gₚμN B/2, so E(+1) = A/4 + (gₑμB − gₚμN)B/2, and |↓↓⟩ is the mirror image. Both are exact for any B: no other state has mF = ±1, so nothing can push on these levels and they stay straight into the strong-field regime. Their slope over h is ±13.991 GHz T⁻¹. Check it against the weak-field route: the projection theorem of the Landé topic, applied to F = 1 built from J = ½ and I = ½, gives gFμB = (gₑμB − gₚμN)/2, the same number, obtained there only for weak fields and here for all of them. Dimensions check, since (J T⁻¹)(T) is an energy; the limit B → 0 returns the F = 1 energy A/4; and at B = 1.0 mT the mF = +1 level has risen by 13.99 MHz.
The mF = 0 block is a two-state problem you have met
In the coupled basis |1, 0⟩ = (|↑↓⟩ + |↓↑⟩)/√2 and |0, 0⟩ = (|↑↓⟩ − |↓↑⟩)/√2 the hyperfine term is diagonal, A/4 and −3A/4. The Zeeman operator multiplies |↑↓⟩ by +(gₑμB + gₚμN)B/2 and |↓↑⟩ by the opposite, so it has no diagonal element in either coupled state and only connects them: ⟨1, 0|HZ|0, 0⟩ = (gₑμB + gₚμN)B/2 = Ax/2. That is the two-state Hamiltonian of QM I with splitting A and coupling Ax/2, so E± = −A/4 ± √((A/2)² + (Ax/2)²) = −A/4 ± (A/2)√(1 + x²). The pair repels and never crosses. The upper state is cos φ |1, 0⟩ + sin φ |0, 0⟩ with tan 2φ = x, and its electron-up probability, (1 + x/√(1 + x²))/2, runs from 0.500 at B = 0 through 0.854 at x = 1 to 0.9994 at 1 T. Checks: x is dimensionless, x → 0 returns A/4 and −3A/4, and the trace of H stays zero at every field.
Which basis is good: F below 50.6 mT, mₛ and mI above
Set x = 1: Bc = A/(gₑμB + gₚμN) = 1420.406 MHz ÷ 28.068 GHz T⁻¹ = 50.6 mT. For x ≪ 1, E± ≈ A/4 + Ax²/4 and −3A/4 − Ax²/4, exactly the second-order shifts (Ax/2)²/(±A) of perturbation theory: F is good and the fan is a weak-field Zeeman pattern. For x ≫ 1 the energies tend to gₑμB B mₛ − gₚμN B mI + A mₛ mI: mₛ and mI are good, the hyperfine term survives only through its diagonal part, and the lines pair off by electron spin. That is how an ESR spectrometer at 9.5 GHz sees free hydrogen, as two lines about 51 mT apart; these levels put them at 0.3116 T and 0.3626 T. The order-α check: the QED anomaly in gₑ = 2(1 + α/2π + …) moves Bc by 0.12%, and binding lowers the 1s electron's g by a further fraction α²/3 = 1.8 × 10⁻⁵, a result of relativistic quantum mechanics. Both are quoted, not derived.
The clock line: why f₀₀ has no linear term
The F = 1, mF = 0 → F = 0 frequency is the gap of the Breit–Rabi pair, f₀₀ = f₀√(1 + x²). At small x it is f₀ + f₀x²/2 = f₀ + KB², with K = (gₑμB + gₚμN)²/(2h²f₀) = (28.068 GHz T⁻¹)² ÷ (2 × 1420.406 MHz) = 2.77 × 10¹¹ Hz T⁻², or 277 kHz mT⁻². The units are (Hz T⁻¹)²/Hz = Hz T⁻², as they must be. The linear term is missing because HZ has no diagonal element in either mF = 0 state. At 1.0 × 10⁻⁷ T, the order of field a maser keeps inside its shields, a wobble of 1.0 × 10⁻⁹ T moves an mF = ±1 ↔ 0 Zeeman line by 14 Hz but moves f₀₀ by only 2KBδB = 5.5 × 10⁻⁵ Hz, 3.9 × 10⁻¹⁴ of f₀. The measured f₀ itself comes from masers run in such fields with KB² subtracted. The quadratic form is a small-x approximation: at 40 mT it is already 3% high, so above a few millitesla use the square root.
Sorting atoms and closing the loop: the hydrogen maser
A maser needs more atoms in the upper mF = 0 level than in F = 0, and a field gradient supplies them. Where x ≫ 1 an atom's energy is U ≈ gₑμB mₛ|B|, so mₛ = +½ atoms are pushed toward weak field and mₛ = −½ atoms toward strong field. An ideal hexapole magnet has |B| = B₀(r/r₀)², zero on its axis: a harmonic lens for mₛ = +½ and a scatterer for mₛ = −½. With B₀ = 0.90 T the well is 5.2 × 10⁻⁵ eV deep, 0.61 K, so atoms from a 300 K source, whose mean thermal speed is 2.5 km s⁻¹, are kept only if their transverse speed is under 100 m s⁻¹, within 0.04 rad of the axis. Leaving the magnet slowly, e↑p↑ follows its line to F = 1, mF = +1, and e↑p↓ follows the upper Breit–Rabi branch to F = 1, mF = 0, since levels of equal mF never cross; F = 0 arrives empty. The atoms bounce for about a second in a coated storage bulb inside a cavity tuned to 1420.406 MHz, and above threshold they sustain their own oscillation: both steps are named here, not developed.
Change one variable at a time
Make the relationship visible.
Step up from B = 0: the mF = ±1 lines fan out at 14 MHz per mT while the slope of f₀₀ starts at zero, because the maser line grows as B². By 25 mT the quadratic rule already runs 9 MHz high, and past the dashed line the levels pair off by electron spin.
MASER LINE f₀₀, EXACT1584.271 MHz
QUADRATIC RULE f₀ + KB²1593.725 MHz
SLOPE df₀₀/dB12.431 MHz mT⁻¹
E-UP WEIGHT, UPPER mF = 00.721
Live interpretationMASER LINE f₀₀, EXACT: 1584.271 MHz. QUADRATIC RULE f₀ + KB²: 1593.725 MHz. SLOPE df₀₀/dB: 12.431 MHz mT⁻¹. E-UP WEIGHT, UPPER mF = 0: 0.721
Catch the common trap
Explain before calculating.
A hydrogen maser runs in a residual field of 2.0 × 10⁻⁷ T along its axis. How far is its F = 1, mF = 0 → F = 0 frequency shifted from the zero-field value f₀ = 1420.405 751 768 MHz?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA beam of ground-state hydrogen crosses a region where B = 0.30 mT. Decide which quantum numbers are good, then find how far the F = 1, mF = +1 level and the F = 1, mF = 0 → F = 0 frequency move from their zero-field values.
- Compare with the crossover: Bc = 1420.406 MHz ÷ 28.068 GHz T⁻¹ = 50.6 mT, so x = 0.30 ÷ 50.6 = 5.93 × 10⁻³. That is far below 1, so F and mF are good and the states are labelled |F, mF⟩.
- mF = +1 is a one-state block, so its shift is exact: ΔE/h = (gₑμB − gₚμN)B/2h = 13.991 GHz T⁻¹ × 3.0 × 10⁻⁴ T = 4.20 MHz.
- The maser line has no linear term: Δf₀₀ = KB² = 2.773 × 10¹¹ Hz T⁻² × (3.0 × 10⁻⁴ T)² = 2.50 × 10⁴ Hz.
- Cross-check with the exact form: f₀(√(1 + x²) − 1) = 1420.406 MHz × 1.757 × 10⁻⁵ = 24.96 kHz, the same to one part in 10⁵, as it should be when x is this small.
Answerx = 5.9 × 10⁻³, so F is good. The mF = +1 level rises 4.20 MHz; f₀₀ rises only 25.0 kHz, 168 times less.
MediumHydrogen atoms sit in a 40.0 mT field. Find the four hyperfine energies as frequencies from the zero-field centroid, the F = 1, mF = 0 → F = 0 frequency, and the probability that the upper mF = 0 state has its electron spin up. How far off is the quadratic rule f₀ + KB² here?
- x = B/Bc = 40.0 ÷ 50.61 = 0.790, so √(1 + x²) = √1.62475 = 1.27466. Neither limit applies, so use the exact blocks.
- mF = ±1: E/h = 355.10 MHz ± 13.991 GHz T⁻¹ × 0.0400 T = 355.10 ± 559.65 MHz, giving +914.7 MHz and −204.5 MHz.
- mF = 0: E/h = −355.10 ± 710.20 × 1.27466 MHz = −355.10 ± 905.26 MHz, giving +550.2 MHz and −1260.4 MHz. The four sum to zero, as they must for a traceless H.
- f₀₀ = 550.2 − (−1260.4) = 1810.5 MHz, which is f₀ × 1.27466.
- Electron-up weight of the upper state: (1 + x/√(1 + x²))/2 = (1 + 0.790/1.27466)/2 = (1 + 0.620)/2 = 0.810.
- The quadratic rule gives 1420.41 MHz + 277.3 kHz mT⁻² × (40.0 mT)² = 1420.41 + 443.70 = 1864.1 MHz, which is 53.6 MHz, or 3.0%, too high because x = 0.79 is not small.
AnswerLevels at +914.7, +550.2, −204.5 and −1260.4 MHz (mF = +1, upper 0, −1, lower 0); f₀₀ = 1810.5 MHz; electron-up probability 0.810; the quadratic rule is 54 MHz (3.0%) high.
HardA hydrogen maser's field is set by driving the F = 1, mF = +1 ↔ mF = 0 Zeeman transition, which resonates at 850 Hz. Find B, the fractional offset of the maser line from f₀, and how steady B must be for that offset to drift by less than 1 × 10⁻¹⁵ of f₀. If the magnetic noise outside is 1 × 10⁻⁷ T, what shielding factor does that demand?
- For x ≪ 1 the Zeeman line is [E(+1) − E₊]/h = (gₑμB − gₚμN)B/2h − f₀x²/4. The second term is about 5 × 10⁻⁴ Hz, so B = 850 Hz ÷ 13.991 GHz T⁻¹ = 6.08 × 10⁻⁸ T, and x = 1.20 × 10⁻⁶.
- Maser-line offset: KB² = 2.773 × 10¹¹ Hz T⁻² × (6.075 × 10⁻⁸ T)² = 1.02 × 10⁻³ Hz, a fractional offset of 1.02 × 10⁻³ ÷ 1.4204 × 10⁹ = 7.2 × 10⁻¹³, which must be corrected for, not ignored.
- Differentiate: δf₀₀ = 2KBδB, so δf₀₀/f₀ = 2(KB²/f₀)(δB/B). Setting this to 1 × 10⁻¹⁵ gives δB/B = 1 × 10⁻¹⁵ ÷ (2 × 7.21 × 10⁻¹³) = 6.9 × 10⁻⁴.
- So δB ≤ 6.9 × 10⁻⁴ × 6.08 × 10⁻⁸ T = 4.2 × 10⁻¹¹ T. Units check: (Hz T⁻²)(T)(T) = Hz, and 2 × 2.773 × 10¹¹ × 6.08 × 10⁻⁸ × 4.2 × 10⁻¹¹ = 1.4 × 10⁻⁶ Hz, which is 1 × 10⁻¹⁵ of f₀.
- Shielding: 1 × 10⁻⁷ T ÷ 4.2 × 10⁻¹¹ T = 2.4 × 10³ at the very least. The same δB moves the 850 Hz Zeeman line by 13.991 GHz T⁻¹ × 4.2 × 10⁻¹¹ T = 0.59 Hz, which is how the field is monitored.
AnswerB = 6.1 × 10⁻⁸ T; the maser line sits 1.0 mHz (7.2 × 10⁻¹³ of f₀) high; B must hold to 4.2 × 10⁻¹¹ T, 0.07%, needing a shielding factor of at least 2.4 × 10³.