University Quantum Mechanics II · Hydrogen in External Fields and Hyperfine Structure · 9.4

The intermediate field: diagonalising the n = 2 matrix

Near 1 T, hydrogen's n = 2 level fits neither textbook limit, and the right move is to stop choosing a basis and diagonalise. Two quantum numbers, l and mⱼ, survive every field; they cut eight states into blocks no larger than 2 × 2, and one square root then gives every level from 0 to 5 T, with both limits inside it.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

In a field near 1 T the Zeeman energy μ_B B, 57.9 μeV per tesla, is comparable with the 45.3 μeV fine-structure splitting Δ of n = 2, so neither perturbation is small beside the other: the fine structure is diagonal in |l, j, mⱼ⟩, the Zeeman term in |l, mₗ, mₛ⟩, and no basis diagonalises both. What rescues the calculation is not a better approximation but a count of symmetries. Both terms commute with L² and with J_z, so the Hamiltonian inside n = 2 never connects states of different l or different mⱼ, and the eight states fall into four 1 × 1 blocks and two 2 × 2 blocks.

Writing the Zeeman term as (μ_B B/ħ)(J_z + S_z) shows that the only off-diagonal element is S_z linking 2P₃/₂ and 2P₁/₂ at equal mⱼ = ±1/2, of size (√2/3)μ_B B. Each block is then solved exactly by the quadratic formula, and expanding its square root in μ_B B/Δ or in Δ/μ_B B returns the Landé slopes and the Paschen–Back energies as its two limits. The geometry follows: the two levels of a block repel and never meet, while every other pair crosses freely, 2s and 2p of equal mⱼ included, because the field cannot change l.

The model stops where what it drops becomes comparable with what it keeps: the 1.06 GHz Lamb shift, which can be put back on the 2s diagonal exactly; hyperfine structure, below about 10 mT; gₛ − 2, at the 0.1% level; and the diamagnetic term, which rivals Δ only above 100 T.

Simple definition: In the intermediate field μ_B B is comparable with the fine-structure splitting Δ, so the n = 2 energies come from diagonalising the fine structure and the Zeeman term together, exactly, inside each block of fixed l and mⱼ.
Example: At B = 1.00 T, μ_B B/h = 14.00 GHz against Δ/h = 10.95 GHz. The 2p pair with mⱼ = −1/2 lies at +5.78 and −8.83 GHz from 2P₁/₂, where the weak-field formula says +1.62 and −4.67 GHz and the Paschen–Back one +3.65 and −6.70 GHz.

The Zeeman term in coupled formH_Z = (μ_B B/ħ)(L_z + 2S_z) = (μ_B B/ħ)(J_z + S_z)

J_z is diagonal in |l, j, mⱼ⟩ and commutes with L², so only S_z can put anything off the diagonal, and only between j = l ± 1/2 at the same l and mⱼ.

μ_B = 9.274 × 10⁻²⁴ J T⁻¹ (57.88 μeV T⁻¹); B in T; L_z, S_z, J_z in J s; gₛ = 2, quoted from Dirac theory

The one off-diagonal element⟨1, 3/2, mⱼ|H_Z|1, 1/2, mⱼ⟩ = −(√2/3)μ_B B, mⱼ = ±1/2

Derived here from S_z between two Clebsch–Gordan combinations. The same S_z puts ±ħ/6 on the diagonal, which rebuilds g_J = 4/3 and 2/3 as a check.

Energy in J; the sign follows the Condon–Shortley Clebsch–Gordan phases; magnitude h × 6.60 GHz at 1 T

Energies of the two mixed 2p levelsE± = (Δ + 2mⱼμ_B B)/2 ± ½√(Δ² + (4/3)mⱼΔμ_B B + (μ_B B)²)

Exact at every field. The root never vanishes while Δ ≠ 0, so the pair never touches; for mⱼ = −1/2 the closest approach is (2√2/3)Δ, at μ_B B = Δ/3.

E in J, measured from 2P₁/₂; Δ = E(2P₃/₂) − E(2P₁/₂) = 45.28 μeV, quoted from Unit 8; mⱼ = ±1/2

The first sign of the intermediate fieldδE = ±2(μ_B B)²/(9Δ)

Second-order perturbation theory through the (√2/3)μ_B B element across the gap Δ: 0.16 GHz at 0.2 T, already 17% of the linear shift of 2P₁/₂.

J; + for the level from 2P₃/₂, − for the level from 2P₁/₂, at mⱼ = ±1/2; valid while μ_B B ≪ Δ

How much 2P₃/₂ the upper level keepsP₃/₂ = ½[1 + (Δ + (2/3)mⱼμ_B B)/√(Δ² + (4/3)mⱼΔμ_B B + (μ_B B)²)]

Falls from 1 at B = 0 to 2/3 for mⱼ = +1/2 and 1/3 for mⱼ = −1/2 as B → ∞: the Clebsch–Gordan weights of the Paschen–Back states |0, ↑⟩ and |−1, ↑⟩.

Dimensionless probability, mⱼ = ±1/2; the lower level of the pair holds the rest, 1 − P₃/₂

The unmixed 2s levelsE(2S₁/₂, mⱼ = ±1/2) = L ± μ_B B

A 1 × 1 block at every field, because H_Z cannot change l: these straight lines pass through the 2p levels, even those of equal mⱼ, without repulsion.

J, from 2P₁/₂; L = h × 1057.8 MHz, the measured Lamb shift, quoted; it moves only these two lines

Two perturbations, neither small, and what still commutes

At B = 1.00 T, μ_B B/h = 14.00 GHz while the 2P₃/₂–2P₁/₂ splitting is Δ/h = 10.95 GHz: a ratio of 1.28, and neither the weak-field nor the Paschen–Back basis diagonalises H_fs + H_Z. The fine structure is a rotational scalar built from r, p and L⋅S, so it commutes with J² and J_z but not with L_z or S_z; the Zeeman term commutes with L_z and S_z but not with J². That is exactly why j and (mₗ, mₛ) both fail. Two operators survive. L² commutes with every piece of both terms, and J_z commutes with both because each is symmetric about the field axis. So l and mⱼ are good at every field, and the matrix can only connect states sharing both. Sort the eight n = 2 states by (l, mⱼ): l = 0 gives mⱼ = ±1/2, one state each; l = 1 gives mⱼ = ±3/2, one state each, since only j = 3/2 reaches them; and mⱼ = ±1/2, two states each, one from each j. Four 1 × 1 blocks and two 2 × 2 blocks: nothing larger than 2 × 2 is ever diagonalised.

Filling in the matrix from Clebsch–Gordan coefficients

Put the fine structure on the diagonal as Unit 8 left it: 2S₁/₂ and 2P₁/₂ at zero, 2P₃/₂ at Δ = α⁴mₑc²/32 = 45.28 μeV, with the 1057.8 MHz Lamb shift set aside for now. For the Zeeman term write L_z + 2S_z = J_z + S_z. J_z returns mⱼħ, so everything off the diagonal comes from S_z. Unit 5's table gives |1, 3/2, +1/2⟩ = √(2/3)|0, ↑⟩ + √(1/3)|1, ↓⟩ and |1, 1/2, +1/2⟩ = −√(1/3)|0, ↑⟩ + √(2/3)|1, ↓⟩. So ⟨S_z⟩ is (2/3 − 1/3)(ħ/2) = +ħ/6 in the first state and −ħ/6 in the second, and the cross element is √(2/3)(−√(1/3))(ħ/2) + √(1/3)√(2/3)(−ħ/2) = −(√2/3)ħ. The diagonal Zeeman entries are (1/2 + 1/6)μ_B B = (2/3)μ_B B and (1/2 − 1/6)μ_B B = (1/3)μ_B B: g_J mⱼ μ_B B with g_J = 4/3 and 2/3, the Landé values back as a check. The mⱼ = −1/2 block has these diagonal entries negated and an off-diagonal element of the same size. At 1 T that element is 6.60 GHz, more than half of Δ, so no perturbative treatment of it is safe.

Diagonalising a 2 × 2 block in closed form

Write b = μ_B B. The mⱼ = −1/2 block is [[Δ − 2b/3, −(√2/3)b], [−(√2/3)b, −b/3]]. Its trace is Δ − b and its determinant −Δb/3, so the eigenvalues are half the trace plus or minus ½√(trace² − 4 det) = ½√(Δ² − (2/3)Δb + b²). For mⱼ = +1/2 the diagonal Zeeman entries change sign but the off-diagonal element does not, so every term linear in b in the trace, the determinant and the root changes sign. Read as a quadratic in b, the expression under the root has discriminant (2Δ/3)² − 4Δ² = −32Δ²/9, negative, so the root never vanishes and the two levels never meet. Its minimum, 8Δ²/9 at b = Δ/3, or 0.26 T, sets the closest approach at (2√2/3)Δ = 10.32 GHz. The weak-field lines Δ − 2b/3 and −b/3 would have crossed at b = 3Δ, 2.35 T; there the exact levels are 2√2Δ = 30.97 GHz apart, twice the off-diagonal element. That is level repulsion in its simplest form: two states joined by a nonzero matrix element cannot be degenerate.

Both limits from one square root

For b ≪ Δ expand the root: ½√(Δ² ± (2/3)Δb + b²) ≈ Δ/2 ± b/6 + 2b²/(9Δ). The pair becomes Δ ± (2/3)b + 2b²/(9Δ) and ±(1/3)b − 2b²/(9Δ): topic 9.2's Landé slopes plus an equal and opposite quadratic push, which is second-order perturbation theory through the (√2/3)b element. For b ≫ Δ expand in Δ/b instead: the root tends to b ± Δ/3, the mⱼ = +1/2 levels to b + 2Δ/3 and Δ/3, and the mⱼ = −1/2 levels to Δ/3 and −b + 2Δ/3. These are b(mₗ + 2mₛ) plus the Paschen–Back fine structure (2Δ/3)(1 + mₗmₛ) of topic 9.3, for (mₗ, mₛ) = (0, ↑), (1, ↓), (−1, ↑) and (0, ↓). Set Δ = 0 and the root is exactly b: the four levels become b, 0, 0 and −b, which is b(mₗ + 2mₛ) at every field. The states interpolate as the energies do: the upper mⱼ = −1/2 level starts as pure 2P₃/₂ and ends as |−1, ↑⟩, only one-third 2P₃/₂.

Which levels cross, and which repel

Only states in the same block are coupled, so only they repel. The 2P₃/₂ levels with mⱼ = ±3/2 sit alone and run as straight lines Δ ± 2b; the 2S₁/₂ levels run as ±b. Straight lines cross the curves and each other wherever the algebra puts them: 2S₁/₂(+1/2) meets 2P₃/₂(−3/2) at b = Δ/3, 0.26 T, and 2S₁/₂(−1/2) meets it at b = Δ, 0.78 T. The telling case is equal mⱼ with different l. Restore the Lamb shift, so the 2s lines become L ± b with L/h = 1.058 GHz, and 2S₁/₂(−1/2) falls through the 2P₁/₂(−1/2) level near 0.12 T: same mⱼ, yet no repulsion, because H_Z commutes with L² and with parity and has no element between l = 0 and l = 1. An electric field does have one, since ez is odd. A field along B joins that equal-mⱼ pair, and the motional field v × B, perpendicular to B, joins 2S₁/₂(−1/2) to 2P₁/₂(+1/2), which cross at 0.057 T. Either opens a gap and quenches metastable 2s atoms, the business of topic 9.6.

Checks, and what the matrix leaves out

Dimensions: every entry is μ_B B or Δ, both energies, and the root of a sum of squared energies is an energy. Sum rule: H_Z has zero trace, so the eight levels always add to 4Δ, plus 2L with the Lamb shift, a one-line test of any numerical diagonalisation. Order in α: Δ = α⁴mₑc²/32 while μ_B B carries no α, so the crossover μ_B B = Δ sits at α²Bₐₜ/16 = 0.78 T, where Bₐₜ = ħ/(ea₀²) = 2.35 × 10⁵ T is the atomic unit of field. Measurement: Δ/h is 10.969 GHz, against 10.949 GHz from Eₙⱼ and 10.943 GHz with the reduced mass; the last 0.23% is the electron's anomalous moment, which scales the spin-orbit splitting by 1 + 2aₑ with aₑ ≈ α/2π, a QED result quoted here. The same aₑ moves each Zeeman slope by at most 0.12%, and not always upward: 2S₁/₂ and 2P₃/₂ steepen, but g_J(2P₁/₂) = 2/3 − 2aₑ/3 flattens, because that level's spin projection on J is negative. Left out: hyperfine structure, measured at 177.6 MHz in 2S₁/₂ and 59.2 MHz in 2P₁/₂, which reshapes the diagram below about 10 mT; and the diamagnetic term e²B²⟨x² + y²⟩/(8mₑ), 0.42 MHz for 2s at 1 T, which rivals Δ only above 100 T.

Change one variable at a time

Make the relationship visible.

Interactive model

Field range, 0 to B2.50 T

Fine-structure splitting Δ/h10.95 GHz

Lamb shift L/h of 2S₁/₂0.00 GHz

At 2.5 T the faint Landé lines meet near the right edge while the two curves they started on never touch. At 0.2 T the curves sit on the Landé lines; at Δ = 0 they turn into straight Paschen–Back lines. With L = 1.06 GHz and a 0.15 T range, the falling dashed 2s line cuts through both 2P₁/₂ curves.

FIELD WHERE μ_B B = Δ0.78 T

UPPER mⱼ = −1/2, RIGHT EDGE4.48 GHz

LOWER mⱼ = −1/2, RIGHT EDGE-28.52 GHz

2P₃/₂ SHARE OF THE UPPER ONE48.9 %

Live interpretationFIELD WHERE μ_B B = Δ: 0.78 T. UPPER mⱼ = −1/2, RIGHT EDGE: 4.48 GHz. LOWER mⱼ = −1/2, RIGHT EDGE: −28.52 GHz. 2P₃/₂ SHARE OF THE UPPER ONE: 48.9 %

Catch the common trap

Explain before calculating.

Misconception check

The tempting reading is that in the intermediate field nothing is conserved: j fails, mₗ and mₛ fail, so the full 8 × 8 matrix must go to a computer, or the energies can be estimated by blending the weak-field and Paschen–Back formulas. Both halves are wrong. l and mⱼ are conserved exactly at every field, because L² and J_z commute with the fine structure and with (μ_B B/ħ)(J_z + S_z) alike, so the problem never exceeds a 2 × 2 block, and those blocks are solved exactly. And no blend of the limits can work, because the exact pair lies outside both: each block's trace is the same in all three treatments, but both limits drop part of the repulsion. At 1.00 T the upper mⱼ = −1/2 level sits at 5.78 GHz, above the weak-field 1.62 GHz and the Paschen–Back 3.65 GHz alike.

With the Lamb shift restored, 2S₁/₂ starts 1.06 GHz above 2P₁/₂ at zero field. As B rises from 0 to 5 T, which statement about the eight n = 2 levels of hydrogen is right?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyWrite the mⱼ = +1/2 block of the n = 2 Zeeman matrix numerically at B = 0.400 T, as E/h in GHz from 2P₁/₂, with Δ/h = 10.95 GHz and μ_B/h = 13.996 GHz T⁻¹. Then estimate how much 2P₁/₂ the upper state has taken on.

Zeeman scale: b = μ_B B/h = 13.996 GHz T⁻¹ × 0.400 T = 5.598 GHz, about half of Δ, so this is already an intermediate field.
Diagonal entries, the weak-field g_J mⱼ values: 2P₃/₂ at Δ + (2/3)b = 10.95 + 3.732 = 14.682 GHz, and 2P₁/₂ at (1/3)b = 1.866 GHz.
Off-diagonal entry: −(√2/3)b = −0.4714 × 5.598 GHz = −2.639 GHz.
First-order estimate: the admixed amplitude is 2.639/(14.682 − 1.866) = 2.639/12.816 = 0.206, so about 0.206² = 4.2% of 2P₁/₂.
Exact check: √(Δ² + (2/3)Δb + b²) = √(119.90 + 40.87 + 31.34) = 13.861 GHz, so P₃/₂ = ½(1 + 12.816/13.861) = 0.962, leaving 3.8% of 2P₁/₂.

Answer[[14.68, −2.64], [−2.64, 1.87]] GHz. The upper state is 96.2% 2P₃/₂ and 3.8% 2P₁/₂; first-order theory overestimates the admixture at 4.2%.

MediumAt B = 1.00 T, find the two 2p levels with mⱼ = −1/2 exactly, as E/h in GHz from 2P₁/₂, with Δ/h = 10.95 GHz. Compare them with the weak-field and Paschen–Back predictions, check the result against the trace, and find the 2P₃/₂ share of the upper level.

b = μ_B B/h = 13.996 GHz, so b/Δ = 1.28: neither limit applies.
Under the root: Δ² − (2/3)Δb + b² = 119.90 − 102.17 + 195.89 = 213.62 GHz², whose root is 14.616 GHz.
E± = (Δ − b)/2 ± 14.616/2 = −1.523 ± 7.308 GHz, giving +5.785 and −8.831 GHz. Their sum, −3.046 GHz, equals the trace Δ − b.
Weak field: Δ − 2b/3 = 1.619 and −b/3 = −4.665 GHz, each 4.166 GHz too close to the middle. Paschen–Back: Δ/3 = 3.650 and 2Δ/3 − b = −6.696 GHz, each 2.135 GHz too close. The errors are equal and opposite because both limits keep the trace.
P₃/₂ = ½[1 + (Δ − b/3)/14.616] = ½(1 + 6.285/14.616) = 0.715.

Answer+5.78 GHz and −8.83 GHz from 2P₁/₂. The weak-field values miss by 4.17 GHz and the Paschen–Back values by 2.13 GHz, both too narrow; the upper level is 71.5% 2P₃/₂.

HardRestore the measured Lamb shift, L/h = 1.0578 GHz, and use the measured Δ/h = 10.969 GHz. Find, within this model, the fields at which 2S₁/₂(mⱼ = −1/2) crosses 2P₁/₂(mⱼ = −1/2) and 2P₁/₂(mⱼ = +1/2), compare each with the linear weak-field estimate, and say why neither is an anticrossing.

2S₁/₂(−1/2) is a 1 × 1 block, E = L − b with b = μ_B B. 2P₁/₂(−1/2) is the lower root of its block, (Δ − b)/2 − ½√(Δ² − (2/3)Δb + b²).
Equate and isolate the root: √(Δ² − (2/3)Δb + b²) = Δ + b − 2L. Squaring cancels the b² terms and leaves b(8Δ/3 − 4L) = 4L(Δ − L), so b = L(Δ − L)/(2Δ/3 − L).
Numbers: b = 1.0578 × 9.9112/6.2549 = 1.6761 GHz, so B = 1.6761/13.996 = 0.1198 T. The linear estimate L − b = −b/3 gives b = 3L/2, 0.1134 T; the 2b²/(9Δ) push on 2P₁/₂ moves the crossing 6.4 mT higher.
For 2P₁/₂(+1/2), the lower root of the other block, the same steps give √(Δ² + (2/3)Δb + b²) = Δ + 3b − 2L, which squares to 2b² + (4Δ/3 − 3L)b + L² − ΔL = 0, with 4Δ/3 − 3L = 11.452 GHz and L² − ΔL = −10.484 GHz².
So b = [−11.452 + √(131.15 + 83.87)]/4 = 0.8029 GHz and B = 0.0574 T, against 3L/4 → 0.0567 T.
H_Z conserves l and parity, so it has no element between 2s and 2p, even at equal mⱼ: both crossings are exact in a pure magnetic field. Hyperfine structure, 177.6 MHz in 2S₁/₂ and 59.2 MHz in 2P₁/₂, splits each into several crossings, spread over roughly 14 mT for the first and 6.5 mT for the second, and any electric field, which does join 2s to 2p, opens a gap.

AnswerEqual mⱼ: B = 0.120 T (linear estimate 0.113 T). Different mⱼ: B = 0.0574 T (0.0567 T). Both are true crossings, since the field cannot change l.