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Advanced Quantum Mechanics course sequence · Interactive lesson

Kramers degeneracy derived

This is the first theorem that Θ² = −1 pays for. Two lines of antiunitary algebra forbid an odd-electron system a single level while H is even under T, and the work of the topic is learning exactly what that forbids: which fields shift a doublet, which split it, and which particles you must count.

UAQM-01.07 · Lesson 7 of 9 in this unit. Position is not a completion score.
Starting model: Kramers degeneracy derivedLevels of one T-even spin Hamiltonian, D(S_z² − S(S+1)/3) + E(Sₓ² − S_y²) with D/h = -5.75 GHz and E/h = 1.50 GHz, against B along z (g = 2.00). Left, S = 1: the pair from m = ±1 is 3.00 GHz apart at B = 0. Right, S = 3/2: two doublets 12.62 GHz apart, each pair of curves meeting at B = 0. Dots mark the S = 3/2 ground pair at the cursor; both panels share one energy scale.S = 1, Θ² = +1 (Ni²⁺)S = 3/2, Θ² = −1 (Cr³⁺)ε/hε/hB along z from −0.2 T to +0.2 T; cursor 0.05 T
Start with a picture. Change the model in the lab below.
1

Build the model

When Θ² = −1, every level comes in pairs that only a T-odd field can part.

The key idea

Kramers' theorem: if the Hamiltonian commutes with time reversal and Θ² = −1, as for any odd number of spin-half particles, every energy level is at least twofold degenerate, |n⟩ and Θ|n⟩ being orthogonal states of one energy.

Example. Cr³⁺ in ruby has 21 electrons and spin 3/2. Its four spin states form two Kramers doublets, m = ±3/2 and m = ±1/2, measured about 11.5 GHz (0.38 cm⁻¹) apart at zero field; the crystal field shifts them, but no T-even field can split either pair.

The complete physical model and assumptions

Time reversal is antiunitary, and its square is a sign no choice of phase can change: Θ² = (−1)(2j) (topic 1.06), which for N spin-half particles is (−1)N. Kramers' theorem turns that sign into a statement about every energy level. If Θ commutes with H, then Θ|n⟩ has the same energy as |n⟩, because the energy is real and survives the complex conjugation; and if Θ² = −1 it is a different state, because the antiunitary rule ⟨Θα|Θβ⟩ = ⟨β|α⟩ forces ⟨n|Θn⟩ = −⟨n|Θn⟩ = 0. Splitting the eigenspace into such orthogonal pairs shows that its dimension is even. Nothing about the shape of H entered: crystal fields of no symmetry, applied electric fields, spin-orbit coupling and electron repulsion are all even under T, so every level of an odd-electron ion stays evenly degenerate however it is squeezed, at most (2J + 1)/2 levels from one J multiplet, and inside each pair a T-even perturbation acts as a multiple of the identity. The theorem stops where its two assumptions stop. A magnetic field or an exchange field is odd under T, breaks [Θ, H] = 0 and generally splits each doublet linearly in B. And the count must include every fermion: hydrogen's proton makes N = 2 and Θ² = +1, which is why its F = 0 ground level may stand alone. What the theorem never supplies is the size of a splitting or the g-factor of a doublet; those need the states themselves.

1Two lines of antiunitary algebra

Units are SI, with small splittings also quoted as frequencies ΔE/h; the lesson is non-relativistic, and the metric (+, −, −, −) is not needed until Unit 3. Two results are quoted, not re-derived: an antiunitary Θ obeys ⟨Θα|Θβ⟩ = ⟨β|α⟩ (topic 1.02), and Θ² = (−1)(2j) whatever phase is chosen (topic 1.06). Now let [Θ, H] = 0 and H|n⟩ = Eₙ|n⟩. Then HΘ|n⟩ = ΘH|n⟩ = Θ(Eₙ|n⟩) = EₙΘ|n⟩: antilinearity conjugates the number, but Eₙ is real. So Θ|n⟩ is an eigenstate of the same energy and the same norm. Is it a new state? Put α = |n⟩ and β = Θ|n⟩ into the rule: ⟨Θn|Θ²n⟩ = ⟨Θn|n⟩. With Θ² = −1 the left side is −⟨Θn|n⟩, so ⟨Θn|n⟩ = 0. A shorter route lands in the same place: if Θ|n⟩ = c|n⟩, then Θ²|n⟩ = c* Θ|n⟩ = |c|²|n⟩, which can never equal −|n⟩. Derived here: the level is at least twofold.

2From one partner to an even count

One orthogonal partner makes a level at least twofold; the theorem claims more, that every eigenspace W of H has even dimension. Θ maps W into itself. Pick |a⟩ in W: the pair |a⟩, Θ|a⟩ spans a plane that Θ also maps into itself, since Θ(Θ|a⟩) = −|a⟩. The rest of W, the states |c⟩ orthogonal to both, is Θ-invariant too: the rule gives ⟨Θc|Θa⟩ = ⟨a|c⟩ = 0 and ⟨Θc|a⟩ = −⟨Θc|Θ²a⟩ = −⟨Θa|c⟩ = 0. Repeat inside that remainder until W is used up: it is a sum of orthogonal pairs, so its dimension is even. For a free-ion multiplet of half-integer J, a crystal field can therefore make at most (2J + 1)/2 levels: 5 from Nd³⁺ ⁴I₉/₂ and 8 from Er³⁺ ⁴I₁₅/₂. Optical spectra agree: in Nd:YAG the ⁴I₉/₂ ground multiplet shows five crystal-field levels and the ⁴F₃/₂ upper laser level shows two, roughly 10 meV apart, never more.

3A test by hand: spin 1 against spin 3/2

The proof never used the form of H, so try the most general T-even H that a small multiplet allows. Since ΘSΘ⁻¹ = −S, a Hermitian product of k spin components carries the T-sign (−1)k, and for S = 1 or 3/2 the T-even operators are a constant plus a rank-2 term, which in its principal axes reads D(Sz² − S(S+1)/3) + E(Sₓ² − Sy²). For S = 1 the real states |x⟩, |y⟩, |z⟩ are eigenstates at D/3 − E, D/3 + E and −2D/3: three singlets once E ≠ 0, which an even-electron ion such as Ni²⁺ is free to show. For S = 3/2, Sₓ² − Sy² = (S₊² + S₋²)/2 links m to m ± 2, so the states fall into two blocks, m = (3/2, −1/2) and m = (−3/2, 1/2), each with diagonal ±D and off-diagonal √3E. Both blocks give ε = ±√(D² + 3E²): two doublets for every D and E, Kramers' theorem checked case by case. Cr³⁺ in ruby, with 21 electrons, shows exactly two doublets, about 11.5 GHz apart.

4Shifted, never split: T-even perturbations

Inside one Kramers pair a T-even perturbation can do only one thing. Let V be Hermitian with ΘVΘ⁻¹ = V. Apply the rule to α = VΘ|n⟩ and β = |n⟩: since ΘVΘ|n⟩ = VΘ²|n⟩ = −V|n⟩, it gives −⟨n|V|Θn⟩ = ⟨n|V|Θn⟩, so the off-diagonal element vanishes. Apply it to α = V|n⟩ and β = |n⟩ and it gives ⟨Θn|V|Θn⟩ = ⟨n|V|n⟩. On the pair, V is ⟨n|V|n⟩ times the identity: it moves the doublet as a block. That is the first-order statement, and nothing changes at higher order, because H + V still commutes with Θ. The T-even list is long: a crystal field V(r); the Stark term −qE⋅r, with r even and the applied E a fixed external number; spin-orbit coupling ξ L⋅S, a product of two odd factors; electron repulsion; strain. So an odd-electron ion in an electric field of any size and direction shows doublets that shift, and never a pair that splits.

5A magnetic field parts the pair, linearly

Run the same algebra with a T-odd operator such as the magnetic moment, ΘμΘ⁻¹ = −μ. Now the diagonal elements come out opposite, ⟨Θn|μz|Θn⟩ = −⟨n|μz|n⟩, and the off-diagonal ones are unconstrained, so −μ⋅B on the pair is a traceless 2 × 2 Hermitian matrix with eigenvalues ±½ geff μB B. The splitting ΔE = geff μB B is linear in B, with geff set by the field direction through the doublet's g-tensor. With μB/h = 13.996 GHz T⁻¹, a doublet with geff = 2 opens 28.0 GHz per tesla, which is why odd-electron ions are the staple of EPR near 9.5 GHz and 0.34 T. Compare a non-degenerate level, possible only when Θ² = +1: there Θ|n⟩ = c|n⟩, the same step gives ⟨n|μ|n⟩ = −⟨n|μ|n⟩ = 0, and the Zeeman shift starts at B². At B = 10 mT an S = 1 pair held 4.0 GHz apart by E/h = 2.0 GHz widens by 0.04 GHz, while an m = ±3/2 Kramers doublet with g = 2 opens by 0.84 GHz.

6Count every fermion

The many-particle Θ is a product of one-particle factors, exp(−iπSy/ħ) for each spin, times one K for the whole system; the factors act on different particles and commute, so Θ² = (−1)N, derived from topic 1.06's one-particle result. “An odd number of electrons” is shorthand that drops the nuclei, and hydrogen tests it. With the proton's spin ignored, 1s is a Kramers pair, mₛ = ±½. Keep the proton and N = 2, Θ² = +1, so nothing stops the hyperfine coupling from leaving a single F = 0 level below the F = 1 triplet, measured 1420.405 751 768 MHz, or 5.87 μeV, apart. Deuterium adds a neutron, N = 3, and its ground levels are F = 1/2 and F = 3/2, two and four states, both even, 327.384 MHz apart. Unit 7's Dirac hydrogen, quoted here, obeys the theorem: its energies depend on n and j only, so every level holds 2j + 1 or 2(2j + 1) states, always even, and the measured 1057.8 MHz Lamb shift parts 2s₁/₂ from 2p₁/₂ while leaving each a pair.

2

Key equations

Read the units, assumptions and limits before substituting.

Kramers' orthogonality

⟨n|Θn⟩ = −⟨n|Θn⟩ = 0 whenever Θ² = −1

Derived from ⟨Θα|Θβ⟩ = ⟨β|α⟩ with α = n and β = Θn. The partner of every state is a new, orthogonal state, so no level can be single.

Θ antiunitary time reversal, |Θn⟩ ≡ Θ|n⟩, |n⟩ any normalised state; the overlap is a pure number

Time-reversal square for N spin-half fermions

Θ² = (−1)N

Derived: Θ is a product of one-particle factors that each square to −1 (topic 1.06). Odd N is the Kramers case; hydrogen, electron plus proton, has N = 2.

N counts electrons, plus protons and neutrons when nuclear spin is kept; a phase-independent sign, dimensionless

Crystal-field levels of a half-integer J multiplet

number of levels ≤ (2J + 1)/2, each of even degeneracy

A site with no point symmetry removes every degeneracy except Kramers', so the bound is generically reached. For integer J the bound is 2J + 1 singlets.

J the multiplet's total angular momentum in units of ħ, half-integer for odd N; Nd³⁺ ⁴I₉/₂ gives 5, Er³⁺ ⁴I₁₅/₂ gives 8

T-even perturbation inside a Kramers pair

P V P = ⟨n|V|n⟩ P, with P = |n⟩⟨n| + |Θn⟩⟨Θn| and ΘVΘ⁻¹ = V

Derived from ⟨n|V|Θn⟩ = 0 and ⟨Θn|V|Θn⟩ = ⟨n|V|n⟩. The pair moves as a block: a Stark shift, never a Stark splitting.

V Hermitian, in joules: a crystal field V(r), −qE⋅r with E in V m⁻¹, spin-orbit ξ L⋅S, strain; P projects on the pair

Zeeman splitting of a Kramers doublet

ΔE = geff μB B

μ is odd under T, so inside the pair it is traceless and the levels move as ±½ geff μB B. ΔE/h is 13.996 geff GHz per tesla.

μB = 9.274 × 10⁻²⁴ J T⁻¹ = 5.788 × 10⁻⁵ eV T⁻¹, B in T, geff dimensionless and set by the field direction

Most general T-even Hamiltonian for spin 3/2

H = D(Sz² − 5/4) + E(Sₓ² − Sy²) has levels ε = ±√(D² + 3E²), each twofold

Derived by hand: blocks m = (3/2, −1/2) and m = (−3/2, 1/2), each with diagonal ±D and off-diagonal √3E. The same form for S = 1 gives three singlets once E ≠ 0.

D, E in joules, quoted as D/h and E/h in GHz; S dimensionless; any rank-2 T-even term takes this form in its principal axes
3

Interactive lab

Predict first. Explore the model. Compare the result.

Explore the model

−5.75 GHz
−6 to 6 GHz; steps of 0.25.
1.50 GHz
0 to 3 GHz; steps of 0.25.
0.05 T
−0.2 to 0.2 T; steps of 0.01.

Try this. Raise E from 0: the left pair parts into two singlets whose curves go flat at B = 0, while on the right every curve still meets a partner at B = 0 for any D and E. Park the cursor at 0.01 T: at the default D and E the S = 3/2 ground gap opens 0.79 GHz, the S = 1 pair moves 0.05 GHz.

Live results

S = 1 PAIR GAP AT B = 0
3.00 GHz
S = 1 PAIR GAP AT CURSOR
4.10 GHz
S = 3/2 GROUND GAP AT CURSOR
3.94 GHz
S = 3/2 ZERO-FIELD SPACING
12.62 GHz

Results follow this lesson’s model and assumptions. Displayed values are rounded.

Interactive Advanced Quantum Mechanics modelLevels of one T-even spin Hamiltonian, D(S_z² − S(S+1)/3) + E(Sₓ² − S_y²) with D/h = -5.75 GHz and E/h = 1.50 GHz, against B along z (g = 2.00). Left, S = 1: the pair from m = ±1 is 3.00 GHz apart at B = 0. Right, S = 3/2: two doublets 12.62 GHz apart, each pair of curves meeting at B = 0. Dots mark the S = 3/2 ground pair at the cursor; both panels share one energy scale.S = 1, Θ² = +1 (Ni²⁺)S = 3/2, Θ² = −1 (Cr³⁺)ε/hε/hB along z from −0.2 T to +0.2 T; cursor 0.05 T
Levels of one T-even spin Hamiltonian, D(Sz² − S(S+1)/3) + E(Sₓ² − Sy²) with D/h = −5.75 GHz and E/h = 1.50 GHz, against B along z (g = 2.00). Left, S = 1: the pair from m = ±1 is 3.00 GHz apart at B = 0. Right, S = 3/2: two doublets 12.62 GHz apart, each pair of curves meeting at B = 0. Dots mark the S = 3/2 ground pair at the cursor; both panels share one energy scale.

Predict → change one variable → check

Plan a change before running it. Every other input stays at the saved baseline.

Saved baseline. Axial term D/h: −5.75 GHz; Rhombic term E/h: 1.50 GHz; Field cursor B along z: 0.05 T

Baseline −5.75; choose −6 to 6, in steps of 0.25.
4. Compared with the saved baseline, this result will…

This experiment is kept only while this page is open. It does not change saved course progress.

4

Catch the common trap

The tempting error is to file Kramers degeneracy with the degeneracies of spatial symmetry, like the ±m pairs of an axial crystal site, and so to expect a site of no symmetry, or a strong enough electric field, to split an odd-electron doublet. Neither can. The proof used nothing about H except [Θ, H] = 0 and Θ² = −1: no rotation, no reflection, no small parameter. A crystal field, an applied electric field, strain and spin-orbit coupling are all even under T, so they keep every pair exactly and to all orders; only a T-odd agent, an applied magnetic field or the exchange field of magnetically ordered neighbours, can part one. The mirror-image slip is to count electrons where nuclear spin matters and then call hydrogen's single F = 0 level a counterexample: with the proton counted, N is even and no pairing is promised.

5

Practice & worked examples

3 authored examples. Make an attempt, then reveal the reasoning one step at a time.

EasyExample 1

Yb³⁺ (Yb has Z = 70; ground multiplet ²F₇/₂) and Tm³⁺ (Z = 69; ground multiplet ³H₆) each sit at a crystal site with no point symmetry. Ignoring nuclear spin, find the largest number of distinct levels each ground multiplet can form, and say which ion can have a single level. Then find the splitting of a Yb³⁺ doublet with geff = 3.40 in B = 0.250 T, in μeV and as a frequency.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 5 steps revealed.

MediumExample 2

V³⁺ (20 electrons, S = 1) and Cr³⁺ (21 electrons, S = 3/2) are each modelled by the T-even spin Hamiltonian D(Sz² − S(S+1)/3) + E(Sₓ² − Sy²) with the same illustrative values D/h = −4.00 GHz and E/h = 1.00 GHz. Find each ion's zero-field levels, then the splitting of each ion's lowest pair in B = 10.0 mT along z with g = 2.00, and say which response is linear in B.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 5 steps revealed.

HardExample 3

Model the ground doublet of Er³⁺ (65 electrons; ⁴I₁₅/₂, gJ = 6/5) at an axial site as |±⟩ = |J = 15/2, m = ±1/2⟩. Show it is a Kramers pair, show that an off-axis strain term c(JₓJz + JzJₓ) cannot split it although it links m to m ± 1, and find g∥ and g⊥. Predict the EPR frequency for B = 50.0 mT perpendicular to the axis, and compare with the measured g-values of Er³⁺ in CaWO₄, g∥ = 1.247 and g⊥ = 8.38.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 5 steps revealed.

6

Quick check

Select an answer and explain it before checking.

An Nd³⁺ ion (57 electrons; ground multiplet ⁴I₉/₂, J = 9/2) sits at a crystal site with no point symmetry, and a static electric field of 1 × 10⁸ V m⁻¹ is applied in an arbitrary direction. Ignoring nuclear spin, what is the largest number of distinct energy levels the ten ⁴I₉/₂ states can form?
7

Review & continue

Kramers' theorem: if the Hamiltonian commutes with time reversal and Θ² = −1, as for any odd number of spin-half particles, every energy level is at least twofold degenerate, |n⟩ and Θ|n⟩ being orthogonal states of one energy.

  • Explain the relationship without looking at the equation.
  • Check a sign, a unit and a model assumption in your working.
  • Return to the lab or worked example if a step is still unclear.

Extra challenging questions

Connect several ideas, explore supplied data, and compare your reasoning with a worked solution.