University Quantum Mechanics II · Time-Independent Perturbation Theory · 6.5
Degenerate levels and the good states
The levels most worth perturbing are degenerate: hydrogen's n = 2 shell, the mₛ = ±1 pair of a spin-1 defect, a rotor's ±m. There the formulas of the last four lessons divide by zero. The repair is a small diagonalisation done first, after which those formulas work again, and Units 8 and 9 run on it.
Build the model
Connect the measurement to the mechanism.
A degenerate level poses a question H₀ cannot answer. Its g states share one energy, so every combination of them is an equally good eigenstate of H₀, and nothing in H₀ says which combination the true state grows out of as the perturbation is switched on. The non-degenerate recipe assumes an answer and then fails on it: its first-order state divides ⟨i|H′|j⟩ by a zero gap. Keep the start state unknown, |a⁽⁰⁾⟩ = Σ cᵢ|i⟩, and project the first-order equation onto the level.
H₀ − E⁽⁰⁾ vanishes there, and what is left says the column c must be an eigenvector of the g × g matrix Wᵢⱼ = ⟨i|H′|j⟩. Its eigenvalues are the first-order shifts and its eigenvectors the good states, fixed by the form of H′ and not by its size, so an arbitrarily weak perturbation still turns the zeroth-order states through a finite angle. In the good basis every zero gap inside the level meets a zero numerator, and the non-degenerate formulas resume with sums over the states outside it. The method stops in two places.
If W has a repeated eigenvalue, first order leaves the good states undetermined and a higher order must choose them. And if two levels are merely close, with a gap comparable to their coupling, neither recipe is right: their 2 × 2 block is diagonalised exactly, the avoided crossing of Quantum Mechanics I, and everything else is added perturbatively.
- Simple definition
- The good states of a degenerate level are the combinations of its states that diagonalise the perturbation inside the level; the first-order shifts are the eigenvalues of that small matrix, Wᵢⱼ = ⟨i|H′|j⟩.
- Example
- On a ring, e(iφ) and e(−iφ) share one level. Add V₀ cos 2φ with V₀ = 0.40 meV and the good states are cos φ, raised by 0.20 meV, and sin φ, lowered by 0.20 meV. Neither running wave is a good state, however small V₀ is made.
Derived by projecting the order-λ equation onto the level, where H₀ − E⁽⁰⁾ vanishes. The shifts are W's eigenvalues and the good states its eigenvectors.
|1⟩ … |g⟩ span the g-fold level of H₀ at E⁽⁰⁾; Wᵢⱼ and E⁽¹⁾ in J; cᵢ dimensionless, with |a⁽⁰⁾⟩ = Σ cᵢ|i⟩
Derived: a trace does not depend on the basis. The mean shift comes out right in any basis; the individual shifts need the good one.
Sum over the g good states on the left, over any orthonormal basis of the level on the right; every term in J
Derived by projecting onto ⟨m|. Components inside the level are not fixed here: they wait for second order, and are fixed then only if W's eigenvalues differ.
|a⁽⁰⁾⟩ a good state; numerator and denominator in J, so each component is dimensionless
Derived. The shifts ignore φ₀ while the states rotate with it: H₀ accepts every basis of the pair, and H′ chooses one.
V₀ in J; φ₀ locates the hills of H′; W₊₋ = (V₀/2)e(−2iφ₀) between e(±iφ)/√(2π)
Quoted from the 2 × 2 problem of Quantum Mechanics I. δ ≫ |W₁₂| gives Ē ± (δ/2 + |W₁₂|²/δ), the non-degenerate result; δ → 0 gives Ē ± |W₁₂|.
Ē and δ ≥ 0 are the mean and difference of the diagonal entries E₁⁽⁰⁾ + W₁₁ and E₂⁽⁰⁾ + W₂₂; all in J
Derived from the 2 × 2 eigenvector. At δ = 0, θ = 45° however small W₁₂ is: the good-state statement in one line.
θ dimensionless, from 0 when |W₁₂| ≪ δ to 45° at δ = 0; the lower state carries weight sin²θ in level 1, the upper
A zero gap, and a start state nobody chose
The non-degenerate first-order state has components ⟨m|H′|n⟩/(Eₙ⁽⁰⁾ − Eₘ⁽⁰⁾). Now let H₀ have g orthonormal states |1⟩ … |g⟩ at one energy E⁽⁰⁾. Between two of them the gap is zero, and unless ⟨i|H′|j⟩ happens to vanish the formula returns an infinite admixture. The fault is not a large correction but a wrong start. Take a pair with H = E⁽⁰⁾ + λVσₓ in the basis |1⟩, |2⟩, where E⁽⁰⁾ multiplies the identity and V = 0.10 meV. For every λ ≠ 0, however small, the eigenstates are (|1⟩ ± |2⟩)/√2 at E⁽⁰⁾ ± λV. As λ → 0 the energies return to E⁽⁰⁾, but the states do not return to |1⟩ and |2⟩: they stay at 45°. At λ = 0 every combination is an eigenstate of H₀, so H₀ cannot say which one the true state grows out of. Split the pair by even 1 μeV and the states do return to |1⟩ and |2⟩ once λV falls well below 0.5 μeV: what decides is the coupling against the gap, never either alone.
Project the first-order equation onto the level
Leave the start state unknown, |a⁽⁰⁾⟩ = Σᵢ cᵢ|i⟩, and keep the order-λ equation from the start of this unit: (H₀ − E⁽⁰⁾)|a⁽¹⁾⟩ = (Eₐ⁽¹⁾ − H′)|a⁽⁰⁾⟩. Project it onto ⟨j| for each j in the level. H₀ is Hermitian, so ⟨j|H₀ = E⁽⁰⁾⟨j| and the left side vanishes whatever |a⁽¹⁾⟩ is. What remains is g equations, Σᵢ Wⱼᵢcᵢ = Eₐ⁽¹⁾cⱼ with Wⱼᵢ = ⟨j|H′|i⟩: the eigenvalue equation of a g × g Hermitian matrix. The first-order equation has a solution only if c is an eigenvector of W, and the shift is then its eigenvalue. So the shifts are the g eigenvalues of W and the good states its eigenvectors, orthonormal because W is Hermitian. Units check: Wⱼᵢ and Eₐ⁽¹⁾ in J, cᵢ dimensionless. A free test follows: the trace of W is the same in every basis, so the shifts must add up to Σᵢ⟨i|H′|i⟩ whichever basis you began in. For a two-fold level the secular equation is a quadratic, with roots (W₁₁ + W₂₂)/2 ± √(((W₁₁ − W₂₂)/2)² + |W₁₂|²).
The good states belong to H′, not to H₀
A particle on a ring, H₀ = Lz²/2I, has a level at ε = ħ²/2I holding e(iφ)/√(2π) and e(−iφ)/√(2π), or just as well cos φ/√π and sin φ/√π; H₀ prefers none of them. Add H′ = V₀ cos 2(φ − φ₀), a potential with hills at φ₀ and φ₀ + 180°. Written as two exponentials it changes m by ±2, so it has no diagonal elements in the pair, and ⟨+1|H′|−1⟩ = (V₀/2)e(−2iφ₀). The eigenvalues of W are ±V₀/2, and its eigenvectors are cos(φ − φ₀)/√π at +V₀/2 and sin(φ − φ₀)/√π at −V₀/2: standing waves locked to the perturbation. The lower one puts its probability in the valleys, where cos 2(φ − φ₀) = −1, which is why it goes down. Rotate the hills and the good states rotate with them while the shifts stay put; halve V₀ and the shifts halve while the states do not move. Neither running wave e(±iφ) is a good state for any non-zero V₀. The figure plots both densities.
In the good basis the old formulas work again
Once W is diagonal, ⟨b⁽⁰⁾|H′|a⁽⁰⁾⟩ = 0 for any two different good states, so every zero gap inside the level now meets a zero numerator, and the non-degenerate formulas resume with their sums restricted to states outside the level. Projecting the first-order equation onto an outside ⟨m| gives ⟨m|a⁽¹⁾⟩ = ⟨m|H′|a⁽⁰⁾⟩/(E⁽⁰⁾ − Eₘ⁽⁰⁾), and the second-order energy is Σₘ |⟨m|H′|a⁽⁰⁾⟩|²/(E⁽⁰⁾ − Eₘ⁽⁰⁾) over the same m. On the ring, cos 2φ cos φ = (cos 3φ + cos φ)/2, so V₀ cos 2φ couples the good state cos φ only to cos 3φ, with element V₀/2, across the gap ε − 9ε = −8ε. The admixture is −V₀/(16ε), and the level drops by V₀²/(32ε) at second order, as does sin φ through sin 3φ. The ratio to watch is (V₀/2)/(8ε), a coupling out of the level over a real gap. The coupling inside the level, however strong, was already handled exactly by diagonalising W.
When W is degenerate too
If two eigenvalues of W coincide, first order has only half finished. The two shifts are known, but any combination of the two eigenvectors satisfies the first-order condition equally well, so the good states are still undetermined, and the choice passes to second order, through the couplings the pair shares with states outside the level. That case is stated here, not developed. The ring shows both ways it can go. H′ = V₀ cos φ changes m by ±1, so W vanishes identically in the m = ±1 pair and first order predicts neither a shift nor a preferred state; yet the route +1 → 0 → −1 joins the pair at second order and splits it into cos φ and sin φ. H′ = V₀ cos 4φ changes m by ±4 and can never join +1 to −1. The reflection φ → −φ, a symmetry of H₀ + H′, maps the states built on +1 onto those built on −1, and the pair stays degenerate to every order.
Merely close: solve the pair, perturb the rest
Two levels a gap δ apart, with a coupling |W₁₂| comparable to δ, defeat both recipes: non-degenerate second order, ±|W₁₂|²/δ, grows without bound as δ shrinks, and degenerate first order ignores δ altogether. Take the 2 × 2 block of H₀ + H′ in those two states and diagonalise it exactly: E± = Ē ± √((δ/2)² + |W₁₂|²), with δ and Ē the difference and mean of the diagonal entries. That is the avoided crossing of Quantum Mechanics I, quoted rather than rederived and now one step of a method: the pair's exact eigenstates become the zeroth order, and every other level is added perturbatively across its own, larger gap. The limits check it. For δ ≫ |W₁₂| the root expands to δ/2 + |W₁₂|²/δ, the non-degenerate answer; as δ → 0 it becomes |W₁₂|, the degenerate one. With δ = 0.40 meV and |W₁₂| = 0.30 meV the exact splitting is 0.72 meV, against 0.85 meV from second order and 0.60 meV from degenerate first order. Hydrogen's 2s₁/₂ and 2p₁/₂, apart only by the measured 1057.8 MHz Lamb shift, form such a pair in a field of a few hundred V cm⁻¹, worked below.
Change one variable at a time
Make the relationship visible.
At δ = 0 the states are standing waves locked to the hills: slide |W| down to 0.05 meV and only the gap shrinks while the waves stay; turn φ₀ and they follow the hills. Then raise δ past 2|W| and the contrast fades toward running waves e(±iφ).
EXACT SPLITTING E₊ − E₋0.600 meV
DEGENERATE FIRST ORDER 2|W|0.60 meV
LOWER STATE: P(m = +1)0.500
DENSITY CONTRAST sin 2θ1.000
Live interpretationEXACT SPLITTING E₊ − E₋: 0.600 meV. DEGENERATE FIRST ORDER 2|W|: 0.60 meV. LOWER STATE: P(m = +1): 0.500. DENSITY CONTRAST sin 2θ: 1.000
Catch the common trap
Explain before calculating.
A two-fold degenerate level of H₀ is spanned by |a⟩ and |b⟩. A perturbation has ⟨a|H′|a⟩ = ⟨b|H′|b⟩ = 0 and ⟨a|H′|b⟩ = 0.30 meV, and the nearest other level is 50 meV away. What does first-order perturbation theory predict for this level?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA nitrogen-vacancy centre in diamond has a spin-1 ground state with H₀ = (D/ħ²)Sz², where D/h = 2.870 GHz is measured, so mₛ = +1 and mₛ = −1 are degenerate. Strain adds H′ = (η/ħ²)(Sₓ² − Sy²) with η/h = 5.0 MHz. Find W in the mₛ = ±1 pair, the good states and their shifts, and the two microwave frequencies that drive mₛ = 0 to them.
- Rewrite H′ with ladder operators: Sₓ² − Sy² = (S₊² + S₋²)/2. S₊² raises mₛ by 2, so its only element is ⟨+1|S₊²|−1⟩ = (√2 ħ)(√2 ħ) = 2ħ², from S₊|−1⟩ = √2 ħ|0⟩ and S₊|0⟩ = √2 ħ|+1⟩; S₋² supplies the mirror element ⟨−1|S₋²|+1⟩ = 2ħ².
- So W has zero diagonal, because H′ changes mₛ by ±2, and ⟨+1|H′|−1⟩ = (η/ħ²)(2ħ²/2) = η, an energy as it must be. Its eigenvalues are ±η, with eigenvectors (|+1⟩ ± |−1⟩)/√2. Trace check: 0 + 0 = η − η.
- The good states are (|+1⟩ ± |−1⟩)/√2 at D ± η, and neither is an eigenstate of Sz. They do not depend on η: make the strain ten times weaker and the splitting shrinks tenfold while the states stay equal mixtures.
- Is first order the whole story? H′ connects nothing outside the pair, since reaching mₛ = 0 needs a change of 1. So H₀ + H′ is block-diagonal, |0⟩ alone at zero and the pair at D times the identity plus W, and D ± η is exact to every order.
- Frequencies from mₛ = 0: (D ± η)/h = 2.870 GHz ± 5.0 MHz, that is 2.865 GHz and 2.875 GHz, 10.0 MHz apart. As η → 0 the two lines merge at D/h, the unperturbed limit.
AnswerW has zero diagonal and off-diagonal η; good states (|+1⟩ ± |−1⟩)/√2 at D ± η; lines at 2.865 and 2.875 GHz, 10.0 MHz apart, exact because H′ couples nothing outside the pair.
MediumA particle on a ring has ε = ħ²/2I = 2.00 meV, so e(iφ) and e(−iφ) share the level ε. A potential V₀ cos 2φ with V₀ = 1.20 meV is added. Find the good states, their energies through second order and the admixture of other states in each, then compare with 2.5766 meV and 1.3783 meV, the levels from diagonalising H numerically in the e(imφ) basis.
- Inside the pair: cos 2φ = (e(2iφ) + e(−2iφ))/2 changes m by ±2, so W has zero diagonal and ⟨+1|H′|−1⟩ = V₀/2 = 0.60 meV. Eigenvalues ±0.60 meV, with eigenvectors cos φ/√π at +0.60 meV and sin φ/√π at −0.60 meV.
- Outside the pair: cos 2φ cos φ = (cos 3φ + cos φ)/2 and cos 2φ sin φ = (sin 3φ − sin φ)/2, so each good state couples to one outside state, cos 3φ or sin 3φ, with element V₀/2 = 0.60 meV, across the gap ε − 9ε = −16.0 meV.
- First-order admixture: 0.60/(−16.0) = −0.0375 in each, a probability of 0.0014. That ratio is the small parameter, not W against the zero gap: the coupling inside the pair has already been treated exactly.
- Second order: E⁽²⁾ = (0.60 meV)²/(−16.0 meV) = −0.0225 meV for both, in meV²/meV = meV. Totals: 2.00 + 0.60 − 0.0225 = 2.5775 meV for cos φ, and 2.00 − 0.60 − 0.0225 = 1.3775 meV for sin φ.
- Against the numerical levels the errors are −0.0009 and +0.0008 meV, the size of the third-order term ∓(V₀/2)³/(8ε)² = ∓0.0008 meV, upper sign for cos φ. As V₀ → 0 both levels return to ε = 2.00 meV.
Answercos φ at 2.5775 meV and sin φ at 1.3775 meV through second order (numerical 2.5766 and 1.3783 meV); each carries a −0.0375 admixture of cos 3φ or sin 3φ.
HardIn hydrogen, 2s₁/₂ lies 1057.8 MHz above 2p₁/₂ (the measured Lamb shift) and 9911.2 MHz below 2p₃/₂. A field F along z couples the mⱼ = ½ states: |⟨2s₁/₂|eFz|2p₁/₂⟩| = √3 ea₀F and |⟨2s₁/₂|eFz|2p₃/₂⟩| = √6 ea₀F. Treat the 2s₁/₂, 2p₁/₂ pair exactly and 2p₃/₂ at second order: find the field at which the pair's coupling equals half its gap, and the 2s-like level's shift there.
- Why exactly: both states have j = ½, so they are degenerate through order α⁴, where the Dirac energy, quoted from relativistic quantum mechanics, depends on n and j only; the Lamb shift, 4.37 μeV ≈ 0.4α⁵mₑc², is a QED correction, quoted here and treated in Unit 8. 2p₃/₂ lies an α⁴ interval above 2p₁/₂, α⁴mₑc²/32 = 10.95 GHz from the same quoted Dirac energy against the measured 10.969 GHz, so it stays perturbative while a weak field swamps the pair's gap.
- Crossover: √3 ea₀F/h = 528.9 MHz gives ea₀F/h = 305.36 MHz, so F = (305.36 × 10⁶ s⁻¹)(4.1357 × 10⁻¹⁵ eV s)/(e × 5.2918 × 10⁻¹¹ m) = 2.386 × 10⁴ V m⁻¹ ≈ 239 V cm⁻¹, in eV/(e m) = V m⁻¹.
- Exact pair: E± = Ē ± √(528.9² + 528.9²) MHz = Ē ± 748.0 MHz about the midpoint Ē. The upper level began at Ē + 528.9 MHz, so it rises by 219.1 MHz. With tan 2θ = 1, θ = 22.5°, and the upper state is cos²θ = 0.854 of 2s and 0.146 of 2p₁/₂.
- Only its 2s part couples to 2p₃/₂: V = cos θ × √6 × 305.36 MHz = 0.9239 × 2.4495 × 305.36 MHz = 691.0 MHz, across a gap of (528.9 + 9911.2 − 748.0) MHz = 9692.1 MHz. Second order gives −691.0²/9692.1 = −49.3 MHz, and the ratio 691.0/9692.1 = 0.071 confirms the rest is perturbative.
- Total: +219.1 − 49.3 = +169.8 MHz above the unperturbed 2s₁/₂; a 3 × 3 numerical diagonalisation gives +170.3 MHz. Treating 2s as non-degenerate instead gives 3(305.36)²/1057.8 − 6(305.36)²/9911.2 = 264.45 − 56.45 = +208.0 MHz, 38 MHz too high.
- Limits: as F → 0 the shift falls as F², the non-degenerate form; well above 239 V cm⁻¹ the pair's splitting approaches 2√3 ea₀F, linear in F, the degenerate form, but only while √6 ea₀F stays small against the 9911.2 MHz gap, F ≪ 3 kV cm⁻¹; nearer that field 2p₃/₂ must join the exact block.
AnswerF ≈ 239 V cm⁻¹ (2.39 × 10⁴ V m⁻¹). There the 2s-like level sits +169.8 MHz above unperturbed 2s₁/₂, +219.1 MHz from the exact pair and −49.3 MHz from 2p₃/₂, against +170.3 MHz from a 3 × 3 diagonalisation.