University Quantum Mechanics II · Time-Independent Perturbation Theory · 6.6

Finding the good states from a commuting operator

Diagonalising W always works, but it hides why the answer comes out as it does. Here the good states arrive before any matrix is written: find an operator that commutes with both halves of the Hamiltonian, and its eigenvalues sort the level into pieces H′ cannot mix. Units 7, 8 and 9 all lean on this move.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

The previous lesson found the good states of a degenerate level by diagonalising W, the matrix of H′ inside the level. Symmetry usually does that job before the matrix exists. If a Hermitian A commutes with H′, then (aᵢ − aⱼ)⟨i|H′|j⟩ = ⟨i|[A, H′]|j⟩ = 0, so H′ never connects eigenstates of A with different eigenvalues. If A also commutes with H₀, it maps the degenerate subspace into itself, so its eigenstates can be chosen inside the level.

In that basis W falls into blocks, one per eigenvalue of A; where A's eigenvalues in the level are all distinct, W is already diagonal, the eigenstates of A are the good states, and each first-order shift is a single expectation value. Both conditions are needed, and the one students forget is H′: a quantum number that labels the unperturbed states is good only if H′ respects it too. The method stops in three places. A repeated eigenvalue leaves a block, which a second commuting operator may split or a small diagonalisation must finish.

A symmetry that H′ obeys only approximately gives states that are good only while the breaking term is small against the splitting. And the labels never give the sizes: how far each state moves, and whether two distinct labels share an energy, still come from the matrix elements.

Simple definition: A good state of a degenerate level can be read off as an eigenstate of any Hermitian operator that commutes with both H₀ and H′, provided no other state of the level shares its eigenvalue; shared eigenvalues leave blocks of W still to diagonalise.
Example: In a 2D oscillator with ħω = 3.00 meV and H′ = εmω²xy, ε = 0.0400, the swap x ↔ y commutes with both terms and has eigenvalues ±1 on (|1,0⟩ ± |0,1⟩)/√2. These are the good states, at 6.00 ± 0.0600 meV, with no matrix diagonalised.

A commuting operator forbids mixing(aᵢ − aⱼ)⟨i|H′|j⟩ = ⟨i|[A, H′]|j⟩ = 0 when [A, H′] = 0

Derived in one line from ⟨i|A = aᵢ⟨i|. If aᵢ ≠ aⱼ, H′ cannot connect the two states; if aᵢ = aⱼ, the relation says nothing.

A Hermitian, A|i⟩ = aᵢ|i⟩ with aᵢ real and in A's own unit; ⟨i|H′|j⟩ in J

Why A must also commute with H₀[A, H₀] = 0 and H₀|ψ⟩ = E⁽⁰⁾|ψ⟩ ⇒ H₀(A|ψ⟩) = E⁽⁰⁾(A|ψ⟩)

Derived. A restricted to D is Hermitian, so its eigenvectors form a basis of D, and in that basis W splits into one block per eigenvalue of A.

E⁽⁰⁾ in J is the degenerate energy; A|ψ⟩ stays inside the level's subspace D

Good states of the 2D oscillator's first excited pairPψ(x, y) = ψ(y, x), P|±⟩ = ±|±⟩ with |±⟩ = (|1,0⟩ ± |0,1⟩)/√2

Derived from P|1,0⟩ = |0,1⟩. The eigenvalues +1 and −1 differ, so W is diagonal in |±⟩ and no 2 × 2 problem is solved.

P, the swap of coordinates, is dimensionless; [P, H₀] = 0 for the isotropic oscillator and [P, xy] = 0

Their first-order energiesE± = 2ħω ± εħω/2 for H′ = εmω²xy

Derived. The exact normal modes, along the diagonals with frequencies ω√(1 ± ε), give ħω[(3/2)√(1 ± ε) + (1/2)√(1 ∓ ε)], equal to this through order ε.

ε dimensionless, ħω in J; uses ⟨1,0|xy|0,1⟩ = ⟨1|x|0⟩⟨0|y|1⟩ = ħ/(2mω), in m²

When a term breaks the swapswap-odd weight of the lower state = [1 + ε/√(ε² + δ²)]/2

Derived from the eigenvectors of W. It equals 1 only at δ = 0, where the swap commutes; at ε = 0.20 it is 0.985 for δ = 0.05 but 0.854 for δ = 0.20.

For H′ = mω²[εxy + δ(x² − y²)/2], where W = (ħω/2)(δσ_z + εσₓ) on |1,0⟩, |0,1⟩; ε and δ dimensionless

Labels for a spin–orbit termE_J = (A/2)[J(J + 1) − L(L + 1) − S(S + 1)] for H′ = (A/ħ²)L⋅S

Derived from L⋅S = (J² − L² − S²)/2 once the |J, M⟩ labels are shown distinct. It gives the interval rule E_J − E_(J−1) = AJ, tested in the Hard example.

A in J or cm⁻¹; J², L², S² and J_z commute with L⋅S and with the central-field H₀

A commutator forbids mixing between different eigenvalues

Let A be Hermitian with A|i⟩ = aᵢ|i⟩, and suppose [A, H′] = 0. Sandwich the commutator between two of A's eigenstates. Because A is Hermitian, ⟨i|A = aᵢ⟨i| with aᵢ real, so ⟨i|[A, H′]|j⟩ = ⟨i|AH′ − H′A|j⟩ = (aᵢ − aⱼ)⟨i|H′|j⟩. The left side is zero, so whenever aᵢ ≠ aⱼ the element ⟨i|H′|j⟩ vanishes: H′ cannot carry a state from one eigenvalue of A to another. This is derived, costs no integral, and is the precise meaning of saying that H′ conserves A. The dimensions agree on both sides, A's unit times joules. Note what it does not say. It is silent about ⟨i|H′|j⟩ when aᵢ = aⱼ, and it says nothing about the diagonal elements, which still have to be computed.

Both halves of the Hamiltonian must commute with A

Degenerate perturbation theory needs a basis of the degenerate subspace D, not of the whole space. If [A, H₀] = 0 and H₀|ψ⟩ = E⁽⁰⁾|ψ⟩, then H₀(A|ψ⟩) = AH₀|ψ⟩ = E⁽⁰⁾(A|ψ⟩), so A maps D into itself. Its restriction to D is a Hermitian matrix with a complete set of eigenvectors inside D, and in that basis W is block-diagonal by the first result, one block per eigenvalue. Drop either condition and the argument fails. For the 2D oscillator with H′ = εmω²xy, the position x commutes with H′, but its eigenstates are points in space, not members of the level. The number operator nₓ = aₓ†aₓ commutes with H₀ and has distinct eigenvalues 1 and 0 on |1,0⟩ and |0,1⟩, yet those states are not good: nₓ does not commute with xy, and W in that basis is purely off-diagonal, with ⟨1,0|H′|0,1⟩ = εħω/2.

The swap x ↔ y hands over the oscillator's good states

From QM I, the isotropic 2D oscillator has E = (nₓ + n_y + 1)ħω, so |1,0⟩ and |0,1⟩ share 2ħω. Let P swap the coordinates, Pψ(x, y) = ψ(y, x). It commutes with H₀, whose x and y parts are identical, and with xy. Since P|1,0⟩ = |0,1⟩, its eigenstates in the level are |±⟩ = (|1,0⟩ ± |0,1⟩)/√2 with eigenvalues ±1. They differ, so W is diagonal there and E⁽¹⁾ = ±⟨1,0|H′|0,1⟩. With x = √(ħ/2mω)(aₓ + aₓ†), ⟨1,0|xy|0,1⟩ = ⟨1|x|0⟩⟨0|y|1⟩ = ħ/(2mω), so E± = 2ħω ± εħω/2. The exact answer checks it: the potential ½mω²(x² + y² + 2εxy) has normal modes along the diagonals with frequencies ω√(1 ± ε), and at ε = 0.10 the pair lies at 2.0476ħω and 1.9474ħω against the first-order 2.05ħω and 1.95ħω. The residue, close to −ε²ħω/4, is second order. The normal modes lie on the diagonals at every ε because P commutes with the whole H, so its labels hold at every order.

Repeated eigenvalues leave blocks for a second operator

When A takes the same value on several states of D, the theorem is silent about W inside that group, and a block survives. The remedy is a second operator B that commutes with H₀, H′ and A: the joint eigenvalues (a, b) label smaller blocks, and a set of such operators whose joint labels are all distinct inside D diagonalises W completely. Hydrogen's n = 2 level with spin is the case Unit 8 needs: eight states, and a fine-structure H′ that commutes with L², J² and J_z. J_z alone gives mⱼ = ±3/2 once each and mⱼ = ±1/2 three times each, blocks of 1, 3, 3 and 1. L² splits each block of 3 into the 2s state, with l = 0, and two 2p states, with l = 1, and J² separates those two into j = 1/2 and j = 3/2. All eight labels (l, j, mⱼ) are now distinct, so W is diagonal in the coupled basis. With J_z alone you would still be diagonalising a 3 × 3 block twice. In Unit 9's strong field the Zeeman term joins H₀ and fine structure is the H′. J² no longer commutes with H₀, but L² and J_z still do, and they part every pair sharing mₗ + 2mₛ: J_z separates (mₗ, mₛ) = (1, −½) from (−1, ½), and L² separates each 2s state from the 2p state with mₗ = 0 and the same mₛ, so the uncoupled states |l, mₗ, mₛ⟩ are already good.

Symmetry fixes the states, never the sizes

The labels decide which combinations are good; the diagonal elements decide how far each moves, and distinct labels do not promise distinct energies. In hydrogen's n = 2 level the fine structure, quoted from Unit 8, puts 2p₃/₂ 4.5 × 10⁻⁵ eV above 2p₁/₂ (measured as 10.97 GHz), a fraction α²/4 of the 3.40 eV binding, yet leaves 2s₁/₂ and 2p₁/₂ together although their l differs; the measured 1057.8 MHz Lamb shift, from quantum electrodynamics at the next order in α, parts them. Helium's 1s2s pair tests the other half against a measurement. Swapping the two electrons commutes with H₀ and with their repulsion, and so does the total L², which keeps 1s2s, with L = 0, apart from the 1s2p states, with L = 1, that share its unperturbed energy; so the good spatial states are (|1s2s⟩ ± |2s1s⟩)/√2, shifted by a common direct term plus or minus the exchange integral K. With the hydrogenic Z = 2 value K = 1.194 eV, quoted ahead of Unit 7, first order predicts a 2¹S − 2³S splitting 2K = 2.39 eV; the measured value is 0.796 eV. The states are right, because the swap and L² are exact symmetries. The size is three times too large, because unscreened orbitals are not exact.

An approximate symmetry gives approximately good states

Add to the oscillator a term that breaks the swap, ½δmω²(x² − y²), which stiffens x and softens y for δ above zero. On |1,0⟩ and |0,1⟩ it has diagonal elements ±δħω/2 and no off-diagonal one, so W = (ħω/2)(δσ_z + εσₓ), with σ the Pauli matrices of QM I. P no longer commutes, the splitting becomes ħω√(ε² + δ²), and the upper good state cos φ|1,0⟩ + sin φ|0,1⟩ rotates away from the diagonal as tan 2φ = ε/δ. The swap-odd weight of the lower state, [1 + ε/√(ε² + δ²)]/2, measures how good the old labels still are: at ε = 0.20 it is 0.985 for δ = 0.05 but 0.854 for δ = 0.20. A symmetry-based state is trustworthy only while the breaking element is small against the splitting the symmetric term produces; when the two are comparable, return to the full W. At ε = 0 the reflection x → −x commutes instead, and |1,0⟩ and |0,1⟩ themselves are good. The figure sweeps between these limits.

Change one variable at a time

Make the relationship visible.

Interactive model

Swap-symmetric coupling ε0.20

Swap-breaking anisotropy δ0.00

At δ = 0 drag ε: the lobes stay on a diagonal while the splitting grows, because the swap commutes with H′. Then step δ away from 0: the swap stops commuting, and the lobes turn toward the axes, reaching them at ε = 0, where x → −x commutes instead. At the corners the dots leave the bars.

FIRST-ORDER SPLITTING0.200 ħω

EXACT SPLITTING0.201 ħω

LOWER LOBES TO x AXIS135 °

SWAP-ODD WEIGHT, LOWER STATE1.000

Live interpretationFIRST-ORDER SPLITTING: 0.200 ħω. EXACT SPLITTING: 0.201 ħω. LOWER LOBES TO x AXIS: 135 °. SWAP-ODD WEIGHT, LOWER STATE: 1.000

Catch the common trap

Explain before calculating.

The first excited level of a 2D isotropic oscillator, |1,0⟩ and |0,1⟩, is perturbed by H′ = εmω²xy. Which operator hands you the good states without diagonalising W, and why?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA particle on a ring has levels Eₘ = m²E₁ with E₁ = ħ²/(2I) = 0.150 eV, so m = +1 and m = −1 are degenerate. Squeezing the ring slightly into an ellipse adds H′ = V₀ cos 2φ with V₀ = 12.0 meV. Find the good states of the |m| = 1 level from a symmetry, give their first-order energies, and check the energy-denominator condition.

Look for an operator that commutes with both terms. The mirror M: φ → −φ sends L_z to −L_z, so it leaves H₀ = L_z²/(2I) unchanged, and cos 2φ is even, so it leaves H′ unchanged too. Both commutators vanish.
M sends e^(iφ) to e^(−iφ), so its eigenstates in the level are cos φ/√π with M = +1 and sin φ/√π with M = −1. The eigenvalues differ, so these are the good states and W is diagonal in them.
Diagonal elements: (V₀/π)∫cos²φ cos 2φ dφ over one turn, with cos²φ = (1 + cos 2φ)/2, gives V₀/2; the same integral with sin²φ = (1 − cos 2φ)/2 gives −V₀/2. So E = 0.150 eV ± 6.0 meV: 0.156 eV for cos φ, whose lobes sit where cos 2φ is largest, and 0.144 eV for sin φ.
Why not the rotation φ → φ + π? It also commutes with H₀ and H′, but it gives −1 on both e^(±iφ), and a repeated eigenvalue separates nothing. Cross-check in the e^(±iφ) basis: W = (V₀/2)σₓ, whose eigenvectors are exactly cos φ and sin φ.
Energy denominators: cos 2φ changes m by ±2, so the pair also couples to |m| = 3, a gap of 8E₁ = 1.20 eV, through an element V₀/2 = 6.0 meV. The ratio is 0.005, and the second-order shift −(6.0 meV)²/(1.20 eV) = −0.030 meV is the same for both states, so the 12.0 meV splitting survives. As V₀ → 0 both states return to 0.150 eV.

AnswerGood states cos φ/√π and sin φ/√π, fixed by the mirror φ → −φ; E = 0.156 eV and 0.144 eV, split by V₀ = 12.0 meV, each lowered a further 0.030 meV at second order.

MediumA hydrogen atom sits at the centre of an electrostatic quadrupole with potential Φ = (G/2)(x² − y²), where G = 2.50 × 10¹⁷ V m⁻². Ignoring spin and fine structure, find the good states of the n = 2 level from symmetry, and their first-order shifts.

The electron's energy is H′ = −eΦ = −(eG/2)(x² − y²). The Coulomb H₀ is unchanged by each reflection Πₓ: x → −x, Π_y and Π_z, and so is x² − y², so all three commute with H₀ and H′. L_z does not: [L_z, x² − y²] = 4iħxy, so the m = ±1 states are not good.
The real orbitals carry reflection signs (Πₓ, Π_y, Π_z) of 2s (+, +, +), 2pₓ (−, +, +), 2p_y (+, −, +) and 2p_z (+, +, −). No pattern repeats, so W is diagonal in the basis 2s, 2pₓ, 2p_y, 2p_z, and each shift is one expectation value.
Angles: the 2s and 2p_z averages of x² − y² vanish by symmetry. For 2pₓ, whose angular part is √(3/4π) x/r, ⟨x² − y²⟩ = 3(1/5 − 1/15)⟨r²⟩ = (2/5)⟨r²⟩, using the sphere averages 1/5 of x⁴/r⁴ and 1/15 of x²y²/r⁴; for 2p_y the sign flips.
Radius: with R₂₁ = r e^(−r/2a₀)/(√24 a₀⁵⁄²) from QM I, ⟨r²⟩ = (1/(24a₀⁵))∫r⁶e^(−r/a₀) dr = 720a₀²/24 = 30a₀². So ΔE(2pₓ) = −(eG/2)(2/5)(30a₀²) = −6eGa₀², ΔE(2p_y) = +6eGa₀², and 2s and 2p_z do not move.
Numbers: eGa₀² = e × (2.50 × 10¹⁷ V m⁻²)(5.29 × 10⁻¹¹ m)² = 0.700 meV, so the shifts are ∓4.20 meV. Checks: C × V m⁻² × m² is J; the four shifts sum to zero, as they must for an H′ whose average over all directions vanishes; 4.20 meV is tiny against the 1.89 eV gap to n = 3, and large against the 0.045 meV fine structure, which justifies dropping spin.

AnswerGood states 2s, 2pₓ, 2p_y and 2p_z, not the m = ±1 pair: 2pₓ moves by −4.20 meV, 2p_y by +4.20 meV, and 2s and 2p_z stay put, degenerate but never mixed, since Π_z tells them apart at every order.

HardCalcium's 4s4p ³P term (L = 1, S = 1) holds nine states, degenerate before spin–orbit coupling, which within the term acts as H′ = (A/ħ²)L⋅S. Use commuting operators to reduce W, find the energies in terms of A, and test them against the measured levels ³P₀, ³P₁ and ³P₂ at 15157.901, 15210.063 and 15315.943 cm⁻¹.

In the |M_L, M_S⟩ basis W is 9 × 9. J_z = L_z + S_z commutes with H₀, since the whole atom can be rotated, and with the scalar L⋅S, so W splits by M = M_L + M_S: M = ±2 once, ±1 twice and 0 three times, blocks of 1, 2, 3, 2 and 1. J_z alone still leaves a 3 × 3 and two 2 × 2 problems.
J² also commutes with H₀, H′ and J_z, and inside each block its eigenvalues differ: J = 2, 1, 0 in the M = 0 block and J = 2, 1 in the M = ±1 blocks. Every (J, M) label is now distinct, so W is diagonal in the coupled basis |J, M⟩ built with the Clebsch–Gordan coefficients of Unit 5.
From J² = L² + S² + 2L⋅S, E_J = (A/2)[J(J + 1) − 4], so E₂ = A, E₁ = −A and E₀ = −2A. Check: the (2J + 1)-weighted sum 5A − 3A − 2A = 0 equals the trace of L⋅S in the uncoupled basis, whose diagonal elements M_L M_S ħ² sum to zero.
The prediction needs no value of A: E₂ − E₁ = 2A and E₁ − E₀ = A, a ratio of exactly 2, the Landé interval rule. Measured: 15315.943 − 15210.063 = 105.880 cm⁻¹ and 15210.063 − 15157.901 = 52.162 cm⁻¹, a ratio of 2.030. The two intervals give A = 52.94 and 52.16 cm⁻¹, 1.5% apart.
Where the 1.5% goes: J² and J_z commute with the full atomic H, so the labels are exact, but 4s4p ¹P₁, measured at 23652.304 cm⁻¹, shares ³P₁'s labels J = 1, M, and spin–orbit coupling mixes them. Repulsion from above pushes ³P₁ down, widening the upper interval: ratio above 2, as seen. A push δ with 2A + δ = 105.880 cm⁻¹ and A − δ = 52.162 cm⁻¹ gives A = 52.68 cm⁻¹ and δ = 0.52 cm⁻¹.
Size check: A = 52.68 cm⁻¹ = 6.53 meV is 0.35% of the term's 15158 cm⁻¹ excitation, small as a relativistic correction of order α² should be, though enlarged by the nuclear charge the 4p electron meets inside the core.

AnswerJ_z cuts the 9 × 9 W to blocks 1, 2, 3, 2, 1 and J² diagonalises it: E_J = A, −A, −2A for J = 2, 1, 0, intervals 2A : A. Measured ratio 2.030; A ≈ 52.7 cm⁻¹, with ³P₁ pushed down about 0.52 cm⁻¹ by ¹P₁, which shares its labels.