University Quantum Mechanics II · Time-Independent Perturbation Theory · 6.7

Higher degeneracy worked: 2D oscillator and cubical box

The recipe for a degenerate level fits on one line; running it is where the marks are lost. Here it is run twice, all the way to numbers: on an oscillator whose exact solution checks every order, and on a box where a 3 × 3 matrix falls apart into a 1 × 1 and a 2 × 2 once you ask which integrals must vanish.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Degenerate perturbation theory comes down to one concrete task: write the matrix Wᵢⱼ = ⟨i|H′|j⟩ among the states that share an unperturbed energy, and diagonalise it. Its eigenvalues are the first-order shifts and its eigenvectors the good states, and most of the work is deciding, before any integral, which entries must vanish. Two systems show the whole method.

In the 2D isotropic oscillator, H′ = εmω²xy is odd in x, so both diagonal entries on |1,0⟩ and |0,1⟩ vanish, and the ladder operators give the one off-diagonal entry, εħω/2. The level splits by ±εħω/2 into (|1,0⟩ ± |0,1⟩)/√2, the combinations the exchange symmetry x ↔ y demanded all along. The same problem is solved exactly by rotating to u, v = (x ± y)/√2, where the frequencies are ω√(1 ± ε): expanding them returns the first-order splitting and a second-order shift of −ε²ħω/4, and shows that at |ε| = 1 the soft frequency reaches zero and every bound state is gone, something no first-order line hints at.

In a cubical box with V₀ added over a quarter-column, the level (1,1,2), (1,2,1), (2,1,1) needs a 3 × 3 matrix, but H′ ignores z, so (1,1,2) decouples and the remaining pair splits by ±16V₀/(9π²) about V₀/4. First order is trustworthy only while couplings out of the level stay small beside the gaps to other levels; inside the level the gap is zero, which is why a matrix, not a formula, is needed.

Simple definition: Working a degenerate level means writing the matrix of H′ among the states that share one unperturbed energy, using symmetry to zero as many entries as possible, and diagonalising the rest: the eigenvalues are the first-order shifts and the eigenvectors the good states.
Example: In a 2D trap with ħω = 40.0 meV, H′ = εmω²xy with ε = 0.050 has zero diagonal entries and an off-diagonal entry of 1.00 meV on |1,0⟩, |0,1⟩, so the 80.0 meV level splits to 79.0 and 81.0 meV; the exact values are 78.97 and 80.98 meV.

The oscillator's degenerate block⟨1,0|H′|0,1⟩ = εmω² × ħ/(2mω) = εħω/2

Derived from x = √(ħ/(2mω)) (a + a†), used from Quantum Mechanics I. This one number fills the 2 × 2 matrix: zeros on the diagonal, εħω/2 off it.

H′ = εmω²xy with ε dimensionless, m in kg, ω in rad s⁻¹; both diagonal entries vanish because ⟨1|x|1⟩ = 0 by parity

First-order doublet and its good statesE = 2ħω ± εħω/2 for |±⟩ = (|1,0⟩ ± |0,1⟩)/√2

Derived by diagonalising the block. Exchange symmetry fixes the good states for every ε; the sign of ε decides only which of them goes up.

E in J; |±⟩ are even and odd under the exchange x ↔ y, which commutes with both H₀ and H′

Exact normal-mode frequenciesω₊ = ω√(1 + ε) along u = (x + y)/√2, ω₋ = ω√(1 − ε) along v = (x − y)/√2

Derived exactly. E = (nᵤ + ½)ħω₊ + (nᵥ + ½)ħω₋ expands to 2ħω ± εħω/2 − ε²ħω/4 for the doublet, a check on first and second order at once.

ω₊ and ω₋ in rad s⁻¹; from x² + y² + 2εxy = (1 + ε)u² + (1 − ε)v²; both real and nonzero only while |ε| stays below 1

Half-box overlap of the first two sines(2/L)∫₀^(L/2) sin(πx/L) sin(2πx/L) dx = 4/(3π) ≈ 0.424

Derived with t = πx/L and sin 2t = 2 sin t cos t. Orthogonality needs the whole interval, so this is not zero, and its square is the box's only coupling.

Dimensionless, L in m; over the same half, sin²(nπx/L) integrates to ½ for every n, being symmetric about L/2

The box's 3 × 3 blockW₁₁ = W₂₂ = W₃₃ = V₀/4, W₂₃ = 16V₀/(9π²), W₁₂ = W₁₃ = 0

Derived. H′ ignores z, so every entry linking n_z = 2 to n_z = 1 carries ⟨2|1⟩ = 0, and the 3 × 3 block falls apart into a 1 × 1 and a 2 × 2.

Basis 1, 2, 3 = |1,1,2⟩, |1,2,1⟩, |2,1,1⟩; V₀ in J; each entry is V₀ times an x, a y and a z overlap

First-order shifts of the box levelE⁽¹⁾ = V₀/4 for |1,1,2⟩ and V₀[¼ ± 16/(9π²)] for (|1,2,1⟩ ± |2,1,1⟩)/√2

Derived. Each shift is V₀ times the good state's probability of lying in the stepped quarter-column: the + state crowds into it and the − state avoids it.

0.250V₀, 0.430V₀ and 0.070V₀, in J; they sum to 3V₀/4, the trace of the block

Set up the oscillator's block before integrating anything

The 2D isotropic oscillator, used as Quantum Mechanics I left it, has E = (nₓ + n_y + 1)ħω, so its first excited level at 2ħω holds |1,0⟩ and |0,1⟩. Add H′ = εmω²xy with ε dimensionless, the coupling a tilted anisotropy in a trap or a quantum dot produces. Because the level is degenerate, the first-order shifts are the eigenvalues of the 2 × 2 matrix Wᵢⱼ = ⟨i|H′|j⟩, not its diagonal entries. Decide what vanishes first. ⟨1,0|xy|1,0⟩ = ⟨1|x|1⟩⟨0|y|0⟩, and x is odd, so ⟨1|x|1⟩ = 0; the same kills ⟨0,1|xy|0,1⟩. The one survivor factorises too: ⟨1,0|xy|0,1⟩ = ⟨1|x|0⟩⟨0|y|1⟩ = ħ/(2mω), from x = √(ħ/(2mω)) (a + a†) and ⟨1|a†|0⟩ = 1. So W₁₂ = εmω² × ħ/(2mω) = εħω/2, and the units check: kg s⁻² times m² is a joule. The whole first-order problem is now εħω/2 times the matrix with zeros on its diagonal and 1 off it.

Diagonalise, then notice symmetry had already chosen

That matrix has eigenvalues ±εħω/2 and eigenvectors (|1,0⟩ ± |0,1⟩)/√2. At ħω = 40.0 meV and ε = 0.050 the 80.0 meV level splits into 79.0 and 81.0 meV. The good states could have been named without the matrix. The exchange P, which swaps x and y, commutes with H₀, which is isotropic, and with H′, since xy = yx. P turns |1,0⟩ into |0,1⟩, so its eigenstates in this level are exactly the symmetric and antisymmetric sums, and a W that commutes with P cannot mix them. The calculation adds only the ordering: xy = (u² − v²)/2 is positive along u = (x + y)/√2, so for ε above 0 the symmetric state, whose one quantum lies along u, goes up. Symmetry fixes the good states for every ε, large or small; only the energies depend on ε. The three-fold level at 3ħω yields to the same move, worked in the second example below: P splits off (|2,0⟩ − |0,2⟩)/√2, and the level becomes three sublevels spaced εħω.

Solve it exactly and compare order by order

This perturbation can be removed exactly. With u = (x + y)/√2 and v = (x − y)/√2, x² + y² = u² + v² and xy = (u² − v²)/2, so V = ½mω²[(1 + ε)u² + (1 − ε)v²]: two independent oscillators of frequency ω₊ = ω√(1 + ε) and ω₋ = ω√(1 − ε), and E = (nᵤ + ½)ħω₊ + (nᵥ + ½)ħω₋. Expanding √(1 ± ε) = 1 ± ε/2 − ε²/8 + … gives, for nᵤ + nᵥ = N, E = (N + 1)ħω + (nᵤ − nᵥ)εħω/2 − (N + 1)ε²ħω/8 + …. For the doublet the linear term is ±εħω/2, the matrix's answer, and the state with nᵤ = 1 is (|1,0⟩ + |0,1⟩)/√2, the good state. The quadratic term, −ε²ħω/4, is what second-order theory builds from outside the level: H′ carries each good state to |2,1⟩ and |1,2⟩ with matrix element εħω/2 across a gap of 2ħω, and 2 × (εħω/2)²/(−2ħω) = −ε²ħω/4. At ε = 0.30 the exact levels are 1.8251ħω and 2.1286ħω; first order gives 1.85 and 2.15, second order 1.8275 and 2.1275.

What first order cannot see: the end of the bound states

Follow the lower sublevel as ε grows. First order says 2ħω − εħω/2, a straight line that reaches a harmless 1.5ħω at ε = 1. The exact energy, ½ħω√(1 + ε) + (3/2)ħω√(1 − ε), falls ever faster: 1.673ħω at ε = 0.5, 1.164ħω at ε = 0.9 and 0.855ħω at ε = 0.99, and at ε = 1 the soft frequency ω₋ is zero. There the potential is flat along v, and for |ε| above 1 it is a saddle: the particle runs away along v and no bound state survives. Nothing in the first-order matrix hints at this, because a first-order shift is linear in H′ and sees only the level's own states. Even the energy-denominator ratio stays modest, a coupling of εħω/2 against a gap of 2ħω, or ε/4. What limits the series is the branch point of √(1 − ε) at ε = 1, the same place the physics collapses: it sets the radius of convergence, each added order helps less as ε approaches it, and beyond it the series diverges. The limiting checks still pass: ε → 0 returns 2ħω, and reversing the sign of ε only swaps the roles of u and v.

The cubical box: let separability empty the matrix

In a cube of side L, ψ = (2/L)³⁄² sin(nₓπx/L) sin(n_yπy/L) sin(n_zπz/L) and E = (nₓ² + n_y² + n_z²)E₁, with E₁ = π²ħ²/(2mL²). The level 6E₁ holds (1,1,2), (1,2,1) and (2,1,1). Raise the potential by V₀ wherever x and y both lie between 0 and L/2, at every z. H′ is a product of a function of x, one of y and none of z, so every entry factorises: ⟨n|H′|n′⟩ = V₀ I(nₓ, nₓ′) I(n_y, n_y′) ⟨n_z|n_z′⟩, where I is the overlap of two sines over the half-box. The z factor is an overlap over the whole box, 1 when n_z is unchanged and 0 otherwise. (1,1,2), the only state with n_z = 2, is therefore coupled to neither partner, and the 3 × 3 problem is a 1 × 1 and a 2 × 2 before any integral is done. Two integrals remain: I(n, n) = ½ for every n, because sin² is symmetric about L/2, and I(1,2) = 4/(3π) = 0.424. So W has V₀/4 on its diagonal and one off-diagonal pair, V₀[4/(3π)]² = 16V₀/(9π²) = 0.180V₀, linking (1,2,1) and (2,1,1).

Diagonalise the pair, and read each shift as a probability

The 2 × 2 block has equal diagonal entries V₀/4 and off-diagonal 16V₀/(9π²), so the shifts are V₀[¼ ± 16/(9π²)] = 0.430V₀ and 0.070V₀, for (|1,2,1⟩ ± |2,1,1⟩)/√2, while |1,1,2⟩ keeps 0.250V₀. The exchange x ↔ y, a symmetry of the quarter-column, predicted those good states again. Because H′ is V₀ inside the step and 0 outside, each shift is V₀ times the good state's probability of being in the step. The + state is proportional to sin(πx/L) sin(πy/L)[cos(πx/L) + cos(πy/L)], with its node on x + y = L, which only touches the step's corner: 43% of its probability sits in the step. The − state has its node on x = y, straight through the step, and puts 7% there. Checks: the shifts sum to 3V₀/4, the trace, and V₀ spread over the whole box would shift all three by V₀. A numerical diagonalisation at V₀ = E₁ over 400 sine products in x and y, n_z being conserved, computed rather than derived, gives 0.064, 0.216 and 0.418 V₀, within 0.034V₀ of first order although V₀ is a third of the 3E₁ gap to the nearest levels.

Change one variable at a time

Make the relationship visible.

Interactive model

Coupling ε in H′ = εmω²xy0.30

Level N = nₓ + n_y1

At ε = 0 the contour is a circle and no direction is special. Raise ε: the contour stretches along v while the dashed lines stay straight. Near ε = 0.99 the lowest exact sublevel dives far below its line and the contour is about three times its ε = 0 width along v. Set N = 2 for three sublevels spaced εħω.

EXACT LOWEST SUBLEVEL1.825 ħω

FIRST ORDER1.850 ħω

THROUGH SECOND ORDER1.828 ħω

SOFT FREQUENCY ω₋0.837 ω

Live interpretationEXACT LOWEST SUBLEVEL: 1.825 ħω. FIRST ORDER: 1.850 ħω. THROUGH SECOND ORDER: 1.828 ħω. SOFT FREQUENCY ω₋: 0.837 ω

Catch the common trap

Explain before calculating.

Misconception check

The first-order shift of a degenerate state is not its diagonal element ⟨n|H′|n⟩. Students find ⟨1,0|H′|1,0⟩ = ⟨0,1|H′|0,1⟩ = 0 for the oscillator and conclude the level does not move at first order, or find V₀/4 on the box's whole diagonal and predict a rigid shift. The full matrix refutes both, and the exact solution and a numerical diagonalisation confirm it: the doublet splits by εħω, and the box level spreads from 0.070V₀ to 0.430V₀. Inside a degenerate level every combination of the basis states is an equally good eigenstate of H₀, so the diagonal entries depend on which basis you happened to write down; only the eigenvalues of the whole block are basis-independent. The diagonal entries are right about one thing only, their sum: the trace, 0 for the oscillator and 3V₀/4 for the box, equals the sum of the first-order shifts. They give the shifts themselves only in a basis that already diagonalises W, which is what finding the good states means.

In a cube of side L, H′ = V₀ wherever x and y both lie between 0 and L/2, at every z. Which statement correctly describes the off-diagonal entries of H′ within the level |1,1,2⟩, |1,2,1⟩, |2,1,1⟩?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyAn electron in a GaAs quantum dot, with the measured effective mass 0.067 mₑ, moves in a 2D isotropic trap with ħω = 3.00 meV. In-plane strain adds H′ = κxy with κ = 1.52 × 10⁻⁷ J m⁻². Find ε in H′ = εmω²xy, the matrix of H′ on |1,0⟩ and |0,1⟩, and the first-order energies and good states of the level at 2ħω, then compare with the exact energies.

Convert: ω = (3.00 meV)/ħ = 4.558 × 10¹² rad s⁻¹ and m = 0.067 × 9.109 × 10⁻³¹ kg = 6.103 × 10⁻³² kg, so mω² = 1.268 × 10⁻⁶ kg s⁻², the same unit as κ, since J m⁻² = kg s⁻². Then ε = κ/(mω²) = 0.120, dimensionless as it must be.
Diagonal entries: ⟨1,0|xy|1,0⟩ = ⟨1|x|1⟩⟨0|y|0⟩ = 0 because x is odd, and likewise for |0,1⟩. Off-diagonal: ⟨1,0|xy|0,1⟩ = ⟨1|x|0⟩⟨0|y|1⟩ = ħ/(2mω) = ℓ²/2, with ℓ² = ħ/(mω) = 3.79 × 10⁻¹⁶ m², an oscillator length ℓ = 19.5 nm.
So W₁₂ = κℓ²/2 = 2.88 × 10⁻²³ J = 0.180 meV, which equals εħω/2 = 0.120 × 3.00 meV/2 as it should. The 2 × 2 matrix, zeros on the diagonal and 0.180 meV off it, has eigenvalues ±0.180 meV.
First order: E = 6.00 ± 0.180 meV, that is 6.180 meV for (|1,0⟩ + |0,1⟩)/√2 and 5.820 meV for (|1,0⟩ − |0,1⟩)/√2. The symmetric state goes up because κ is positive, making xy positive along the line x = y where its quantum lies.
Exact: E = ħω[(3/2)√1.120 + ½√0.880] = 6.169 meV and ħω[½√1.120 + (3/2)√0.880] = 5.809 meV. Both lie 0.011 meV below first order, the common second-order shift −ε²ħω/4 = −0.0108 meV. The splitting, 0.360 meV at first order and 0.361 meV exactly, is untouched at second order because that shift moves the two levels alike; it changes only at third order, by ε³ħω/8 = 0.0006 meV.

Answerε = 0.120 and W₁₂ = 0.180 meV; first order gives 5.820 and 6.180 meV for (|1,0⟩ ∓ |0,1⟩)/√2, against exact values of 5.809 and 6.169 meV.

MediumA particle in a 2D isotropic trap with ħω = 20.0 meV feels H′ = εmω²xy with ε = 0.100. Find the first-order energies and good states of the three-fold level at 3ħω, using the exchange x ↔ y to reduce the 3 × 3 matrix, and compare them with the exact energies.

The level holds |2,0⟩, |1,1⟩ and |0,2⟩. Write xy = [ħ/(2mω)](aₓ + aₓ†)(a_y + a_y†). Only aₓ†a_y + a_y†aₓ keeps nₓ + n_y fixed, so inside the level W = (εħω/2)(aₓ†a_y + a_y†aₓ). Its diagonal vanishes, and so does ⟨2,0|W|0,2⟩, which would need two quanta moved at once.
On the basis: (aₓ†a_y + a_y†aₓ)|2,0⟩ = √2|1,1⟩ and (aₓ†a_y + a_y†aₓ)|1,1⟩ = √2|2,0⟩ + √2|0,2⟩. The only non-zero entries are ⟨1,1|W|2,0⟩ = ⟨1,1|W|0,2⟩ = √2εħω/2.
Exchange swaps |2,0⟩ and |0,2⟩ and keeps |1,1⟩. The antisymmetric A = (|2,0⟩ − |0,2⟩)/√2 cannot mix with the symmetric states, and W A = 0, so its shift is zero. With S = (|2,0⟩ + |0,2⟩)/√2, W|1,1⟩ = εħω S, so the symmetric block is εħω times the matrix with zeros on the diagonal and 1 off it: shifts ±εħω for (S ± |1,1⟩)/√2.
With εħω = 2.00 meV the 60.0 meV level becomes 58.0, 60.0 and 62.0 meV, three sublevels spaced by εħω. The trace of W is 0, and so is the sum of the shifts.
Exact: E = (nᵤ + ½)ħω√1.100 + (nᵥ + ½)ħω√0.900 with nᵤ + nᵥ = 2 gives 61.927, 59.925 and 57.922 meV. The second-order term of the expansion, −3ε²ħω/8 = −0.075 meV for all three, brings first order within 0.003 meV of each.

Answer58.0, 60.0 and 62.0 meV at first order, for (S − |1,1⟩)/√2, A and (S + |1,1⟩)/√2; the exact values are 57.922, 59.925 and 61.927 meV.

HardA GaAs cube, electron effective mass 0.067 mₑ (measured), has side L = 20.0 nm. A gate raises the potential by V₀ = 5.00 meV over the octant where x, y and z all lie between 0 and L/2. Find the first-order shifts of the level (1,1,2), (1,2,1), (2,1,1), explain why a degeneracy survives, test the energy-denominator condition, and compare with a numerical diagonalisation.

E₁ = π²ħ²/(2mL²) = 14.03 meV with m = 6.103 × 10⁻³² kg, so the level sits at 6E₁ = 84.2 meV. Its nearest neighbours, (1,1,1) at 3E₁ and (1,2,2) and partners at 9E₁, are each 3E₁ = 42.09 meV away.
H′ now depends on z too, so each entry is V₀ times a half-box overlap in every coordinate. Diagonal: V₀(½)³ = 0.625 meV. Off-diagonal, for example ⟨1,1,2|H′|1,2,1⟩ = V₀ × ½ × 4/(3π) × 4/(3π) = 8V₀/(9π²) = 0.4503 meV, and relabelling the axes shows all three off-diagonal entries are equal.
W has 0.625 meV on the diagonal and 0.4503 meV everywhere off it. The vector (1,1,1)/√3 gives 0.625 + 2 × 0.4503 = 1.526 meV; any vector orthogonal to it gives 0.625 − 0.4503 = 0.175 meV, twice. Check: 1.5256 + 2 × 0.1747 = 1.875 meV = 3V₀/8, the trace. The shifts depend on neither m nor L.
The octant, like the cube, is unchanged by every permutation of x, y and z, so W is unchanged by any relabelling of the three states and can single out only (1,1,1)/√3. The quarter-column respected only the x ↔ y permutation, which is why it split all three. This pair stays degenerate at every order, protected by the permutations (stated, not derived).
Energy denominators: the largest couplings out of the level, such as ⟨1,1,1|H′|1,1,2⟩ = V₀ × ½ × ½ × 4/(3π) = 0.531 meV, sit against gaps of 42.09 meV, a ratio of 0.013, so second order should move each shift by a few hundredths of a meV at most.
A numerical diagonalisation in a basis of 9³ sine states, computed rather than derived, gives shifts of 0.170 meV, exactly twice, and 1.482 meV: first order is high by 0.005 and 0.044 meV, as the second-order estimate predicted, and the degeneracy is exact.

AnswerShifts of 0.175 meV (two states, still degenerate) and 1.526 meV above 6E₁ = 84.2 meV, split by 1.351 meV; numerics give 0.170 meV twice and 1.482 meV.