University Quantum Mechanics II · Time-Independent Perturbation Theory · 6.9
Convergence, and when perturbation theory fails
Every answer in this unit has been a truncated series, and this lesson prices the truncation. On the x⁴ oscillator, whose coefficients are known to any order, you will see one more term help and then hurt, learn why Dyson's argument forbids convergence, and learn where to stop and what error bar to quote.
Build the model
Connect the measurement to the mechanism.
Perturbation theory hands you a power series in the coupling and quietly invites you to believe that summing all of it gives the answer. The anharmonic oscillator H = p²/(2m) + ½mω²x² + λx⁴ is where that belief can be tested completely. With g = λħ/(m²ω³), ladder operators give E₀ = ħω[½ + (3/4)g − (21/8)g² + …], and the coefficients are known to any order: from first order on they alternate in sign and grow like 3ⁿΓ(n + ½) (quoted from Bender and Wu), so at every positive g the terms shrink only until n ≈ 1/(3g) and then grow without limit.
Dyson's argument says this had to happen. At negative g the potential falls to −∞ beyond a barrier of height ħω/(16|g|), no bound state exists, and so E₀(g) cannot be analytic at g = 0: the radius of convergence is zero, a physical argument whose conclusion Bender and Wu proved (quoted). What survives is an asymptotic series.
At fixed order it improves as g → 0; at fixed g it has a best order, near its smallest term, and an accuracy floor of about e(−1/3g) ħω, the same exponential that sets the tunnelling decay at negative g, which no power series can represent. At g = 0.01 that floor is 4 × 10⁻¹⁵ ħω and three terms already match numerical diagonalisation to 4 parts in 10⁵; at g = 0.2 the smallest term is the second-order one, and nothing beyond first order helps.
- Simple definition
- A perturbation series is asymptotic when, for every fixed number of terms N, its error vanishes faster than gN as g → 0; it can still diverge at every positive g, as the x⁴ oscillator's does, and is then best stopped near its smallest term.
- Example
- At g = 0.05 the x⁴ oscillator's terms shrink to 1.0 × 10⁻³ ħω at sixth order and then grow; S₅ = 0.53315 ħω and S₆ = 0.53215 ħω bracket the diagonalised 0.53264 ħω, and no later partial sum comes closer.
The quartic energy at the zero-point length, in units of the level spacing. Every statement about convergence and accuracy is a statement about g, not about λ alone.
For H′ = λx⁴ added to the oscillator: λ in J m⁻⁴, m in kg, ω in s⁻¹, ħ in J s, x₀ in m; g is a pure number
c₁ is positive because x⁴ ≥ 0, and c₂ negative, as every ground-state second-order shift must be. From here on the signs alternate and the sizes grow.
E₀ in J; write E₀ = ħω Σ cₙgⁿ. c₁ and c₂ derived with ladder operators; c₃ quoted
Successive terms have ratio cₙ₊₁g/cₙ ≈ −3ng, which passes 1 in size at n ≈ 1/(3g) for every positive g: the ratio test gives a radius of convergence of zero.
Quoted from Bender and Wu; cₙ is dimensionless; the exact c₃₀ is 0.95 of this, closing on it as n grows
A finite barrier over a bottomless fall holds no bound state, so E₀(g) has no real value just below g = 0 and cannot be analytic there: Dyson's argument.
Derived for negative g, where V = ½mω²x² − |λ|x⁴; Vₘₐₓ in J, x₀ in m; beyond the barrier V falls without limit
Stop the sum near n*: the error is then below this term. At g = 0.03 it predicts 1.6 × 10⁻⁵ ħω near n = 11; the actual smallest term is 1.4 × 10⁻⁵ ħω, at n = 11.
Derived from the quoted large-order law with Γ(n + ½) ≈ √(2π) nⁿ e⁻ⁿ; √(12/π²) = 1.10; energies in J
Every derivative of it vanishes at g = 0, so no coefficient cₙ contains it: invisible at every order, yet it carries the same exponential as the accuracy floor.
Γ in J, lifetime ħ/Γ in s; quoted at leading order: exponent from Unit 11's WKB barrier integral, prefactor from Bender and Wu
Three terms from ladder operators, checked at g = 0.01
Quote x = x₀(a + a†)/√2 from Quantum Mechanics I. Then (a + a†)⁴|0⟩ = 3|0⟩ + 6√2|2⟩ + 2√6|4⟩, so H′ = λx⁴ has ⟨0|H′|0⟩ = (3/4)għω, ⟨2|H′|0⟩ = (3√2/2)għω and ⟨4|H′|0⟩ = (√6/2)għω, with g = λx₀⁴/(ħω). First order is the diagonal element, +(3/4)għω. Second order divides the squared off-diagonal elements by the gaps 2ħω and 4ħω: −(4.5/2 + 1.5/4)g²ħω = −(21/8)g²ħω. The third coefficient, 333/16, is quoted. Checks: [λħ/(m²ω³)] = (J m⁻⁴)(J s)/(kg² s⁻³) = 1; g → 0 returns ħω/2; the first-order shift is positive because x⁴ ≥ 0, and the second is negative, as every ground-state second-order shift must be. At g = 0.01 the three terms give 0.5072375 ħω, while diagonalising H in the lowest 70 even oscillator states gives 0.5072562 ħω: an error of 1.9 × 10⁻⁵ ħω, or 3.7 × 10⁻⁵ of the energy, just under the next term, 2.1 × 10⁻⁵ ħω.
At g = 0.2, every order after the first makes it worse
Raise the coupling to g = 0.2 and compare the partial sums with the diagonalised 0.60241 ħω. S₀ = 0.5000, S₁ = 0.6500, S₂ = 0.5450, S₃ = 0.7115 and S₄ = 0.3254 ħω have errors of −0.102, +0.048, −0.057, +0.109 and −0.277 ħω. First order helped; each later order made the answer worse, and the swings grow. The energy-denominator test did not warn you: |⟨2|H′|0⟩|/(2ħω) = 0.21 and |⟨4|H′|0⟩|/(4ħω) = 0.06 both look comfortably small, because that test only judges the first-order state. The warning is in the ratio of successive coefficients, cₙ₊₁/cₙ = −3.5, −7.9, −11.6, −14.8, −17.9, which grows in size by 3 to 4.5 per order. So the ratio of successive terms grows roughly as 3ng, and at g = 0.2 it is already 1.59 between second and third order. Once that ratio passes 1, each extra order adds a bigger swing than the one before.
Dyson's argument: why the radius of convergence is zero
A power series that converges at some positive g₀ converges for every complex g with |g| smaller than g₀, and its sum is analytic there. If the x⁴ series did so, E₀(g) would be analytic in that disc and, the coefficients being real, real on the negative axis inside it, for instance at −g₀/2. Now look at negative g. V = ½mω²x² − |λ|x⁴ rises to ħω/(16|g|) at x = ±x₀/(2√|g|) and then falls to −∞, so a particle in the well tunnels out: there is no true bound state, only a decaying resonance, defined by outgoing waves beyond the barrier, whose energy carries an imaginary part −iΓ/2. That is the contradiction: the continued E₀ is complex at −g₀/2, so the series cannot converge at g₀, and since g₀ was arbitrary, the radius of convergence is zero. This is Dyson's physical argument, made here from the shape of the potential alone; it is not a proof. The rigorous result is quoted from Bender and Wu, whose large-order law |cₙ| ≈ √(6/π³) 3ⁿ Γ(n + ½) gives, by the ratio test, a radius of lim 1/(3n) = 0. At n = 30 the exact coefficient is 0.95 of that formula, and the gap keeps closing as n grows.
Asymptotic, not convergent: stop at the smallest term
Divergent does not mean useless. At each fixed order N the error |E₀ − SN| shrinks like g(N+1) as g → 0, which is what makes the series asymptotic. The price is that at fixed g there is a best N. The terms shrink while 3ng stays below 1, so the smallest sits near n* ≈ 1/(3g). Putting Stirling's Γ(n + ½) ≈ √(2π) nⁿ e⁻ⁿ into the quoted law at that n gives a smallest term of √(12/π²) e(−1/3g) ħω ≈ 1.10 e(−1/3g) ħω, derived here from the quoted law. Stop there. For this oscillator the exact energy lies between any two consecutive partial sums, a property proved for this series (quoted), so the smallest term is an honest error bar. At g = 0.05 the estimate gives n* ≈ 6.7 and a floor of 1.4 × 10⁻³ ħω; the actual smallest term is at n = 6, 1.0 × 10⁻³ ħω, and S₅ = 0.53315 ħω and S₆ = 0.53215 ħω bracket the exact 0.53264 ħω. At g = 0.02 the floor falls to 6 × 10⁻⁸ ħω, near n = 17.
What no power series can see
The accuracy floor measures what the series is blind to. At negative g the state leaks through the barrier with a width Γ ≈ √(8/π|g|) e(−1/3|g|) ħω, quoted at leading order ahead of Unit 11's WKB integral. Every derivative of e(−1/3|g|) vanishes as g → 0, so it adds zero to every coefficient: no truncation of the series contains it, and two functions differing by that much share the same series. That is why the partial sums at positive g stall at a floor carrying the same exponential. The information is not wholly lost, since the factorial growth of the cₙ is the tunnelling's fingerprint, and Borel summation of the whole series does return E₀ for this oscillator (quoted). At g = −0.01 the barrier stands 6.25 ħω high and e(−1/3|g|) = 3 × 10⁻¹⁵, so the state is bound for every practical purpose; at g = −0.2 it is 0.31 ħω, below the zero-point energy. Strong coupling is out of reach too: rescaling x forces E₀ → C g¹⁄³ ħω as g → ∞, with C = 0.668 from a numerical solution (quoted), a growth no polynomial in g can follow.
The same argument in QED, and a measurement it passes
Dyson first made this argument in 1952 for quantum electrodynamics, and there its conclusion is quoted, not derived: if e² changed sign, like charges would attract, and the vacuum could lower its energy without limit by creating a cluster of electrons and a separate cluster of positrons, so no series in the fine-structure constant α can converge. Yet the series for the electron's anomalous magnetic moment is among the most successful in physics. Its terms, quoted from QED calculations, are +1.161 × 10⁻³, −1.77 × 10⁻⁶, +1.48 × 10⁻⁸, −5.6 × 10⁻¹¹ and a few times 10⁻¹³, each 100 to 1000 times smaller than the last, because each order brings one more factor of α/π = 0.00232 and the coefficients stay of order one. Five orders, with small pieces from muons, hadrons and the weak interaction, give aₑ = 0.00115965218, and the Penning-trap measurement agrees in every digit shown. The factorial growth is estimated to win only near order 1/α ≈ 137. Usefulness depends on where the smallest term lies: n ≈ 33 for the oscillator at g = 0.01, but n = 2 at g = 0.2.
Change one variable at a time
Make the relationship visible.
Set g = 0.20 and step N from 0 to 4: the ring drops once, then climbs with the dots, as S₁ to S₄ swing 0.6500, 0.5450, 0.7115, 0.3254 ħω around the exact 0.6024 ħω. Drop g to 0.05 and the dip moves out to n = 6, where the error bottoms out near 5 × 10⁻⁴ ħω.
PARTIAL SUM SN0.5487 ħω
EXACT E₀ (NUMERICAL)0.5591 ħω
SMALLEST TERM0.0208 ħω
TURNING ORDER 1/(3g)3.3
Live interpretationPARTIAL SUM SN: 0.5487 ħω. EXACT E₀ (NUMERICAL): 0.5591 ħω. SMALLEST TERM: 0.0208 ħω. TURNING ORDER 1/(3g): 3.3
Catch the common trap
Explain before calculating.
At g = 0.1 the x⁴ oscillator's ground-state series gives, in units of ħω, the terms ½, +0.0750, −0.0263, +0.0208, −0.0241 and +0.0358. Using only these terms, which is the most defensible statement of E₀?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyAn electron in a GaAs nanowire dot (effective mass 0.067 mₑ) sits in a potential with ħω = 4.0 meV plus a quartic term λx⁴ that alone would reach 1.0 meV at x = 30 nm. Find x₀ and g, show that g is dimensionless, estimate the ground-state energy to second order, and judge whether the series can be trusted.
- Oscillator length: m = 0.067 × 9.109 × 10⁻³¹ kg = 6.103 × 10⁻³² kg and ω = (4.0 meV)/ħ = 6.077 × 10¹² s⁻¹, so x₀ = √(ħ/mω) = √(1.0546 × 10⁻³⁴ / (6.103 × 10⁻³² × 6.077 × 10¹²)) m = 16.86 nm.
- Coupling: λ = 1.0 meV/(30 nm)⁴ = 1.978 × 10⁸ J m⁻⁴, and g = λx₀⁴/(ħω) = (1.0/4.0) × (16.86/30)⁴ = 0.02495. Dimensions: [λħ/(m²ω³)] = (J m⁻⁴)(J s)/(kg² s⁻³) = J² s⁴ m⁻⁴ kg⁻², and since J = kg m² s⁻² this is 1.
- Energy: E₀ ≈ ħω[½ + 0.75g − 2.625g²] = 4.0 meV × [0.5 + 0.018713 − 0.001634] = 2.0683 meV. The quartic term raises the level by 74.9 μeV at first order and lowers it by 6.5 μeV at second.
- Trust: the next term is (333/16)g³ħω = 1.3 μeV, which sets the error. The terms keep shrinking until n ≈ 1/(3g) = 13, where the floor 1.10 e(−1/3g) ħω is about 7 neV, far below anything measurable. Limit check: λ → 0 gives ħω/2 = 2.000 meV.
Answerx₀ = 16.9 nm, g = 0.025 and E₀ ≈ 2.068 meV, uncertain by about the next term, 1.3 μeV; numerical diagonalisation gives 2.0693 meV. Here the series is excellent.
MediumReverse the sign of the quartic term: take g = −0.05, so V = ½mω²x² − |λ|x⁴. In units of x₀ and ħω, find the height and position of the barrier, the classical turning points at the unperturbed energy ħω/2, and the coupling at which the barrier sinks to that energy. Then say what the state is, and why this stops the series converging at any positive g.
- Barrier: with y = x/x₀ and energies in ħω, V = ½y² − |g|y⁴. Setting dV/dy = y − 4|g|y³ = 0 gives y² = 1/(4|g|) = 5, so y = ±2.24, where V = 1/(16|g|) = 1.25 ħω. Limit check: as g → 0 the barrier moves out to infinite height and distance, and the plain oscillator returns.
- Turning points at E = ½: |g|y⁴ − ½y² + ½ = 0 is a quadratic in y², so y² = [½ ∓ √(¼ − 2|g|)]/(2|g|) = (0.5 ∓ 0.3873)/0.1, giving y² = 1.127 or 8.873 and y = ±1.06 or ±2.98. The particle oscillates between y = −1.06 and +1.06, meets a forbidden region from 1.06 to 2.98, and beyond 2.98 V falls without limit.
- Critical coupling: the barrier top sinks to ½ when 1/(16|g|) = ½, at |g| = 1/8 = 0.125, the same coupling at which ¼ − 2|g| vanishes and the two turning points merge. Beyond it nothing holds even the zero-point motion.
- The state: at g = −0.05 it is a resonance with complex energy Eᵣ − iΓ/2. The quoted leading-order width Γ ≈ √(8/π|g|) e(−1/3|g|) ħω = 7.14 × 1.27 × 10⁻³ ħω = 9.1 × 10⁻³ ħω gives a lifetime ħ/Γ ≈ 110/ω, about 18 oscillation periods.
- The consequence: at every negative g, however small, E₀ has an imaginary part and no real stationary value, so E₀(g) cannot be analytic at g = 0. A series converging at some positive g would also converge, to a real sum, at every negative g of smaller size, so it converges at no positive g at all.
AnswerBarrier 1.25 ħω at x = ±2.24 x₀; turning points ±1.06 x₀ and ±2.98 x₀; the barrier falls to ħω/2 at |g| = 0.125. The state is a resonance with Γ ≈ 9 × 10⁻³ ħω at leading order (quoted), which no real power series in g can represent.
HardA student needs the x⁴ oscillator's ground-state energy at g = 0.04 to 1 part in 10⁵. The series through sixth order sums to 0.5266189 ħω, and the next coefficients are c₇ = 1 329 734, c₈ = −31 448 215 and c₉ = 833 541 603. Decide whether any partial sum can deliver, and explain the limit you find.
- Where the terms turn: the quoted large-order law makes the term ratio about 3ng, which reaches 1 at n* ≈ 1/(3g) = 8.3. The given coefficients agree: |c₈/c₇|g = 23.65 × 0.04 = 0.95 and |c₉/c₈|g = 26.5 × 0.04 = 1.06, so the terms shrink through n = 8 and grow after.
- The terms: c₇g⁷ = 1 329 734 × 1.638 × 10⁻¹⁰ = +2.179 × 10⁻⁴ ħω, c₈g⁸ = −31 448 215 × 6.554 × 10⁻¹² = −2.061 × 10⁻⁴ ħω, and c₉g⁹ = +2.185 × 10⁻⁴ ħω. The smallest is the eighth-order term.
- The bracket: S₇ = 0.5266189 + 0.0002179 = 0.5268368 ħω and S₈ = 0.5268368 − 0.0002061 = 0.5266307 ħω. The exact energy lies between them, so the series gives E₀ = 0.52673 ± 0.00010 ħω, about 2 parts in 10⁴, and no later order narrows it.
- The requirement: 1 part in 10⁵ of 0.527 ħω is 5.3 × 10⁻⁶ ħω, twenty times finer than the bracket's half-width of 1.03 × 10⁻⁴ ħω. No partial sum can meet it.
- The limit: the floor estimate 1.10 e(−1/3g) = 2.65 × 10⁻⁴ ħω sits beside the actual smallest term, 2.06 × 10⁻⁴ ħω, and both carry e(−1/3g) = 2.4 × 10⁻⁴, the exponential of the tunnelling width at g = −0.04 that no partial sum contains. Diagonalising H in 70 even oscillator states gives 0.5267340 ħω, inside the bracket.
AnswerNo partial sum can. The terms bottom out at n = 8 (2.06 × 10⁻⁴ ħω), so the series brackets E₀ between 0.52663 and 0.52684 ħω, twenty times too wide; diagonalisation gives 0.526734 ħω.