University Quantum Mechanics II · Time-Independent Perturbation Theory · 6.4

Second-order energy, and why a ground state always goes down

A perturbation that averages to zero in a state, such as a uniform field acting on an oscillator or on an atom's ground state, has no first-order effect, so the change it makes starts at second order. Here we derive that term, read its sign before doing any integral, and test it where it is the whole answer.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

The second-order energy is what a level learns from being mixed with the others. Projecting the λ² equation onto ⟨n⁽⁰⁾| gives Eₙ⁽²⁾ = ⟨n⁽⁰⁾|H′|n⁽¹⁾⟩, and inserting the first-order state turns it into a sum over every other state m of |⟨m⁽⁰⁾|H′|n⁽⁰⁾⟩|² divided by the gap Eₙ⁽⁰⁾ − Eₘ⁽⁰⁾. No numerator is negative, so the ordering of the levels alone sets each sign: a state above pushes level n down, a state below pushes it up, and since the (m, n) term enters both levels' sums with opposite signs, a pair coupled mainly to each other is pushed apart.

A ground state has nothing below it, so its second-order shift is never positive, whatever the sign of H′. That is the small-λ shadow of an exact statement: E⁽⁰⁾ + E⁽¹⁾ is the expectation value of the full H in the unperturbed ground state, and no expectation value lies below the true ground energy. The formula stops working in three places.

It diverges when a state that H′ couples to the level shares its energy, the degenerate case of the next lesson. It needs the whole spectrum, and for hydrogen's ground state in a field the continuum supplies nearly a fifth of the sum. And it is only the leading term of a series, to be trusted while each coupling is small against its gap, unless the series ends, as for a charged oscillator in a uniform field, where completing the square shows second order is exact.

Simple definition: The second-order energy shift Eₙ⁽²⁾ is the part of a level's energy quadratic in the perturbation: the sum, over every other unperturbed state m, of |⟨m⁽⁰⁾|H′|n⁽⁰⁾⟩|² divided by the energy gap Eₙ⁽⁰⁾ − Eₘ⁽⁰⁾.
Example: A ⁴⁰Ca⁺ ion trapped at ω = 2π × 1.00 MHz in a stray field of 0.100 V m⁻¹: only n = 1 couples to the ground state, with |⟨1|x|0⟩| = 11.2 nm, and the one term (qF × 11.2 nm)² over the gap E₀⁽⁰⁾ − E₁⁽⁰⁾ = −ħω gives E₀⁽²⁾ = −4.90 × 10⁻²⁹ J, a drop of h × 73.9 kHz.

Second-order energy as a sum over statesEₙ⁽²⁾ = Σₘ |⟨m⁽⁰⁾|H′|n⁽⁰⁾⟩|²/(Eₙ⁽⁰⁾ − Eₘ⁽⁰⁾), m ≠ n

Derived by projecting the λ² equation onto ⟨n⁽⁰⁾| and inserting the first-order state; Hermiticity turns each numerator into a modulus squared.

Numerator in J², gap in J, so Eₙ⁽²⁾ in J; m runs over every other eigenstate of H₀, bound and continuum

A ground state is never raised at second orderE₀⁽²⁾ = −Σₘ |⟨m⁽⁰⁾|H′|0⁽⁰⁾⟩|²/(Eₘ⁽⁰⁾ − E₀⁽⁰⁾) ≤ 0, m ≠ 0

Derived. The sign is set by the ordering of the levels, not by H′: replacing H′ with −H′ leaves every term unchanged. Zero only if H′ couples |0⁽⁰⁾⟩ to nothing.

Every gap Eₘ⁽⁰⁾ − E₀⁽⁰⁾ is positive for a non-degenerate ground state; energies in J

Two levels coupled mainly to each otherE₂⁽²⁾ = −E₁⁽²⁾ = |V|²/Δ

Derived: one term enters both sums with opposite gaps. The pair's mean stays put and its splitting grows by 2|V|²/Δ, which is level repulsion.

V = ⟨2⁽⁰⁾|H′|1⁽⁰⁾⟩ and Δ = E₂⁽⁰⁾ − E₁⁽⁰⁾, positive, both in J; couplings to other states neglected

Second-order shifts sum to zeroΣₙ Eₙ⁽²⁾ = 0, over a finite non-degenerate basis

Derived. A check on any truncated calculation: whatever the low levels lose at second order, the high ones gain, and the top state of the basis never falls.

Because Σₙ Eₙ = Tr H₀ + λ Tr H′ exactly, with no λ² term; energies in J; can fail in an infinite space

First order never underestimates a ground energyE₀⁽⁰⁾ + E₀⁽¹⁾ = ⟨0⁽⁰⁾|H|0⁽⁰⁾⟩ ≥ E₀

Derived by expanding |0⁽⁰⁾⟩ in eigenstates of the full H. The sign rule is its small-λ shadow, and Unit 10 builds the variational method on it.

E₀ is the exact ground energy of H = H₀ + H′, in J; holds at any strength of H′

Charged oscillator in a uniform fieldEₙ⁽²⁾ = −q²F²/(2mω²), the same for every n

Derived from the terms m = n ± 1 with ladder operators, and exact: completing the square lowers the whole ladder by this amount, so every higher order vanishes.

H′ = −qFx; q in C, F in V m⁻¹, m in kg, ω in rad s⁻¹, result in J; E⁽¹⁾ = 0 by parity

Project the λ² equation, then insert the first-order state

Collecting λ² in (H₀ + λH′)|n⟩ = Eₙ|n⟩ gives (H₀ − Eₙ⁽⁰⁾)|n⁽²⁾⟩ = (Eₙ⁽¹⁾ − H′)|n⁽¹⁾⟩ + Eₙ⁽²⁾|n⁽⁰⁾⟩. Project onto ⟨n⁽⁰⁾|. The left side vanishes because ⟨n⁽⁰⁾|H₀ = Eₙ⁽⁰⁾⟨n⁽⁰⁾|, and the Eₙ⁽¹⁾ term vanishes because the first-order state was chosen orthogonal to |n⁽⁰⁾⟩, leaving Eₙ⁽²⁾ = ⟨n⁽⁰⁾|H′|n⁽¹⁾⟩. The previous lesson's |n⁽¹⁾⟩ = Σₘ |m⁽⁰⁾⟩⟨m⁽⁰⁾|H′|n⁽⁰⁾⟩/(Eₙ⁽⁰⁾ − Eₘ⁽⁰⁾), m ≠ n, turns this into a sum of products ⟨n⁽⁰⁾|H′|m⁽⁰⁾⟩⟨m⁽⁰⁾|H′|n⁽⁰⁾⟩, and since H′ is Hermitian the first factor is the complex conjugate of the second. So Eₙ⁽²⁾ = Σₘ |⟨m⁽⁰⁾|H′|n⁽⁰⁾⟩|²/(Eₙ⁽⁰⁾ − Eₘ⁽⁰⁾). Dimensions: J² over J is J. Scaling: multiply H′ by s and the shift grows as s², so reversing a field cannot reverse it. And note what was not needed: the second-order energy uses the state only to first order.

Read the sign from the gaps before computing anything

Every numerator |⟨m⁽⁰⁾|H′|n⁽⁰⁾⟩|² is zero or positive, so each term carries the sign of Eₙ⁽⁰⁾ − Eₘ⁽⁰⁾. A state above level n contributes a negative term and pushes it down; a state below contributes a positive term and pushes it up; and a near state pushes harder than a far one with the same coupling, in proportion to the inverse gap. A ground state has nothing below it, so E₀⁽²⁾ ≤ 0, with equality only when H′ couples |0⁽⁰⁾⟩ to no other state, in which case |0⁽⁰⁾⟩ is already an exact eigenstate of H. The rule does not care about the sign of H′: a bump and a dip of the same shape give identical second-order shifts, because H′ enters squared. The top state of a finite spectrum obeys the mirror rule and never goes down at second order. An excited level can go either way, in a contest of couplings and gaps: with equal couplings, a close neighbour below beats a distant one above, and the level rises.

A pair coupled to each other is pushed apart

The term linking states m and n appears twice: in Eₙ⁽²⁾ as |V|²/(Eₙ⁽⁰⁾ − Eₘ⁽⁰⁾) and in Eₘ⁽²⁾ as |V|²/(Eₘ⁽⁰⁾ − Eₙ⁽⁰⁾), equal and opposite. When two levels are coupled far more strongly to each other than to anything else, the lower one falls by |V|²/Δ and the upper one rises by the same amount: the levels repel and their mean stays put. With Δ = 50.0 meV and |V| = 5.00 meV each moves 0.500 meV, so the splitting grows to 51.0 meV, against 50.99 meV from the exact 2 × 2 root √(Δ² + 4|V|²). Summed over all the levels of a finite basis every pair cancels, so Σₙ Eₙ⁽²⁾ = 0, which also follows exactly because the trace of H₀ + λH′ is linear in λ. Now shrink Δ towards zero at fixed V: |V|²/Δ grows without bound, which is the formula announcing that the pair must be diagonalised together, the degenerate case of the next lesson.

The exact statement behind the sign

At λ = 1 the first-order estimate is E₀⁽⁰⁾ + E₀⁽¹⁾ = ⟨0⁽⁰⁾|H₀|0⁽⁰⁾⟩ + ⟨0⁽⁰⁾|H′|0⁽⁰⁾⟩ = ⟨0⁽⁰⁾|H|0⁽⁰⁾⟩. Expand |0⁽⁰⁾⟩ = Σₖ cₖ|k⟩ in the exact eigenstates of H, with Σₖ |cₖ|² = 1; then ⟨0⁽⁰⁾|H|0⁽⁰⁾⟩ = Σₖ |cₖ|²Eₖ ≥ E₀ Σₖ |cₖ|² = E₀. This is exact, not perturbative: first-order theory never underestimates a ground energy, however strong H′ is. Run the same argument at every λ and the exact ground energy E₀(λ) is the lowest of a family of straight lines ⟨ψ|H₀ + λH′|ψ⟩, so it is concave in λ and lies below its tangent at λ = 0, E₀⁽⁰⁾ + λE₀⁽¹⁾. A concave function has non-positive curvature, and the curvature at λ = 0 is 2E₀⁽²⁾: the sign rule again. What carries no guarantee is the second-order sum E⁽⁰⁾ + E⁽¹⁾ + E⁽²⁾. For levels 0 and Δ coupled by V = 0.300Δ it gives −0.0900Δ, below the exact −0.0831Δ. Unit 10 turns the bound into the variational method.

The sum runs over everything, and a measurement tests it

The sum needs every eigenstate of H₀, and for an atom that includes the unbound ones, so it becomes a sum plus an integral over the continuum. For hydrogen's 1s level in a field F along z, H′ = eFz reaches only p states with m = 0, and writing E⁽²⁾ = −½α_d F² defines the polarisability α_d. The bound np terms, from quoted closed-form radial integrals, give α_d = 3.66(4πε₀)a₀³; the exact total, also quoted, is 4.50(4πε₀)a₀³. The continuum supplies 19%, which is why Unit 9 bounds the sum by closure instead. No power of the fine-structure constant enters; the small number is eFa₀ against the 10.2 eV gap to n = 2. Every term of a ground state's sum has one sign, so any truncated sum underestimates the size of the shift, and a measurement can test that. Sodium's 3s level couples almost entirely to 3p, 2.104 eV up. With the tabulated oscillator strength f = 2mₑω|⟨3p|z|3s⟩|²/ħ = 0.961, that one term, 2e²|⟨3p|z|3s⟩|²/ħω = e²f/(mₑω²), gives α_d = 2.65 × 10⁻³⁹ C m² V⁻¹, just below the 2.68 × 10⁻³⁹ C m² V⁻¹ measured by atom interferometry, as the sign rule demands.

A charged oscillator, where second order is the whole answer

Add H′ = −qFx to an oscillator. Quote x = x₀(a + a†)/√2 with x₀ = √(ħ/mω): ⟨n|x|n⟩ = 0 by parity, so E⁽¹⁾ = 0, and x links n only to n ± 1, with ⟨n+1|x|n⟩ = x₀√((n+1)/2) and ⟨n−1|x|n⟩ = x₀√(n/2). The level above contributes −(n+1)(qFx₀)²/(2ħω) and the level below +n(qFx₀)²/(2ħω); at high n they nearly cancel, and for every n they leave −(qFx₀)²/(2ħω) = −q²F²/(2mω²). Completing the square, ½mω²x² − qFx = ½mω²(x − d)² − q²F²/(2mω²) with d = qF/(mω²), so the exact ladder is the old one lowered by exactly that, and every higher order vanishes. ħ has cancelled: the shift is the classical energy −½qFd of a spring settling at its new equilibrium. For a ⁴⁰Ca⁺ ion at ω = 2π × 1.00 MHz in 1.00 V m⁻¹, x₀ = 15.9 nm, d = 61.2 nm and the shift is 7.39 ħω, exact although the ground state's coupling to n = 1 is 2.72 times its gap: a failed energy-denominator test warns that the series may be poor, not that it must be.

Change one variable at a time

Make the relationship visible.

Interactive model

Field as displacement d/x₀ = qF/(mω²x₀)0.80

Level n examined0

Start at n = 0 with d = 0.80 x₀: only the level above acts, and the ground state drops 0.32 ħω. Step n up to 3: the bars grow to −1.28 and +0.96 ħω, their sum stays −0.32 ħω, and the coupling-to-gap ratio reaches 1.13, yet the shift is still exact. Then slide d: the whole ladder moves as one.

PUSH FROM LEVEL n + 1-0.320 ħω

PUSH FROM LEVEL n − 10.000 ħω

NET E⁽²⁾, EXACT, ANY n-0.320 ħω

COUPLING/GAP TO n + 10.57

Live interpretationPUSH FROM LEVEL n + 1: −0.320 ħω. PUSH FROM LEVEL n − 1: 0.000 ħω. NET E⁽²⁾, EXACT, ANY n: −0.320 ħω. COUPLING/GAP TO n + 1: 0.57

Catch the common trap

Explain before calculating.

Misconception check

“A ground state always goes down” is a statement about E₀⁽²⁾, not about the total shift. First order can raise a ground level, and usually dominates when it is non-zero: a small positive bump in the middle of a well raises the ground energy by ⟨0⁽⁰⁾|H′|0⁽⁰⁾⟩, and adding a constant c raises every level by c with no second-order term at all. What is always true is weaker and exact: the true ground energy lies at or below the first-order estimate E₀⁽⁰⁾ + E₀⁽¹⁾. The companion error is to give E⁽²⁾ the sign of H′, expecting a repulsive perturbation to push a level up at second order. H′ enters every term squared, so H′ and −H′ give the same E⁽²⁾; its sign comes from where the other levels sit, which is why the ground state and the top state of a finite system move in opposite directions under the same perturbation.

A level n has ⟨n⁽⁰⁾|H′|n⁽⁰⁾⟩ = 0 and couples to only two other states: one 10.0 meV below it and one 30.0 meV above it, each with |⟨m⁽⁰⁾|H′|n⁽⁰⁾⟩| = 3.00 meV. What is Eₙ⁽²⁾?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyAn impurity's ground level couples through a perturbation to three excited levels lying 2.00, 3.50 and 5.00 eV above it, with |⟨m⁽⁰⁾|H′|0⁽⁰⁾⟩| = 0.100, 0.200 and 0.150 eV respectively, and ⟨0⁽⁰⁾|H′|0⁽⁰⁾⟩ = 0. Find E₀⁽²⁾, say which level contributes most, and judge whether second order can be trusted.

Every other level lies above, so every gap E₀⁽⁰⁾ − Eₘ⁽⁰⁾ is negative and every term lowers the level: E₀⁽²⁾ = −Σₘ |⟨m⁽⁰⁾|H′|0⁽⁰⁾⟩|²/(Eₘ⁽⁰⁾ − E₀⁽⁰⁾).
The three terms: (0.100 eV)²/(2.00 eV) = 5.00 meV, (0.200 eV)²/(3.50 eV) = 11.4 meV and (0.150 eV)²/(5.00 eV) = 4.50 meV. Dimensions: eV² over eV is eV.
With the sign: E₀⁽²⁾ = −(5.00 + 11.43 + 4.50) meV = −20.9 meV.
The level 3.50 eV up gives more than half of it: its coupling is twice that of the nearest level, and squaring the coupling (×4) outweighs its 1.75 times larger gap.
Trust: the coupling-to-gap ratios are 0.050, 0.057 and 0.030, all far below 1, so higher orders should change the answer by only a few per cent. Independently of that, the exact ground energy lies at or below E₀⁽⁰⁾ + E₀⁽¹⁾ = E₀⁽⁰⁾, consistent with a negative shift.

AnswerE₀⁽²⁾ = −20.9 meV, of which 11.4 meV comes from the level 3.50 eV up; every coupling is at most 6% of its gap, so second order is reliable.

MediumAn electron in an InSb quantum dot is modelled as an oscillator along x with ħω = 2.00 meV and the tabulated effective mass 0.0140 mₑ. A field F = 3.50 × 10⁴ V m⁻¹ is applied along x, so H′ = eFx. For the level n = 2, find the second-order contribution from each neighbouring level, the net shift, and compare with the exact result.

Scales: m = 0.0140 × 9.109 × 10⁻³¹ kg = 1.275 × 10⁻³² kg and ω = 2.00 meV/ħ = 3.039 × 10¹² s⁻¹, so x₀ = √(ħ/mω) = 52.2 nm and the coupling scale is eFx₀ = 1.826 meV.
E⁽¹⁾ = eF⟨2|x|2⟩ = 0 by parity. From n = 2, x reaches only n = 3 and n = 1: |⟨3|H′|2⟩| = eFx₀√(3/2) = 2.24 meV and |⟨1|H′|2⟩| = eFx₀√(2/2) = 1.83 meV. The electron's negative charge drops out once squared.
Both terms are multiples of (eFx₀)²/ħω = (1.826 meV)²/(2.00 meV) = 1.667 meV. Level above, gap −ħω: −(3/2)(1.667 meV) = −2.50 meV. Level below, gap +ħω: +(2/2)(1.667 meV) = +1.67 meV. Net: E₂⁽²⁾ = −0.833 meV.
Exact: e²F²/(2mω²) = (5.608 × 10⁻¹⁵ N)²/(2 × 1.275 × 10⁻³² kg × (3.039 × 10¹² s⁻¹)²) = 1.335 × 10⁻²² J = 0.833 meV, a rigid drop shared by every level, while the well's minimum moves d = eF/(mω²) = 47.6 nm against the field.
Trust: the coupling to the level above is 2.24/2.00 = 1.12 times its gap, so the energy-denominator condition fails, yet the answer is exact, because x links only neighbours and the field merely translates the well. For a general H′ at this ratio, second order would be a guess.

Answer−2.50 meV from n = 3 and +1.67 meV from n = 1, net E₂⁽²⁾ = −0.833 meV, equal to the exact shift e²F²/(2mω²) that every level shares.

HardA polar molecule is modelled as a rigid rotor with E_J = BJ(J + 1), B = h × 1.114 GHz and dipole moment μ = 0.570 D (1 D = 3.336 × 10⁻³⁰ C m), close to the measured values for ⁴⁰K⁸⁷Rb. A field F = 1.00 × 10⁵ V m⁻¹ along z adds H′ = −μF cos θ. Find the second-order shifts of the ground state and of the J = 1, m = 0 level in frequency units, and compare with a numerical diagonalisation.

Quoted from the spherical harmonics of Unit 1: ⟨J+1, m|cos θ|J, m⟩ = √[((J+1)² − m²)/((2J+1)(2J+3))]. cos θ commutes with L_z, so m is conserved and the m = ±1 partners of J = 1 never couple to m = 0; non-degenerate theory applies inside the m = 0 block, where ⟨1, 0|cos θ|0, 0⟩ = 1/√3 and ⟨2, 0|cos θ|1, 0⟩ = 2/√15.
Scale: μF/h = 0.570 × 3.336 × 10⁻³⁰ C m × 1.00 × 10⁵ V m⁻¹/(6.626 × 10⁻³⁴ J s) = 287 MHz, against 2B/h = 2228 MHz. The largest coupling-to-gap ratio is (287/√3)/2228 = 0.074, and odd orders vanish by parity, so the error should be of order 0.074², about 0.5%.
Ground state: only J = 1 lies above and couples, so E₀⁽²⁾ = −(μF)²(1/3)/(2B) = −(μF)²/(6B), and E₀⁽²⁾/h = −(287 MHz)²/(6 × 1114 MHz) = −12.3 MHz. It goes down, as a ground state must.
J = 1, m = 0: the ground state 2B below pushes it up by (μF)²(1/3)/(2B) = h × 12.3 MHz, and J = 2, 4B above, pushes it down by (μF)²(4/15)/(4B) = h × 4.93 MHz. Net: E⁽²⁾/h = +7.39 MHz. The (0, 1) term enters both levels with opposite signs: level repulsion.
Diagonalising H numerically in the m = 0 basis J = 0 to 19 gives shifts of −h × 12.27 MHz and +h × 7.34 MHz. The differences of h × 0.05 MHz are the fourth-order terms, 0.4% of the ground-state shift, as step 2 predicted.

AnswerE₀⁽²⁾/h = −12.3 MHz (numerical −12.27 MHz) and, for J = 1, m = 0, E⁽²⁾/h = +7.39 MHz (numerical +7.34 MHz): the ground state falls while the excited level is pushed up.