University Quantum Mechanics II · Time-Independent Perturbation Theory · 6.3

The first-order state and the energy-denominator condition

The first-order energy says how far a level moves; the first-order state says what the level turns into, and whether any of the calculation can be believed. Here we build that state, read it level by level as a coupling divided by a gap, and test the verdict on an oscillator we can also solve exactly.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

The first-order state answers a question the first-order energy cannot: what does the perturbation do to the state itself? Each other unperturbed state is mixed in with an amplitude equal to how strongly H′ couples it to the level followed, divided by how far away it sits in energy, so |n⁽¹⁾⟩ = Σ (m ≠ n) |m⁽⁰⁾⟩⟨m⁽⁰⁾|H′|n⁽⁰⁾⟩/(Eₙ⁽⁰⁾ − Eₘ⁽⁰⁾). Two consequences drive everything else.

Nearby levels dominate, even when their couplings are weak, because their denominators are small. And the expansion is only as good as the largest of these ratios, so the working test is |⟨m⁽⁰⁾|H′|n⁽⁰⁾⟩| ≪ |Eₙ⁽⁰⁾ − Eₘ⁽⁰⁾| for every m, not a small λ, which is a label set to 1 at the end. A charge in an oscillator with a uniform field shows the whole argument in one term: −qFx reaches only |1⟩, the ratio is d/(√2 x₀), the classical shift d of the equilibrium measured against the zero-point width x₀, and the first-order state is exactly the first Taylor term of the displaced Gaussian ψ₀(x − d).

The test fails when a partner level is degenerate or nearly so, where no field is weak enough, and when d grows towards x₀, where the state has moved by its own width and a single correction can no longer describe it.

Simple definition: The first-order state is the part of the corrected eigenstate linear in H′: every other unperturbed state mixed in with amplitude ⟨m⁽⁰⁾|H′|n⁽⁰⁾⟩/(Eₙ⁽⁰⁾ − Eₘ⁽⁰⁾), trustworthy only while each amplitude is much smaller than 1.
Example: A charge e in an oscillator with ħω = 5.0 meV and x₀ = 15.1 nm, in a field of 1.0 × 10⁵ V m⁻¹: the coupling to |1⟩ is 1.07 meV, so |0⟩ gains 0.213|1⟩, and the exact probability of staying in |0⟩ is 0.956.

The first-order state|n⁽¹⁾⟩ = Σ (m ≠ n) |m⁽⁰⁾⟩ ⟨m⁽⁰⁾|H′|n⁽⁰⁾⟩/(Eₙ⁽⁰⁾ − Eₘ⁽⁰⁾)

Derived: the λ¹ components found by projection, reassembled with the completeness relation. Each level enters with weight coupling ÷ gap, and a zero numerator removes it.

Matrix elements of H′ and energies in J, so every coefficient is dimensionless; no |n⁽⁰⁾⟩ term, by the orthogonality choice

The energy-denominator condition|⟨m⁽⁰⁾|H′|n⁽⁰⁾⟩| ≪ |Eₙ⁽⁰⁾ − Eₘ⁽⁰⁾| for every m ≠ n

Keeps each admixture small, so the state is still mostly |n⁽⁰⁾⟩. The nearest coupled level usually fails first; at a ratio of 0.5 a two-level amplitude is already 21% too large.

Both sides in J; unchanged by moving the zero of energy or rescaling λ; a degenerate partner fails it at any strength

First-order change of an observableδ⟨A⟩ = 2 Re Σ (m ≠ n) ⟨n⁽⁰⁾|A|m⁽⁰⁾⟩⟨m⁽⁰⁾|H′|n⁽⁰⁾⟩/(Eₙ⁽⁰⁾ − Eₘ⁽⁰⁾)

Derived from ⟨n|A|n⟩ with |n⟩ = |n⁽⁰⁾⟩ + |n⁽¹⁾⟩. Only partners that both H′ and A connect to n contribute: an induced dipole, a bond's stretch.

In the units of A; the norm changes only at order λ², so it does not enter at this order

Charged oscillator in a field: the one amplitude|0⁽¹⁾⟩ = (qFx₀/(√2 ħω))|1⟩ = (d/(√2 x₀))|1⟩

Derived from x = x₀(a + a†)/√2: −qFx reaches only |1⟩, across the gap ħω. The criterion becomes d ≪ x₀, the equilibrium's classical shift small beside the zero-point width.

q in C, F in V m⁻¹, ħω in J; d = qF/(mω²) and x₀ = √(ħ/(mω)), both in m

Exact ground state in the fieldψ(x) = ψ₀(x − d) = ψ₀(x) + (d/(√2 x₀))ψ₁(x) + O(d²)

Derived by completing the square in ½mω²x² − qFx. The Taylor step uses ψ₀′ = −(x/x₀²)ψ₀ and xψ₀ = (x₀/√2)ψ₁, and returns the first-order state term for term.

ψ₀ and ψ₁ are the unperturbed oscillator states, in m^(−1/2); d = qF/(mω²) in m

Probability of staying in |0⟩|⟨0⁽⁰⁾|ψ⟩|² = e^{(−d²/(2x₀²))}

The normalised first-order state gives 1/(1 + d²/(2x₀²)) instead. They agree to order d² and part at d⁴: 0.956 against 0.957 at d = 0.30x₀, 0.368 against 0.500 at d = √2 x₀.

Dimensionless; d and x₀ in m; exact, from a Gaussian overlap integral

Reassemble the components into one state

The λ¹ rung has already been projected: onto ⟨m⁽⁰⁾| with m ≠ n it gave ⟨m⁽⁰⁾|n⁽¹⁾⟩ = ⟨m⁽⁰⁾|H′|n⁽⁰⁾⟩/(Eₙ⁽⁰⁾ − Eₘ⁽⁰⁾), and the component along |n⁽⁰⁾⟩ was set to zero. Insert the completeness relation Σ|m⁽⁰⁾⟩⟨m⁽⁰⁾| = 1 and the correction becomes one superposition, |n⁽¹⁾⟩ = Σ (m ≠ n) cₘ|m⁽⁰⁾⟩. Read each amplitude cₘ as a ratio. The numerator, in J, says whether H′ connects the two states at all, and parity or a conserved quantum number usually makes most numerators zero. The denominator, also in J, says how far away the partner sits. Their quotient is a pure number. Its sign carries information too: a level above n enters with the opposite sign to its coupling, a level below with the same sign. Take two partners, one 0.020 eV above coupled by 0.004 eV and one 1.0 eV above coupled by 0.10 eV: the amplitudes are −0.20 and −0.10. The weaker coupling wins, because its level is fifty times closer.

What the first-order state moves, and what it leaves alone

Amplitudes change at first order; probabilities do not. The probability of finding |m⁽⁰⁾⟩ is |cₘ|², second order in H′, and the norm is 1 + Σ|cₘ|², so the probability of staying in |n⁽⁰⁾⟩ falls only by Σ|cₘ|². What does move at first order is any expectation value that links |n⁽⁰⁾⟩ to the states mixed in: ⟨A⟩ shifts by 2 Re Σ ⟨n⁽⁰⁾|A|m⁽⁰⁾⟩cₘ, the interference between the old state and its admixture. With A = qx and H′ = −qFx it is the dipole a field induces, 2q²F Σ |⟨m⁽⁰⁾|x|n⁽⁰⁾⟩|²/(Eₘ⁽⁰⁾ − Eₙ⁽⁰⁾) in C m, linear in F; for a ground state every denominator is positive, so a ground state always polarises along the field. The sum asks two things of each partner: that H′ mixes it in, and that A connects it back. A cubic term x³ mixes |1⟩ and |3⟩ into an oscillator's ground state, yet ⟨x⟩ hears only |1⟩. That is how the Hard example turns a molecule's zero-point bond stretch into a measured rotational constant.

The working criterion: each coupling against its own gap

Everything so far assumed each amplitude is small, so the state is still mostly |n⁽⁰⁾⟩. That is the test: |⟨m⁽⁰⁾|H′|n⁽⁰⁾⟩| ≪ |Eₙ⁽⁰⁾ − Eₘ⁽⁰⁾| for every m. It is not the size of λ, a label set to 1 at the end, and not H′ against Eₙ⁽⁰⁾, which moves when a constant is added to H₀ while the gaps stay put. A solved pair shows how fast the amplitude degrades. For coupling V across a gap Δ the exact admixture is tan θ with tan 2θ = 2V/Δ, quoted from Quantum Mechanics I, against V/Δ at first order: 1% too large at V/Δ = 0.1, 21% at 0.5, 62% at 1. Couplings change slowly from level to level while gaps can be arbitrarily small, so the nearest levels break the test first. Hydrogen shows both extremes: relativistic corrections mix different n with amplitudes of order α² ≈ 5 × 10⁻⁵, but inside one n the gap is zero and the formula divides by nothing unless H′ has no matrix element there. That case belongs to the lessons on degenerate levels.

One term: a charged oscillator in a uniform field

Put a charge q in the oscillator of Quantum Mechanics I and switch on a uniform field F along x, so H′ = −qFx. Quote x = x₀(a + a†)/√2 with x₀ = √(ħ/(mω)): x changes n by exactly one, so ⟨m|x|0⟩ survives only for m = 1, where it is x₀/√2. The sum has one term, with coupling −qFx₀/√2 across E₀ − E₁ = −ħω, so c₁ = qFx₀/(√2 ħω). Writing ħω = mω²x₀² turns this into d/(√2 x₀), where d = qF/(mω²) is where the spring force balances the field force. The condition therefore reads d ≪ x₀: the classical shift of the equilibrium must be small beside the zero-point width. For an electron in a GaAs quantum dot (effective mass 0.067 mₑ, quoted from a table) with ħω = 5.0 meV, x₀ = 15.1 nm, and a field of 1.0 × 10⁵ V m⁻¹ gives d = 4.5 nm and c₁ = 0.21. A level n has two partners, n + 1 with amplitude (d/x₀)√((n + 1)/2) and n − 1 with −(d/x₀)√(n/2), so the test tightens as √n. And ħ sits in it: as ħ → 0, x₀ shrinks while d stays put, so first order fails in the classical limit.

Check it against the exact answer: the displaced Gaussian

This perturbation can be removed exactly. Complete the square: ½mω²x² − qFx = ½mω²(x − d)² − ½mω²d², so the exact ground state is ψ₀(x − d), the same Gaussian moved by d. Expand in d: ψ₀(x − d) = ψ₀(x) − dψ₀′(x) + O(d²), and since ψ₀′ = −(x/x₀²)ψ₀ and xψ₀ = (x₀/√2)ψ₁, the first-order term is (d/(√2 x₀))ψ₁, which is c₁|1⟩ exactly. Gaussian overlaps give ⟨0|ψ⟩ = e^{(−c₁²/2)} and ⟨1|ψ⟩ = c₁e^{(−c₁²/2)}, so first order even gets their ratio right; what it misses is |2⟩, |3⟩, …, which start in the d² term. The probability of staying in |0⟩ shows the cost: e^(−c₁²) exactly against 1/(1 + c₁²), 0.9555 against 0.9565 at c₁ = 0.2133, but 0.368 against 0.500 at c₁ = 1, where d = √2 x₀. The energy is kinder. Its whole shift, −½mω²d², is second order, so its series stops there even when c₁ exceeds 1, as the next lesson shows. The state's series never stops: ψ₀(x − d) holds every power of d.

Change one variable at a time

Make the relationship visible.

Interactive model

Field F100 kV m⁻¹

Level spacing ħω5.0 meV

At ħω = 5.0 meV and F = 100 kV m⁻¹ the coupling bar is a fifth of the gap and dashed lies on solid. Then set ħω = 2.0 meV and F = 200 kV m⁻¹: the coupling outgrows the gap, the exact Gaussian slides 57 nm, while the first-order curve stalls near 19 nm and grows a node.

COUPLING |⟨1|H′|0⟩|1.07 meV

RATIO c = COUPLING / GAP0.213

P(STAY IN |0⟩), EXACT0.9555

P(STAY IN |0⟩), FIRST ORDER0.9565

Live interpretationCOUPLING |⟨1|H′|0⟩|: 1.07 meV. RATIO c = COUPLING / GAP: 0.213. P(STAY IN |0⟩), EXACT: 0.9555. P(STAY IN |0⟩), FIRST ORDER: 0.9565

Catch the common trap

Explain before calculating.

Misconception check

A zero first-order energy is not evidence that the perturbation is weak or that the state is unchanged. The diagonal element ⟨n⁽⁰⁾|H′|n⁽⁰⁾⟩ and the off-diagonal ⟨m⁽⁰⁾|H′|n⁽⁰⁾⟩ are independent numbers: the first moves the level, the others reshape the state and decide whether any of it can be trusted. The charged oscillator makes the point sharply. By parity E⁽¹⁾ = 0 for every level at every field, yet at d = 2x₀ the first-order amplitude is √2 ≈ 1.41, larger than the state it corrects, and the exact ground state has only e⁽⁻²⁾ = 0.14 probability of being found in |0⟩. The converse misleads as well: a constant added to H′ shifts E⁽¹⁾ by any amount you like and leaves the state exactly as it was. Judge a perturbation by its off-diagonal couplings, each against its own gap, never by its diagonal element.

A level at Eₙ⁽⁰⁾ = −4.00 eV is coupled by H′ to three others: one at −3.98 eV with |⟨m⁽⁰⁾|H′|n⁽⁰⁾⟩| = 0.010 eV, one at −3.00 eV with 0.20 eV, and one at −9.00 eV with 0.60 eV. Can its first-order state be trusted, and which level decides?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA ⁹Be⁺ ion (mass 1.4964 × 10⁻²⁶ kg, quoted) is cooled to the ground state of a trap axis with ω/2π = 2.00 MHz, and a stray field F = 0.10 V m⁻¹ lies along that axis. Find x₀, the displacement d and the first-order amplitude of |1⟩, decide whether first order can be trusted, and find what it misses: the probability the exact state puts in n ≥ 2.

ω = 2π × 2.00 × 10⁶ s⁻¹ = 1.257 × 10⁷ s⁻¹, so x₀ = √(ħ/(mω)) = √(1.0546 × 10⁻³⁴ J s ÷ (1.4964 × 10⁻²⁶ kg × 1.257 × 10⁷ s⁻¹)) = 23.7 nm. The units are J s/(kg s⁻¹) = m², as a squared length should be.
The spring and field forces balance at d = eF/(mω²) = (1.602 × 10⁻¹⁹ C × 0.10 V m⁻¹) ÷ (1.4964 × 10⁻²⁶ kg × 1.579 × 10¹⁴ s⁻²) = 6.78 nm.
Only |1⟩ is coupled, with amplitude c₁ = d/(√2 x₀) = 6.78 nm ÷ 33.5 nm = 0.202. The criterion itself gives the same number: the coupling eFx₀/√2 = 1.67 × 10⁻⁹ eV sits across a gap ħω = 8.27 × 10⁻⁹ eV.
c₁ = 0.20 ≪ 1, so first order holds. The normalised first-order state gives P(n = 1) = c₁²/(1 + c₁²) = 0.0394, and the exact displaced Gaussian gives c₁²e^(−c₁²) = 0.0393.
What first order cannot see is the exact weight in n ≥ 2: 1 − (1 + c₁²)e^(−c₁²) = 8.2 × 10⁻⁴, close to c₁⁴/2, a fourth-order effect. It reaches 0.026 at c₁ = 0.5, which takes F = 0.25 V m⁻¹, where d = 16.7 nm = 0.71x₀. As F → 0 all of it vanishes.

Answerx₀ = 23.7 nm, d = 6.78 nm and c₁ = 0.202 ≪ 1, so first order holds; P(n = 1) = 0.039, and the exact state puts only 8.2 × 10⁻⁴ in n ≥ 2.

MediumIn ammonia's rotational level (J, K) = (3, 3) with M = 3, inversion splits a lower state |s⟩ from an upper |a⟩ by Δ = h × 23.870 GHz, and a field F along z couples them through μ_eff = 0.75μ, with μ = 1.47 D = 4.90 × 10⁻³⁰ C m (all quoted). The rotational levels the field reaches lie about 8B = h × 2.385 THz away. For F = 1.00 × 10⁵ V m⁻¹ find the first-order state of |s⟩, the induced dipole, and the field at which first order fails.

Parity first: |s⟩ and |a⟩ have opposite inversion parity, so ⟨s|μ_z|s⟩ = 0 and E⁽¹⁾ = 0; only the off-diagonal element acts. Scales: Δ = (6.626 × 10⁻³⁴ J s)(2.387 × 10¹⁰ s⁻¹) = 1.582 × 10⁻²³ J = 98.7 μeV, and the coupling μ_eff F = (3.68 × 10⁻³⁰ C m)(1.00 × 10⁵ V m⁻¹) = 3.68 × 10⁻²⁵ J = 2.30 μeV, since C V = J.
With the phase chosen so that ⟨a|μ_z|s⟩ = +μ_eff, H′ = −μ_zF gives ⟨a|H′|s⟩ = −μ_eff F across Eₛ − Eₐ = −Δ, so the state is |s⟩ + c|a⟩ with c = μ_eff F/Δ = 2.30/98.7 = 0.0233.
Every other partner is a rotational level or lies farther still. No matrix element of −μ_zF can exceed μF = 3.06 μeV, and the J = 4 levels it reaches sit about 8B = 9.86 meV away, so their amplitudes are at most 3.1 × 10⁻⁴: 75 times smaller than c. The inversion partner, 100 times closer, decides.
Induced dipole: ⟨μ_z⟩ = 2c⟨s|μ_z|a⟩ = 2μ_eff²F/Δ = 2(0.0233)(1.10 D) = 0.0513 D along the field, linear in F; with μ_eff in C m, F in V m⁻¹ and Δ in J the units are C m.
c reaches 0.5 at F = Δ/(2μ_eff) = 2.15 × 10⁶ V m⁻¹ = 21.5 kV cm⁻¹, where the rotational amplitudes are still at most 0.0067. There the exact pair, quoted from Quantum Mechanics I, has tan 2θ = 1: the admixture is tan θ = 0.414 and the dipole μ_eff sin 2θ = 0.707μ_eff, where first order says 1.00μ_eff, 41% too high.
Far above that field the exact dipole saturates at μ_eff and the shift becomes linear in F. The remedy is to solve the pair exactly and perturb only the distant rotational levels, the plan of the lessons on degenerate levels.

Answer|s⟩ + 0.0233|a⟩, with every rotational amplitude below 3.1 × 10⁻⁴; induced dipole 0.0513 D; first order fails near 21.5 kV cm⁻¹, where it overstates the dipole by 41%.

HardModel N₂ as a Morse oscillator, V = D(1 − e^(−βx))² in the stretch x = r − rₑ, with the measured ωₑ = 2358.57 cm⁻¹, ωₑxₑ = 14.324 cm⁻¹, Bₑ = 1.99824 cm⁻¹ and rₑ = 1.09768 Å (quoted). Take the cubic term of V as H′. Find the first-order state of v = 0 and its bond stretch ⟨x⟩, then predict αₑ in Bᵥ = Bₑ − αₑ(v + ½) and compare with the measured 0.017318 cm⁻¹.

Expand: V = Dβ²x² − Dβ³x³ + …, so k = 2Dβ² = μω² and H′ = −κx³ with κ = Dβ³ = kβ/2. As energies, the Morse spectrum (quoted) has ωₑxₑ = ħ²β²/(2μ) and Bₑ = ħ²/(2μrₑ²), so βrₑ = √(ωₑxₑ/Bₑ) = √(14.324/1.99824) = 2.677 and β = 2.439 Å⁻¹. Likewise x₀²/rₑ² = ħ/(μωrₑ²) = 2Bₑ/ωₑ = 1.694 × 10⁻³, so x₀ = 0.04518 Å.
Ladder operators: x³ = (x₀/√2)³(a + a†)³ takes |0⟩ to 3|1⟩ + √6|3⟩. Since ħω = kx₀², the natural size is κx₀³/ħω = βx₀/2 = 0.0551. So c₁ = 0.0551 × 3/(2√2) = 0.0584 across ħω, and c₃ = 0.0551 × √6/(2√2)/3 = 0.0159 across 3ħω. Both are far below 1: first order holds.
The stretch: x connects |0⟩ only to |1⟩, so ⟨x⟩₀ = 2⟨0|x|1⟩c₁ = √2 x₀c₁ = (3/4)βx₀² = 3.73 × 10⁻³ Å, 0.34% of rₑ; |3⟩ is in the state but invisible to ⟨x⟩. Check by force balance: a stationary state has ⟨V′⟩ = 0, so k⟨x⟩ = 3κ⟨x²⟩ = 3κx₀²/2, the same (3/4)βx₀².
Level v: the same balance with ⟨x²⟩ = x₀²(v + ½) gives ⟨x⟩ = (3/2)βx₀²(v + ½). Bᵥ is ħ²/(2μ) times ⟨1/r²⟩, and 1/(rₑ + x)² = rₑ⁻²(1 − 2x/rₑ + 3x²/rₑ² − …), so Bᵥ = Bₑ[1 − 3(v + ½)(x₀²/rₑ²)(βrₑ − 1)]. The stretch lowers B; the spread in x, through the curvature of 1/r², raises it.
So αₑ = 3Bₑ(x₀²/rₑ²)(βrₑ − 1) = 3(1.99824 cm⁻¹)(1.6945 × 10⁻³)(1.6774) = 0.01704 cm⁻¹, in the units of B as it must be. The stretch alone gives 0.02720 cm⁻¹ and the spread takes back 0.01016 cm⁻¹. Against the measured 0.017318 cm⁻¹ the prediction is 1.6% low, the price of the Morse shape and of terms of higher order in x₀/rₑ.
Limits: a harmonic bond, β → 0, gives αₑ = −3Bₑ x₀²/rₑ² and a B that rises with v, the opposite of N₂ and of almost every measured diatomic; the classical limit x₀ → 0 gives αₑ → 0. No power of the fine-structure constant enters: the small number is x₀/rₑ = 0.041.

Answerc₁ = 0.058 and c₃ = 0.016, so first order holds; ⟨x⟩₀ = 3.73 × 10⁻³ Å; αₑ = 3Bₑ(x₀²/rₑ²)(βrₑ − 1) = 0.0170 cm⁻¹ against 0.01732 cm⁻¹ measured, 1.6% low.