University Quantum Mechanics II · Time-Independent Perturbation Theory · 6.2

The first-order energy shift E⁽¹⁾ = ⟨n|H′|n⟩

This is the formula you will reach for most often in the rest of the course, and it asks only for the state you already have. We derive it by projection, read it as an average over |ψ|², and spend it on a spike in a well, an x⁴ oscillator and the finite size of the proton.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

For a non-degenerate level, projecting the order-λ equation onto ⟨n⁽⁰⁾| wipes out H₀ − Eₙ⁽⁰⁾, because H₀ is Hermitian, and leaves Eₙ⁽¹⁾ = ⟨n⁽⁰⁾|H′|n⁽⁰⁾⟩. How the state bends in response never enters, because an energy is stationary at an eigenstate. For a potential, the shift is the perturbation averaged over the unperturbed density |ψₙ⁽⁰⁾|², so the work is knowing where the particle is: a spike at a node does nothing, a polynomial in x becomes a count of ladder-operator paths, and a nucleus tens of thousands of times smaller than the atom is sampled at the single value |ψ(0)|².

The model stops in three places. When symmetry makes ⟨H′⟩ vanish, as for a uniform field acting on an oscillator, the true shift starts at second order. When other states share the level, ⟨n|H′|n⟩ depends on which of them you call n, and the good states must be found first.

And when ⟨H′⟩ is not small beside the gaps to the levels H′ couples, the linear answer fails: a spike's first-order shift grows without limit, while the exact level can rise no further than the next one.

Simple definition: The first-order shift of a non-degenerate level is the expectation value of the perturbation in the unperturbed state, Eₙ⁽¹⁾ = ⟨n⁽⁰⁾|H′|n⁽⁰⁾⟩; for a potential it is H′ averaged over the unperturbed probability density.
Example: An electron in a 1.00 nm well with a spike of strength β = 0.0200 eV nm at the centre: the ground state, with |ψ₁(L/2)|² = 2/L = 2.00 nm⁻¹, rises by 2β/L = 40.0 meV, while n = 2, which has a node there, does not move.

Projecting the order-λ rungEₙ⁽¹⁾ = ⟨n⁽⁰⁾|H′|n⁽⁰⁾⟩

Derived: ⟨n⁽⁰⁾|(H₀ − Eₙ⁽⁰⁾) = 0 because H₀ is Hermitian, so the unknown |n⁽¹⁾⟩ drops out. No gap and no other state is needed.

H′ in J; |n⁽⁰⁾⟩ a non-degenerate eigenket of H₀ with ⟨n⁽⁰⁾|n⁽⁰⁾⟩ = 1; the bookkeeping λ set to 1 at the end

A potential perturbation as an averageEₙ⁽¹⁾ = ∫ |ψₙ⁽⁰⁾(r)|² H′(r) d³r

Derived by inserting position completeness. H′ is weighted by where the particle already is, so the shift is ⟨H′(r)⟩, never H′(⟨r⟩).

|ψₙ⁽⁰⁾|² in m⁻³, H′(r) in J, d³r in m³; in one dimension |ψ|² is in m⁻¹ and dx in m

Spike in an infinite wellEₙ⁽¹⁾ = (2β/L) sin²(nπx₀/L)

Derived: a δ function samples |ψₙ⁽⁰⁾|² = (2/L) sin²(nπx/L) at one point. A state with a node at x₀ never feels the spike, to any order.

β in J m, since δ(x) carries m⁻¹; L in m; x₀ the spike's position. At x₀ = L/2: 2β/L for odd n, 0 for even n

Quartic term in the oscillatorEₙ⁽¹⁾ = (3λħ²/(4m²ω²))(2n² + 2n + 1)

Derived from x = √(ħ/2mω)(a + a†): six orderings of two a and two a† return n to n, and their weights sum to 6n² + 6n + 3.

λ in J m⁻⁴, m in kg, ω in rad s⁻¹, so ħ/mω is in m²; n = 0 gives 3λħ²/(4m²ω²)

Finite nuclear size, any spherical chargeΔE = (2π/3)(Ze²/4πε₀)|ψ(0)|²⟨r²⟩

Derived with ψ held constant across the nucleus: Green's identity and Gauss's law leave only ⟨r²⟩, so the nucleus's shape cannot be seen.

⟨r²⟩ the mean-square charge radius in m², |ψ(0)|² in m⁻³, e²/4πε₀ = 2.31 × 10⁻²⁸ J m; s states only

Hydrogen 1s and the proton's radiusΔE₁ₛ = (2/3)(e²/(4πε₀a))(rₚ/a)²

Derived from the card above with |ψ₁₀₀(0)|² = 1/(πa³): 4.6 × 10⁻⁹ eV, or 1.1 MHz, equal to (2/3)α⁴mc²(rₚmc/ħ)².

a = 5.29 × 10⁻¹¹ m, e²/(4πε₀a) = 27.2 eV; rₚ = 0.84 fm is measured and quoted

Project the order-λ rung, and the new state drops out

Quote the order-λ rung from the previous lesson: (H₀ − Eₙ⁽⁰⁾)|n⁽¹⁾⟩ = (Eₙ⁽¹⁾ − H′)|n⁽⁰⁾⟩. Act on both sides with ⟨n⁽⁰⁾|. Because H₀ is Hermitian, ⟨n⁽⁰⁾|H₀ = Eₙ⁽⁰⁾⟨n⁽⁰⁾|, so the left side is zero whatever |n⁽¹⁾⟩ turns out to be, and with ⟨n⁽⁰⁾|n⁽⁰⁾⟩ = 1 the right side gives Eₙ⁽¹⁾ = ⟨n⁽⁰⁾|H′|n⁽⁰⁾⟩. No other state, no energy gap and no convention about |n⁽¹⁾⟩ enters. The reason is that an energy is stationary at an eigenstate: |n⁽⁰⁾⟩ differs from the true eigenstate of H by order λ, so ⟨n⁽⁰⁾|H|n⁽⁰⁾⟩ misses the true eigenvalue only at order λ². Test it where the exact answer is known. A constant H′ = V₀ gives V₀, exactly. Stiffen an oscillator's spring from k to k + κ, so H′ = ½κx²: with ⟨x²⟩ = (ħ/2mω)(2n + 1) from Quantum Mechanics I, Eₙ⁽¹⁾ = (κ/2k)(n + ½)ħω, the first Taylor term of the exact (√(1 + κ/k) − 1)(n + ½)ħω, since ω grows as √k. At κ/k = 0.10 they give 0.0500 and 0.0488 in units of (n + ½)ħω, and the gap, −0.0012, is close to −(κ/k)²/8, the second-order term.

Read it as H′ averaged over where the particle already is

For a perturbation that depends only on position, insert ∫|x⟩⟨x| dx on each side of H′: Eₙ⁽¹⁾ = ∫ |ψₙ⁽⁰⁾(x)|² H′(x) dx, with d³r in three dimensions. The shift is H′ weighted by the unperturbed probability density, and that settles three things before any integral is done. The sign: an H′ that is nowhere negative can only raise the level. The size: at first order, a perturbation confined to a region holding probability P moves the level by about H′ there times P, so even a large H′ does little where the particle seldom goes. And symmetry: an H′ odd under x → −x, added to a potential even under it, gives Eₙ⁽¹⁾ = 0, because |ψₙ|² is even. A charge in an oscillator with a uniform field F, H′ = −qFx, therefore has no first-order shift, although completing the square gives the exact shift −q²F²/(2mω²), quadratic in F: first order is silent there, not wrong. What is averaged is H′, not x.

A spike samples the density at a single point

Put a thin barrier βδ(x − x₀) into an infinite well of width L; β is in J m, because δ carries m⁻¹, and the letter α is kept for the fine-structure constant. The δ function does the integral: Eₙ⁽¹⁾ = β|ψₙ⁽⁰⁾(x₀)|² = (2β/L) sin²(nπx₀/L). At the centre, odd n rise by 2β/L, twice what a uniform smear of the same strength would give, because sin² peaks there; even n sit on a node and do not move, and that zero is exact, since a state vanishing at the spike remains an eigenstate. For an electron in a 1.00 nm well with β = 0.0200 eV nm, 2β/L = 40.0 meV. The exact condition for odd n, tan(kL/2) = −ħ²k/(mβ), from the jump in dψ/dx across the spike, gives 39.0 meV for n = 1, 2.6% lower. First order grows linearly in β for ever, but the exact n = 1 level can rise no further than E₂⁽⁰⁾, 1.13 eV up, as the spike becomes a wall that splits the well; first order is already 10% high at β = 0.073 eV nm.

Polynomials in x: count the ladder paths

For H′ = λx⁴ in an oscillator, quote x = √(ħ/2mω)(a + a†) and let the ladder operators replace the integral. Of the 16 products in (a + a†)⁴, only the six with two a and two a† return |n⟩ to itself. Acting from the right, a†a†aa gives n(n − 1), aaa†a† gives (n + 1)(n + 2), and the mixed orderings aa†aa†, a†aa†a, aa†a†a and a†aaa† give (n + 1)², n², n(n + 1) and n(n + 1). The sum is 6n² + 6n + 3, so Eₙ⁽¹⁾ = (3λħ²/(4m²ω²))(2n² + 2n + 1) and the ground state rises by 3λħ²/(4m²ω²). Units: λ in J m⁻⁴ times ħ²/m²ω² in m⁴ is J. With g = λħ/(m²ω³), E₀⁽¹⁾ = (3g/4)ħω; at g = 0.010 that is 0.0075 ħω, against 0.0073 ħω from a numerical diagonalisation. The same count shows that λx³ has no first-order shift, since three steps cannot return to n. The growth as n² is the warning: at g = 0.050, n = 3 gets 0.94 ħω at first order against a numerically exact 0.68 ħω.

A nucleus too small to resolve: hold ψ fixed across it

Model the proton as a uniformly charged sphere of radius R. Inside it the electron's potential energy is −(e²/4πε₀)(3R² − r²)/(2R³) rather than −e²/(4πε₀r), so H′ = (e²/4πε₀)[1/r − (3R² − r²)/(2R³)] for r ≤ R and zero outside. H′ is positive, since inside the sphere the electron is held less tightly, and for R = 1.08 fm reaches 0.83 MeV at r = R/2, yet the level barely moves, because |ψ₁₀₀|² ∝ e^(−2r/a) changes by only about 2R/a ≈ 4 × 10⁻⁵ across the proton. Hold |ψ₁₀₀|² at its value 1/(πa³) from Unit 2, so ΔE₁ₛ = |ψ₁₀₀(0)|² ∫H′ d³r. The integral is 4π(e²/4πε₀)(R²/2 − R²/2 + R²/10) = (2π/5)(e²/4πε₀)R². The rms radius of a uniform sphere is rₚ = √(3/5) R, so ΔE₁ₛ = (2/3)(e²/(4πε₀a))(rₚ/a)². The measured rₚ = 0.84 fm (quoted) gives (rₚ/a)² = 2.52 × 10⁻¹⁰ and, with e²/(4πε₀a) = 27.2 eV, ΔE₁ₛ = 4.57 × 10⁻⁹ eV, or 1.11 MHz after dividing by h: about 1.1 MHz, since rₚ carries two figures.

The shape drops out, and the checks: α, n and a measured line

The sphere was a model, but the answer does not depend on it. Write H′ = −e δφ, where δφ is the change in the nucleus's potential; for a spherical nucleus it vanishes outside, since Gauss's law on a concentric sphere makes the field there depend only on the charge enclosed. Green's identity gives ∫r²∇²δφ d³r = 6∫δφ d³r, and Gauss's law in the form ∇²δφ = −δρ/ε₀ turns the left side into −Ze⟨r²⟩/ε₀, so ∫H′ d³r = (2π/3)(Ze²/4πε₀)⟨r²⟩ for any spherical nucleus. Only the mean-square radius survives. Order in α: e²/(4πε₀a) = α²mc² and rₚ/a = αrₚmc/ħ, so ΔE₁ₛ = (2/3)α⁴mc²(rₚmc/ħ)², an energy times a pure number: the α⁴mc² of fine structure, suppressed by (rₚmc/ħ)² = 4.7 × 10⁻⁶. Only s states feel it, since ψ ∝ rˡ near the origin, and Unit 2's |ψₙ₀₀(0)|² = 1/(πn³a³) makes it fall as 1/n³: 2s rises by 0.14 MHz, lowering the 1S–2S frequency by 0.97 MHz. That line is measured to about 10 Hz (quoted); with a second line to fix the Rydberg constant, and quantum electrodynamics, quoted not derived, for all else, it measures rₚ.

Change one variable at a time

Make the relationship visible.

Interactive model

Level n1

Bump centre x₀0.50 nm

Bump width w0.10 nm

Narrow the bump to 0.05 nm and the bar's top sits level with the dot, because a spike reads |ψₙ|² at one point. Set n = 2 at the centre and the shift nearly vanishes at the node; widen the bump to 0.50 nm and it climbs to exactly β/L.

PROBABILITY IN BUMP0.198

FIRST-ORDER SHIFT E⁽¹⁾39.7 meV

SPIKE LIMIT β|ψₙ(x₀)|²40.0 meV

LEVEL Eₙ⁽⁰⁾ + E⁽¹⁾0.416 eV

Live interpretationPROBABILITY IN BUMP: 0.198. FIRST-ORDER SHIFT E⁽¹⁾: 39.7 meV. SPIKE LIMIT β|ψₙ(x₀)|²: 40.0 meV. LEVEL Eₙ⁽⁰⁾ + E⁽¹⁾: 0.416 eV

Catch the common trap

Explain before calculating.

Misconception check

The tempting shortcut is to put the particle at its average position and evaluate the perturbation there, writing Eₙ⁽¹⁾ = H′(⟨x⟩) instead of ⟨H′(x)⟩. The two agree only when H′ is linear in x, and every example in this lesson breaks the shortcut. In the oscillator ⟨x⟩ = 0, so λ⟨x⟩⁴ = 0, yet the ground state rises by 3λħ²/(4m²ω²); even the subtler λ⟨x²⟩² gives λħ²/(4m²ω²), a third of the truth, because the Gaussian's tails weigh heavily in x⁴. In the well, n = 1 and n = 2 both have ⟨x⟩ = L/2, exactly where the centred spike sits, yet one rises by 2β/L and the other not at all. In hydrogen ⟨r⟩ = 1.5a, nearly 10⁵ proton radii out, where H′ = 0, yet 1s moves by 1.1 MHz. The shift averages the perturbation over the density, and the density's shape, not its centre, decides it.

A thin barrier βδ(x − L/4) is added to an infinite well with walls at x = 0 and x = L. To first order, how do the four lowest levels shift?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyAn electron of effective mass 0.067 mₑ is held in a GaAs well of width L = 10.0 nm with infinite walls, where E₁⁽⁰⁾ = 56.1 meV. A gate raises the potential by V₀ = 20.0 meV over the right-hand quarter, from x = 3L/4 to x = L. Find the first-order shifts of n = 1, 2 and 3, and compare them with the shift a step over the whole right half would give.

The step is constant where it acts, so Eₙ⁽¹⁾ = ∫|ψₙ⁽⁰⁾|²H′ dx = V₀Pₙ, where Pₙ is the probability of finding the electron under the step. Units: meV times a pure number.
With |ψₙ⁽⁰⁾|² = (2/L) sin²(nπx/L) and 2 sin²θ = 1 − cos 2θ, the integral from 3L/4 to L is Pₙ = 1/4 + sin(3nπ/2)/(2nπ).
n = 1: sin(3π/2) = −1, so P₁ = 0.2500 − 0.1592 = 0.0908 and E₁⁽¹⁾ = 1.82 meV. n = 2: sin 3π = 0, so P₂ = 0.2500 and E₂⁽¹⁾ = 5.00 meV. n = 3: sin(9π/2) = +1, so P₃ = 0.2500 + 0.0531 = 0.3031 and E₃⁽¹⁾ = 6.06 meV.
The ground state barely feels the step because its density falls to zero at the wall. As n grows the density evens out and Pₙ tends to 1/4, the classical fraction, so the shift tends to V₀/4 = 5.00 meV.
Over the right half, every |ψₙ⁽⁰⁾|² is symmetric about x = L/2, so Pₙ = ½ exactly and each level rises by V₀/2 = 10.0 meV. Limiting case: a step over the whole well gives V₀ for every n, which is exact. Size: every shift is small beside E₂⁽⁰⁾ − E₁⁽⁰⁾ = 168 meV.

AnswerE⁽¹⁾ = 1.82 meV, 5.00 meV and 6.06 meV for n = 1, 2 and 3; a step over the right half would shift every level by 10.0 meV.

MediumA ⁸⁷Rb atom sits in one well of a deep optical lattice V(x) = V₀ sin²(kx), with k = 2π/(1064 nm), recoil energy E_R = ħ²k²/2m = h × 2.028 kHz and V₀ = 40.0 E_R. Take the x² term of V into H₀ and the x⁴ term as H′. Find the first-order shifts of the two lowest levels and the 0 → 1 transition frequency, and compare with a full numerical solution whose band centres lie 6.063 E_R and 17.609 E_R above the well bottom.

Expand about the well bottom with sin²u = u² − u⁴/3 + …: V ≈ V₀k²x² − (V₀k⁴/3)x⁴. So ½mω² = V₀k², giving ħω = 2√(V₀E_R) = 2 × √40.0 × E_R = 12.649 E_R, and H′ = λx⁴ with λ = −V₀k⁴/3 in J m⁻⁴. The lattice softens the well, so λ is negative.
Quote the oscillator result Eₙ⁽¹⁾ = (3λħ²/(4m²ω²))(2n² + 2n + 1). With m²ω² = 2mV₀k² this becomes −(ħ²k²/8m)(2n² + 2n + 1) = −(E_R/4)(2n² + 2n + 1): the depth V₀ cancels, and the result is an energy, as it must be.
n = 0: E₀⁽¹⁾ = −0.250 E_R, so E₀ = 6.325 − 0.250 = 6.075 E_R. n = 1: E₁⁽¹⁾ = −1.250 E_R, so E₁ = 18.974 − 1.250 = 17.724 E_R.
The 0 → 1 spacing is ħω − E_R = 11.649 E_R. Multiplying by E_R/h = 2.028 kHz gives 23.62 kHz, against 25.65 kHz for the harmonic well.
The numerical band centres, 6.063 E_R and 17.609 E_R, give a spacing of 11.546 E_R, or 23.41 kHz. Anharmonicity lowers the line by 1.103 E_R in all, and first order supplies 1.000 E_R of it, 91%.
The remainder comes from second order in H′ and from the next Taylor term, +2V₀k⁶x⁶/45, both smaller by a further factor of order √(E_R/V₀) = 0.16. A deeper lattice therefore makes first order better, not worse.

AnswerE₀⁽¹⁾ = −0.250 E_R and E₁⁽¹⁾ = −1.250 E_R, whatever V₀ is; the 0 → 1 line falls from 25.65 kHz to 23.62 kHz at first order, against 23.41 kHz from the full solution.

HardIn muonic hydrogen a muon, m_μ = 206.77 mₑ, replaces the electron. The measured 2S–2P Lamb shift is 202.371 meV, and quantum electrodynamics sums every contribution except the finite-radius term to 206.067 meV (both quoted). Treating the proton's finite size at first order, with ψ₂₀₀ held constant across the proton, infer rₚ and test the assumption.

Reduced mass: μ = m_μmₚ/(m_μ + mₚ) with mₚ = 1836.15 mₑ gives μ = 185.84 mₑ, so the muon's Bohr radius is a_μ = a₀mₑ/μ = 52 918 fm/185.84 = 284.7 fm, 186 times closer to the proton than an electron's.
Unit 2's R₂₀ gives |ψ₂₀₀(0)|² = 1/(8πa_μ³). The finite-size result ΔE = (2π/3)(e²/4πε₀)|ψ(0)|²rₚ² then gives ΔE₂ₛ = (e²/4πε₀)rₚ²/(12a_μ³). With e²/4πε₀ = 1.440 × 10⁶ eV fm this is 5.197 meV fm⁻² × rₚ².
The 2P states barely move: near the origin ψ₂₁ₘ ∝ r, which costs a further factor of order (rₚ/a_μ)² ≈ 10⁻⁵. Raising 2S alone narrows the 2S–2P interval, so the deficit 206.067 − 202.371 = 3.696 meV is ΔE₂ₛ.
rₚ² = 3.696 meV/(5.197 meV fm⁻²) = 0.711 fm², so rₚ = 0.843 fm. Dimensions: meV divided by meV fm⁻² is fm².
Test the assumption: across the uniform-sphere radius R = √(5/3) rₚ = 1.09 fm, |ψ₂₀₀|² changes by about 2R/a_μ = 0.8%, against 4 × 10⁻⁵ in ordinary hydrogen. The published analysis, with relativistic and higher-order terms (5.2275 meV fm⁻², quoted), gives 0.8409 fm, and first order lands 0.3% above it.
Scale: |ψ(0)|² grows as μ³, so the muonic 2s shift is (185.84)³ = 6.4 × 10⁶ times hydrogen's 5.7 × 10⁻¹⁰ eV. That factor is why the muonic measurement pins rₚ to about 0.05% (quoted).

AnswerΔE₂ₛ = 5.197 meV fm⁻² × rₚ², so rₚ = 0.843 fm, 0.3% above the published 0.8409 fm; holding ψ fixed across the proton is good to better than 1% even for the muon.