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Advanced Quantum Mechanics course sequence · Interactive lesson

The atom-field Hamiltonian ordered in powers of e

This is the Hamiltonian the rest of the course calculates with. We assemble it from pieces already derived, sort its terms by the powers of the charge they carry and the photons they move, learn which ones a given problem may drop, and see exactly why Quantum Mechanics II's classical light was the right first step.

UAQM-11.08 · Lesson 8 of 9 in this unit. Position is not a completion score.
Starting model: The atom-field Hamiltonian ordered in powers of eLog-log plot of each interaction term's size at the field peak against intensity, for hydrogen 1s at 1064 nm. Solid: order-e A⋅p, slope ½. Dashed: order-e² A², slope 1. Faint: order-e spin term σ⋅B. Dashed level: binding energy 13.61 eV. At 10^15.0 W m⁻² the ratio A² ÷ A⋅p is 0.0341 and Uₚ is 0.0106 eV.term size at the field peak / eV, logEb = 13.61 eVe² A²e A⋅pe σ⋅B10⁴110⁻⁴10⁻⁸intensity, 10¹⁰ to 10²⁰ W m⁻², log
Start with a picture. Change the model in the lab below.
1

Build the model

Count the powers of e in a term and you have counted its photon operators.

The key idea

The atom-field Hamiltonian ordered in powers of e splits the non-relativistic Hamiltonian of an atom in the quantised field into the uncoupled atom and field, two terms of order e that change the photon number by one, and one term of order e² that changes it by zero or two.

Example. With 3 photons in one mode at the atom, A ∝ a + a† reaches only 2 or 4 photons, while A² ∝ a² + a†² + 2a†a + 1 reaches 1, 3 or 5, with matrix elements √6 = 2.45, 7 and √20 = 4.47.

The complete physical model and assumptions

Three results already derived fit together. Unit 6's Pauli Hamiltonian is the non-relativistic limit of the Dirac equation with g = 2 built in; topic 11.01's Coulomb gauge turns the scalar potential into the instantaneous Coulomb attraction; topic 11.03 makes A an operator. Expanding the square gives one operator for all non-relativistic atom-photon physics, H = Hₐₜₒₘ + Hrad − (q/m)A⋅p − (qħ/2m)σ⋅B + (q²/2m)A², and ordering it in the charge orders it in photons. The two order-e terms are linear in a and a†, so each absorbs or emits exactly one photon, the spin term trailing A⋅p by the atom's size over the reduced wavelength. The order-e² term is quadratic and moves zero or two photons. Taken at the nucleus it contains no atomic operator and shifts every level alike at first order, yet as a†ₖ′aₖ it is the whole of Thomson scattering. Displacing a coherent field turns H exactly into Quantum Mechanics II's semiclassical Hamiltonian plus a quantised field that starts empty. The ordering works while its small numbers stay small: eA₀/p for a driven atom, α = 1/137.036 for the vacuum's own field, ka for the multipoles. It fails when the quiver energy Uₚ rivals the binding energy, and the Hamiltonian itself fails when photon energies approach mc² = 511 keV or the electron's quiver speed approaches c, where the Dirac theory and pair creation take over.

1Assembling H from parts already derived

SI units throughout, with metric (+, −, −, −) wherever a four-vector appears. Quoted from Unit 6: for an electron of charge q = −e, H = [σ⋅(p − qA)]²/2m + qφ. The identity (σ⋅a)(σ⋅b) = a⋅b + iσ⋅(a × b), with p × A + A × p = −iħ∇ × A, turns it into (p − qA)²/2m − (qħ/2m)σ⋅B + qφ, derived, with g = 2 built in. In Coulomb gauge (topic 11.01) qφ is the instantaneous Coulomb attraction VC, and A is topic 11.03's A(r) = Σₖₛ √(ħ/(2ε₀ωₖV)) [εₖₛaₖₛe(ik⋅r) + h.c.], taken at the electron's position operator. Expanding the square, p⋅A = A⋅p − iħ∇⋅A = A⋅p, since every εₖₛ is perpendicular to k. Adding Hrad = Σₖₛ ħωₖ(a†ₖₛaₖₛ + ½) gives H = Hₐₜₒₘ + Hrad − (q/m)A⋅p − (qħ/2m)σ⋅B + (q²/2m)A², derived. Dimensions: V s m⁻¹ times C is kg m s⁻¹, so qA is a momentum and every term an energy. Limit: as e → 0 only Hₐₜₒₘ + Hrad remain, with product eigenstates |j⟩|n₁, n₂, …⟩, the basis everything after expands in.

2Counting photons: what each order can do

A is linear in a and a†, and so is B = ∇ × A, so both order-e terms change the total photon number by exactly one: they absorb or emit. A² is quadratic, and its products aₖaₖ′, a†ₖa†ₖ′, a†ₖ′aₖ and aₖa†ₖ′ change the number by −2, +2 or 0. A process is therefore classified by its power of e, not by how many times the interaction acts. At order e, one photon is absorbed or emitted. At order e², scattering, two-photon absorption and level shifts each have two routes, A² once or an order-e term twice, and a calculation at that order is wrong unless it keeps both. For the vacuum field, e and ε₀ enter only through α = e²/(4πε₀ħc) = 1/137.036, so vacuum effects are ordered in α. Measured check: hydrogen's 2p level, whose decay rate is the square of an order-e amplitude, lives 1.60 ns, so its rate is only Γ = 4.0 × 10⁻⁸ ω for Lyman-α: effects of order α are small fractions of the energies they modify.

3How big each term is: two small parameters

Take an atom in a classical field E₀ cos ωt along z, so A₀ = E₀/ω. At the field's peak the order-e orbital term acting on a state has size (e/m)A₀√⟨pz²⟩; for hydrogen ns the virial theorem gives ⟨p²⟩ = ħ²/(n²a₀²), so this is (e/m)A₀ħ/(√3na₀). The order-e² term is (e²/2m)A₀², and the ratio eA₀/(2√⟨pz²⟩) grows as E₀λ: the strength of the applied field, not the smallness of e, decides whether a driven atom can be expanded. The spin term is also of order e, but B₀ = kA₀, so it trails A⋅p by (ħk/2)/√⟨pz²⟩ = (√3/2)kna₀, an atom-size-over-wavelength factor, the same one that weakens magnetic-dipole and electric-quadrupole lines. For Lyman-α, ka₀ = 2.7 × 10⁻³. The spin term wins only where A⋅p is forbidden outright: hydrogen's 1420 MHz line is a pure spin flip, driven by σ⋅B alone, with a rate of 2.9 × 10⁻¹⁵ s⁻¹ (quoted).

4The dipole approximation, and a shift every level shares

When ka ≪ 1 the phase e(ik⋅r) is 1 across the atom and A(r) can be taken at the nucleus. Then (q²/2m)A(0)² contains no atomic operator: its matrix elements are δⱼⱼ′⟨F′|(q²/2m)A(0)²|F⟩, so at first order it shifts every level j by the same amount and moves no transition frequency, derived. In a classical field that shift is Uₚ = q²E₀²/(4mω²), 0.0106 eV at 1.0 × 10¹⁵ W m⁻² and 1064 nm. Order e² does not stop there. Second-order A⋅p shifts 1s by −Uₚ − ¼αd(ω)E₀²: writing ⟨n|p|1s⟩ = imωₙ₁⟨n|r|1s⟩, the Thomas–Reiche–Kuhn sum rule (quoted) splits off the −Uₚ exactly. So 1s moves by only −¼αd(ω)E₀² ≈ −9 × 10⁻⁵ eV, with polarisability αd ≈ 4.5 × 4πε₀a₀³. A free electron, whose p commutes with A(0), has no second-order A⋅p shift and keeps all of Uₚ. Rydberg levels, nearly free, rise with the threshold, which short-pulse above-threshold ionisation observes (Freeman et al., 1987).

5What A² does that no other term does

Beyond that uniform shift, A² leads two familiar effects. Its a†ₖ′ₛ′aₖₛ part moves a photon from mode ks to k′s′ with amplitude (2πrₑħc²/(V√(ωω′)))ε*ₖ′ₛ′⋅εₖₛ whatever the electron's state, derived: Thomson scattering, which Unit 12 turns into σT = (8π/3)rₑ² = 6.652 × 10⁻²⁹ m². For a free electron it is the whole answer, since in the dipole approximation A⋅p cannot change a free electron's state, and it dominates for any electron once ħω far exceeds the binding energy. Its 2a†a part retunes every mode, raising ω by e²/(2mε₀ωV) for one electron in volume V; diagonalised exactly together with the a² and a†² parts, that becomes e²/(mε₀V) added to ω², and a density nₑ gives ω² = c²k² + ωₚ², the dispersion relation of a plasma. For nₑ = 1.0 × 10¹² m⁻³, fₚ = ωₚ/2π = 8.98 MHz, and the ionosphere's reflection of short-wave radio below such critical frequencies is this term measured. Both results are non-relativistic: Klein–Nishina corrections to Thomson scattering (quoted) have relative size 2ħω/mc², already 3% for 8.05 keV copper Kα X-rays.

6Displacing a coherent field: semiclassical light, derived

Let one mode start in topic 11.07's coherent state |α⟩ = D(α)|0⟩, with D(β) = exp(βa† − β*a) and D†aD = a + β. Write |Ψ(t)⟩ = D(β(t))|Φ(t)⟩ with β(t) = αe(−iωt), the classical motion. Then iħ∂ₜ|Φ⟩ = [D†HD − iħD†∂ₜD]|Φ⟩. In D†HD every a becomes a + β, so A becomes A + Acl, the classical wave the coherent state carries, and −iħD†∂ₜD = −ħω(βa† + β*a + |β|²) cancels exactly the linear and constant pieces that a + β adds to ħωa†a. What remains is Hsc(t) + Hrad + V(t): Quantum Mechanics II's semiclassical Hamiltonian Hsc = (p − qAcl)²/2m + VC − (qħ/2m)σ⋅Bcl, plus the quantised field and its coupling V, which includes (q²/m)Acl⋅A. And |Φ(0)⟩ = |ψ⟩|0⟩: the quantised field starts in its vacuum. Nothing was approximated, so semiclassical theory is exact except for V acting on an empty field, the source of spontaneous emission (Unit 12), the Lamb shift (Unit 13) and topic 11.09's photon-number effects.

2

Key equations

Read the units, assumptions and limits before substituting.

The Hamiltonian, term by term in e

H = Hₐₜₒₘ + Hrad − (q/m)A⋅p − (qħ/2m)σ⋅B + (q²/2m)A²

Derived from Unit 6's Pauli Hamiltonian in Coulomb gauge, where p⋅A = A⋅p. Hₐₜₒₘ = p²/2m + VC holds the Coulomb attraction, and as e → 0 atom and field decouple.

q = −e for an electron of mass m in kg; A(r) in V s m⁻¹ at the electron's position; B = ∇ × A in T; σ the Pauli matrices.

One mode at the atom: what A² contains

(q²/2m)A(0)² = (q²ħ/(4mε₀ωV))(a² + a†² + 2a†a + 1)

Derived: a² and a†² change n by −2 and +2, 2a†a + 1 by zero. The 2a†a part raises the mode frequency by q²/(2mε₀ωV), the origin of a plasma's cutoff; the 1 is a vacuum constant.

One mode: angular frequency ω in s⁻¹, real polarisation, mode volume V in m³, dipole approximation; a, a† its ladder operators.

First-order A² shift: the same for every level

ΔEⱼ⁽¹⁾ = (q²/2m)⟨F|A(0)²|F⟩, independent of the atomic level j

Derived: with A at the nucleus, A² is the identity on the atom. Second-order A⋅p, also of order e², cancels Uₚ for a tightly bound level, so the ionisation threshold rises by about Uₚ.

|F⟩ is the field's state. A classical field E₀ cos ωt gives the cycle average Uₚ = q²E₀²/(4mω²), in J.

Size of the order-e² term against the order-e term

(q²A₀²/2m) ÷ (|q|A₀√⟨pz²⟩/m) = |q|A₀/(2√⟨pz²⟩)

Derived: this ratio, not e alone, is a driven atom's expansion parameter. It is 0.034 for 1s at 1064 nm and 1.0 × 10¹⁵ W m⁻², grows as λ√I, and nears 1 where Uₚ rivals the binding energy.

A₀ = E₀/ω, the vector-potential amplitude along z in V s m⁻¹. Hydrogen ns: ⟨pz²⟩ = ħ²/(3n²a₀²), from the virial theorem.

The A² matrix element that scatters a photon

⟨k′s′|(q²/2m)A²|ks⟩ = (2πrₑħc²/(V√(ωω′))) ε*ₖ′ₛ′⋅εₖₛ e(i(k − k′)⋅r)

Derived: the two orderings of a†ₖ′ₛ′aₖₛ in A⋅A give a factor 2. Unit 12 turns this into Thomson's σT = (8π/3)rₑ² = 6.652 × 10⁻²⁹ m², the whole cross-section of a free electron.

rₑ = e²/(4πε₀mc²) = 2.818 fm; V the box volume in m³; ω, ω′ in s⁻¹. In the dipole approximation the phase is 1.

Displacing a coherent field into the vacuum

|Ψ(t)⟩ = D(αe(−iωt))|Φ(t)⟩ ⇒ iħ∂ₜ|Φ⟩ = [Hsc(t) + Hrad + V(t)]|Φ⟩, |Φ(0)⟩ = |ψ⟩|0⟩

Derived and exact: Quantum Mechanics II's semiclassical Hamiltonian is the leading term, and V acting on a field that starts empty supplies spontaneous emission and the Lamb shift.

D(β) = exp(βa† − β*a). Hsc = (p − qAcl)²/2m + VC − (qħ/2m)σ⋅Bcl; V(t) is H's coupling plus (q²/m)Acl⋅A.
3

Interactive lab

Predict first. Explore the model. Compare the result.

Explore the model

15.0
10 to 20; steps of 0.5.
1064 nm
300 to 2000 nm; steps of 2.
1
1 to 6; steps of 1.

Try this. Start at 1064 nm and 10¹⁵ W m⁻², where A² sits about 30 times below A⋅p. Raise the intensity and the steeper dashed line closes in, crossing near 10¹⁸ W m⁻²; lengthen λ or raise n and the crossing moves left, while σ⋅B never moves.

Live results

A² ÷ A⋅p (EXPANSION)
0.0341
A⋅p TERM ÷ SPIN TERM
3695
Uₚ, SHIFT OF EVERY LEVEL
0.0106 eV
QUIVER SPEED, % OF c
0.0288 %

Results follow this lesson’s model and assumptions. Displayed values are rounded.

Interactive Advanced Quantum Mechanics modelLog-log plot of each interaction term's size at the field peak against intensity, for hydrogen 1s at 1064 nm. Solid: order-e A⋅p, slope ½. Dashed: order-e² A², slope 1. Faint: order-e spin term σ⋅B. Dashed level: binding energy 13.61 eV. At 10^15.0 W m⁻² the ratio A² ÷ A⋅p is 0.0341 and Uₚ is 0.0106 eV.term size at the field peak / eV, logEb = 13.61 eVe² A²e A⋅pe σ⋅B10⁴110⁻⁴10⁻⁸intensity, 10¹⁰ to 10²⁰ W m⁻², log
Log-log plot of each interaction term's size at the field peak against intensity, for hydrogen 1s at 1064 nm. Solid: order-e A⋅p, slope ½. Dashed: order-e² A², slope 1. Faint: order-e spin term σ⋅B. Dashed level: binding energy 13.61 eV. At 10¹⁵ W m⁻² the ratio A² ÷ A⋅p is 0.0341 and Uₚ is 0.0106 eV.

Predict → change one variable → check

Plan a change before running it. Every other input stays at the saved baseline.

Saved baseline. log₁₀ of intensity in W m⁻²: 15.0; Wavelength: 1064 nm; Hydrogen ns state, n: 1

Baseline 15.0; choose 10 to 20, in steps of 0.5.
4. Compared with the saved baseline, this result will…

This experiment is kept only while this page is open. It does not change saved course progress.

4

Catch the common trap

The tempting shortcut is to drop (q²/2m)A² outright: it is of order e² and, in the dipole approximation, only shifts every level alike. Both reasons fail. Order e² is exactly where scattering and level shifts live, and there A² competes on equal terms with A⋅p acting twice; drop it and both limits of light scattering go wrong. A free electron, with no other route to elastic scattering, would stop scattering light, yet Thomson scattering at σT = 6.65 × 10⁻²⁹ m² is how plasmas are diagnosed. A bound electron errs the other way: as ω → 0 the A⋅p amplitude alone tends to minus the Thomson amplitude, so atoms would scatter radio waves with the full σT. Kept, the two cancel through the Thomas–Reiche–Kuhn sum rule and leave Rayleigh's ω⁴ law, by which air scatters 450 nm light 5.9 times more strongly than 700 nm light. Uniform means it moves no line at first order, not that it does nothing.

5

Practice & worked examples

3 authored examples. Make an attempt, then reveal the reasoning one step at a time.

EasyExample 1

Hydrogen in its 1s state sits in a 1064 nm laser beam of intensity 1.0 × 10¹⁵ W m⁻², polarised along z. At the peak of the field, find the size of the order-e term (e/m)A⋅p acting on 1s, of the order-e² term (e²/2m)A², and of the order-e spin term (eħ/2m)σ⋅B. Rank them, and check that ordering in e is justified.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 5 steps revealed.

MediumExample 2

An atom with a 780 nm transition of dipole moment d = 2.5 × 10⁻²⁹ C m sits at the antinode of one resonant, linearly polarised cavity mode of mode volume 1.0 × 10⁻¹⁵ m³. In the dipole approximation, find the order-e coupling ħg, find what the A² term does to |g, n⟩ and |e, n⟩, and decide whether topic 11.09's model may drop it. Then displace a coherent state with |α|² = 100.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 6 steps revealed.

HardExample 3

In the photon vacuum the A² term gives an electron in state ψ the first-order shift (e²/2m)⟨ψ, 0|A(r)²|ψ, 0⟩. Evaluate it as a mode sum cut off at |k| = K, show it is the same for every ψ, bound or free, with no dipole approximation, and evaluate it for K = 1/a₀ and K = mc/ħ. Compare with hydrogen's measured 2S₁/₂–2P₁/₂ splitting, 1057.8 MHz, and say which order-e² route must produce that splitting.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 6 steps revealed.

6

Quick check

Select an answer and explain it before checking.

Hydrogen in its 1s state sits in a 1064 nm laser beam of intensity 1.0 × 10¹⁵ W m⁻², where Uₚ = e²E₀²/(4mω²) = 0.011 eV. Working in the dipole approximation, what happens to the 1s ionisation energy at order e², and why?
7

Review & continue

The atom-field Hamiltonian ordered in powers of e splits the non-relativistic Hamiltonian of an atom in the quantised field into the uncoupled atom and field, two terms of order e that change the photon number by one, and one term of order e² that changes it by zero or two.

  • Explain the relationship without looking at the equation.
  • Check a sign, a unit and a model assumption in your working.
  • Return to the lab or worked example if a step is still unclear.

Extra challenging questions

Connect several ideas, explore supplied data, and compare your reasoning with a worked solution.