Advanced Quantum Mechanics course sequence · Interactive lesson
Bound electron's second-order energy in the quantised field
Unit 12 left a two-level atom's radiative shift as a principal-value sum and stopped. This topic writes the same second-order energy for a real atom, every intermediate state and both kinds of term kept, and lands on the one formula the rest of Unit 13 renormalises and evaluates.
UAQM-13.01 · Lesson 1 of 9 in this unit. Position is not a completion score.Build the model
Emit a photon, pass through any atomic state, take the photon back: every path counts.
The key idea
The second-order energy of a bound state |n⟩ in the quantised field is the shift δEₙ from virtual emission and reabsorption: the electron emits a photon into any mode, passes through any atomic state m, and takes the photon back, each path weighted by its squared coupling over Eₙ − Eₘ − ħω.
Example. For hydrogen 2s every path runs through a p state at or above 3p, so every denominator is negative: through 3p, 1.89 eV up, with a 5.00 eV photon it is −6.89 eV. For 2p the denominator through 1s vanishes at E = 10.20 eV, and that pole carries its 6.27 × 10⁸ s⁻¹ decay.
The complete physical model and assumptions
Put a hydrogen atom in the photon vacuum and ask for its energy to order e². Unit 11's coupling (e/m)A⋅p, with A taken at the nucleus, does nothing at first order, because ⟨0|a|0⟩ = 0, and its A² partner adds one constant to every level alike. At second order A⋅p lifts |n, 0⟩ to every |m, 1ₖₛ⟩ and brings it back, each path weighted by its squared coupling over Eₙ − Eₘ − ħω. The two polarisations, averaged over directions, leave (2/3)|pₘₙ|²; the ω² density of modes times the 1/ω in each mode's squared amplitude leaves a weight ω; and α gathers the constants into δEₙ = (2α/(3πm²c²)) ∫dE E Σₘ|pₘₙ|²/(Eₙ − Eₘ − E), derived. Three things are in it at once. States below n produce poles whose residues give Unit 12's spontaneous-emission rates. The principal value of one such term is Unit 12's Weisskopf-Wigner shift, and every state above n contributes a counter-rotating term that approximation threw away; for 2s, whose dipole partners all lie above it, those terms are the whole shift. And each term tends to −|pₘₙ|² per unit photon energy, so the integral grows with its cutoff K. The model, nonrelativistic and in the dipole approximation, holds only well below ħc/a₀ = 3.73 keV and fails outright near mc²; what the cutoff hides is the business of the next three topics.
1Only one-photon states can be reached
SI units throughout; the metric (+, −, −, −) never enters, since everything here is in the atom's rest frame with the transverse field in Coulomb gauge. Quoted from Unit 11: H = HA + HF + (e/m)A⋅p + (e²/2m)A² for the electron's charge −e, with A(r) = Σₖₛ √(ħ/(2ε₀ωₖV)) êₖₛ(aₖₛe(ik⋅r) + aₖₛ†e(−ik⋅r)). The dipole approximation sets r = 0 in A, fair while the wavelength dwarfs the orbit: ħω ≪ ħc/a₀ = 3.73 keV for the ground state. Start in |n, 0⟩, the atom in |n⟩ and no photons. First order in A⋅p gives ⟨0|aₖₛ + aₖₛ†|0⟩ = 0. One action of A⋅p can only create a photon, since aₖₛ|0⟩ = 0, so the intermediate states are |m, 1ₖₛ⟩ for every atomic state m, every k and both s, reached with amplitude (e/m)√(ħ/(2ε₀ωV)) êₖₛ⋅pₘₙ: derived. The path with m = n is closed, because pₙₙ = (im/ħ)⟨n|[HA, r]|n⟩ = 0 for a bound stationary state.
2Summing over modes, polarisations and directions
Second-order perturbation theory, quoted from Quantum Mechanics II, gives δEₙ = Σₘ Σₖₛ |⟨m, 1ₖₛ|HI|n, 0⟩|²/(Eₙ − Eₘ − ħωₖ): start energy minus intermediate energy, atom in m plus one photon. Polarisation: Σₛ|êₖₛ⋅pₘₙ|² = |pₘₙ|² − |k̂⋅pₘₙ|², since no transverse photon carries the component along k. Direction: k̂ᵢk̂ⱼ averages to δᵢⱼ/3 over the sphere, leaving (2/3)|pₘₙ|². Magnitude: V∫d³k/(2π)³ becomes V∫ω² dω/(2π²c³). Times ħ/(2ε₀ωV), V cancels and ω² × 1/ω leaves ω: δEₙ = (e²ħ/(6π²ε₀m²c³)) Σₘ|pₘₙ|² ∫ω dω/(Eₙ − Eₘ − ħω). With E = ħω and e² = 4πε₀αħc the prefactor becomes 2α/(3πm²c²), 1.5485 × 10⁻³/(m²c²): derived. Dimensions: |pₘₙ|²/(m²c²) is a number and ∫E dE over an energy is an energy. No rotating-wave approximation was made anywhere.
3Poles: the decay rate is already inside
When Eₘ lies below Eₙ, the denominator Eₙ − Eₘ − E vanishes at E = Eₙ − Eₘ: an intermediate state with the energy of the start, a real emitted photon. A perturbation switched on adiabatically writes it with + i0, and 1/(x + i0) = P(1/x) − iπδ(x) splits δEₙ into a principal value, the shift, and Im δEₙ = −(2α/(3m²c²)) Σ|pₘₙ|²(Eₙ − Eₘ) over the lower m: derived. Read as Eₙ − iħΓₙ/2, that gives Γₙ = (4α/(3m²c²ħ)) Σ|pₘₙ|²(Eₙ − Eₘ), and with pₘₙ = imωₘₙrₘₙ, quoted, each term is ω³|d|²/(3πε₀ħc³), Unit 12's A coefficient. For 2p the one lower partner is 1s, 10.20 eV down, with the quoted |r|² = 0.5549 a₀²: Γ = (2/3)⁸α⁵mc²/ħ = 6.27 × 10⁸ s⁻¹, a lifetime of 1.60 ns, as measured. Keep only that term and only its principal value and you have Unit 12's Weisskopf-Wigner shift exactly: the two-level limit of this formula.
4What the rotating-wave approximation dropped
Unit 12's two-level atom kept only |e, 0⟩ → |g, 1ₖₛ⟩, the atom falling as the photon appears. Every term here with Eₘ above Eₙ has the atom rising while a photon is emitted: its denominator −(Eₘ − Eₙ) − E is negative at every E, and it pushes level n down. These counter-rotating terms are not small. The weight E grows with photon energy, so the integral is collected mostly where E ≫ |Eₘ − Eₙ|, and there a state below n and a state above it give the same −|pₘₙ|² per unit E. Now take hydrogen's 2s. Its dipole partners are p states: 1s is out because ⟨1s|p|2s⟩ joins two s states, and 2p is out because pₘₙ = imωₘₙrₘₙ vanishes between degenerate levels. The rest, 3p, 4p, … and the continuum, all lie above. So 2s has no pole and no one-photon dipole decay (it lives 0.12 s by two-photon emission at order e⁴, and a relativistic magnetic-dipole decay at 2.5 × 10⁻⁶ s⁻¹ is slower still, both quoted), and its whole second-order energy is counter-rotating: a two-level rotating-wave model would not shift it at all.
5The A² term, and what is left out
At first order the A² term gives (e²/2m)⟨0|A(0)²|0⟩. Each mode contributes ħ/(2ε₀ωV) times Σₛ|êₖₛ|² = 2, and the same mode sum makes this αK²/(2πmc²): derived, and the same for every n, because A at the nucleus never touches the electron's state. With K = mc² it is 593 eV for every level, 1.4 × 10⁸ times the 4.37 µeV (1057.8 MHz) Lamb interval, and it cancels from that interval exactly. Its second-order terms are of order e⁴ and go with everything beyond e². Three omissions are deliberate. The spin coupling −μ⋅B is left out, so this is Bethe's spinless calculation; the anomalous moment's share of the shift returns in topic 13.09. The dipole approximation drops e(ik⋅r), so the photon's momentum ħk never reaches the electron: no recoil. And the proton is held fixed; a reduced mass would rescale every level by about 1 − 5.4 × 10⁻⁴.
6Where the formula stops
Every term's integrand tends to −|pₘₙ|² once E ≫ |Eₘ − Eₙ|. For one term the principal-value integral is D ln|1 + K/D| − K with D = Eₘ − Eₙ, derived, so each grows as −K: with D = 1.89 eV and K = 1.00 keV it is −1000 eV + 11.9 eV = −988.1 eV. That growth is the next topic's subject; here it marks the model's edges. The dipole approximation needs ħω ≪ 3.73 keV for 1s, the nonrelativistic coupling (e/m)A⋅p fails as photon energies approach mc² = 511 keV, and nothing in the formula stops the integral before either. At the low end the weight E makes the integrand vanish as E → 0, so soft photons contribute little and there is no infrared trouble. One more limit tests the whole structure. A free electron's p commutes with its Hamiltonian, so pₘₙ = p δₘₙ and only m = n survives, with denominator −E: δE = −(2α/(3πm²c²))p²K, proportional to p²/2m, a change of mass that topic 13.03 takes up.
Key equations
Read the units, assumptions and limits before substituting.
One-photon matrix element of A⋅p
⟨m, 1ₖₛ|(e/m)A(0)⋅p|n, 0⟩ = (e/m)√(ħ/(2ε₀ωV)) êₖₛ⋅pₘₙ
Derived from Unit 11's A, quoted. a† makes the photon and a|0⟩ = 0, so this is the only exit from |n, 0⟩; pₙₙ = 0 for a bound state removes m = n.
e in C, m in kg, ω = ck in s⁻¹, V the box volume in m³, êₖₛ a real unit polarisation ⊥ k, pₘₙ = ⟨m|p|n⟩ in kg m s⁻¹.Polarisation sum, averaged over directions
Σₛ |êₖₛ⋅pₘₙ|² = |pₘₙ|² − |k̂⋅pₘₙ|², with average (2/3)|pₘₙ|²
Derived: transverse photons miss only the component along k, and ⟨k̂ᵢk̂ⱼ⟩ = δᵢⱼ/3 over the sphere. It is Unit 12's 8π/3 divided by 4π.
k̂ the unit vector along k; |pₘₙ|² = Σᵢ |⟨m|pᵢ|n⟩|² over x, y, z, in kg² m² s⁻².Second-order energy of level n
δEₙ = (2α/(3πm²c²)) ∫₀ᴷ dE E Σₘ |pₘₙ|²/(Eₙ − Eₘ − E)
Derived, with no rotating-wave approximation: every m, above n or below, is in the sum. Dimensions: |pₘₙ|²/(m²c²) is a number and ∫E dE/E is an energy.
E = ħω the photon energy and K its cutoff, in J; Eₙ, Eₘ atomic energies in J; α = e²/(4πε₀ħc) = 1/137.036.Imaginary part: the decay rate
Γₙ = −2 Im δEₙ/ħ = (4α/(3m²c²ħ)) Σₘ |pₘₙ|²(Eₙ − Eₘ), lower m only
Derived from 1/(x + i0) = P(1/x) − iπδ(x). Each term is Unit 12's ω³|d|²/(3πε₀ħc³): 6.27 × 10⁸ s⁻¹ for 2p, the measured 1.60 ns lifetime.
Γₙ in s⁻¹; only states m below n contribute, those whose denominator vanishes at a photon energy inside the integral.Photon integral for one intermediate state
P∫₀ᴷ E dE/(Eₙ − Eₘ − E) = D ln|1 + K/D| − K, with D = Eₘ − Eₙ
Derived from E/(−D − E) = −1 + D/(D + E). The −K is common to every m; the logarithm remembers where m lies. D = +1.89 eV, K = 1.00 keV give −988.1 eV.
D and K in eV or J; the principal value P matters only when D is negative, a state below n.The A² term at first order
(e²/2m)⟨0|A(0)²|0⟩ = αK²/(2πmc²)
Derived from the same mode sum with Σₛ|êₖₛ|² = 2. At K = mc² it is 593 eV for every level, so it cancels exactly from the 2s–2p interval.
K the photon-energy cutoff; the result is an energy in J, the same for every atomic state n.Interactive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. At Eₘ − Eₙ = −10.2 eV, the 2p → 1s path, drag E past 10.2 eV: the virtual state crosses |n, 0⟩, the integrand flips through its pole and Γ is 6.26 × 10⁸ s⁻¹. Then set Eₘ − Eₙ to +1.9 eV, 2s → 3p: the band lifts clear, the pole and Γ vanish, and every E pushes n down.
Live results
- DENOMINATOR Eₙ − Eₘ − E
- 4.95 eV
- INTEGRAND AT E
- 1.06
- POLE AT E (0 IF NONE)
- 10.2 eV
- Γ FROM THE POLE
- 6.26 10⁸ s⁻¹
Results follow this lesson’s model and assumptions. Displayed values are rounded.
Predict → change one variable → check
Plan a change before running it. Every other input stays at the saved baseline.
Saved baseline. Eₘ − Eₙ, intermediate state: −10.2 eV; Photon energy E: 5.25 eV
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
The tempting step is to carry over Unit 12's verdict that counter-rotating terms are negligible. For decay that verdict is right: a rate is fixed at the resonance E = Eₙ − Eₘ, where a kept term's denominator vanishes while a counter-rotating one, −(Eₘ − Eₙ) − E, is about twice the transition energy in size. A shift is fixed at no single photon energy. The weight E grows, the integral is collected almost entirely where E ≫ |Eₘ − Eₙ|, and there both kinds of denominator are close to −E, so a state above n counts as much as one below it. For a two-level atom 10.2 eV apart, with photons to 1 keV, the dropped ground-state term is 91% of the kept excited-state one, and the rotating-wave transition shift comes out 11 times too large. For hydrogen's 2s, with no dipole partner below, the rotating-wave model predicts no shift at all.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
Hydrogen's 3p level has two dipole partners below it. Given the quoted |⟨1s|r|3p⟩|² = (3⁶/2¹³)a₀² = 0.08899 a₀² and |⟨2s|r|3p⟩|² = 3.131 a₀², each for any one 3p substate, locate the poles of δE₃ₚ, use its imaginary part to find the decay rate and lifetime of 3p, and find the share of decays that end in 2s.
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.
- 3p's dipole partners are s and d states. Two lie below it: 1s, (8/9) × 13.606 eV = 12.094 eV down, and 2s, (5/36) × 13.606 eV = 1.8897 eV down. 2p is no partner, since ⟨2p|p|3p⟩ joins two p states; 3s and 3d are degenerate with 3p, so pₘₙ = imωₘₙrₘₙ = 0; and 4s, 4d, … lie above, with denominators negative at every E.
- So δE₃ₚ has two poles, at E = 12.094 eV and E = 1.8897 eV, and each one's −iπδ adds to Im δE₃ₚ. With |p|² = m²ω²|r|² and ω = E/ħ, Γ = −2 Im δE/ħ = Σ 4αω³|r|²/(3c²) = (4αc/(3(ħc)³)) Σ E³|r|²: one Unit 12 A coefficient per pole, derived.
- With c = 5.6653 × 10¹⁸ a₀ s⁻¹ and ħc = 3728.9 eV a₀, the constant is 1.0631 × 10⁶ s⁻¹ eV⁻³ a₀⁻². To 1s: 1.0631 × 10⁶ × (12.094)³ × 0.08899 = 1.673 × 10⁸ s⁻¹. To 2s: 1.0631 × 10⁶ × (1.8897)³ × 3.131 = 2.246 × 10⁷ s⁻¹. Dimensions: s⁻¹ eV⁻³ a₀⁻² × eV³ × a₀² = s⁻¹.
- Γ₃ₚ = 1.673 × 10⁸ s⁻¹ + 0.2246 × 10⁸ s⁻¹ = 1.90 × 10⁸ s⁻¹, so Im δE₃ₚ = −ħΓ/2 = −0.0625 µeV and τ = 1/Γ = 5.27 ns, the tabulated 3p lifetime. 2s receives 2.246/18.98 = 11.8% of the decays, as Balmer-α photons at 656 nm. Check: the same constant with 2p's 10.204 eV and 0.5549 a₀² gives 6.27 × 10⁸ s⁻¹, the measured 1.60 ns.
- The 2s pole carries 3.131/0.08899 = 35.18 times the |r|² of the 1s pole yet only 0.134 of its rate. The weight E in the integrand makes each residue scale as E³|r|², and (1.8897/12.094)³ = (5/32)³ = 3.81 × 10⁻³, so 35.18 × 3.81 × 10⁻³ = 0.134. One-partner limit: drop either pole and Γ is Unit 12's two-level rate for the other.
Answer. Poles at E = 12.094 eV (1s) and 1.8897 eV (2s); Γ₃ₚ = 1.90 × 10⁸ s⁻¹ and τ = 5.27 ns, with 11.8% of decays to 2s: that pole has 35 times the |r|², but E³ leaves it 0.134 of the 1s rate.
MediumExample 2
A model atom has two levels, g and e, 10.20 eV apart, joined by |peg|² = 4.155 × 10⁻⁶ m²c², hydrogen's value for 1s and one 2p substate. Keep photons up to K = 1.00 keV. From the full second-order formula find the energies of |g, 0⟩ and |e, 0⟩ and the shift of the transition, then say what a two-level rotating-wave model gives for each.
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.
- Each level has one partner. For e it is g, 10.20 eV below: Unit 12's term, with a pole. For g it is e, 10.20 eV above: the atom rises while a photon is emitted, a counter-rotating term. Both carry C = (2α/(3π))|peg|²/(m²c²) = 1.5485 × 10⁻³ × 4.155 × 10⁻⁶ = 6.434 × 10⁻⁹.
- With D = Eₘ − Eₙ the photon integral is D ln|1 + K/D| − K. For e, D = −10.20 eV: −1000 eV − 10.20 eV × ln(97.04) = −1000 eV − 46.67 eV = −1046.67 eV, a principal value. For g, D = +10.20 eV: −1000 eV + 10.20 eV × ln(99.04) = −953.13 eV.
- δE(e) = 6.434 × 10⁻⁹ × (−1046.67 eV) = −6.734 µeV, or −1628 MHz at 241.8 MHz per µeV. δE(g) = 6.434 × 10⁻⁹ × (−953.13 eV) = −6.132 µeV, or −1483 MHz: 91% of the kept term.
- The transition moves by δE(e) − δE(g) = 6.434 × 10⁻⁹ × (−93.54 eV) = −0.602 µeV, −145.5 MHz. The −K pieces cancel because both levels share one |peg|², leaving close to −2|D| ln(K/|D|). The rotating-wave model sets δE(g) = 0 and reports −1628 MHz, 11.2 times too large.
- Checks: the pole gives Im δE(e) = −π|D|C = −0.2062 µeV, so Γ = 2|Im δE(e)|/ħ = 6.26 × 10⁸ s⁻¹, within 0.1% of 2p's 6.27 × 10⁸ s⁻¹ (the gap is rounding), which the rotating-wave model gets right because a rate is decided at resonance. Dimensions: C is a pure number and both integrals are in eV.
Answer. δE(e) = −6.73 µeV (−1.63 GHz), δE(g) = −6.13 µeV (−1.48 GHz), transition shift −0.602 µeV (−146 MHz). The rotating-wave model gives δE(g) = 0 and a −1.63 GHz transition shift, 11 times too large, but the right rate, 6.26 × 10⁸ s⁻¹.
HardExample 3
An electron is bound in a three-dimensional isotropic harmonic well with ħω₀ = 2.00 eV, and photons are kept up to K = 500 eV, below ħc/b = 1.01 keV for the oscillator length b = √(ħ/(mω₀)). Find the ground state's second-order energy and the share of it from photons below ħω₀. Then find the decay rate of the first excited level |1, 0, 0⟩ and compare it with a Lorentz oscillator's classical damping rate e²ω₀²/(6πε₀mc³).
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.
- pₓ = i√(mħω₀/2)(aₓ† − aₓ) changes one Cartesian quantum number by one. From |0, 0, 0⟩ the three components reach only |1, 0, 0⟩, |0, 1, 0⟩ and |0, 0, 1⟩, all ħω₀ above, each with |pₘₙ|² = mħω₀/2, so 3mħω₀/2 in all. No partner lies below: no pole, and every term is one the rotating-wave model drops.
- With D = ħω₀: δE₀ = (2α/(3πm²c²))(3mħω₀/2)[ħω₀ ln(1 + K/ħω₀) − K] = (αħω₀/(πmc²))[ħω₀ ln(1 + K/ħω₀) − K], derived. The prefactor is 7.2974 × 10⁻³ × 2.00 eV ÷ (π × 5.1100 × 10⁵ eV) = 9.091 × 10⁻⁹, and the bracket is 2.00 eV × ln 251 − 500 eV = 11.05 eV − 500 eV = −488.95 eV.
- δE₀ = 9.091 × 10⁻⁹ × (−488.95 eV) = −4.445 µeV, or −1.075 GHz. The −K term alone is 102% of it, trimmed by the logarithm: the cutoff rules the integral, the next topic's subject. Photons below ħω₀ supply −ħω₀(1 − ln 2) = −0.614 eV of the −488.95 eV, 0.13%.
- |1, 0, 0⟩ has |0, 0, 0⟩ below it with |p|² = mħω₀/2, a pole at E = ħω₀. So Γ = (4α/(3m²c²ħ))(mħω₀/2)ħω₀ = (2α/3)(ħω₀)²/(mc²ħ) = (2/3) × 7.2974 × 10⁻³ × (2.00 eV)² ÷ (5.1100 × 10⁵ eV) × 1.5193 × 10¹⁵ s⁻¹ eV⁻¹ = 5.79 × 10⁷ s⁻¹, τ = 17.3 ns.
- Classically, e²ω₀²/(6πε₀mc³) becomes (2αħ/(3mc²))ω₀² once e² = 4πε₀αħc: the same expression, 5.79 × 10⁷ s⁻¹, with ħ gone from the result. This correspondence check holds at every ω₀. The level's other partners, |2, 0, 0⟩, |1, 1, 0⟩ and |1, 0, 1⟩, lie above and only shift it.
Answer. δE₀ = −4.45 µeV (−1.07 GHz) with K = 500 eV, all of it counter-rotating and 99.87% from photons above ħω₀; Γ(1, 0, 0) = 5.79 × 10⁷ s⁻¹ (τ = 17.3 ns), equal to the classical damping rate.
Quick check
Select an answer and explain it before checking.
Review & continue
The second-order energy of a bound state |n⟩ in the quantised field is the shift δEₙ from virtual emission and reabsorption: the electron emits a photon into any mode, passes through any atomic state m, and takes the photon back, each path weighted by its squared coupling over Eₙ − Eₘ − ħω.
- 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.