University Quantum Mechanics II · Emission and Absorption of Radiation · 13.4

The 2p lifetime of hydrogen computed

This is where two courses of machinery finally produce a number you can hold against an experiment. The work is short but unforgiving: split one matrix element into an angle and a radius, keep all three components of the dipole, do one γ-function integral, then test the answer by units, by powers of α and against the clock.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Topic 13.03 left a formula, A = ω³|⟨f|d|i⟩|²/(3πε₀ħc³), obtained from B by detailed balance with Planck's law and reproduced by quantum electrodynamics, both quoted here. It turns a lifetime into two atomic inputs: the transition frequency, which enters cubed, and the dipole matrix element, a vector whose three components all count because a spontaneous photon may leave in any direction with either polarisation. For hydrogen 2p → 1s both come from Unit 2's functions.

The matrix element splits into an angular integral, 1/3 for every m, and a radial integral of R₁₀R₂₁r³ worth (128√6/243) a₀, so |⟨100|r|21m⟩|² = (2¹⁵/3¹⁰) a₀² = 0.555 a₀². With ħω = 10.2 eV every constant collapses to A = (2/3)⁸α⁵mc²/ħ = 6.27 × 10⁸ s⁻¹, τ = 1.595 ns, against 1.60 ns measured. The model is nonrelativistic hydrogen in a classical field at dipole order, spin carried along untouched and the spontaneous rate imported; fine structure, reduced mass, relativity and retardation each move τ by under 0.1%.

It stops working where an input fails: a state with no dipole-allowed partner below it, like 2s; a radio-frequency channel, which ω³ silences, so 2p → 2s runs 10⁻¹⁵ as fast; heavy ions, where Zα is not small; and an atom outside free space, since A inherits the free-space mode density that Planck's law carries.

Simple definition: The 2p lifetime is τ = 1/A, the inverse of the spontaneous-emission rate to 1s, which is fixed by the cube of the Lyman-α frequency and by the squared dipole matrix element e²|⟨100|r|21m⟩|², the same for all three m.
Example: With |⟨100|r|21m⟩|² = 0.555 a₀² and ħω = 10.2 eV, A = 6.27 × 10⁸ s⁻¹, so a cloud of 2p atoms falls to 1/e of its number in 1.595 ns and to 1% in 7.35 ns, ringing about 3.9 million optical cycles per lifetime.

Spontaneous rate (quoted from 13.03)A = ω³|⟨f|d|i⟩|² / (3πε₀ħc³)

13.03 got it from B by detailed balance, borrowing Planck's law; quantum electrodynamics agrees (quoted). The 3 already averages over direction and polarisation.

ω in s⁻¹, d = −er in C m, ε₀ in F m⁻¹, ħ in J s, c in m s⁻¹; A in s⁻¹. |⟨f|d|i⟩|² adds all three Cartesian parts.

Angular factor, equal for every m (derived)|⟨Y₀⁰|r̂|Y₁ᵐ⟩|² = 1/3 for m = −1, 0, +1

Completeness gives Σₘ|⟨Y₀⁰|r̂|Y₁ᵐ⟩|² = ⟨Y₀⁰|r̂⋅r̂|Y₀⁰⟩ = 1, and a rotation-invariant sum cannot favour one m, so each takes 1/3 and the lifetime ignores the choice of axis.

Dimensionless, with r̂ = r/r. For m = 0 it is all z; for m = ±1 it is 1/6 from x plus 1/6 from y.

Radial integral (derived)∫ R₁₀R₂₁ r³ dr (r from 0 to ∞) = (128√6/243) a₀ = 1.290 a₀

One gamma integral, ∫ r⁴ e^(−βr) dr = 4!/β⁵ with β = 3/(2a₀). One power of r is the dipole operator; the other two are the volume element.

a₀ = 5.29 × 10⁻¹¹ m. Each R carries m^(−3/2) and r³ dr carries m⁴, so the result is a length.

Squared dipole element (derived)|⟨100|r|21m⟩|² = (2¹⁵/3¹⁰) a₀² = 0.5549 a₀²

The radial integral squared, 2¹⁵/3⁹, times the angular 1/3. The transition dipole has magnitude 0.745 ea₀ = 6.32 × 10⁻³⁰ C m.

1.554 × 10⁻²¹ m²; times e², |⟨f|d|i⟩|² = 3.989 × 10⁻⁵⁹ C² m², the same for m = −1, 0 and +1.

The 2p rate in closed form (derived)A = (2/3)⁸ α⁵ mc²/ħ = 6.27 × 10⁸ s⁻¹

Fifth order in α. A/ω = (2/3)⁸(8/3)α³ = 4.0 × 10⁻⁸, so ω/(2πA) = 3.9 million cycles pass before it decays; other hydrogen-like systems follow A ∝ Z⁴μ.

α = 1/137.036 and mc²/ħ = 7.763 × 10²⁰ s⁻¹; τ = 1/A = 1.595 ns, against 1.60 ns measured (quoted).

Suppression of the 2s branch (derived, gap quoted)A(2p→2s)/A(2p→1s) = (ω′/ω)³ × 9a₀²/(0.5549 a₀²) = 1.05 × 10⁻¹⁵

A squared dipole 16 times larger cannot beat ω³: this branch alone would take 18 days. The same cube is why radio-frequency transitions are so slow.

ω′/ω = 9.91 GHz ÷ 2.467 × 10¹⁵ Hz = 4.02 × 10⁻⁶ for 2p₃/₂ → 2s₁/₂; 9a₀² is Unit 9's Stark element 3a₀, squared.

What the rate needs, and what is quoted

The rate formula is not derived here. Topic 13.03 obtained A = ω³|⟨f|d|i⟩|²/(3πε₀ħc³) from the absorption coefficient B by Einstein's detailed balance, which borrows Planck's law, and quantum electrodynamics gives the same result; both are quoted. A lifetime therefore needs two atomic numbers. The first is ω, from Unit 2's levels: ħω = (13.606 eV)(1 − 1/4) = 10.204 eV, so ω = 1.5503 × 10¹⁶ s⁻¹ and λ = 121.5 nm. The second is d = −e⟨100|r|21m⟩, and it is a vector. All three Cartesian components count, because the isotropic average behind B, carried into A by 13.03, already covered both polarisations and every direction, which is where the 3 in the denominator came from. Spin rides along: r does not act on spin, so spin-up 2p reaches spin-up 1s with overlap 1, and there is exactly one final state. The units close: s⁻³ × C² m² over (C² J⁻¹ m⁻¹)(J s)(m³ s⁻³) is s⁻¹.

Split the matrix element into radius and angle

Each state is a radial function times a spherical harmonic, ψ₁₀₀ = R₁₀Y₀⁰ and ψ₂₁ₘ = R₂₁Y₁ᵐ, quoted from Units 1 and 2, where they were derived. Write the operator the same way, r = r r̂: a length times a unit vector that depends on direction only. With d³r = r² dr dΩ the integral separates, ⟨100|r|21m⟩ = [∫ R₁₀R₂₁ r³ dr] × [∫ Y₀⁰* r̂ Y₁ᵐ dΩ], the radial integral running from 0 to ∞. Pause on the r³. One power is the dipole operator and two are the volume element, so the radial integrand leans further out, towards the 2p cloud, than the bare overlap r²R₁₀R₂₁ does: it peaks at 8a₀/3 rather than 2a₀. The angular factor is a vector of pure numbers and the radial factor is a length, so the product is a length, as a dipole moment divided by a charge must be. Everything that depends on m and on polarisation lives in the angles; everything that depends on the size of the atom lives in the radius, and the two can be done separately.

The angular factor is one-third for every m

For m = 0 the z-component of r̂ is cos θ, and with Y₀⁰ = 1/√(4π) and Y₁⁰ = √(3/4π) cos θ the integral is (√3/4π) ∫ cos²θ dΩ = (√3/4π)(4π/3) = 1/√3; the x and y parts vanish in the φ integral. For m = +1, Y₁¹ = −√(3/8π) sin θ e^(iφ): now z vanishes, x and y each have magnitude 1/√6, and 1/6 + 1/6 is again 1/3. A reason beats both integrals. Each component of r̂ acting on Y₀⁰ makes only l = 1 functions, so summing over m inserts a complete set: Σₘ |⟨Y₀⁰|r̂|Y₁ᵐ⟩|² = ⟨Y₀⁰|r̂⋅r̂|Y₀⁰⟩ = 1. Each term is the expectation value in Y₁ᵐ of Σᵢ r̂ᵢ|Y₀⁰⟩⟨Y₀⁰|r̂ᵢ, an operator rotations leave unchanged, so on the three l = 1 states it is a multiple of the identity: no m is favoured, and each takes 1/3 (derived). The lifetime is therefore the same for all three states, as it must be: m = 0 about z is a superposition of m = ±1 about x, and empty space has no preferred axis. Which direction and polarisation the photon favours does depend on m; that is topic 13.05.

The radial integral is one gamma function

Unit 2's functions are R₁₀ = 2a₀^(−3/2) e^(−r/a₀) and R₂₁ = a₀^(−3/2)(r/a₀)e^(−r/2a₀)/√24, quoted. Their product with r³ is (2/√24) a₀⁻⁴ r⁴ e^{(−3r/2a₀)}: the exponentials merge into one with β = 3/(2a₀), and the polynomial is a single power. The only integral needed is ∫ rⁿ e^(−βr) dr = n!/βⁿ⁺¹ from 0 to ∞, so the result is (2/√24) a₀⁻⁴ × 4! × (2a₀/3)⁵ = (128√6/243) a₀ = 1.2903 a₀ (derived). The integrand peaks at r = 4/β = 8a₀/3, between the peaks of rR₁₀ at a₀ and rR₂₁ at 4a₀. Half the integral has built up by 3.11 a₀, but half of its square, which is what the rate uses, only by 3.96 a₀: the outer atom matters more than the inner region where the 1s density sits. Squaring gives (2¹⁵/3⁹) a₀² = 1.665 a₀². Units: each R is m^(−3/2) and r³ dr is m⁴, so the integral is a length.

Assemble the rate, then test it three ways

The product is |⟨100|r|21m⟩|² = (1/3)(2¹⁵/3⁹) a₀² = (2¹⁵/3¹⁰) a₀² = 0.5549 a₀² = 1.554 × 10⁻²¹ m², so |⟨f|d|i⟩|² = 3.989 × 10⁻⁵⁹ C² m². Then A = (1.5503 × 10¹⁶ s⁻¹)³(3.989 × 10⁻⁵⁹ C² m²)/(3πε₀ħc³) = 6.27 × 10⁸ s⁻¹ and τ = 1.595 ns (derived), against 1.60 ns measured (quoted). Order in α: write e²/4πε₀ = αħc, ħω = (3/8)α²mc² and a₀ = ħ/αmc, and every constant collapses to A = (2/3)⁸α⁵mc²/ħ, fifth order in α times mc²/ħ = 7.76 × 10²⁰ s⁻¹. Validity: A/ω = (2/3)⁸(8/3)α³ = 4.0 × 10⁻⁸, so the atom oscillates ω/(2πA) = 3.9 million times per lifetime, the separation of time scales the golden rule needs. Limiting case: energies go as Z²μ and lengths as 1/(Zμ), so A ∝ Z⁴μ, and He⁺ 2p lives 1.595/16 = 0.0997 ns.

The other exit, and everything left out

Besides 1s, the only dipole-allowed partner of 2p that is not above it is 2s, and the matrix element to it is larger: Unit 9's Stark element 3ea₀ gives |⟨200|r|21m⟩|² = 9a₀², 16 times the 1s value. But the two levels are degenerate in this model, and with Unit 8's fine structure and the Lamb shift, which that unit quoted from QED, only 2p₃/₂ lies above 2s₁/₂, by 9.91 GHz. Against Lyman-α's 2.467 × 10¹⁵ Hz that is 4.0 × 10⁻⁶, which ω³ turns into 6.5 × 10⁻¹⁷: the branch carries 1.05 × 10⁻¹⁵ of the decay. The same cube is why radio-frequency transitions are so slow. The rest of what was neglected is smaller than the precision of the three-figure measurement: the reduced mass multiplies A by μ/m = 0.99946, so τ = 1.596 ns; fine structure moves ω by about 1.5 × 10⁻⁵; relativistic corrections to the matrix element are of order α² = 5 × 10⁻⁵ (quoted); and 13.01's retardation terms are of order (ka₀)² = 7.5 × 10⁻⁶.

Change one variable at a time

Make the relationship visible.

Interactive model

Upper limit R of the radial integral4.0 a₀

Nuclear charge Z (hydrogen-like ion)1

At Z = 1 drag R outward: the integral is half built near 3.1 a₀, but its square, which sets A, only near 4.0 a₀, out where the 2p radial density peaks. Raise Z and the curve contracts towards r = 0 as 1/Z while A grows as Z⁴.

RADIAL INTEGRAL FROM 0 TO R0.922 a₀

SHARE OF |d|² INSIDE R51.1 %

FULL RATE A6.27 × 10⁸ s⁻¹

LIFETIME τ = 1/A1.5953 ns

Live interpretationRADIAL INTEGRAL FROM 0 TO R: 0.922 a₀. SHARE OF |d|² INSIDE R: 51.1 %. FULL RATE A: 6.27 × 10⁸ s⁻¹. LIFETIME τ = 1/A: 1.5953 ns

Catch the common trap

Explain before calculating.

Misconception check

Fresh from the Δm rules, students often compute the 2p decay with ⟨100|z|21m⟩ alone. For m = 0 that happens to give the right number, but for m = ±1 it gives zero, and the student concludes those states are metastable, or that the three m states have different lifetimes. Both conclusions are wrong. A single component belongs to absorption from a beam polarised along z, where ε̂⋅d picks out one direction. A spontaneous photon has no prescribed polarisation or direction: A sums over both polarisations and integrates over all directions, which is what put the 3 into 3πε₀ħc³ and left the full |⟨f|d|i⟩|² = e²(|x_fi|² + |y_fi|² + |z_fi|²). For m = ±1 the x and y parts carry (2¹⁵/3¹⁰) a₀² between them, exactly what z carries for m = 0. It has to be so: m = 0 about z is a superposition of m = ±1 about x, and free space has no preferred axis, so no rate can depend on which axis was called z.

A hydrogen atom is prepared in 2p with m = +1, and a student finds ⟨100|z|211⟩ = 0. Which value of |⟨100|r|211⟩|² belongs in A = ω³e²|⟨100|r|211⟩|²/(3πε₀ħc³), and what lifetime follows?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyIn muonic hydrogen a negative muon, m_μ = 206.768 mₑ, is bound to a proton, mₚ = 1836.15 mₑ. Scale the hydrogen result A = (2/3)⁸α⁵mc²/ħ = 6.268 × 10⁸ s⁻¹ to find the 2p → 1s photon energy, the 2p lifetime, and the chance that the muon, whose own lifetime is 2.197 μs, decays before the 2p state does.

Scale first. Replacing m by the reduced mass μ multiplies every energy by μ/m, so ω ∝ μ, and divides every length, so a ∝ 1/μ and |⟨100|r|21m⟩|² ∝ 1/μ². The pure numbers 2¹⁵/3¹⁰ and 1/3 are untouched, so A ∝ ω³|r|² ∝ μ³/μ² = μ.
Reduced mass: μ = (206.768 × 1836.15)/(206.768 + 1836.15) mₑ = 185.84 mₑ.
Photon: ħω = 185.84 × 10.204 eV = 1896 eV, a 1.90 keV X-ray at 0.654 nm. The size parameter k a = 3α/8 = 2.7 × 10⁻³ contains no mass, so the dipole approximation is exactly as good as in hydrogen.
Rate: A = 185.84 × 6.268 × 10⁸ s⁻¹ = 1.165 × 10¹¹ s⁻¹, so τ = 1/A = 8.58 ps.
Competition: the muon decays at 1/(2.197 μs) = 4.55 × 10⁵ s⁻¹, so the chance it goes first is 4.55 × 10⁵/(1.165 × 10¹¹ + 4.55 × 10⁵) = 3.9 × 10⁻⁶.

Answerħω = 1.90 keV; A = 1.17 × 10¹¹ s⁻¹ and τ = 8.6 ps, 186 times shorter than hydrogen's; only about 4 muons in a million decay before the X-ray leaves.

MediumIn hydrogen, 2p₃/₂ lies 9.91 GHz above 2s₁/₂ (fine structure minus Lamb shift, quoted). Using R₂₀ = a₀^(−3/2)(1 − r/2a₀)e^(−r/2a₀)/√2 and R₂₁, compute |⟨200|r|21m⟩|² by hand, then find the 2p₃/₂ → 2s₁/₂ rate, its branching ratio against 2p → 1s, and how long that channel alone would take.

Angles: 2s carries Y₀⁰, exactly like 1s, so the angular factor is again 1/3 for every m, and spin rides along as before.
Radius: R₂₀R₂₁r³ = (1/√48) a₀⁻⁴ (1 − r/2a₀) r⁴ e^(−r/a₀). With ∫ rⁿ e^(−r/a₀) dr = n! a₀ⁿ⁺¹, the integral is (1/√48)(4! − 5!/2) a₀ = (24 − 60) a₀/√48 = −3√3 a₀ = −5.196 a₀. The 2s node at 2a₀ lets the outer, negative lobe win.
So |⟨200|r|21m⟩|² = (1/3)(27 a₀²) = 9 a₀², Unit 9's Stark element 3ea₀ squared, and 9/0.5549 = 16.22 times the 1s value.
Frequencies: Lyman-α is 10.204 eV/h = 2.467 × 10¹⁵ Hz, so ω′/ω = (9.91 × 10⁹)/(2.467 × 10¹⁵) = 4.02 × 10⁻⁶ and (ω′/ω)³ = 6.48 × 10⁻¹⁷.
Rate: A′ = 6.268 × 10⁸ s⁻¹ × 16.22 × 6.48 × 10⁻¹⁷ = 6.6 × 10⁻⁷ s⁻¹, a branching ratio of 1.05 × 10⁻¹⁵. Alone it would need 1/A′ = 1.5 × 10⁶ s, about 18 days, to empty the state.

Answer|⟨200|r|21m⟩|² = 9 a₀² (radial integral −3√3 a₀); A(2p₃/₂ → 2s₁/₂) = 6.6 × 10⁻⁷ s⁻¹, a branch of 1.05 × 10⁻¹⁵ and about 18 days on its own, so the 2p lifetime is the 1s channel alone.

HardHydrogen's 3d level decays only to 2p. With R₃₂ = 4a₀^(−3/2)(r/a₀)²e^(−r/3a₀)/(81√30) and R₂₁, compute the radial integral, get the angular factor by starting from m = 2, and find A(3d → 2p) and the 3d lifetime. Explain why 1s and 2s are not open to it.

Channels: Δl = ±1 allows only l = 1 below 3d, which is 2p. 3d → 1s and 3d → 2s have Δl = 2 and vanish at dipole order, and 3d → 3p is degenerate here, ω³-suppressed like 2p → 2s. So A(3d) is a single rate.
Radius: R₃₂R₂₁r³ = [4/(81√720)] a₀⁻⁶ r⁶ e^{(−5r/6a₀)}, with 4/(81√720) = 1.8404 × 10⁻³. The gamma integral gives 6!(6a₀/5)⁷ = 2579.9 a₀⁷, so the radial integral is 4.748 a₀, exactly (2¹¹3⁴/(5⁶√5)) a₀, and its square is 22.54 a₀².
Angles: the total rate is the same for every m, so take m = 2. r̂ acting on Y₂² reaches l = 1 only through Y₁¹; with Y₂² = √(15/32π) sin²θ e^(2iφ), the x and y components each give |⟨Y₁¹|r̂|Y₂²⟩|² = 1/5 and z gives 0. The factor 2/5 is the l/(2l+1) (quoted) that also gives 1/3 for l = 1.
So |⟨2p|r|3d⟩|², summed over the final 2p states, is (2/5)(22.54 a₀²) = 9.017 a₀², 16.25 times the 2p → 1s value.
Frequency: ħω = 13.606 eV × (1/4 − 1/9) = 1.8897 eV, Balmer-α at 656.1 nm (measured 656.5 nm in vacuum; the 0.05% gap is the reduced mass). Scaling from 2p: A = 6.268 × 10⁸ s⁻¹ × (1.8897/10.204)³ × 16.25 = 6.268 × 10⁸ s⁻¹ × 6.351 × 10⁻³ × 16.25 = 6.47 × 10⁷ s⁻¹.
Lifetime: τ = 1/A = 15.5 ns, 9.7 times the 2p value, because a 16 times larger squared dipole loses to a 157-fold smaller ω³. The tabulated 6.465 × 10⁷ s⁻¹ (quoted) is this number times μ/m = 0.99946.

AnswerRadial integral 4.748 a₀, angular factor 2/5, |r|² = 9.02 a₀²; A(3d → 2p) = 6.47 × 10⁷ s⁻¹ and τ(3d) = 15.5 ns.