University Quantum Mechanics II · The Hydrogen Radial Equation Solved · 2.1

Centre of mass, relative coordinates, and the reduced mass

Every hydrogen result in Quantum Mechanics I pinned the proton in place. Here the pin comes out with no approximation: two bodies become a free centre of mass plus one particle of reduced mass μ, and μ alone sets isotope shifts, positronium's halved spectrum and the first correction to every hydrogen level.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Two particles that interact only through their separation have a Hamiltonian that cares how the pair is arranged, not where it is. Change variables to the mass-weighted centre R = (m₁r₁ + m₂r₂)/M and the separation r = r₁ − r₂, and that indifference becomes algebra: the chain rule turns the two kinetic energies into one for the total mass M and one for the reduced mass μ = m₁m₂/M, and the cross terms cancel exactly because R is weighted by mass. The centre then moves as a free particle and, apart from Doppler and recoil shifts, drops out of every spectral line, leaving Quantum Mechanics I's one-body problem with m replaced by μ.

Nothing has been approximated, for any mass ratio and any V(r₁ − r₂). The approximation was the habit of pinning the nucleus, and it costs a fraction m₁/M of every energy: 5.4 × 10⁻⁴ in hydrogen, ten times the α² ≈ 5.3 × 10⁻⁵ that sets the scale of fine structure, and half the answer in positronium. Since energies scale as μ and lengths as 1/μ, the spectra of two isotopes are exact rescalings of each other, which is how a 0.18 nm split in Balmer-α announced deuterium.

The separation stops being exact with a third body, where a coupling survives every change of variables; in a magnetic field, which couples the centre's motion to the internal motion; and in relativity, where kinetic energy is no longer quadratic in momentum. In hydrogen the last is smaller than the fine structure by a further factor of order mₑ/mₚ.

Simple definition: The reduced mass μ = m₁m₂/(m₁ + m₂) is the inertia of the separation r₁ − r₂ of two interacting particles: once the free centre-of-mass motion is split off, the pair's internal problem is exactly a one-body problem of mass μ in the potential V(r).
Example: Hydrogen has μ = 0.999 456 mₑ, so its ground state lies at −13.598 eV rather than the pinned-proton −13.606 eV, 7.4 meV higher; positronium has μ = mₑ/2, and its ground state lies at −6.80 eV.

Centre-of-mass and relative coordinatesR = (m₁r₁ + m₂r₂)/M, r = r₁ − r₂, M = m₁ + m₂

Inverted, r₁ = R + (m₂/M)r and r₂ = R − (m₁/M)r. The Jacobian has unit magnitude, so d³r₁ d³r₂ = d³R d³r and a normalised state stays normalised.

r₁, r₂, R, r: positions in m; m₁, m₂, M: masses in kg; r points from particle 2 to particle 1

The kinetic energy splits with no cross term∇₁²/m₁ + ∇₂²/m₂ = ∇_R²/M + ∇ᵣ²/μ

The chain rule's mixed terms, +(2/M)∇_R⋅∇ᵣ and −(2/M)∇_R⋅∇ᵣ, cancel identically. That cancellation is the whole separation, and it asks nothing of the masses.

∇ in m⁻¹, masses in kg; multiplied by −ħ²/2, with ħ in J s, every term is a kinetic energy in J

Reduced massμ = m₁m₂/(m₁ + m₂)

Hydrogen 0.999 455 7 mₑ, deuterium 0.999 727 6 mₑ, positronium 0.5 mₑ, muonic hydrogen 185.84 mₑ: only equal or comparable masses move μ far from the lighter mass.

kg; between m₁/2 and m₁ when m₂ ≥ m₁, reaching m₁ only as m₂ → ∞, the pinned-centre limit

The internal equation−(ħ²/2μ)∇ᵣ²ψ(r) + V(r)ψ(r) = Eψ(r)

Quantum Mechanics I's one-body equation with m → μ, so every one-body result carries over by that single substitution; the centre's free energy enters a line only through Doppler and recoil shifts.

ψ in m^(−3/2); E and V in J; the full state is e^(iK⋅R)ψ(r), K in m⁻¹, with total energy ħ²K²/2M + E

Hydrogen-like levels with the reduced massEₙ = −μc²(Zα)²/(2n²)

Every level carries the same factor μ/mₑ = 1/(1 + mₑ/m₂), so hydrogen's spectrum is the pinned-proton one shrunk by 1 part in 1837: the ground state rises 7.41 meV.

μc² in J; α ≈ 1/137.036, Z and n = 1, 2, 3, … dimensionless; quoted from QM I, derived in topic 2.5

Isotope shift of one lineλ_H − λ_D = λ∞(mₑ/mₚ − mₑ/m_d)

Each wavelength is λ∞(1 + mₑ/m₂), so Balmer-α splits by 656.112 nm × 2.7217 × 10⁻⁴ = 0.1786 nm, the displacement that revealed deuterium.

λ∞: the line for an infinitely heavy nucleus, in m (656.112 nm for Balmer-α); mass ratios dimensionless

Change variables, and the cross terms cancel

Start from H = −(ħ²/2m₁)∇₁² − (ħ²/2m₂)∇₂² + V(r₁ − r₂), acting on Ψ(r₁, r₂) = ⟨r₁, r₂|Ψ⟩. Work one Cartesian component at a time with X = (m₁x₁ + m₂x₂)/M and x = x₁ − x₂. The chain rule gives ∂/∂x₁ = (m₁/M)∂/∂X + ∂/∂x and ∂/∂x₂ = (m₂/M)∂/∂X − ∂/∂x. Square each and divide by its mass. The ∂²/∂X² pieces add to (m₁ + m₂)/M² = 1/M, the ∂²/∂x² pieces to 1/m₁ + 1/m₂ = 1/μ, and the mixed pieces, +(2/M)∂²/∂X∂x and −(2/M)∂²/∂X∂x, cancel. Nothing about the masses was assumed. In momentum language P = p₁ + p₂ and p = (m₂p₁ − m₁p₂)/M = μ(v₁ − v₂); one line of commutators gives [X, Pₓ] = [x, pₓ] = iħ and [X, pₓ] = [x, Pₓ] = 0, so the new pairs are canonical and p₁²/2m₁ + p₂²/2m₂ = P²/2M + p²/2μ holds as an operator identity. The Jacobian of (x₁, x₂) → (X, x) has determinant −1, so d³r₁ d³r₂ = d³R d³r and normalisation carries over. Every step here is derived.

The centre of mass drops out, for any V(r₁ − r₂)

Because V depends only on r₁ − r₂, H commutes with P: translation invariance, stated as an operator fact. The product Ψ(R, r) = e^(iK⋅R)ψ(r) then separates the equation exactly. The centre moves freely with energy ħ²K²/2M, which the atom carries as a whole. It reaches a spectral line only through the photon's momentum: as a Doppler shift from the atom's motion, and as a recoil shift, a fraction hf/2Mc² of the line, 1.0 × 10⁻⁹ for Balmer-α. The internal part obeys −(ħ²/2μ)∇²ψ + V(r)ψ = Eψ. Notice what was not used. V need not be Coulomb and need not even be central; being central matters only for the next step, the separation in angles. The same algebra fixes a diatomic molecule's vibration at ω = √(k/μ), with k the curvature of the internuclear potential, which is the same for two isotopes in the Born–Oppenheimer picture (quoted). The tabulated harmonic constants of H³⁵Cl and D³⁵Cl, ωₑ = 2990.9 and 2145.2 cm⁻¹, stand in the ratio 1.394, and √(μ(DCl)/μ(HCl)) from the atomic masses is also 1.394.

Where the two particles actually are

Invert the definitions: r₁ = R + (m₂/M)r and r₂ = R − (m₁/M)r. The relative wavefunction ψ(r) is the amplitude for the separation, not for the electron, and the two particles sit on opposite sides of the centre at fixed fractions of r: in hydrogen the electron at 0.999 456 r and the proton at 5.44 × 10⁻⁴ r. Both move; the proton moves little only because it is heavy. Every one-body length now carries 1/μ. QM I's Bohr radius becomes a_μ = 4πε₀ħ²/(μe²) = a₀(mₑ/μ): 52.947 pm for hydrogen against 52.918 pm for a pinned proton, and 2a₀ = 105.8 pm for positronium. Its dimensions are (C² J⁻¹ m⁻¹)(J² s²)/(kg C²) = J s² kg⁻¹ m⁻¹ = m, a length. Two limits check the algebra. As m₂ → ∞, R lands on the heavy particle and μ → m₁, returning Quantum Mechanics I's fixed-centre problem. With m₁ = m₂, R sits at the midpoint, μ = m₁/2, and each particle is r/2 from the centre.

Every level rises by the same fraction

Because the internal equation is the one-body equation with m → μ, dimensional analysis alone fixes how energies scale: the only energy that can be built from μ, e²/4πε₀ and ħ is μ(e²/4πε₀ħ)² = μc²α², so every bound energy is proportional to μ (derived). Quantum Mechanics I quoted the coefficient, Eₙ = −mc²α²/2n², here with m → μ; topic 2.5 derives it. Every hydrogen level is therefore multiplied by μ/mₑ = 1 − 1/1837.15: the ground state moves from −13.605 69 eV to −13.598 29 eV, up 7.41 meV. The measured ionisation energy is 13.598 43 eV (tabulated), so the reduced mass carries the prediction across the 7.26 meV gap and 0.15 meV beyond it. That remainder is the fine structure, 0.18 meV deeper at order α², less the Lamb shift, 0.03 meV back at order α³, both quoted from later units. As a fraction of the level the reduced-mass correction is mₑ/M = 5.4 × 10⁻⁴, about ten times α² = 5.3 × 10⁻⁵. It is the first correction to make, and it carries no power of α: it is a mass ratio, not relativity.

Hydrogen against deuterium

Every wavelength scales as 1/μ, so a line computed for an infinitely heavy nucleus, λ∞, becomes λ∞(1 + mₑ/m₂) for a nucleus of mass m₂. For Balmer-α, 3 → 2, λ∞ = 36/(5R∞) = 656.112 nm. Hydrogen, with mₚ = 1836.153 mₑ, gives 656.470 nm; deuterium, with m_d = 3670.483 mₑ, gives 656.291 nm. The split is 0.179 nm, the displacement Urey, Brickwedde and Murphy measured in 1932 and read as a hydrogen isotope of mass 2. Hydrogen's measured vacuum wavelength, 656.46 nm, sits about 0.01 nm below the reduced-mass value, a fraction near 10⁻⁵ set by the order-α² fine structure, which is nearly identical in the two isotopes and so drops out of the split. The sharpest test is the 1S–2S interval, measured by two-photon laser spectroscopy in both atoms: the reduced mass alone predicts an isotope shift of 671.007 GHz, against 670.994 GHz measured. The 12.7 MHz remainder, 1.9 × 10⁻⁵ of the shift, is of order α²: the fine structure's own mass dependence, relativistic recoil, the Lamb shift and the deuteron's larger radius, all quoted.

Positronium, and where the model stops

Equal masses make the correction as large as it can be: μ = mₑ/2, so Eₙ = −6.803 eV/n², the size doubles, and Lyman-α moves from 121.568 nm in hydrogen to 243.00 nm, where positronium's Lyman-α was first observed in 1975. Positronium also exposes what the separation leaves out. Both particles are light, so relativistic corrections are not suppressed by a heavy partner, and the pair can annihilate, a coupling to photons that only QED supplies. Each assumption can be sized by scaling. Relativistic recoil beyond μ is smaller than the fine structure by a further factor of order m₁/m₂: 5 × 10⁻⁴ in hydrogen, 1 in positronium. The point-nucleus assumption worsens as μ grows, because an S level's finite-size shift goes as μ³r_N², or (μr_N)² relative to the binding: muonic hydrogen, with μ = 185.84 mₑ and a_μ = 285 fm, feels it, relative to its binding, 3.5 × 10⁴ times more than hydrogen. A third body breaks the separation itself: in helium, with both electrons measured from the nucleus, the kinetic energy keeps a mass-polarisation term −(ħ²/M_N)∇₁⋅∇₂, and Jacobi coordinates that remove it only move the coupling into the potential, so no change of variables separates the problem.

Change one variable at a time

Make the relationship visible.

Interactive model

Partner mass, log₁₀(m₂/mₑ)3.2639

Start at hydrogen, 3.2639, where the two ladders look identical although E₁ has risen 7.4 meV; step to 3.5647, deuterium, and Balmer-α shortens by 0.18 nm; then slide to 0, positronium, where the centre sits midway and both levels halve.

REDUCED MASS μ/mₑ0.9995

GROUND STATE E₁-13.5983 eV

BOHR RADIUS a_μ52.947 pm

BALMER-α (VACUUM)656.470 nm

Live interpretationREDUCED MASS μ/mₑ: 0.9995. GROUND STATE E₁: −13.5983 eV. BOHR RADIUS a_μ: 52.947 pm. BALMER-α (VACUUM): 656.470 nm

Catch the common trap

Explain before calculating.

Muonium is an electron bound to a positive muon of mass 206.77 mₑ. How does its ground-state energy compare with hydrogen's, −13.598 eV?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyIn hydrogen's ground state the mean electron–proton separation is ⟨r⟩ = 3a_μ/2, quoted from Quantum Mechanics I and derived in topic 2.8. With a₀ = 52.9177 pm for a pinned proton and mₚ = 1836.15 mₑ, find a_μ, ⟨r⟩ and the proton's mean distance from the centre of mass, and compare that distance with the proton's charge radius, 0.841 fm.

Lengths scale as 1/μ, and mₑ/μ = 1 + mₑ/mₚ = 1.000 5446, so a_μ = 52.9177 pm × 1.000 5446 = 52.9465 pm.
Mean separation: ⟨r⟩ = 1.5 × 52.9465 pm = 79.420 pm. This is the average length of r = r₁ − r₂, the argument of ψ(r), not the electron's distance from anything fixed.
The proton sits at r₂ − R = −(mₑ/M)r, so its distance from the centre is exactly |r|/(1 + mₚ/mₑ) = |r|/1837.15. Averaging, 79.420 pm/1837.15 = 0.043 23 pm = 43.2 fm. The electron's mean distance is the rest, 79.377 pm.
Compare: 43.2 fm/0.841 fm = 51. Checks: a ratio of two lengths is dimensionless, and as mₚ → ∞ the proton's excursion goes to zero, the pinned proton of Quantum Mechanics I.

Answera_μ = 52.95 pm and ⟨r⟩ = 79.42 pm; the proton's mean distance from the centre of mass is 43.2 fm, about 51 times its own charge radius.

MediumThe ⁴He⁺ ion is one electron bound to a Z = 2 nucleus of mass 7294.30 mₑ. Find its reduced mass and its ground-state energy with the nucleus pinned and with it free, and compare both with the measured value, −54.417 765 eV. Which is larger here: the reduced-mass correction, or the fine-structure shift of 1s, (Zα)²/4 of the level?

The nuclear charge sits in V(r) = −Ze²/(4πε₀r), which still depends only on r₁ − r₂, so the separation goes through unchanged: μ/mₑ = 7294.30/7295.30 = 0.999 862 93.
Energies scale as Z²μ. Pinned: E₁ = −4 × 13.605 693 eV = −54.422 772 eV. Free: E₁ = −54.422 772 eV × 0.999 862 93 = −54.415 312 eV, higher by 7.460 meV.
Against −54.417 765 eV measured, the pinned value is 5.007 meV too deep and the reduced-mass value 2.453 meV too shallow, a fraction 4.5 × 10⁻⁵.
Order in α: the Dirac equation (quoted) lowers 1s by (Zα)²/4 of the level, (2/137.036)²/4 × 54.4153 eV = 2.898 meV. That overshoots the residual by 0.445 meV, the size of the Lamb shift that raises 1s at order α(Zα)² of the level (quoted from QED).
Compare: the reduced-mass correction, 7.460 meV, is 2.6 times the fine structure, 2.898 meV. In hydrogen it is 41, 7.41 meV against 0.18 meV, because mₑ/M falls with nuclear mass while (Zα)² grows with nuclear charge.

Answerμ = 0.999 863 mₑ; E₁ = −54.4153 eV free against −54.4228 eV pinned. The reduced mass (7.46 meV) still beats the fine structure (2.90 meV), but by a factor of 2.6, against 41 in hydrogen.

HardPositronium's 1S–2S interval is driven by two counter-propagating laser photons of equal frequency, each supplying half the energy; the measured interval is 1 233 607 216 MHz. Predict the interval and the laser wavelength from the reduced mass, express the disagreement as a multiple of α², and find what the prediction would have been with the positron treated as a fixed centre.

μ = mₑmₑ/(2mₑ) = mₑ/2, so every level is half hydrogen's pinned-nucleus value: Eₙ = −(13.605 693 eV/2)/n² = −6.802 847 eV/n².
Interval: E₂ − E₁ = 6.802 847 eV × (1 − 1/4) = 5.102 135 eV. As a frequency, f = ΔE/h = 5.102 135 eV/(4.135 668 × 10⁻¹⁵ eV s) = 1.233 6907 × 10¹⁵ Hz.
Each photon carries f/2, so λ = 2c/f = 2 × 2.997 925 × 10⁸ m s⁻¹/(1.233 6907 × 10¹⁵ Hz) = 486.009 nm. The measured interval corresponds to 486.042 nm.
Disagreement: (1.233 6907 − 1.233 6072)/1.233 6072 = 6.77 × 10⁻⁵, and α² = 5.325 × 10⁻⁵, so the residual is 1.27 α². Order α² is what relativity and QED, both quoted, should leave once the non-relativistic two-body problem is solved exactly.
A pinned positron means μ → mₑ: the interval would be 10.204 eV, or 2.467 × 10¹⁵ Hz, twice the measured value. Here the reduced mass is not a correction; it is half the answer.

AnswerΔE = 5.1021 eV, f = 1.233 69 × 10¹⁵ Hz and a 486.01 nm laser; the measurement lies 6.8 × 10⁻⁵ lower, 1.27 α². Pinning the positron would double the prediction.