University Physics IV · Limits of Classical Physics · 1.9
Old quantum theory and correspondence
Between Planck and Schrödinger sits a decade of rules that half-worked. Quantise the loop integral of momentum, then force the result to agree with Newton at large quantum numbers. Here is the recipe, Bohr and Planck in three lines each from it, and the exact point where it collapsed.
Build the model
Connect the measurement to the mechanism.
Between 1913 and 1925 physics ran on a hybrid: keep Newton's orbits and Maxwell's radiation, but admit only those orbits whose action integral ∮p dq closes on a whole number of Planck constants. The choice of the action was not arbitrary — Ehrenfest had shown that ∮p dq is an adiabatic invariant, so squeezing a system slowly moves its energy and its frequency while leaving the loop area alone, and only such a quantity can carry an integer label that a slow change of the apparatus cannot alter. What the condition itself cannot supply, Bohr's correspondence principle does: at large n the photon from n → n−k must approach the k-th harmonic of the classical orbital frequency, which fixes the constant in Eₙ = −A/n² and lets the Fourier amplitudes of the classical orbit stand in for transition intensities nobody could otherwise compute.
The cost is that the rule is only defined for separable, multiply periodic motion. Einstein pointed out in 1917 that a trajectory not confined to an invariant torus has no ∮p dq to quantise at all, and the helium atom is exactly that case — so the theory that fixed hydrogen's spectrum and Sommerfeld's fine structure could not compute the ground state of the next element in the table, and had to be replaced rather than patched.
- Simple definition
- Old quantum theory keeps Newtonian orbits but allows only those whose action integral ∮p dq around one period equals a whole number of Planck constants, and it settles what that rule leaves open by demanding agreement with classical physics at large quantum numbers.
- Example
- For a circular hydrogen orbit the loop integral is 2πL, so 2πL = nh gives L = nħ; the Coulomb force balance then returns Eₙ = −13.606 eV/n², and n = 2 → 1 emits 10.20 eV — the measured Lyman-α line.
Selects a countable set of orbits from the classical continuum; the dynamics stay Newtonian, only the occupancy is restricted.
One condition per separable coordinate; h = 6.626 × 10⁻³⁴ J s, so the loop area carries units of J s
Delivers the whole Bohr model from the one condition — no separate standing-wave postulate is needed.
ħ = h/2π = 1.055 × 10⁻³⁴ J s; n = 1, 2, 3 … for the circular orbit
Returns Planck's E = nhν, and exposes what is missing: the ½hν appears only once nh becomes (n + ½)h.
Ellipse semi-axes √(2E/mω²) and √(2mE); ν = ω/2π in Hz, E in joules
The reason the integer sits on J and not on E: a slow squeeze moves E and ν together and cannot change n.
Slow means the parameter alters little in one period; in general ν = dE/dJ, and only for a harmonic oscillator does that integrate to J = E/ν
Fixes the constant in Eₙ = −A/n², and hands each line's intensity to the k-th Fourier amplitude of the classical orbit.
k = 1, 2, 3 … is the harmonic index; forb is the classical orbital frequency in Hz
The number correspondence is checked against: the n = 100 → 99 line comes out at 6.68 GHz, 1.5 per cent above it.
n = 1 gives 6.58 PHz (extreme ultraviolet); n = 100 gives 6.58 GHz (microwave)
Quantise the area enclosed in the phase plane
For any coordinate whose motion repeats, draw the trajectory in the phase plane (q, p). The Wilson-Sommerfeld condition says the enclosed area must be a whole number of Planck constants, ∮p dq = nh, with one such condition for every separable coordinate. Nothing about the dynamics changes — the orbits are still Newton's — only which of them are occupied, and since h = 6.626 × 10⁻³⁴ J s the restriction bites only when the enclosed action is within a few h. Run it on the simplest case. A particle of mass m between rigid walls a distance L apart crosses with momentum +p and returns with −p, so one closed loop gives ∮p dx = 2pL = nh, hence p = nh/2L and Eₙ = n²h²/8mL². For an electron in a 0.50 nm box that is E₁ = 1.50 eV, and the result is not an approximation: it is exactly what the Schrödinger equation returns. That coincidence is what made the rule look like a law rather than a recipe.
Why the integer sits on the action, not the energy
At the 1911 Solvay meeting Lorentz posed a trap: quantise a pendulum as E = nhν, then slowly shorten the string. The frequency ν changes, so either E jumps or n does — yet nothing sudden has happened, and a quantum number has no business changing during a smooth pull. Einstein's answer, which Ehrenfest developed into a general theorem, is that E/ν is unchanged by a slow change, and more generally that ∮p dq is invariant under any variation slow compared with one period. Watch it work on a mass on a spring whose stiffness is slowly quadrupled. The angular frequency goes as the square root of the stiffness, so ν doubles; J = E/ν is held fixed, so E doubles with it, the extra energy supplied by whatever agent is tightening the spring; and because A² = 2E/stiffness, the amplitude falls to 1/√2 of its old value. Frequency, energy and amplitude all move, and E/ν alone does not. That is the whole argument for putting the integer on the action rather than on the energy: it is the one quantity a slow experimenter cannot disturb, so it can carry a label that survives the experiment. The pendulum itself is worked through in the Hard example.
Bohr, Sommerfeld, and the ellipse that fitted too well
The circular orbit of the definition example is the one-coordinate case: the angle φ repeats, its momentum pφ = L is constant, and 2πL = nh gives L = nħ, after which the force balance mv²/r = e²/4πε₀r² fixes rₙ = n²a₀ with a₀ = 4πε₀ħ²/me² = 52.9 pm and the Bohr energies follow. Sommerfeld then allowed ellipses, imposing one condition on the radial motion and one on the angular, with n = nᵣ + nφ; non-relativistically all the ellipses of a given n share an energy. Use the relativistic momentum p = γmv on the fast inner leg of an eccentric orbit and the degeneracy splits by of order α² = 5.33 × 10⁻⁵ of the binding energy: E = Eₙ[1 + (α²/n²)(n/nφ − 3/4)] puts the n = 2 splitting at 4.53 × 10⁻⁵ eV, or 10.95 GHz, against the 10.97 GHz that spectroscopy finds. It was the theory's most persuasive success and its most misleading one — spin was entirely absent, and two errors cancelled to leave the right spacing attached to the wrong quantum numbers.
Correspondence as a working constraint, not a slogan
Bohr's principle is quantitative: as n grows, the frequency radiated in n → n−k must approach k times the frequency at which the electron actually goes round. Set the hydrogen line ν = Rc[1/(n−k)² − 1/n²] against k forb using the orbital-frequency card above. For k = 1 the ratio is n(2n−1)/2(n−1)², which is 3.00 at n = 2, 1.080 at n = 20 and 1.0152 at n = 100 — the mismatch dying away as 3/2n, always from above. The principle earns its keep by being run backwards. Bohr demanded agreement in the limit and used it to fix the coefficient in Eₙ = −A/n², obtaining A = me⁴/8ε₀²h² and so predicting the Rydberg constant from m, e and h rather than fitting it. With the values of e and h available in 1913 the prediction landed within about 6 per cent of the measured constant; with today's values it agrees to the 0.05 per cent reduced-mass correction, which the model can also supply. A statement about a limit had produced a number.
Intensities and selection rules from a Fourier series
The quantisation condition says which levels exist and is silent about which lines are bright, which is a serious gap when your evidence is a photographic plate of spectral lines of wildly different strength. Correspondence closes it. Expand the classical orbit over one period, x(t) = Σₖ aₖ cos(2πk forb t + φₖ). Classically the k-th harmonic radiates dipole power proportional to (k forb)⁴ aₖ², so identify that harmonic with the transition n → n−k: a vanishing aₖ means a forbidden line. A circular orbit is a single harmonic, so only Δn = 1 survives; a Kepler ellipse of eccentricity 0.6 carries second and third harmonics at roughly 0.27 and 0.11 of the fundamental amplitude, so Δn = 2 and 3 appear and are weak. Heisenberg took this literally in 1925. Keep the array of amplitudes aₖ(n), relabel each one by the pair of states it connects as x(n, n−k), and the rule for multiplying such arrays turns out to be matrix multiplication — matrix mechanics grew straight out of this section of the old theory.
Where it stops: helium, and what survived
The condition needs one closed loop per coordinate, so it presupposes separable, multiply periodic motion. Einstein saw in 1917 that this is a strong dynamical assumption rather than a formality: unless the trajectory winds on an invariant torus there is no well-defined ∮p dq to set equal to nh, and generic many-body motion is not of that kind. Helium is the decisive case — two electrons, and no choice of coordinates separates them. Bohr's 1913 picture of both electrons sharing one circular orbit binds the pair by 83.3 eV against the measured 79.0 eV (24.59 eV to remove the first electron, 54.42 eV the second); Kramers's far more careful 1923 calculation of tilted orbits, the best the old theory ever managed, gives a first ionisation energy of 20.7 eV against the measured 24.59 eV, with no free parameter left to adjust. The list lengthened: hydrogen's ground state was assigned L = ħ instead of 0, the oscillator lost its ½hν, the anomalous Zeeman effect demanded half-integer quantum numbers, and Pauli's 1922 thesis on H₂⁺ disagreed with experiment. Three things survived intact: ∮p dq = (n + ½)h as the WKB rule, the adiabatic theorem, and correspondence as a permanent constraint on any successor.
Change one variable at a time
Make the relationship visible.
Hold k = 1 and drag n up: the gap falls from 8.0 per cent at n = 20 to 1.3 per cent at n = 120, and always from above. Then set k = 3 — the Δn = 3 line still misses its harmonic by 28 per cent at n = 20, so the higher the harmonic the larger n has to be.
QUANTUM ν(n → n−k)106.8 GHz
CLASSICAL k forb102.8 GHz
MISMATCH3.88 %
3k/2n ESTIMATE3.75 %
Live interpretationQUANTUM ν(n → n−k): 106.8 GHz. CLASSICAL k forb: 102.8 GHz. MISMATCH: 3.88 %. 3k/2n ESTIMATE: 3.75 %
Catch the common trap
Explain before calculating.
A particle of mass m bounces elastically between rigid walls a distance L apart, with no other forces acting. Quantise it with the Wilson-Sommerfeld condition ∮ p dx = nh, where the integral runs once around the closed path in the phase plane. Which spectrum follows, and why?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA 1.0 kg mass on a spring of stiffness 100 N m⁻¹ oscillates with amplitude 0.10 m. Apply ∮p dx = nh to find which rung of the old quantum ladder it occupies, and the size of one rung.
- ω = √(stiffness/m) = √(100/1.0) = 10.0 rad s⁻¹, so ν = ω/2π = 1.5915 Hz, and the energy of the swing is E = ½ × stiffness × A² = ½(100)(0.10)² = 0.50 J.
- The phase-space path is an ellipse with semi-axes A = 0.10 m and pₘₐₓ = mωA = 1.0 kg m s⁻¹, so its area is ∮p dx = πApₘₐₓ = π(0.10)(1.0) = 0.3142 J s — equal to E/ν = 0.50/1.5915 = 0.3142 J s, as the harmonic-oscillator result requires.
- Set that area equal to nh: n = E/(hν) = 0.50/(6.626 × 10⁻³⁴ × 1.5915) = 0.50/(1.0546 × 10⁻³³) = 4.7 × 10³².
- One rung is hν = 1.05 × 10⁻³³ J, a relative step of 1/n = 2.1 × 10⁻³³. No instrument resolves that, which is the correspondence principle's own account of why the mass simply looks classical.
Answern ≈ 4.7 × 10³², with rung spacing hν = 1.05 × 10⁻³³ J — quantised, and unobservably so.
MediumHydrogen has Eₙ = −13.606 eV/n². Find the frequency of the n = 100 → 99 photon and the classical orbital frequency of the n = 100 orbit, compare them, then repeat the comparison for n = 2 → 1.
- Write the level formula as a frequency: Rc = 13.6057 eV/h = 3.2898 × 10¹⁵ Hz, so ν(n → n−1) = Rc[1/(n−1)² − 1/n²].
- At n = 100: ν = 3.2898 × 10¹⁵ × (1/9801 − 1/10000) = 3.2898 × 10¹⁵ × 2.0304 × 10⁻⁶ = 6.68 × 10⁹ Hz = 6.68 GHz.
- The classical orbital frequency is forb = 2|Eₙ|/(nh) = 6.580 × 10¹⁵/n³ Hz, so at n = 100 it is 6.580 × 10¹⁵/10⁶ = 6.58 GHz. The two agree to 1.5 per cent: the exact ratio is n(2n − 1)/2(n − 1)² = 19900/19602 = 1.0152.
- At n = 2: ν = 3.2898 × 10¹⁵ × (1 − 1/4) = 2.47 × 10¹⁵ Hz, while forb = 6.580 × 10¹⁵/8 = 8.22 × 10¹⁴ Hz. The ratio is exactly 3, so the classical estimate is out by a factor of three.
- The ratio expands as 1 + 3/(2n), which tracks the exact value closely at n = 100 (1.50 per cent predicted against 1.52 per cent actual) and not at all at n = 2. Correspondence is asymptotic: 'large n' means large compared with 3/2 divided by the accuracy you need.
Answer6.68 GHz against 6.58 GHz, 1.5 per cent apart at n = 100; 2.47 × 10¹⁵ Hz against 8.22 × 10¹⁴ Hz, a factor of 3 at n = 2. The mismatch closes as 3/2n.
HardA 0.200 kg bob hangs on a 1.00 m string and swings with angular amplitude 0.100 rad. The string is pulled slowly through a hole in the ceiling until the length is 0.250 m, the pull taking many periods. Find the new frequency, energy and amplitude from the adiabatic invariant, and show that the quantum number is unchanged.
- ν = (1/2π)√(g/L): ν₁ = (1/2π)√(9.81/1.00) = 0.4985 Hz and ν₂ = (1/2π)√(9.81/0.250) = 0.9970 Hz, exactly double, since ν ∝ L(−1/2).
- Initial energy at small amplitude: E₁ = ½mgLθ₀² = ½(0.200)(9.81)(1.00)(0.100)² = 9.81 × 10⁻³ J.
- At small amplitude the pendulum is harmonic, so the adiabatic invariant is J = ∮p dq = E/ν, unchanged provided the pull spans many periods. Hence E₂ = E₁(ν₂/ν₁) = 2E₁ = 1.962 × 10⁻² J; the extra 9.81 mJ is work done by the hand against the string tension, whose time average exceeds mg.
- Amplitude from E₂ = ½mgL₂θ₂²: θ₂ = √(2E₂/(mgL₂)) = √(3.924 × 10⁻²/(0.200 × 9.81 × 0.250)) = √0.0800 = 0.283 rad, or 16.2°. Equivalently θ₀ ∝ L(−3/4), and 4³⁄⁴ = 2.83.
- Now the quantum number, n = J/h = E/(hν). Before: 9.81 × 10⁻³/(6.626 × 10⁻³⁴ × 0.4985) = 2.97 × 10³¹. After: 1.962 × 10⁻²/(6.626 × 10⁻³⁴ × 0.9970) = 2.97 × 10³¹. Identical to every figure.
- That is Ehrenfest's argument in one calculation: E and ν both doubled, yet n did not move, so the integer can safely be attached to J. Yank the string in a fraction of a period instead and the invariance fails — the bob has then simply been kicked, and n does change.
Answerν doubles to 0.997 Hz and E to 19.6 mJ, while θ₀ grows to 0.283 rad (16.2°); J = E/ν and hence n = 2.97 × 10³¹ are unchanged.