University Quantum Mechanics II · The Hydrogen Radial Equation Solved · 2.9

The l-degeneracy explained: the quantum Runge-Lenz vector

Topic 2.5 found that hydrogen's energy depends on n alone; this topic finds the reason. The vector that keeps a Kepler ellipse from turning survives quantisation, commutes with H and carries 2s into 2p at fixed energy. It also picks out the Stark states, and in Pauli's hands it returns the Bohr levels and their n² count.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Rotational symmetry explains only why the 2l + 1 values of m share an energy. It cannot link different l, and in a generic central potential they do split: 1/r and the isotropic oscillator are the exceptions, each with a hidden conserved quantity. The Coulomb problem conserves one more quantity, inherited from Kepler: the Laplace-Runge-Lenz vector, which points to perihelion with length k times the eccentricity and stays fixed only because the force falls as 1/r².

Quantum mechanically it must be symmetrised, A = (p × L − L × p)/(2μ) − k r̂, and commutator algebra then gives [H, A] = 0 and A⋅L = 0. A commutes with H but not with L², so it moves a state to l ± 1 without changing its energy: A_z turns 2s into (k/2) times 2p₀, which forces 2s and 2p to be degenerate, and the combinations (2s ± 2p₀)/√2 that diagonalise A_z are the states a weak electric field selects. With two quoted identities, Pauli turned A and L into two commuting angular momenta and read off Eₙ = −μk²/(2ħ²n²) and the n² degeneracy.

Everything rests on an exact, non-relativistic 1/r potential. Screening in alkali atoms breaks it by electronvolts; the order-α² fine structure bends it, keeping 2s₁/₂ with 2p₁/₂; and the order-α³ Lamb shift finally splits those two, by a measured 1057.8 MHz.

Simple definition: The quantum Runge-Lenz vector A = (p × L − L × p)/(2μ) − k r̂ is a Hermitian vector operator that commutes with the hydrogen Hamiltonian but not with L², so it connects states of the same n whose l differ by one.
Example: Acting on hydrogen's 2s state, A_z gives (k/2)|2p₀⟩, with k = e²/(4πε₀) = 1.440 eV nm. Because A_z commutes with H, that 2p₀ state must share 2s's energy, −3.40 eV.

The quantum Runge-Lenz vectorA = (p × L − L × p)/(2μ) − k r̂ = (p × L − iħp)/μ − k r̂

The Hermitian average of both orderings of the classical p × L/μ − k r̂; the second form uses L × p = −p × L + 2iħp. Classically |A| is k times the eccentricity.

A and k in J m, with k = e²/(4πε₀) = 2.307 × 10⁻²⁸ J m; p in kg m s⁻¹, L and ħ in J s, μ in kg; r̂ a unit vector

It commutes with the Coulomb Hamiltonian (derived)[H, A] = 0, H = p²/(2μ) − k/r

A cannot carry a state out of its energy level: whatever A makes from a state of energy E is again a state of energy E, or zero.

H in J, r in m; true only for an exact 1/r potential. The unsymmetrised p × L/μ − k r̂ gives −(ħ²k/μ) r̂/r² instead

A acting on an s state f(r) (derived)A f(r) = −r̂ [(ħ²/μ) f′(r) + k f(r)]

Zero for 1s, where f′ = −f/a with a = ħ²/(μk); for 2s it gives A_z|2s⟩ = (k/2)|2p₀⟩, which forces E(2p) = E(2s).

f in m^(−3/2), f′ in m^(−5/2); the factor r̂ makes the result pure l = 1

Pauli's identity for A² (quoted)A² = k² + (2H/μ)(L² + ħ²)

A quantum squared eccentricity: as ħ → 0 it returns the classical e² = 1 + 2EL²/(μk²). 2s has 3k²/4, 2p has k²/4, and 1s has 0.

Every term in J² m²; on |n l m⟩ it takes the value k²[1 − (l(l + 1) + 1)/n²]

Pauli's levels from two angular momentaEₙ = −μk²/(2ħ²n²), n = 2j + 1

The Bohr levels, and their n² = (2j + 1)² states, follow from the quoted algebra [Aᵢ, Aⱼ] = −(2iħH/μ)εᵢⱼₖLₖ with no differential equation.

E in J; j = 0, ½, 1, … labels both J₊ and J₋ = (L ± K)/2, with K = √(μ/(−2E)) A in J s

Position inside a level: the Stark linkz → −(3n²a/(2k)) A_z within level n

The quantum form of the classical orbit average. A field F along z picks A_z's eigenstates: for n = 2, (2s ± 2p₀)/√2 shift by ∓3eaF.

a = ħ²/(μk) = 52.95 pm for hydrogen; derived here for n = 2 from ⟨2p₀|z|2s⟩ = −3a, quoted for general n

Why a Kepler ellipse never turns

Any central force f(r) r̂ conserves L, and d(p × L)/dt = f(r) r̂ × L. Since dr̂/dt = (L × r̂)/(μr²), that is −μf(r)r² dr̂/dt, a total derivative only when f(r)r² is constant: the inverse-square force f = −k/r². Then A = p × L/μ − k r̂ is conserved (derived). It points from the centre to perihelion with length ke, e the eccentricity, and its square is A² = k² + 2EL²/μ, the orbit's e² = 1 + 2EL²/(μk²). A fixed A means an ellipse that closes and never turns. Any other force rotates A: general relativity's small 1/r⁴ addition turns Mercury's perihelion by the 43″ per century that planetary perturbations leave unexplained (measured). The energy E = −k/(2d) depends only on the semi-major axis d, so ellipses of every eccentricity, and so every L, share one E: the classical form of the l-degeneracy. Dimensions: (kg m s⁻¹)(J s)/kg = J m, the unit of k.

Symmetrise, then check [H, A] = 0

Carried over literally, p × L is not Hermitian: its adjoint is −L × p, and L × p = −p × L + 2iħp because L and p do not commute. The Hermitian choice averages the orderings, A = (p × L − L × p)/(2μ) − k r̂ = (p × L − iħp)/μ − k r̂. With H = p²/(2μ) − k/r, two pieces of [H, A] vanish at once, since p² commutes with p and L, and 1/r with r̂, leaving [H, A] = −(k/(2μ))([p², r̂] + [1/r, p × L − L × p]). Use [p², f] = −ħ²∇²f − 2iħ(∇f)⋅p and [1/r, pⱼ] = −iħxⱼ/r³. Component i of the first bracket is 2ħ²xᵢ/r³ − 2iħr⁻¹pᵢ + 2iħ(xᵢ/r³)(r⋅p); with r × L = r(r⋅p) − r²p, the second is −2ħ²xᵢ/r³ + 2iħr⁻¹pᵢ − 2iħ(xᵢ/r³)(r⋅p). They cancel term by term (derived). The ħ² terms cancel only because of the symmetrising: the naive A′ = p × L/μ − k r̂ = A + iħp/μ leaves [H, A′] = −(ħ²k/μ) r̂/r².

A lies in the orbit plane and changes l, never E

r⋅L and p⋅L vanish because εᵢⱼₖ is antisymmetric while xᵢxⱼ, or pᵢpₖ, is symmetric. In (p × L)⋅L = εᵢⱼₖpⱼLₖLᵢ only the commutator part of LₖLᵢ survives, giving iħp⋅L = 0; (L × p)⋅L vanishes the same way, and r̂⋅L = 0 at once. So A⋅L = 0: A lies in the plane perpendicular to L, like the perihelion vector (derived). Built from r and p, A obeys [Lᵢ, Aⱼ] = iħεᵢⱼₖAₖ like every vector operator, so A_z commutes with L_z but not with L², and A is odd under parity. [H, A] = 0 gives H(A|ψ⟩) = A(H|ψ⟩) = E(A|ψ⟩): A never leaves an energy level. Inside it, A reverses parity and, as Unit 5 proves for any vector operator (quoted), changes l by at most one. So whenever A|ψ⟩ is not zero, a state with l ± 1 sits at exactly the energy of |ψ⟩. Rotations could never do this: L commutes with L², so it only reshuffles m within one l.

A_z carries 2s into 2p₀

On an s state f(r), Lf = 0 removes p × L, and with p = −iħ∇ the rest of A = (p × L − iħp)/μ − k r̂ gives, in one line, Af = −r̂[(ħ²/μ)f′ + kf]. The output is r̂ times a radial function: pure l = 1, with m = 0 for A_z and m = ±1 for Aₓ ± iA_y. For 1s, f ∝ e^(−r/a) with a = ħ²/(μk), the bracket vanishes: A|1s⟩ = 0, as it must with no 1p to go to. For 2s, f ∝ (1 − r/2a)e^(−r/2a), the bracket is −k(r/4a)e^(−r/2a) times the same constant, exactly 2p's radial shape. Matching normalisations, A_z|2s⟩ = (k/2)|2p₀⟩ and (Aₓ ± iA_y)|2s⟩ = ∓(k/√2)|2p, ±1⟩ in Condon-Shortley phases (derived). Because A_z commutes with H, 2p₀ has 2s's energy, with no appeal to the Bohr formula. Size check: A_z|2s⟩ has norm k/2 = 0.720 eV nm, and the quoted A² = k² + (2H/μ)(L² + ħ²) gives ⟨2s|A²|2s⟩ = k²(1 − 1/4) = 3k²/4, a third of it, k²/4, in each component.

The Stark states are the eigenstates of A_z

In the n = 2, m = 0 pair, A_z is k/2 times a Pauli σₓ, so its eigenstates are (2s ± 2p₀)/√2 with A_z = ±k/2. Position follows the same pattern: ⟨2p₀|z|2s⟩ = (1/√3)∫R₂₁R₂₀r³dr = (1/√3)(−3√3a) = −3a (derived). Inside the pair, then, z = −(6a/k)A_z, the quantum form of the classical orbit average ⟨r⟩ = −(3/2)dA/k with semi-major axis d = n²a = 4a: the electron lingers at aphelion, opposite A. A field F along z adds eFz to the electron's energy, so it selects A_z's eigenstates and shifts them by ∓3eaF, while 2p±1 stay put to first order. At F = 1.00 × 10⁶ V m⁻¹ that is 0.159 meV, or 38.4 GHz. This is Unit 9's linear Stark effect, and it leans on the degeneracy: hydrogen's measured 2s₁/₂–2p₁/₂ Lamb shift, 1057.8 MHz, and 2p fine structure, 10.97 GHz, are both smaller than that, but at 10⁴ V m⁻¹, where 3eaF is 0.38 GHz, the Stark term would lose to both.

Pauli's count, and where the symmetry stops

Pauli (1926) finished the argument with two identities quoted here: [Aᵢ, Aⱼ] = −(2iħH/μ)εᵢⱼₖLₖ and A² = k² + (2H/μ)(L² + ħ²). On a level of negative energy E, K = √(μ/(−2E)) A obeys [Kᵢ, Kⱼ] = iħεᵢⱼₖLₖ, and J₊ = (L + K)/2 and J₋ = (L − K)/2 are two commuting angular momenta. A⋅L = 0 makes J₊² = J₋², one quantum number j for both, and the A² identity becomes (2j + 1)²ħ² = μk²/(−2E). So E = −μk²/(2ħ²n²) with n = 2j + 1, topic 2.5's Bohr levels without a differential equation; the level holds (2j + 1)² = n² states, and adding J₊ to J₋ (Unit 5, quoted) gives l = 0 to n − 1. All of it needs an exact 1/r potential. Screening in sodium puts 3p 2.10 eV above 3s, the 589 nm D lines. The Dirac equation keeps 2s₁/₂ with 2p₁/₂ but puts 2p₃/₂ 45 μeV above them, order α² of the level (quoted). QED splits 2s₁/₂ from 2p₁/₂ by the measured 1057.8 MHz, 1.3 × 10⁻⁶ of E₂: order α³.

Change one variable at a time

Make the relationship visible.

Interactive model

Mixing angle β30 °

Field F along z20 kV cm⁻¹

Start at β = 0, pure 2s: the cloud is symmetric and ⟨A_z⟩ = 0. At β = 45° it piles up at −z, ⟨z⟩ = −3a, while ⟨A_z⟩ = +k/2 points the other way and the energy dot lands on the lower Stark level. Set F = 0 and every β has the same energy.

EXPECTED ⟨A_z⟩ / k0.433

EXPECTED ⟨z⟩-137.6 pm

FIELD ENERGY ⟨eFz⟩-275.1 μeV

STARK SPLITTING 6eaF/h153.6 GHz

Live interpretationEXPECTED ⟨A_z⟩ / k: 0.433. EXPECTED ⟨z⟩: −137.6 pm. FIELD ENERGY ⟨eFz⟩: −275.1 μeV. STARK SPLITTING 6eaF/h: 153.6 GHz

Catch the common trap

Explain before calculating.

Misconception check

Spherical symmetry does not explain why 2s and 2p share an energy. Rotational invariance means [H, L] = 0, and since L commutes with L² it can only turn a state into others of the same l: that is the 2l + 1 degeneracy in m, present in every central potential. A degeneracy between different l needs a conserved operator that changes l. The 1/r potential has one, a vector that moves l by one; the isotropic oscillator, the other textbook exception, has a conserved tensor that moves l by two. In sodium the valence electron feels a screened, non-Coulomb potential, A is not conserved, and 3s lies 2.10 eV below 3p. In hydrogen, 2s₁/₂ and 2p₁/₂ differ by only 4.4 μeV, the QED Lamb shift. “Accidental” means not required by rotations, not coincidental: the Runge-Lenz vector enforces it exactly for as long as the potential is exactly 1/r.

Pauli's identity A² = k² + (2H/μ)(L² + ħ²), quoted in this lesson, is diagonal in |n l m⟩. What is A²/k² for hydrogen's 3d state, the quantum counterpart of its squared eccentricity?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyHydrogen's ground state is ψ₁₀₀ ∝ e^(−r/a) with a = ħ²/(μk) = 52.95 pm. Apply the symmetrised A and the naive A′ = p × L/μ − k r̂ to it, use the difference to show that A′ cannot be conserved, and give the norm of A′ψ₁₀₀ in eV nm.

ψ₁₀₀ depends on r alone, so Lψ₁₀₀ = 0 and the p × L term contributes nothing to either operator. For A′ that leaves A′ψ₁₀₀ = −k r̂ψ₁₀₀.
A = A′ − iħp/μ, and −(iħ/μ)(−iħ∇ψ) = −(ħ²/μ)ψ′r̂, so Aψ₁₀₀ = −r̂[(ħ²/μ)ψ′ + kψ]. With ψ′ = −ψ/a and ħ²/(μa) = k, the bracket is −kψ + kψ = 0.
Consistency with the quoted identity: A² = k² + (2E₁/μ)(0 + ħ²) with E₁ = −μk²/(2ħ²) gives k² − k² = 0, so A|1s⟩ had to vanish.
A′ψ₁₀₀ is r̂ times a radial function, an l = 1 state. Since l ≤ n − 1, no l = 1 state has energy E₁, so A′ has moved ψ₁₀₀ out of its level and cannot commute with H; indeed [H, A′] = −(ħ²k/μ) r̂/r².
Norm: ‖A′ψ₁₀₀‖² = k²⟨ψ₁₀₀|r̂⋅r̂|ψ₁₀₀⟩ = k², so ‖A′ψ₁₀₀‖ = k = 1.440 eV nm = 2.307 × 10⁻²⁸ J m, in the units of A. The missing iħp/μ is not a small correction: on 1s its size is ħ²/(μa) = k.

AnswerAψ₁₀₀ = 0 exactly, while A′ψ₁₀₀ = −k r̂ψ₁₀₀ is an l = 1 function of norm k = 1.440 eV nm with no partner at E₁, so A′ is not conserved.

MediumApply A_z to hydrogen's 3s state, R₃₀ ∝ (1 − 2r/3a + 2r²/27a²)e^(−r/3a), using A f = −r̂[(ħ²/μ)f′ + kf]. Show that the result is proportional to 3p₀, find the coefficient, and check it against the quoted A² identity.

Since ħ²/μ = ka, the bracket is k(af′ + f). With x = r/a and f ∝ P(x)e^(−x/3), P = 1 − 2x/3 + 2x²/27, this is k(P′ + 2P/3)e^(−x/3) times ψ₃₀₀'s constant.
P′ + 2P/3 = −2/3 + 4x/27 + 2/3 − 4x/9 + 4x²/81 = −(8/27)x(1 − x/6). The constant cancels, so no s wave survives, and x(1 − x/6)e^(−x/3) is exactly R₃₁'s shape, node at r = 6a.
So A_zψ₃₀₀ = (8/27)k cosθ x(1 − x/6)e^(−x/3) × 2a^(−3/2)/(3√3√(4π)). Since ψ₃₁₀ = 8√3a^(−3/2) x(1 − x/6)e^(−x/3) cosθ/(27√6√(4π)), the ratio is (8/27) × 2/(3√3) × 27√6/(8√3) = 2√6/9 = 0.544.
Check: 3s is spherically symmetric, so ⟨A_z²⟩ = ⟨A²⟩/3 = (k²/3)(1 − 1/9) = 8k²/27, and A_z|3s⟩ lies wholly in 3p₀, so its coefficient must be √(8/27) k = 0.544k. The two routes agree.
In numbers, √(8/27) × 1.440 eV nm = 0.784 eV nm. Limit: the same check gives (k/n)√((n² − 1)/3) for ns → np₀, which is 0 for n = 1 and k/2 for n = 2, as found in the lesson.

AnswerA_z|3s⟩ = √(8/27) k|3p₀⟩ = 0.544k|3p₀⟩, a coefficient of 0.784 eV nm, confirmed by ⟨A²⟩/3 = 8k²/27.

HardIn hydrogen's n = 3, m = 0 subspace, spanned by 3s, 3p₀ and 3d₀, find the eigenvalues of A_z from Pauli's two angular momenta, deduce ⟨3d₀|A_z|3p₀⟩ given ⟨3p₀|A_z|3s⟩ = √(8/27) k, and predict the first-order Stark shifts of these states in a field of 1.00 × 10⁵ V m⁻¹ along z, in μeV and GHz.

n = 2j + 1 = 3 gives j = 1 for both J₊ and J₋. On this level A = √(−2E₃/μ)K = (k/(3ħ))K, and K = J₊ − J₋, so K_z has eigenvalues (m₊ − m₋)ħ. Since m = m₊ + m₋ = 0 forces m₋ = −m₊, K_z = 2ħm₊ = 2ħ, 0 or −2ħ, and A_z is +2k/3, 0 or −2k/3.
In the basis 3s, 3p₀, 3d₀, A_z keeps m, flips parity and moves l by one, so its matrix is tridiagonal with zero diagonal and off-diagonal elements b₁ = √(8/27)k and b₂. Its eigenvalues are 0 and ±√(b₁² + b₂²).
Matching ±2k/3: b₂² = 4k²/9 − 8k²/27 = 4k²/27, so ⟨3d₀|A_z|3p₀⟩ = b₂ = 2k/(3√3) = 0.385k.
Within n = 3 the quoted z → −(27a/(2k))A_z gives ⟨z⟩ = −9a, 0, +9a for A_z = +2k/3, 0, −2k/3. Check: −(27a/(2k))b₁ = −3√6a = −7.35a, the value the direct integral (1/√3)∫R₃₁R₃₀r³dr gives for ⟨3p₀|z|3s⟩.
The field adds eFz, so the shifts are −9eaF, 0 and +9eaF. With a = 52.95 pm, 9eaF = 9 × 52.95 × 10⁻¹² m × 1.00 × 10⁵ V m⁻¹ = 47.7 μeV, or 11.5 GHz. Dimensions: e × m × V m⁻¹ is an energy. This beats n = 3's whole order-α² fine-structure spread, 4.3 GHz (Dirac, quoted), so the linear picture is fair; as F → 0 the threefold degeneracy returns.

AnswerA_z = 0 or ±2k/3; ⟨3d₀|A_z|3p₀⟩ = 2k/(3√3) = 0.385k; the m = 0 states shift by 0 and ∓47.7 μeV (∓11.5 GHz), the lowest having A_z = +2k/3 and ⟨z⟩ = −9a.