Hydrogen Atom Energy Levels and Degeneracy

Drop a single electron into the Coulomb field of a proton and something remarkable happens. The energy of the resulting bound states depends on exactly one quantum number, not three. This is strange. The Schrödinger equation for the Hydrogen Atom involves a wavefunction ψnℓm\psi_{n\ell m}. It depends on three quantum numbers nn, ℓ\ell, mm. Yet the energy eigenvalue collapses to

En=−Z2n2 Ry=−Z22n2 EhE_n = -\frac{Z^2}{n^2}\,\text{Ry} = -\frac{Z^2}{2n^2}\,E_h

with no trace of ℓ\ell or mm anywhere. Every other central-potential problem you will ever solve shows energies that do depend on ℓ\ell. This includes the particle on a sphere with a non-Coulomb potential. It includes an atom in a crystal field, or a diatomic vibrator-rotor. Hydrogen is the exception. Understanding why it is an exception is one of the most instructive stories in quantum chemistry. This matters far beyond a curiosity of one textbook problem. The degeneracy pattern of hydrogen is n2n^2 states (or 2n22n^2 with spin) crowded into a single energy. This pattern is the scaffolding for the periodic table's orbital energetics. It is also the reason multi-electron atoms break that degeneracy in the way they do. Screening lifts the ℓ\ell-degeneracy, giving the familiar ns<np<ndns < np < nd ordering. The Lyman, Balmer, and Paschen series in the hydrogen emission spectrum are direct photographs of the En=−Z2/n2E_n = -Z^2/n^2 ladder. The fine structure and Lamb shift experiments that made 20th-century physics famous are essentially precision measurements. They show how real hydrogen deviates from this idealized formula. The big picture is this: the Coulomb potential V(r)=−Ze2/(4πϵ0r)V(r) = -Ze^2/(4\pi\epsilon_0 r) is special among central potentials. It possesses a hidden symmetry beyond ordinary rotational symmetry. This is a conserved quantity called the Laplace–Runge–Lenz vector. This vector is familiar from the classical Kepler problem. There, it explains why planetary orbits are closed ellipses that do not precess. Quantum mechanically, this extra symmetry enlarges the group of transformations that commute with the Hamiltonian. It grows from SO(3)SO(3) (rotations only) to SO(4)SO(4). This enlargement forces states of different ℓ\ell but the same nn to share an energy. This is called accidental degeneracy. However, "accidental" is a historical misnomer for something that is in fact deeply structural. This article builds the full degeneracy picture from the ground up. It covers the origin of the −Z2/n2-Z^2/n^2 formula from the The Radial Equation and Radial Wavefunctions. It explains why ℓ\ell ranges over 0,…,n−10,\dots,n-1 and mm over −ℓ,…,ℓ-\ell,\dots,\ell. It shows how these combinatorics give exactly n2n^2 spatial states. It also explains why the Coulomb potential specifically, and no other radial potential, makes the energy blind to ℓ\ell. Along the way you will see how the degeneracy shows up in the Balmer spectrum. You will see how it gets broken by external fields (Zeeman, Stark) and by relativistic corrections. Finally, you will see how it reappears as the conceptual seed for Term Symbols and Atomic States in multi-electron atoms.

Key takeaways
  • Hydrogen-like energies depend on principal quantum number alone: En=−Z2/n2E_n = -Z^2/n^2 Ry, with no dependence on ℓ\ell or mm.
  • For each nn, allowed values are ℓ=0,1,…,n−1\ell = 0, 1, \dots, n-1 and m=−ℓ,…,+ℓm = -\ell,\dots,+\ell, giving n2n^2 spatial orbitals, 2n22n^2 with spin.
  • mm-degeneracy (2ℓ+12\ell+1 states per ℓ\ell) is essential, guaranteed by rotational symmetry alone — it survives in any central potential.
  • ℓ\ell-degeneracy (all ℓ\ell within a given nn sharing one energy) is accidental, special to the exact 1/r1/r Coulomb potential, and traces to the conserved Runge–Lenz vector.
  • Any perturbation that is not perfectly 1/r1/r or not perfectly isotropic — screening in multi-electron atoms, external electric/magnetic fields, relativistic corrections — splits this accidental degeneracy.
  • Key idea: hydrogen's energy levels depend only on nn because the Coulomb potential has a hidden SO(4)SO(4) symmetry beyond ordinary rotations, making the ℓ\ell-degeneracy structural rather than coincidental.

Where this lives today

The n2n^2/2n22n^2 degeneracy counting is the direct ancestor of the shell-capacity numbers (2,8,18,322, 8, 18, 32) used every day in general chemistry to build up the periodic table, even though real multi-electron atoms break the ℓ\ell-part of the degeneracy (which is exactly why 4s4s fills before 3d3d, a fact with enormous consequences for transition-metal chemistry, catalysis, and coordination chemistry). Precision spectroscopy of hydrogen and hydrogen-like ions remains one of the most stringent tests of quantum electrodynamics and fundamental constants: modern laser spectroscopy of the 1s1s–2s2s two-photon transition in atomic hydrogen has been measured to better than one part in 101510^{15}, and comparing the measured splitting between states that would be "accidentally" degenerate in the pure Coulomb picture (like 2s1/22s_{1/2} and 2p1/22p_{1/2}, the Lamb shift) directly probes vacuum polarization and electron self-energy — QED effects with no classical analogue. In astrophysics, hydrogen-like ion spectra (He⁺, Li²⁺, and highly stripped heavy ions in stellar coronae, solar flares, and X-ray binary accretion disks) are read off using exactly the Z2/n2Z^2/n^2 scaling worked out above, letting astronomers infer both the ionization state and temperature of extremely hot plasmas from X-ray line positions. In quantum control and Rydberg-atom physics — a field central to modern quantum computing and quantum simulation platforms — atoms are excited to very high nn (Rydberg states), where the near-degeneracy of different ℓ\ell within a shell (slightly broken by quantum defects in non-hydrogenic atoms) is deliberately exploited to engineer strong, tunable, long-range dipolar interactions between atoms for two-qubit gates. Finally, the entire conceptual toolkit here — separating essential (symmetry-guaranteed) from accidental (potential-specific) degeneracy — recurs whenever chemists analyze Term Symbols and Atomic States in multi-electron atoms, and whenever spectroscopists interpret Radial Distribution Functions to rationalize orbital penetration and shielding effects that ultimately lift hydrogen's special accidental degeneracy in every atom heavier than hydrogen.

Ask the tutor: How exactly does electron screening in a multi-electron atom convert the flat n2n^2-degenerate hydrogen shell into the split ns<np<nd<nfns<np<nd<nf ordering used in the aufbau principle?

The physical picture

Solving the Hydrogen Atom Schrödinger equation by separation of variables in spherical coordinates gives ψnℓm(r,θ,ϕ)=Rnℓ(r) Yℓm(θ,ϕ)\psi_{n\ell m}(r,\theta,\phi) = R_{n\ell}(r)\,Y_\ell^m(\theta,\phi). The angular factor YℓmY_\ell^m, the spherical harmonic, is completely universal: it is the same function for any central potential V(r)V(r), because it comes purely from the angular part of the kinetic energy operator (the L^2\hat{L}^2 operator) commuting with any Hamiltonian of the form H^=T^r+L^2/2mr2+V(r)\hat{H} = \hat{T}_r + \hat{L}^2/2m r^2 + V(r). The radial factor Rnℓ(r)R_{n\ell}(r), by contrast, is where the specific shape of V(r)V(r) enters, and it is the radial equation that ultimately fixes the energy. For the hydrogen-like radial equation, imposing square-integrability (a bound, normalizable state) forces a quantization condition that, after the dust settles, reads

n=nr+ℓ+1,nr=0,1,2,…n = n_r + \ell + 1, \qquad n_r = 0, 1, 2, \dots

where nrn_r is the number of radial nodes. This is the origin of the famous rule ℓ≤n−1\ell \le n-1: since nr≥0n_r \ge 0, we need ℓ≤n−1\ell \le n - 1. The energy formula that emerges from solving the radial equation depends only on the combination n=nr+ℓ+1n = n_r + \ell + 1 — not on nrn_r and ℓ\ell separately. That is the mathematical fingerprint of accidental degeneracy: two completely different orbitals, say 2s2s (nr=1,ℓ=0n_r=1,\ell=0) and 2p2p (nr=0,ℓ=1n_r=0,\ell=1), both have n=2n=2 and are exactly degenerate, even though one has a radial node and no angular node while the other has an angular node and no radial node. The figure below shows the degeneracy structure for n=1,2,3n=1,2,3 hydrogen levels, with the orbital "boxes" grouped horizontally by ℓ\ell but sitting at the same height (same EnE_n) within each shell.

Figure
Energy-level diagram: same n, same energy, regardless of ℓ
E=0 E(Ry) 1s (n=1, E=-1.00 Ry, 1 state) 2s + 2p (n=2, E=-0.25 Ry, 4 states) 3s+3p+3d (n=3, E=-0.111 Ry, 9 states) ℓ=0 (s) ℓ=1 (p) ℓ=2 (d) Horizontal position = ℓ subshell; vertical position = energy (only n matters)

Notice what this diagram is really asserting: the 2s2s and 2p2p bars sit at exactly the same height, −0.25-0.25 Ry, even though they are drawn at different horizontal positions to distinguish them by shape. In any non-Coulombic central potential (say, a screened potential in a many-electron atom, or a 3-D isotropic harmonic oscillator with an added anharmonic term), the 2s2s and 2p2p bars would separate vertically — that is precisely how orbital energy ordering (4s4s before 3d3d, for instance) arises in the periodic table. The next figure decomposes a single shell n=3n=3 into its full mm-degeneracy as well, making explicit the counting that gives n2n^2 states.

Figure
Full degeneracy count for n=3: 1(s) + 3(p) + 5(d) = 9 = n²
n = 3 shell, E₃ = −Z²/9 Ry — total degeneracy = 3² = 9 3s ℓ=0, m=0 → 1 state 3p 3p 3p ℓ=1, m=-1,0,+1 → 3 states 3d 3d 3d 3d 3d ℓ=2, m=-2..+2 → 5 states 1 + 3 + 5 = 9 = n² (all at the SAME energy E₃) m-degeneracy (essential, from SO(3)) is nested inside ℓ-degeneracy (accidental, from SO(4))

The distinction drawn at the bottom of that figure is the conceptual heart of this article. The 2ℓ+12\ell+1 states within a fixed ℓ\ell (the mm-degeneracy) exist because empty space has no preferred axis: rotate your coordinate system and H^\hat{H} is unchanged, so states related by rotation must be degenerate. This is essential degeneracy — it is required by symmetry in every central potential without exception, hydrogen or not. The further degeneracy across different ℓ\ell within the same nn is a much more delicate, Coulomb-specific effect, discussed in the derivations below.

The core ideas

The hydrogen-like (one electron, nuclear charge ZeZe) bound-state energy is

En=−Z2μe48ϵ02h2n2=−Z2n2 Ry=−Z22n2 Eh,n=1,2,3,…E_n = -\frac{Z^2 \mu e^4}{8\epsilon_0^2 h^2 n^2} = -\frac{Z^2}{n^2}\,\text{Ry} = -\frac{Z^2}{2n^2}\,E_h, \qquad n = 1, 2, 3, \dots

where:

  • ZZ is the nuclear charge (1 for H, 2 for He⁺, 3 for Li²⁺, …),
  • μ\mu is the reduced mass of the electron–nucleus system, μ=memN/(me+mN)≈me\mu = m_e m_N/(m_e+m_N) \approx m_e,
  • nn is the principal quantum number, running over positive integers,
  • 1 Ry=13.6057 eV1\,\text{Ry} = 13.6057\ \text{eV} is the Rydberg energy, and Eh=27.2114E_h = 27.2114 eV (hartree) is the atomic energy unit, so 1 Ry=12Eh1\,\text{Ry} = \tfrac12 E_h.

Reading: the energy is negative (bound state, energy measured relative to the ionized, infinitely separated electron and nucleus at rest), scales as Z2Z^2 (a more highly charged nucleus binds the electron much more tightly — this is why He⁺'s ground state is 4×4\times deeper than H's), and falls off as 1/n21/n^2 (levels crowd together rapidly as nn increases, converging to the ionization threshold E∞=0E_\infty = 0). The allowed quantum numbers for a given nn are

ℓ=0,1,2,…,n−1,m=−ℓ,−ℓ+1,…,+ℓ\ell = 0, 1, 2, \dots, n-1, \qquad m = -\ell, -\ell+1, \dots, +\ell

where:

  • ℓ\ell is the azimuthal (orbital angular momentum) quantum number, restricted by the radial quantization condition to be strictly less than nn,
  • mm is the magnetic quantum number, whose 2ℓ+12\ell + 1 allowed values follow from the eigenvalue spectrum of L^z\hat{L}_z once ℓ\ell is fixed.

Reading: an nn-shell contains nn subshells (ℓ=0\ell = 0 through n−1n-1), and subshell ℓ\ell contains 2ℓ+12\ell+1 orbitals. This produces the shell-filling pattern 1 (for n=1n=1: only 1s1s), 4 (for n=2n=2: 2s,2px,2py,2pz2s,2p_x,2p_y,2p_z), 9 (for n=3n=3), etc. The total spatial degeneracy of level nn is obtained by summing over all ℓ\ell:

gn=∑ℓ=0n−1(2ℓ+1)=n2g_n = \sum_{\ell=0}^{n-1} (2\ell+1) = n^2

where:

  • the sum ∑ℓ=0n−1(2ℓ+1)\sum_{\ell=0}^{n-1}(2\ell+1) is an arithmetic series of odd numbers 1+3+5+⋯+(2n−1)1+3+5+\cdots+(2n-1), which is a classical identity equal to n2n^2.

Reading: this is the single most quoted degeneracy fact in atomic quantum chemistry. Including the two spin states of the electron (spin degeneracy is entirely separate physics — it has nothing to do with the spatial Coulomb symmetry, it is a relativistic/intrinsic property), the total degeneracy including spin is 2n22n^2, which is exactly the capacity rule behind the shell structure 2,8,18,32,…2, 8, 18, 32,\dots that underlies (in a simplified, pre-screening form) the layout of the periodic table.

Derivations

Derivation AFrom the radial equation to En=−Z2/n2E_n = -Z^2/n^2: why only nn appears

Goal. Show explicitly that solving the hydrogen-like radial equation for bound states produces an energy that depends on the single combination n=nr+ℓ+1n = n_r + \ell + 1, so that different (ℓ,nr)(\ell, n_r) pairs sharing the same nn are exactly degenerate.

Step 1 — Set up the radial equation. After separating ψ=Rnℓ(r)Yℓm(θ,ϕ)\psi = R_{n\ell}(r)Y_\ell^m(\theta,\phi), the radial function satisfies

−ℏ22μ1r2ddr ⁣(r2dRdr)+[ℏ2ℓ(ℓ+1)2μr2−Ze24πϵ0r]R=ER.-\frac{\hbar^2}{2\mu}\frac{1}{r^2}\frac{d}{dr}\!\left(r^2\frac{dR}{dr}\right) + \left[\frac{\hbar^2\ell(\ell+1)}{2\mu r^2} - \frac{Ze^2}{4\pi\epsilon_0 r}\right]R = E R.

The ℓ(ℓ+1)\ell(\ell+1) term is the centrifugal barrier coming purely from the angular kinetic energy; it enters the radial equation as an effective addition to the potential, which is why ℓ\ell could, in principle, affect the energy — as it does for any potential other than pure Coulomb.

Step 2 — Substitute u(r)=rR(r)u(r) = rR(r) and go dimensionless. Defining u=rRu=rR turns the equation into a 1-D Schrödinger-like form,

−ℏ22μd2udr2+[ℏ2ℓ(ℓ+1)2μr2−Ze24πϵ0r]u=Eu.-\frac{\hbar^2}{2\mu}\frac{d^2u}{dr^2} + \left[\frac{\hbar^2 \ell(\ell+1)}{2\mu r^2} - \frac{Ze^2}{4\pi\epsilon_0 r}\right]u = Eu.

Introduce the Bohr radius a0=4πϵ0ℏ2/(μe2)a_0 = 4\pi\epsilon_0\hbar^2/(\mu e^2) and the dimensionless variable ρ=r/a0\rho = r/a_0 together with λ2=−2μEa02/ℏ2\lambda^2 = -2\mu E a_0^2/\hbar^2 (positive for bound states, since E<0E<0). The equation becomes

d2udρ2=[λ2+ℓ(ℓ+1)ρ2−2Zρ]u.\frac{d^2u}{d\rho^2} = \left[\lambda^2 + \frac{\ell(\ell+1)}{\rho^2} - \frac{2Z}{\rho}\right]u.

Step 3 — Extract asymptotic behavior and posit a series solution. As ρ→∞\rho\to\infty the equation reduces to u′′≈λ2uu''\approx \lambda^2 u, giving u∼e−λρu\sim e^{-\lambda\rho} (the decaying solution is kept for normalizability). As ρ→0\rho\to 0, the centrifugal term dominates and u∼ρℓ+1u \sim \rho^{\ell+1} (again, keeping only the solution regular at the origin). Writing

u(ρ)=ρℓ+1e−λρL(ρ)u(\rho) = \rho^{\ell+1} e^{-\lambda\rho} L(\rho)

and substituting into the differential equation produces a recursion for the power-series coefficients of L(ρ)=∑kckρkL(\rho) = \sum_k c_k \rho^k:

ck+1=2λ(k+ℓ+1)−2Z(k+1)(k+2ℓ+2) ck.c_{k+1} = \frac{2\lambda(k+\ell+1) - 2Z}{(k+1)(k+2\ell+2)}\,c_k.

Step 4 — Impose normalizability: the series must terminate. If the series for L(ρ)L(\rho) does not terminate, its coefficients behave like those of e2λρe^{2\lambda\rho} for large kk, which would make u(ρ)∼ρℓ+1e+λρu(\rho)\sim \rho^{\ell+1}e^{+\lambda\rho} blow up at infinity — unacceptable for a bound, square-integrable state. The only way to avoid this is for the numerator to vanish at some non-negative integer k=nrk = n_r, i.e.

λ(nr+ℓ+1)=Z⟹λ=Znr+ℓ+1.\lambda(n_r + \ell + 1) = Z \quad\Longrightarrow\quad \lambda = \frac{Z}{n_r+\ell+1}.

Define n≡nr+ℓ+1n \equiv n_r + \ell + 1 (so n=1,2,3,…n = 1, 2, 3,\dots since nr=0,1,2,…n_r = 0,1,2,\dots and ℓ≥0\ell \ge 0). Then simply λ=Z/n\lambda = Z/n.

Step 5 — Convert back to energy. Recall λ2=−2μEa02/ℏ2\lambda^2 = -2\mu E a_0^2/\hbar^2, so

E=−ℏ2λ22μa02=−ℏ22μa02⋅Z2n2.E = -\frac{\hbar^2 \lambda^2}{2\mu a_0^2} = -\frac{\hbar^2}{2\mu a_0^2}\cdot\frac{Z^2}{n^2}.

Using ℏ2/(μa02)=e2/(4πϵ0a0)=2 Ry\hbar^2/(\mu a_0^2) = e^2/(4\pi\epsilon_0 a_0) = 2\,\text{Ry} (a standard identity relating the Bohr radius to the Rydberg energy), this becomes

En=−Z2n2 Ry.E_n = -\frac{Z^2}{n^2}\,\text{Ry}.

Conclusion. The crucial point is in Step 4: the termination condition involves nrn_r and ℓ\ell only through their sum nr+ℓ+1n_r+\ell+1. Nowhere does the derivation produce a term depending on ℓ\ell alone. That is a direct consequence of the potential being exactly −Ze2/(4πϵ0r)-Ze^2/(4\pi\epsilon_0 r) — an inverse-first-power law — which is what allows the −2Z/ρ-2Z/\rho term in Step 3 to combine so cleanly with the exponential ansatz. Change the power of rr in the potential (e.g., add a 1/r21/r^2 term, or use a screened Yukawa-type potential), and the termination condition will involve ℓ\ell independently, immediately lifting the ℓ\ell-degeneracy.

Go deeper · Derivation B · 2 minWhy the Coulomb ℓ-degeneracy is not an accident: the Runge–Lenz vector
Derivation BWhy the Coulomb ℓ-degeneracy is not an accident: the Runge–Lenz vector

Goal. Show, using the symmetry argument (rather than brute-force solving the radial equation), why the Coulomb potential specifically produces extra conserved quantities that connect different ℓ\ell values at fixed nn, explaining the term "accidental degeneracy" and why it is really a hidden dynamical symmetry.

Step 1 — Ordinary rotational symmetry and its consequence. In any central potential V(r)V(r), the Hamiltonian commutes with all three components of orbital angular momentum, [H^,L^i]=0[\hat H, \hat L_i] = 0. This algebra closes among the L^i\hat L_i alone: [L^i,L^j]=iℏϵijkL^k[\hat L_i, \hat L_j] = i\hbar\epsilon_{ijk}\hat L_k, generating rotations, i.e. the group SO(3)SO(3). Because H^\hat H, L^2\hat L^2, and L^z\hat L_z can be simultaneously diagonalized, states of the same ℓ\ell but different mm are automatically degenerate — this is the essential (2ℓ+1)(2\ell+1)-fold degeneracy, guaranteed for every central potential with no further assumption.

Step 2 — The classical clue: the Runge–Lenz vector. In the classical Kepler problem with V(r)=−k/rV(r) = -k/r, in addition to angular momentum L=r×p\mathbf{L} = \mathbf{r}\times\mathbf{p}, there is a second conserved vector,

A=p×L−mk r^,\mathbf{A} = \mathbf{p}\times\mathbf{L} - m k\,\hat{\mathbf{r}},

which points along the major axis of the classical elliptical orbit and whose constancy is exactly the statement that Kepler orbits do not precess — a special feature of the 1/r1/r potential (Bertrand's theorem shows only 1/r1/r and r2r^2 potentials give closed orbits for all bound trajectories). No other central-force law has a similarly conserved vector.

Step 3 — Quantum Runge–Lenz operator. The quantum-mechanical (symmetrized, Hermitian) analogue is

A^=12μ(p^×L^−L^×p^)−Ze24πϵ0r^/r.\hat{\mathbf{A}} = \frac{1}{2\mu}\left(\hat{\mathbf{p}}\times\hat{\mathbf{L}} - \hat{\mathbf{L}}\times\hat{\mathbf{p}}\right) - \frac{Ze^2}{4\pi\epsilon_0}\hat{\mathbf{r}}/r.

Direct (if lengthy) commutator algebra shows [H^,A^]=0[\hat H, \hat{\mathbf{A}}] = 0: the Runge–Lenz operator is conserved for the Coulomb Hamiltonian, and only for the Coulomb Hamiltonian among central potentials. Crucially, A^\hat{\mathbf{A}} does not commute with L^2\hat L^2: applying A^\hat{\mathbf A} to a state of angular momentum ℓ\ell produces a superposition containing ℓ\ell and ℓ±1\ell\pm1 components (much as a raising/lowering operator connects adjacent quantum numbers). Because [H^,A^]=0[\hat H, \hat{\mathbf A}]=0, states connected by A^\hat{\mathbf{A}} must share the same energy eigenvalue — so states of different ℓ\ell (but the same nn) are forced into degeneracy by this extra symmetry.

Step 4 — The enlarged algebra: SO(4)SO(4). Suitably rescaling A^\hat{\mathbf A} (dividing by −2μE\sqrt{-2\mu E}, valid within a fixed bound-state energy shell) and combining it with L^\hat{\mathbf L}, one finds that the six operators {L^i,A^i}\{\hat L_i, \hat A_i\} close under commutation into the Lie algebra of SO(4)SO(4) (equivalently, two commuting copies of the SO(3)SO(3)/SU(2)SU(2) algebra after forming J^1=(L^+A^)/2\hat{\mathbf J}_1 = (\hat{\mathbf L}+\hat{\mathbf A})/2 and J^2=(L^−A^)/2\hat{\mathbf J}_2 = (\hat{\mathbf L}-\hat{\mathbf A})/2). Each irreducible representation of SO(4)SO(4) built this way has dimension n2n^2 and corresponds exactly to one principal quantum number shell — reproducing, by pure representation theory (no radial equation needed), both the correct energy spectrum En∝−1/n2E_n\propto -1/n^2 and the degeneracy n2n^2.

Conclusion. The ℓ\ell-degeneracy of hydrogen is not a coincidence of the algebra in the derivation above (which could easily have failed to simplify) — it is a direct signature of the dynamical symmetry group being SO(4)SO(4) rather than merely SO(3)SO(3). This is exactly analogous to (and historically inspired by) the 3-D isotropic harmonic oscillator, whose potential V(r)=12mω2r2V(r) = \tfrac12 m\omega^2 r^2 is the other potential satisfying Bertrand's theorem and which shows its own accidental degeneracy with symmetry group SU(3)SU(3). Any deviation of the potential from pure 1/r1/r — however small — breaks the Runge–Lenz conservation and splits the ℓ\ell-degeneracy; this is exactly what happens physically inside multi-electron atoms.

Worked examples

Example 1 (foundational): Full quantum-number enumeration and energy for n=4n=4, hydrogen atom (Z=1Z=1). List every allowed (ℓ,m)(\ell, m) combination for n=4n=4 and verify the total count equals n2n^2. Allowed ℓ\ell: since ℓ=0,…,n−1\ell = 0,\dots,n-1, we have ℓ=0,1,2,3\ell = 0,1,2,3 (spectroscopic labels s,p,d,fs,p,d,f). For each ℓ\ell, the number of mm values is 2ℓ+12\ell+1:

ℓ=0: 2(0)+1=1 state (m=0)\ell=0:\ 2(0)+1=1 \text{ state } (m=0)
ℓ=1: 2(1)+1=3 states (m=−1,0,+1)\ell=1:\ 2(1)+1=3 \text{ states } (m=-1,0,+1)
ℓ=2: 2(2)+1=5 states (m=−2,−1,0,1,2)\ell=2:\ 2(2)+1=5 \text{ states } (m=-2,-1,0,1,2)
ℓ=3: 2(3)+1=7 states (m=−3,…,+3)\ell=3:\ 2(3)+1=7 \text{ states } (m=-3,\dots,+3)

Sum:

g4=1+3+5+7=16=42✓g_4 = 1+3+5+7 = 16 = 4^2 \checkmark

The energy for every one of these 16 states is identical:

E4=−(1)2(4)2 Ry=−116(13.606 eV)=−0.8504 eV.E_4 = -\frac{(1)^2}{(4)^2}\,\text{Ry} = -\frac{1}{16}(13.606\ \text{eV}) = -0.8504\ \text{eV}.

Limiting-case check: as n→∞n\to\infty, En→0−E_n\to 0^- (states pile up just below the ionization limit — this is the hallmark of a Coulomb potential's infinite bound-state ladder, in contrast to, e.g., a finite square well which has only finitely many bound states). Including spin doubles every count: 2g4=322g_4 = 32 total spin-orbitals in the n=4n=4 shell, matching the familiar "2,8,18,322,8,18,32" shell-capacity sequence.

Example 2 (applied): Comparing He⁺ (Z=2Z=2) and H (Z=1Z=1) at n=2n=2, and connecting to the Balmer-series-like transition. He⁺ is a hydrogen-like ion (one electron, Z=2Z=2) — a system of real spectroscopic importance (its n=4→3n=4\to3 transition mimics part of the hydrogen Balmer series and was historically confused with it in early stellar spectra, the "Pickering series"). For He⁺, using the exact formula (ignoring the small reduced-mass correction, which shifts things by less than 0.03%):

En(He+)=−Z2n2 Ry=−4n2 Ry.E_n(\text{He}^+) = -\frac{Z^2}{n^2}\,\text{Ry} = -\frac{4}{n^2}\,\text{Ry}.

At n=2n=2:

E2(He+)=−44 Ry=−1.0000 Ry=−13.606 eV.E_2(\text{He}^+) = -\frac{4}{4}\,\text{Ry} = -1.0000\,\text{Ry} = -13.606\ \text{eV}.

Compare with hydrogen's ground state, E1(H)=−1.000 Ry=−13.606E_1(\text{H}) = -1.000\,\text{Ry} = -13.606 eV. These are numerically coincident (up to the reduced-mass correction) — this is a direct manifestation of the Z2/n2Z^2/n^2 scaling: He⁺ at n=2n=2 mimics H at n=1n=1 because Z2/n2=4/4=1/1Z^2/n^2 = 4/4 = 1/1. Now consider the n=4→n=3n=4\to n=3 emission transition in He⁺:

ΔE=E4−E3=−416Ry−(−49Ry)=−0.25 Ry+0.4444 Ry=0.1944 Ry.\Delta E = E_4 - E_3 = -\frac{4}{16}\text{Ry} - \left(-\frac{4}{9}\text{Ry}\right) = -0.25\,\text{Ry} + 0.4444\,\text{Ry} = 0.1944\,\text{Ry}.

Converting to a wavelength using 1 Ry=2.1799×10−181\ \text{Ry} = 2.1799\times10^{-18} J and λ=hc/ΔE\lambda = hc/\Delta E:

ΔE=0.1944×2.1799×10−18 J=4.238×10−19 J,\Delta E = 0.1944 \times 2.1799\times10^{-18}\ \text{J} = 4.238\times10^{-19}\ \text{J},
λ=(6.626×10−34)(2.998×108)4.238×10−19=4.686×10−7 m=468.6 nm.\lambda = \frac{(6.626\times10^{-34})(2.998\times10^8)}{4.238\times10^{-19}} = 4.686\times10^{-7}\ \text{m} = 468.6\ \text{nm}.

This falls in the blue-green visible region — one member of the He⁺ Pickering series, historically mistaken for a hydrogen line before it was correctly assigned to He⁺. Limiting-case check: as Z→Z\to larger values (say a hydrogen-like Fe²⁵⁺ ion, Z=26Z=26), the same transition would scale as ΔE∝Z2\Delta E \propto Z^2, pushing the emission from visible light all the way into the X-ray region — precisely how astrophysicists identify highly ionized species in solar-flare and accretion-disk X-ray spectra.

Example 3 (exam / competition level): Constructing an explicit degenerate pair at n=3n=3 and demonstrating the accidental degeneracy numerically from the radial wavefunctions. Show explicitly, using the normalized radial functions, that R30(r)R_{30}(r) (3s3s) and R32(r)R_{32}(r) (3d3d) — which look completely different in shape — are eigenfunctions of the same radial Hamiltonian eigenvalue E3=−Z2/9E_3 = -Z^2/9 Ry, by evaluating the expectation value of the Hamiltonian directly (rather than trusting the general formula) at one representative sample point argument, and cross-checking with the general quantization condition. The hydrogen-like radial functions (in units where a0/Z→aa_0/Z \to a; here consider Z=1 for simplicity so a=a0a=a_0) are:

R30(r)=2813(1a0)3/2(27−18ra0+2r2a02)e−r/3a0R_{30}(r) = \frac{2}{81\sqrt{3}}\left(\frac{1}{a_0}\right)^{3/2}\left(27 - 18\frac{r}{a_0} + 2\frac{r^2}{a_0^2}\right)e^{-r/3a_0}
R32(r)=48130(1a0)3/2r2a02 e−r/3a0R_{32}(r) = \frac{4}{81\sqrt{30}}\left(\frac{1}{a_0}\right)^{3/2}\frac{r^2}{a_0^2}\,e^{-r/3a_0}

Both functions decay with the same exponential rate e−r/3a0e^{-r/3a_0} — note that this exponential decay constant is set only by n=3n=3 (recall from the derivation, λ=Z/n\lambda = Z/n, and u∼e−λρu\sim e^{-\lambda \rho}), not by ℓ\ell. This is already a strong hint of the shared energy, since the asymptotic decay rate of a bound-state wavefunction is directly tied to EE via λ=−2μEa02/ℏ2\lambda = \sqrt{-2\mu E a_0^2/\hbar^2}; identical decay constants can only occur for identical ∣E∣|E|. To make this quantitative, apply the quantization condition from the derivation directly to each case: For 3s3s: nr=2n_r = 2 (two radial nodes — matching the two roots of the quadratic factor 27−18x+2x227-18x+2x^2), ℓ=0\ell=0, so

n=nr+ℓ+1=2+0+1=3.n = n_r+\ell+1 = 2+0+1 = 3.

For 3d3d: nr=0n_r=0 (no radial nodes — the function r2e−r/3a0r^2 e^{-r/3a_0} is nodeless for r>0r>0), ℓ=2\ell=2, so

n=nr+ℓ+1=0+2+1=3.n = n_r+\ell+1 = 0+2+1 = 3.

Both give n=3n=3, hence identically

E3=−1232Ry=−19 Ry=−1.5118 eV.E_3 = -\frac{1^2}{3^2}\text{Ry} = -\frac{1}{9}\,\text{Ry} = -1.5118\ \text{eV}.

Despite 3s3s having two radial nodes and zero angular nodes while 3d3d has zero radial nodes and two angular nodes (the Y2±2Y_2^{\pm2} or real dxy,dx2−y2d_{xy}, d_{x^2-y^2} angular parts), the total node count is identical: nr+ℓ=2n_r + \ell = 2 in both cases. This total-node invariant, n−1=nr+ℓn - 1 = n_r+\ell, is exactly what the Coulomb potential conserves, and is the operational, calculation-level manifestation of the SO(4)SO(4) symmetry argument. Limiting-case check: if we instead used a modified potential V(r)=−Ze2/(4πϵ0r)+δ/r2V(r) = -Ze^2/(4\pi\epsilon_0 r) + \delta/r^2 with small δ≠0\delta\ne 0 (a Coulomb-plus-centrifugal perturbation, as occurs approximately for penetrating orbitals in alkali-metal valence electrons), the effective ℓ(ℓ+1)\ell(\ell+1) in the radial equation would be replaced by ℓ(ℓ+1)+2μδ/ℏ2\ell(\ell+1)+2\mu\delta/\hbar^2, giving an effective non-integer ℓ′\ell' and a quantization condition n=nr+ℓ′+1n = n_r+\ell'+1 that depends on ℓ\ell separately — immediately splitting 3s3s below 3d3d (lower ℓ\ell penetrates the core more and is stabilized), exactly the origin of the orbital energy ordering used in Term Symbols and Atomic States and the aufbau principle for real multi-electron atoms.

Limiting cases, scope & common traps

SituationWhat happens to the degeneracyWhy
Exact one-electron Coulomb potential (H, He⁺, Li²⁺,…)Full n2n^2-fold degeneracy (2n22n^2 with spin)Exact 1/r1/r potential, hidden SO(4)SO(4)/Runge–Lenz symmetry
Multi-electron atoms (Na, Fe, …)ℓ\ell-degeneracy destroyed; ns<np<nd<nfns<np<nd<nfCore electrons screen the nuclear charge non-uniformly; effective potential is no longer pure 1/r1/r
External magnetic field (Zeeman effect)mm-degeneracy lifted; 2ℓ+12\ell+1 sublevels splitBreaks rotational (SO(3)SO(3)) symmetry itself by picking out a preferred axis
External electric field (Stark effect)Mixes states of different ℓ\ell, same nn; can partially lift degeneracy at first orderCouples states already accidentally degenerate via A^\hat{\mathbf A}-like operator (z^\hat z has matrix elements between ℓ\ell and ℓ±1\ell\pm1)
Relativistic corrections (fine structure, Lamb shift)Splits degeneracy within same n,ℓn,\ell (spin–orbit) and even 2s1/22s_{1/2} vs 2p1/22p_{1/2} (Lamb shift)Relativistic kinetic energy and QED effects are not exactly 1/r1/r-symmetric under SO(4)SO(4)
n→∞n\to\inftyLevels crowd toward E=0E=0; degeneracy n2→∞n^2\to\inftyCoulomb potential supports infinitely many bound states (unlike a finite well)
Z→Z\to large (hydrogen-like heavy ions)Degeneracy pattern unchanged, but ∣En∣|E_n| scales as Z2Z^2The SO(4)SO(4) symmetry argument is ZZ-independent; only overall energy scale changes

The chemist behind it

Origins · 1 minErwin Schrödinger and how this was discovered
ES
Erwin Schrödinger
1887–1961 · Vienna and Zürich, 1926

Erwin Schrödinger solved the hydrogen atom wave equation in his landmark 1926 papers, obtaining the −Z2/n2-Z^2/n^2 (Rydberg-scaled) spectrum directly from the newly proposed wave mechanics, matching the empirical Rydberg–Ritz formula that had summarized hydrogen's spectral lines for decades before any theoretical derivation existed. His derivation used the same separation-of-variables and Laguerre-polynomial machinery that underlies the radial equation treatment above. Schrödinger himself, along with Wolfgang Pauli (using matrix mechanics and the algebraic Runge–Lenz-vector method) and later Pauli's independent 1926 letter to Jordan, recognized that the extra ℓ\ell-degeneracy pointed to a symmetry beyond simple rotational invariance — foreshadowing the group-theoretic (SO(4)SO(4)) understanding fully worked out later by Vladimir Fock (1935) and Valentine Bargmann (1936), who made the connection to the four-dimensional rotation group rigorous.

History
- 1885–1888: Balmer and then Rydberg empirically fit the hydrogen spectral lines to what is now recognized as the 1/n21/n^2 formula, with no theoretical derivation.
- 1913: Niels Bohr derives En=−Z2/n2E_n = -Z^2/n^2 Ry from a semiclassical quantized-orbit model, correctly predicting the spectrum but without explaining orbital shapes or degeneracy structure.
- 1926: Schrödinger and, independently via matrix mechanics, Pauli solve the full quantum hydrogen atom, recovering En=−Z2/n2E_n=-Z^2/n^2 Ry and identifying the n2n^2-fold degeneracy including ℓ\ell and mm.
- 1935–1936: Fock and Bargmann formalize the SO(4)SO(4) hidden-symmetry explanation of hydrogen's accidental ℓ\ell-degeneracy using the Runge–Lenz vector, connecting it rigorously to representation theory.
- 1947: Lamb and Retherford experimentally measure the Lamb shift, showing that real hydrogen's 2s1/22s_{1/2} and 2p1/22p_{1/2} levels are not exactly degenerate — direct experimental proof that the accidental degeneracy is a feature of the idealized non-relativistic Coulomb Hamiltonian only, broken by quantum electrodynamic effects.
Why doesn't the hydrogen energy depend on ℓ at all?

Because the Coulomb potential −Ze2/(4πϵ0r)-Ze^2/(4\pi\epsilon_0 r) has an extra conserved quantity beyond ordinary angular momentum: the quantum Runge–Lenz vector. This vector commutes with the Hamiltonian but not with L^2\hat L^2, so it connects states of different ℓ\ell while forcing them to share the same energy. Solving the radial equation directly shows the same thing algebraically: the quantization condition only involves the combination n=nr+ℓ+1n = n_r+\ell+1, never nrn_r and ℓ\ell separately, so different (ℓ,nr)(\ell,n_r) pairs adding to the same nn are exactly degenerate. This is a special feature of the exact 1/r1/r potential, not a general property of central potentials.

Is the m-degeneracy the same kind of thing as the ℓ-degeneracy?

No, and this distinction is essential. The mm-degeneracy (the 2ℓ+12\ell+1 states for a given ℓ\ell) is guaranteed by ordinary rotational symmetry, SO(3)SO(3): since no direction in space is special for an isolated atom, states related by rotation must share the same energy. This holds for any central potential — a screened atomic potential, a 3-D harmonic oscillator, anything. The ℓ\ell-degeneracy (different ℓ\ell's at the same nn) is a much rarer, Coulomb-specific effect tied to the hidden SO(4)SO(4) symmetry, and it disappears the moment the potential deviates from exact 1/r1/r.

Why do real multi-electron atoms show 4s4s filling before 3d3d if hydrogen's energies only depend on n?

In hydrogen, only one electron feels a pure, unscreened 1/r1/r nuclear potential, so the accidental ℓ\ell-degeneracy holds exactly. In multi-electron atoms, each electron feels the nucleus plus the averaged repulsion of every other electron — an effective potential that is no longer pure Coulomb. Lower-ℓ\ell orbitals (like 4s4s) penetrate closer to the nucleus through the electron cloud more than higher-ℓ\ell orbitals of similar nn (like 3d3d), so they feel less screening and are stabilized. This penetration/screening effect breaks the accidental degeneracy and produces the empirical orbital energy ordering used in the aufbau principle.

What happens to hydrogen's degeneracy under a magnetic field?

An external magnetic field along, say, the zz-axis picks out a preferred spatial direction, which breaks the rotational symmetry SO(3)SO(3) responsible for the mm-degeneracy itself (not just the accidental ℓ\ell-degeneracy). The energy gains a term proportional to mm (the Zeeman term ∝mℓB\propto m_\ell B), so the 2ℓ+12\ell+1 states of a given ℓ\ell split into 2ℓ+12\ell+1 distinct, evenly spaced sublevels. This is the normal Zeeman effect, and it is a direct experimental demonstration that the mm-degeneracy, unlike the deeper structure of quantum mechanics, is entirely a consequence of the absence of any external symmetry-breaking field.

Does the accidental degeneracy survive relativistic corrections?

Not exactly. The non-relativistic Schrödinger treatment gives exact ℓ\ell-degeneracy within a shell, but once you include relativistic kinetic-energy corrections and spin–orbit coupling (the Dirac equation treatment), states are labeled by total angular momentum jj rather than ℓ\ell and mm separately, and levels with the same nn and jj but different ℓ\ell (such as 2s1/22s_{1/2} and 2p1/22p_{1/2}) remain degenerate in the pure Dirac theory — but this residual degeneracy is finally broken by quantum electrodynamic effects, the Lamb shift, measured by Lamb and Retherford in 1947. So the Coulomb accidental degeneracy is progressively (and instructively) lifted as you add more complete physics.

How is this degeneracy related to Bertrand's theorem?

Bertrand's theorem is a classical mechanics result stating that among all possible central-force laws, only the inverse-square force (V∝−1/rV\propto -1/r, giving Kepler ellipses) and the Hooke's-law force (V∝r2V\propto r^2, the isotropic harmonic oscillator) produce closed orbits for every bound trajectory, regardless of initial conditions. This classical non-precession is the direct classical shadow of the extra conserved quantities (Runge–Lenz vector for Coulomb, a symmetric tensor for the oscillator) that, upon quantization, produce accidental degeneracy. So Bertrand's theorem essentially predicts in advance which two quantum central-potential problems will show extra (ℓ\ell-independent) degeneracy: exactly the hydrogen atom and the 3-D isotropic harmonic oscillator.

Why is the degeneracy called "n² " and not something else?

Because summing the number of magnetic sublevels (2ℓ+1)(2\ell+1) over every allowed ℓ\ell from 00 to n−1n-1 is precisely the sum of the first nn odd numbers, 1+3+5+⋯+(2n−1)1+3+5+\cdots+(2n-1), a classical arithmetic identity that equals n2n^2 exactly. This is not a coincidence of hydrogen specifically — it's simply the combinatorial consequence of the rule ℓ≤n−1\ell\le n-1 (itself derived from the radial quantization condition) combined with the universal rule that each ℓ\ell carries 2ℓ+12\ell+1 magnetic sublevels. Including electron spin doubles it to 2n22n^2, matching the shell capacities familiar from general chemistry.

Before this

Where this leads