core34 min studyLesson 375 in Physics

Hydrogenic energies and degeneracy

Use the Coulomb spectrum with reduced mass and separate orbital multiplicity from spin and physical level splittings.

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What you’ll understand

  • Calculate Coulomb binding and transition energies with a stated reduced mass.
  • Enumerate allowed n, ℓ, and m values and their multiplicities.
  • Explain the idealizations behind hydrogen's additional orbital degeneracy.

The nonrelativistic two-body Coulomb problem predicts a striking spectrum: several states with different orbital angular momenta share one energy. This equality is stronger than ordinary rotational degeneracy. It makes hydrogen-like ions an excellent test of quantum-number counting, but also a place where neglected spin, relativistic effects, finite nuclear mass, and external fields must be stated before a degeneracy is called exact.

Reduce the two-body problem first

For an electron of mass mem_e and nucleus of mass MM and charge +Ze+Ze, separate the centre-of-mass motion. The internal relative-coordinate Hamiltonian contains reduced mass μ=meM/(me+M)\mu=m_eM/(m_e+M) and potential −Ze2/(4πϵ0r)-Ze^2/(4\pi\epsilon_0r). Its bound energies are measured below the zero-energy ionization threshold at infinite separation. A negative eigenvalue gives binding; the emitted photon in a downward transition carries a positive difference between two negative level energies.

The characteristic Coulomb length is aμ=4πϵ0ℏ2/(μe2)a_\mu=4\pi\epsilon_0\hbar^2/(\mu e^2). Increasing nuclear charge contracts hydrogenic orbitals roughly by a factor Z, while increasing reduced mass contracts them further. The binding scale is proportional to μZ2\mu Z^2. Replacing reduced mass by electron mass is usually a useful heavy-nucleus approximation, but it is not exact and fails badly for equal-mass constituents. The energy and length conventions must be changed together.

En=−μZ2e42(4πϵ0)2ℏ2n2≃−13.6 eV μmeZ2n2,n=1,2,3,…E_n=-\frac{\mu Z^2e^4}{2(4\pi\epsilon_0)^2\hbar^2n^2}\simeq-13.6\,\mathrm{eV}\,\frac{\mu}{m_e}\frac{Z^2}{n^2},\qquad n=1,2,3,\ldots
This is the spin-independent nonrelativistic point-Coulomb internal spectrum. Z is a positive integer nuclear charge number, n is dimensionless, and μ is the reduced mass. The numerical 13.6 eV scale uses electron mass; finite-size, relativistic, radiative, and spin interactions are omitted.↔ Scroll wide equations horizontally.

Count orbitals rather than energy values

For a given principal quantum number n, allowed orbital angular momenta are ℓ=0,1,…,n−1\ell=0,1,\ldots,n-1. Each has m=−ℓ,…,ℓm=-\ell,\ldots,\ell and therefore 2ℓ+12\ell+1 magnetic substates. Adding these multiplicities yields n2n^2 spatial states with the same Coulomb energy. These are distinct orthogonal eigenvectors, not n2n^2 different energy values. Each fixed-ℓ radial function is shared by its m states because the central radial equation contains no m.

Rotational symmetry explains equality between m states at fixed angular momentum. Equality between different ℓ values in this ideal Coulomb spectrum reflects an additional dynamical symmetry, often expressed through the Runge–Lenz operator. A generic central potential preserves m degeneracy but makes energy depend on radial and angular labels separately. Consequently, the degeneracy formula for hydrogen must not be transplanted to an arbitrary screened atomic potential, finite spherical well, or many-electron atom.

ℓ=0,…,n−1,m=−ℓ,…,ℓ,gnorbital=∑ℓ=0n−1(2ℓ+1)=n2\ell=0,\ldots,n-1,\qquad m=-\ell,\ldots,\ell,\qquad g_n^{\rm orbital}=\sum_{\ell=0}^{n-1}(2\ell+1)=n^2
The count is for spatial eigenstates in the ideal spinless Coulomb Hamiltonian. Including an uncoupled electron spin doubles it to 2n². Nuclear spin is not counted, and including interactions that resolve fine or hyperfine structure changes which states remain at exactly equal energy.↔ Scroll wide equations horizontally.

Use differences and label the approximation

A transition energy follows by subtracting final from initial internal energy. Its photon wavelength is hc/ΔEhc/\Delta E when recoil is negligible at the precision of the estimate. The existence of two levels alone does not guarantee a strong radiative line: transition amplitudes and selection rules determine whether a particular coupling connects the states. An energy-level calculation provides possible energy differences, not their intensities or decay lifetimes.

Real spectra resolve corrections to the ideal model. Electron spin and relativistic dynamics produce fine structure, radiative effects can distinguish otherwise degenerate states, and external electric or magnetic fields remove further equalities. Finite nuclear mass already changes the gross binding scale through reduced mass. A useful statement therefore names both the Hamiltonian that has the degeneracy and the resolution at which omitted effects matter, instead of asserting that every state with the same principal number is exactly equal in an actual atom.

A hydrogen-like ion and its n = 2 shell

Approximate the reduced mass by mem_e for a one-electron ion with Z=2Z=2. Find its n = 1 and n = 2 energies, the energy difference between them, and the n = 2 orbital multiplicity. Use hc=1240 eV nmhc=1240\,\mathrm{eV\,nm} to estimate the corresponding photon wavelength when an allowed transition is selected.

  1. The Coulomb scale is 13.6Z2=54.4 eV13.6Z^2=54.4\,\mathrm{eV}. Thus E1=−54.4 eVE_1=-54.4\,\mathrm{eV} and E2=−54.4/4=−13.6 eVE_2=-54.4/4=-13.6\,\mathrm{eV}. Both are relative to the same exterior ionization threshold.

  2. The downward energy difference is E2−E1=40.8 eVE_2-E_1=40.8\,\mathrm{eV}. Therefore λ=1240/40.8=30.4 nm\lambda=1240/40.8=30.4\,\mathrm{nm}. This is a positive photon energy even though both internal energies are negative; recoil is neglected.

  3. For n = 2, ℓ = 0 gives one m state and ℓ = 1 gives three. The orbital count is 1+3=41+3=4; including an otherwise uncoupled electron spin gives eight states in the same idealized shell, before fine structure or external fields are resolved.

The result

The levels lie at −54.4 and −13.6 eV, separated by 40.8 eV, corresponding to about 30.4 nm for an allowed transition. The ideal n = 2 shell contains four spatial orbitals, or eight electron-spin states if spin interactions are omitted.

Explore the model

Interactive lab

Interactive lab · quantum

Degeneracy is not a fixed mixing angle

Treat the lab basis as two chosen orthogonal members of a degenerate subspace at a fixed instant. Reset and set both controls to θ = 0° and φ = 0°; predict basis probability 1 for the first state. Reset and set both controls to θ = 90° and φ = 0°; predict equal probabilities. The pure two-state lab does not count a hydrogen shell or implement its Hamiltonian; its playback phase precession is not evolution within an exactly degenerate hydrogen subspace.

Degeneracy is not a fixed mixing angleA normalized two-level pure state, not an interacting field simulation. Playback adds one radian per second to the relative phase; basis probabilities stay fixed. Current readings: Probability |0⟩: 71.1%; Probability |1⟩: 28.9%; Bloch (x, y, z): 0.78, 0.45, 0.42. All parameters are adjustable with the labeled controls below.xyz|0⟩|1⟩Basis probabilitiesP(0)71.1%P(1)28.9%
Bloch vector / P(0)P(1)Phase latitude
Probability |0⟩
71.1%
Probability |1⟩
28.9%
Bloch (x, y, z)
0.78, 0.45, 0.42
t = 0 s

A normalized two-level pure state, not an interacting field simulation. Playback adds one radian per second to the relative phase; basis probabilities stay fixed. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forDifferent normalized superpositions can belong to one degenerate energy; counting that subspace requires the quantum-number calculation outside the lab.

Check your understanding

What is the spatial degeneracy of the n = 3 shell in the ideal spin-independent point-Coulomb Hamiltonian?

Try it yourself

Consider an ideal Coulomb pair of opposite charges of equal magnitude e and equal constituent masses mem_e, ignoring spin, relativity, radiation corrections, and annihilation. Find the reduced mass, n = 2 binding energy, and the n = 3 to n = 2 energy difference. Count the spatial n = 3 multiplicity.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Equal constituent masses halve the electron-mass binding scale; subtract energies only after including that factor.

Compare with the solution

The reduced mass is me2/(2me)=me/2m_e^2/(2m_e)=m_e/2, so En=−6.8 eV/n2E_n=-6.8\,\mathrm{eV}/n^2. The n = 2 energy is −1.70 eV, meaning binding energy 1.70 eV. The downward difference is 6.8(1/4−1/9)=6.8(5/36)=0.944 eV6.8(1/4-1/9)=6.8(5/36)=0.944\,\mathrm{eV}. The spatial n = 3 shell still has 1+3+5=91+3+5=9 states, because changing reduced mass rescales energies without changing the allowed orbital labels. No spin multiplicity is included for either constituent in this deliberately spinless comparison.

Take it with you

  • Reduced mass and Z² determine the Coulomb binding scale.
  • Hydrogen's n² spatial multiplicity counts eigenvectors, not distinct energy values.
  • Real-atom splittings and transition strengths require physics beyond the ideal energy formula.

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