Degeneracy of the Hydrogen Atom

The nonrelativistic hydrogen atom has more degeneracy than a generic central-potential problem. For a fixed principal quantum number $n$, the ideal spinless Coulomb Hamiltonian has

$$g_n = \sum_{\ell=0}^{n-1}(2\ell+1) = n^2$$

linearly independent spatial bound states with the same energy. Part of this degeneracy is the ordinary rotational degeneracy of angular-momentum multiplets. The rest is special to the Coulomb potential.

This page explains what is being counted, which part follows from rotations, which part is the special Coulomb degeneracy, and why real hydrogen spectra do not preserve the full ideal degeneracy.

Helpful background. Hydrogen Atom to use the Coulomb quantum numbers and bound-state energies.

Ideal Coulomb Hamiltonian

The model is the spinless, nonrelativistic, point-Coulomb Hamiltonian after center-of-mass separation:

$$\hat H = \frac{\hat p^2}{2\mu} - \frac{e^2}{4\pi\epsilon_0r}.$$

Here $\mu$ is the electron-proton reduced mass. The bound-state energies are

$$E_n = - \frac{\mu e^4}{2(4\pi\epsilon_0)^2\hbar^2} \frac{1}{n^2}, \qquad n=1,2,\ldots .$$

The exact wavefunctions and radial normalization conventions are developed in Hydrogen Atom. The present page focuses on why the energy depends only on $n$ in the ideal model.

For the shell-by-shell angular multiplet structure, see Hydrogen Atom Angular Structure.

The claim does not include electron spin, relativistic corrections, radiative corrections, nuclear spin, finite nuclear size, or external fields. If spin is included but all spin-dependent interactions are artificially switched off, each spatial state acquires two spin states and the degeneracy becomes $2n^2$.

The Quantum Numbers Being Counted

The bound states may be labeled

$$\psi_{n\ell m}(r,\theta,\phi) = R_{n\ell}(r)Y_\ell^m(\theta,\phi),$$

with

$$n=1,2,\ldots, \qquad \ell=0,1,\ldots,n-1, \qquad m=-\ell,-\ell+1,\ldots,\ell .$$

For fixed $\ell$, there are $2\ell+1$ possible values of $m$. For fixed $n$, the allowed $\ell$ values run from $0$ through $n-1$. Therefore

$$\begin{aligned} g_n &= \sum_{\ell=0}^{n-1}(2\ell+1) \\ &= 2\sum_{\ell=0}^{n-1}\ell+n \\ &= n(n-1)+n =n^2 . \end{aligned}$$

For example, the $n=3$ shell contains

$$3s: 1, \qquad 3p: 3, \qquad 3d: 5,$$

for a total of $1+3+5=9$ spatial states.

Rotational Degeneracy

Every central potential satisfies

$$V(\mathbf r)=V(r),$$

so the Hamiltonian is rotationally invariant. For a rotationally invariant domain, $[\hat H,\hat L_i]=0$ for all three components. In particular,

$$[\hat H,\hat{\mathbf L}^2]=0, \qquad [\hat H,\hat L_z]=0 .$$

For any fixed $\ell$, the $m$ states form one irreducible angular-momentum multiplet. Rotational invariance prevents the Hamiltonian from assigning different energies to different orientations inside the same multiplet. Thus the $2\ell+1$ degeneracy within a fixed $\ell$ sector is not peculiar to hydrogen; it is the standard degeneracy of a central potential.

This statement has a useful physical reading. The number $m$ labels the projection of orbital angular momentum on a chosen axis. In the absence of an external axis, no direction in space is physically preferred, so different $m$ values cannot be distinguished by the Hamiltonian.

The same conclusion appears directly in the radial Schrödinger equation. Once $\ell$ is fixed, the radial equation contains $\ell(\ell+1)$ but not $m$. The magnetic quantum number changes the angular orientation, not the radial spectral problem.

The Special Coulomb Degeneracy

A generic central potential does not make different $\ell$ sectors degenerate. The reduced radial equation contains the effective potential

$$V_{\mathrm{eff}}(r) = V(r) + \frac{\hbar^2\ell(\ell+1)}{2\mu r^2}.$$

Changing $\ell$ changes the centrifugal term, and therefore usually changes the radial eigenvalues. In a generic central potential, one expects energies of the form

$$E=E_{n_r\ell},$$

where $n_r$ counts radial nodes and $\ell$ labels orbital angular momentum.

The Coulomb potential is exceptional. Its bound-state energies can be written as

$$E_n = - \frac{\hbar^2}{2\mu a_0^2} \frac{1}{n^2}, \qquad n=n_r+\ell+1 .$$

The energy depends on the sum $n_r+\ell+1$, not on $n_r$ and $\ell$ separately. Thus states with different radial-node counts and different angular momenta can share the same energy. For instance, $2s$ and $2p$ are degenerate in the ideal nonrelativistic Coulomb problem even though they have different $\ell$.

This is the part often called the accidental degeneracy of hydrogen. The word “accidental” does not mean mysterious or approximate. It means that the degeneracy is not forced by the manifest rotational symmetry alone. It is a signal that the Coulomb problem has more structure than an arbitrary central potential.

Hidden Symmetry Preview

Classically, the inverse-square force has a conserved Laplace–Runge–Lenz vector. Quantum mechanically, an operator version of this conserved quantity can be constructed for the Coulomb Hamiltonian. Together with angular momentum, it generates a larger algebra for the bound-state sector. The general operator-algebra lesson is Hidden Symmetry.

For negative energies, this structure is often described as an $\mathrm{SO}(4)$ symmetry. The angular momentum operators account for rotations in physical space, while the additional conserved quantities relate states with different $\ell$ inside the same principal shell. This is the symmetry explanation for why the bound-state energy depends only on $n$.

This hidden symmetry is special. The harmonic oscillator has its own enlarged symmetry and its own pattern of degeneracies, but a generic central potential has only the ordinary rotational multiplets. That is why the hydrogen $n^2$ pattern should not be treated as a universal feature of spherically symmetric systems.

How Real Effects Lift the Degeneracy

Real hydrogen is close to the ideal Coulomb problem, but precision spectroscopy sees many effects that split the ideal levels.

Relativistic kinetic-energy corrections, spin–orbit coupling, and the Darwin term produce fine structure. In the exact Dirac-Coulomb theory for a point nucleus, the levels depend on $n$ and total angular momentum $j$, not simply on $\ell$ and $m$. This already changes the nonrelativistic degeneracy pattern.

Radiative corrections produce the Lamb shift. The most famous example is the splitting between $2S_{1/2}$ and $2P_{1/2}$, which are degenerate in the point-nucleus Dirac-Coulomb spectrum but not in real quantum electrodynamics.

Nuclear spin produces hyperfine structure. External magnetic fields produce Zeeman splitting, and external electric fields produce Stark splitting. Finite nuclear size and nuclear recoil further modify high-precision spectra. These corrections are small on the scale of the gross Coulomb binding energy, but they are not conceptually small if the question is degeneracy.

The hierarchy is therefore:

• the gross nonrelativistic Coulomb model gives $E_n$ and $n^2$ spatial degeneracy;
• rotations explain the $m$ degeneracy inside each $\ell$ multiplet;
• hidden Coulomb symmetry explains the additional $\ell$ degeneracy;
• real interactions and external fields split the ideal degeneracy in characteristic ways.

Common Mistakes

• Treating the full $n^2$ degeneracy as a consequence of rotations alone.
• Assuming every central potential has hydrogen-like $\ell$ degeneracy.
• Forgetting that $m$ degeneracy and $\ell$ degeneracy have different explanations.
• Counting spin degeneracy without stating whether spin-dependent interactions are present.
• Calling the degeneracy “accidental” as if it were a numerical coincidence rather than a hidden-symmetry result.
• Using precision hydrogen spectra as if they were described by the spinless Schrödinger Coulomb Hamiltonian alone.

Exercises

1. Count the spatial degeneracy of the $n=4$ shell.

Solution

For $n=4$, the allowed orbital angular momenta are

$$\ell=0,1,2,3 .$$

Their $m$ degeneracies are

$$1,\quad 3,\quad 5,\quad 7 .$$

Therefore

$$g_4 = 1+3+5+7 =16 =4^2 .$$

2. Explain why $m$ degeneracy is expected for any central potential, but $\ell$ degeneracy is not.

Solution

For a central potential, rotational invariance implies that no spatial direction is preferred. For a fixed radial label, the states with the same $\ell$ span one irreducible rotation multiplet. A rotation-invariant Hamiltonian acts as a scalar on this multiplet, so all its $m$ states have the same energy. They need not be pairwise related by a single spatial rotation.

Changing $\ell$ changes the radial equation through the centrifugal term

$$\frac{\hbar^2\ell(\ell+1)}{2\mu r^2}.$$

For a generic central potential this changes the radial eigenvalues. Hydrogen is special because the Coulomb problem has an additional hidden symmetry that makes the bound-state energy depend on $n=n_r+\ell+1$.

3. List the spatial states in the $n=2$ shell and identify which degeneracies are rotational and which are special to the Coulomb problem.

Solution

For $n=2$, the allowed states are

$$2s: \ell=0,\quad m=0,$$

and

$$2p: \ell=1,\quad m=-1,0,1 .$$

The three $2p$ states are degenerate with one another because of rotational symmetry. The degeneracy between $2s$ and $2p$ is not forced by rotations, because they have different $\ell$. It is the special Coulomb degeneracy.

References

• D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018.
• L. D. Landau and E. M. Lifshitz, Quantum Mechanics: Non-Relativistic Theory, 3rd ed., Pergamon, 1977.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.