07 Jun 2026

Hydrogen Atom

Hydrogen atom as a central potential problem, quantum numbers, energy spectrum, and degeneracy.

msc semester-i quantum-mechanics hydrogen-atom central-potential

The hydrogen atom is the most important exactly solvable central-potential problem in quantum mechanics. It consists of an electron moving in the Coulomb field of a proton.

Coulomb potential

The potential energy is

\[V(r)=-\frac{e^2}{4\pi\epsilon_0r}.\]

The Hamiltonian is

\[H=-\frac{\hbar^2}{2\mu}\nabla^2 -\frac{e^2}{4\pi\epsilon_0r},\]

where $\mu$ is the reduced mass:

\[\mu=\frac{m_em_p}{m_e+m_p}.\]

Since the potential is central,

\[[H,L^2]=0,\qquad [H,L_z]=0.\]

Therefore the eigenfunctions can be written as

\[\psi_{nlm}(r,\theta,\phi) = R_{nl}(r)Y_l^m(\theta,\phi).\]

Energy spectrum

Solving the radial equation gives the bound-state energy levels

\[\boxed{ E_n=-\frac{\mu e^4}{2(4\pi\epsilon_0)^2\hbar^2} \frac{1}{n^2}. }\]

In electron-volt units,

\[\boxed{ E_n=-\frac{13.6\ \mathrm{eV}}{n^2} }\]

when the reduced-mass correction is neglected.

Here

\[n=1,2,3,\dots\]

is the principal quantum number.

Allowed quantum numbers

For a given $n$,

\[l=0,1,2,\dots,n-1,\]

and for each $l$,

\[m=-l,-l+1,\dots,l.\]

Thus the states are labeled by

\[\lvert n,l,m\rangle.\]

The total number of orbital states for a given $n$ is

\[\sum_{l=0}^{n-1}(2l+1)=n^2.\]

Including electron spin, the degeneracy becomes

\[2n^2.\]

Bohr radius

The natural length scale of the hydrogen atom is the Bohr radius:

\[\boxed{ a_0=\frac{4\pi\epsilon_0\hbar^2}{\mu e^2}. }\]

The ground-state wavefunction is

\[\boxed{ \psi_{100}(r)= \frac{1}{\sqrt{\pi a_0^3}}e^{-r/a_0}. }\]

It is spherically symmetric because $l=0$.

Physical meaning of quantum numbers

The principal quantum number $n$ determines the energy in the nonrelativistic Coulomb problem.

The orbital quantum number $l$ determines the magnitude of orbital angular momentum:

\[L^2=\hbar^2l(l+1).\]

The magnetic quantum number $m$ determines the $z$-component:

\[L_z=\hbar m.\]

The spin quantum number $m_s=\pm1/2$ labels the two spin states of the electron.

Degeneracy

The energy depends only on $n$:

\[E=E_n.\]

It does not depend on $l$ or $m$ in the nonrelativistic Coulomb problem. This produces a high degeneracy. Some of this degeneracy comes from rotational symmetry, and the additional degeneracy is related to a hidden symmetry of the Coulomb potential.

In more accurate treatments, fine structure, spin-orbit coupling, Lamb shift, and external fields partially remove this degeneracy.

Main points

• The hydrogen atom is a central-potential problem with $V(r)\propto -1/r$.
• The wavefunction separates into radial and spherical-harmonic parts.
• The energy spectrum is $E_n=-13.6\,\mathrm{eV}/n^2$.
• For each $n$, $l=0,\dots,n-1$ and $m=-l,\dots,l$.
• The nonrelativistic energy depends only on $n$.

Practice questions

1. Write the Hamiltonian of the hydrogen atom using the reduced mass.
2. Derive the allowed values of $l$ and $m$ for $n=3$.
3. How many orbital states correspond to a fixed $n$?
4. Why is the ground state spherically symmetric?
5. Explain why the hydrogen spectrum is degenerate in $l$ and $m$ in the nonrelativistic theory.