07 Jun 2026
Hydrogen Atom
Hydrogen atom with one proton and one electron, Coulomb potential, quantum numbers, energy spectrum, and degeneracy.
The hydrogen atom contains one proton as the nucleus and one electron bound to it. The proton has charge $+e$ and the electron has charge $-e$, so the force between them is attractive. Since this electrostatic interaction depends only on the separation $r$ between the two particles, it is modelled by a Coulomb potential.
This makes the hydrogen atom a central-potential problem. The potential depends only on $r$, so the Schrodinger equation can be separated into radial and angular parts. The solution gives quantum numbers, allowed energies, and degeneracies.
Coulomb potential
Choose the zero of potential energy when the electron and proton are infinitely far apart. The Coulomb potential energy is then
\[V(r)=-\frac{e^2}{4\pi\epsilon_0r}.\]The negative sign shows attraction. As $r$ decreases, the potential energy becomes more negative, meaning that the electron is bound more strongly to the proton.
Strictly, hydrogen is a two-body problem. It is reduced to an equivalent one-body problem by using the reduced mass $\mu$ for the relative motion of the electron-proton system.
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
\[\psi_{nlm}(r,\theta,\phi) = R_{nl}(r)Y_l^m(\theta,\phi).\]This separation is possible because the angular part is governed by $L^2$ and $L_z$, while the radial part contains the Coulomb potential.
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.
The negative sign means that the electron is in a bound state. The energy approaches zero from below as $n$ becomes large, corresponding to the ionization limit.
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.
States belonging to $n=3$
For $n=3$, the allowed orbital quantum numbers are
\[l=0,1,2.\]For these values, the possible $m$ values are:
| $l$ | allowed $m$ values | number of states |
|---|---|---|
| 0 | $0$ | 1 |
| 1 | $-1,0,1$ | 3 |
| 2 | $-2,-1,0,1,2$ | 5 |
The total number of orbital states is
\[1+3+5=9=n^2.\]Including spin, the number becomes
\[2n^2=18.\]The energy is
\[E_3=-\frac{13.6}{9}\,\mathrm{eV} \approx -1.51\,\mathrm{eV}.\]Ground-state probability density
For the ground state,
\[\psi_{100}(r)= \frac{1}{\sqrt{\pi a_0^3}}e^{-r/a_0}.\]The probability density is
\[|\psi_{100}|^2= \frac{1}{\pi a_0^3}e^{-2r/a_0}.\]It depends only on $r$, not on $\theta$ or $\phi$, so the ground-state charge distribution is spherically symmetric.
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
- Write the Hamiltonian of the hydrogen atom using the reduced mass.
- Derive the allowed values of $l$ and $m$ for $n=3$.
- How many orbital states correspond to a fixed $n$?
- Why is the ground state spherically symmetric?
- Explain why the hydrogen spectrum is degenerate in $l$ and $m$ in the nonrelativistic theory.
Discussion