The Hydrogen Atom

The potential for the hydrogen atom is the Coulomb potential:

where $e$ is the elementary charge and $\epsilon_0$ is the vacuum permittivity. We use the reduced mass $\mu \approx m_e$ (since $m_p \gg m_e$).

The radial equation for $u(r) = r R(r)$ becomes:

The Coulomb potential is singular at $r = 0$ but not so singular as to cause problems. The boundary conditions are $u(0) = 0$ (normalizability) and $u(\infty) \to 0$ (bound state).

For large $r$, the Coulomb term becomes negligible and the equation behaves like that of a free particle with negative energy. The asymptotic solution is $u(r) \sim e^{-\kappa r}$, where $\kappa = \sqrt{-2\mu E}/\hbar$.

For small $r$, the centrifugal term dominates and $u(r) \sim r^{l+1}$. We factor out both asymptotic behaviors by writing:

Substituting this into the radial equation gives a differential equation for $v(\rho)$. A power series solution $v(\rho) = \sum a_j \rho^j$ leads to a recursion relation:

For the series to terminate (giving a polynomial), we require $j_{\max} + l + 1 = n$ for some integer $n = 1, 2, 3, \ldots$ This is the principal quantum number. The terminating polynomial is the associated Laguerre polynomial.

The termination condition gives the famous Bohr energy formula:

The principal quantum number $n = 1, 2, 3, \ldots$ determines the energy. The orbital quantum number $l$ can be $0, 1, \ldots, n-1$. For each $l$, $m$ ranges from $-l$ to $+l$.

The total degeneracy of level $n$ (ignoring spin) is:

The ground state ($n=1$, $l=0$, $m=0$) has energy $E_1 = -13.6\text{ eV}$. This is the ionization energy of hydrogen. The wave function is:

The characteristic length scale is the Bohr radius $a_0 \approx 0.529\text{ \AA}$. The ground state wave function is spherically symmetric and peaks at $r = 0$, but the radial probability density $P(r) = 4\pi r^2 |\psi|^2$ peaks at $r = a_0$.

The full normalized wave functions are:

where $L_{n-l-1}^{2l+1}(\rho)$ are the associated Laguerre polynomials and $Y_l^m(\theta, \phi)$ are the spherical harmonics.

The radial quantum number $n_r = n - l - 1$ counts the number of radial nodes (excluding the node at $r = 0$ for $l > 0$ and the node at $r = \infty$). The total number of nodes is $n_r + l = n - 1$.

Some important wave functions: $\psi_{100}$ (1s), $\psi_{200}$ (2s), $\psi_{210}$ ($2p_0$), $\psi_{21\pm1}$ ($2p_{\pm1}$). The $2p$ states have angular dependence: $Y_1^0 \sim \cos\theta$ and $Y_1^{\pm1} \sim \sin\theta\, e^{\pm i\phi}$.

Calculate $\langle r \rangle$ and $\langle r^2 \rangle$ for the ground state of hydrogen.

Step 1 — $\langle r \rangle$. The radial probability density is $P(r) = 4\pi r^2 |\psi_{100}|^2 = (4/a_0^3) r^2 e^{-2r/a_0}$. Then $\langle r \rangle = \int r P(r)\, dr$ from $0$ to $\infty$ = $(4/a_0^3) \int r^3 e^{-2r/a_0}\, dr$.

Using $\int x^n e^{-ax}\, dx = n! / a^{n+1}$, we get $\langle r \rangle = (4/a_0^3) \cdot 3! / (2/a_0)^4 = (4/a_0^3) \cdot 6 \cdot a_0^4 / 16 = (3/2) a_0$.

Step 2 — $\langle r^2 \rangle$. $\langle r^2 \rangle = (4/a_0^3) \int r^4 e^{-2r/a_0}\, dr = (4/a_0^3) \cdot 4! / (2/a_0)^5 = (4/a_0^3) \cdot 24 \cdot a_0^5 / 32 = 3 a_0^2$.

Step 3 — Uncertainty. $\Delta r = \sqrt{\langle r^2 \rangle - \langle r \rangle^2} = \sqrt{3 a_0^2 - 9 a_0^2/4} = \sqrt{3/4}\, a_0 = a_0 \sqrt{3}/2 \approx 0.866 a_0$.