Hydrogen atom

The Lyman-alpha line is the hydrogen $2p\leftrightarrow 1s$ electric-dipole transition, with photon energy about $10.2\mathrm{eV}$ and vacuum wavelength about $1216$ angstroms. Its emission cools collisionally excited gas, while absorption traces neutral hydrogen along a background-source sightline.

A neutral hydrogen atom absorbs a photon near the Lyman-alpha resonance by excitation from $1s$ to $2p$. Absorption lines in a quasar spectrum can therefore locate hydrogen along the sightline by their redshift, subject to peculiar-velocity and line-profile effects.

For a population with comoving number density $n_c$ and proper interception area ${\sigma }_p$, the absorption incidence is the proper density multiplied by cross-section and proper light-path length $cdz/[H(z)(1+z)]$. The displayed relation assumes one counted absorption system per interception. Completeness, covering fraction, evolution and the distinction between observed and rest-frame wavelength separation must be stated in an inference from a spectrum.

A circular, opaque, geometrically thin disk has projected area $\pi R^2|\cos i|$. For isotropically oriented normals, $|\cos i|$ is uniform on $[0,1]$, so the mean interception area is half the face-on area. A nonunit covering fraction multiplies this cross-section. Finite thickness, orientation-dependent absorption thresholds and incomplete coverage can change it; this is an average over randomly placed disks, not the inclination distribution conditional on an observed interception.

Radiative decay from the hydrogen $2p$ level to $1s$ produces a Lyman-alpha photon. In an optically thin collisionally excited gas, escape of this photon removes the excitation energy deposited by an Electron collision. The excitation threshold is responsible for the rapid temperature sensitivity of this cooling channel.

The Bohr model assumes circular electron orbits with quantized orbital angular momentum $N\hslash$. It reproduces the hydrogen energy levels, but wavefunctions in quantum mechanics describe spatial probability distributions rather than definite trajectories. In particular, spherically symmetric Coulomb bound states have zero orbital angular momentum.

For a hydrogenic bound state, the principal quantum number is the positive integer $n$ that determines the unperturbed energy. Its allowed orbital quantum numbers are $\ell =0,\dots ,n-1$.

The ionization energy is the minimum energy needed to remove an electron from an isolated atom or molecule in a specified state. Ground-state hydrogen has ionization energy $13.6\mathrm{eV}$.

For two particles of masses $m_1,m_2$ interacting through a central Coulomb potential, separation of the centre of mass leaves the relative-coordinate Schrödinger equation with

For reduced mass $\mu$, the Coulomb length scale is

A no-radial-node Coulomb eigenstate has principal quantum number $n=l+1$ and radial factorIts energy is $-e^2/[8\pi {ϵ}_{0}a(l+1{)}^2]$.

For $R(r)\propto r^l{e}^{-r/[a(l+1)]}$, the radial probability density is proportional to $r^{2l+2}{e}^{-2r/[a(l+1)]}$ and

Neglecting spin, bound hydrogen eigenstates are labelled $|n,\ell ,m⟩$, where $\ell =0,\dots ,n-1$ and $m=-\ell ,\dots ,\ell$. Their unperturbed energy depends only on $n$.

Neglecting spin, the hydrogen level with principal quantum number $n$ has degeneracy

For the $z$ component of position,requires $\Delta m=0$ and $\Delta \ell =\pm 1$. The first condition follows from $[L_z,z]=0$; the second follows from the vector-operator angular-momentum relation together with odd parity.

The even, spherically symmetric hydrogen ground state has no permanent electric dipole, so a uniform electric field gives no first-order energy shift. Odd-parity excited states mix in at second order, yielding a negative shift quadratic in the field.

The $n=2$ eigenspace has basis

A perturbation $g{L}^2$ preserves the hydrogen eigenbasis and shifts $|n,\ell ,m⟩$ byIn the first excited eigenspace, the $2s$ state is unshifted and the three $2p$ states receive the common shift $2g{\hslash }^2$.

The operators $X_3$ and $P_3$ commute with $L_3$, so they preserve $m$. Both are odd under parity, so they connect only states of opposite orbital parity. Within the $n=2$ eigenspace, only the pair $|2,0,0⟩$ and $|2,1,0⟩$ can therefore have nonzero matrix elements.

Forset $a=U(t)b$ withThen $b$ evolves under the constant Hamiltonian

For zero orbital angular momentum and Coulomb potential $-q^2/r$, the radial equation is

For a normalized maximal-angular-momentum hydrogen state, the probability inside a small radius $r_p$ is the integral of $|R{|}^2{r}^2$. Since $R$ behaves as $r^l$ near zero, it equals $2^{2l+3}(r_p/a{)}^{2l+3}/[(2l+3)!(l+1{)}^{2l+3}]$ times $1+O(r_p/a)$. Increasing angular momentum strongly suppresses penetration toward the Coulomb centre.

A hydrogen atom state with principal quantum number $n=l+1$ has no radial nodes and radial factor $R(r)=A{r}^l{e}^{-r/[a(l+1)]}$, where $a$ is the Bohr radius. Its energy is $-{\hslash }^2/[2m{a}^2(l+1{)}^2]$. With unit-normalized spherical harmonics, the radial normalization gives $|A{|}^2=2^{2l+3}/[a^{2l+3}(l+1{)}^{2l+3}(2l+2)!]$.

After the logarithmic radial coordinate $r=e^x$, the Langer replacement $l(l+1)\mapsto (l+1/2{)}^2$ turns the radial Coulomb problem into a two-turning-point WKB approximation problem. Its action integral giveswhich in this Coulomb case agrees with the exact dependence on the principal quantum number.

For a radial solution regular at the origin, writing $R=e^{-\gamma r}{\sum }_{n\ge 0}{a}_n{r}^n$ givesNormalizability at infinity forces the series to terminate, so $2\gamma N=\beta$ for a positive integer $N$ and $E_N=-m{q}^4/(2{\hslash }^2{N}^2)$.

The lowest Coulomb bound-state energy is

With normalized $Y_{00}$, the radial ground state $R(r)=C{e}^{-r/a_0}$ hasIf the angular factor is absorbed into the spherically symmetric wavefunction, its normalization constant is instead $(\pi a_0^3{)}^{-1/2}$.