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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.

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 $|A{|}^2=2^{2l+3}/[a^{2l+3}(l+1{)}^{2l+3}(2l+2)!]$.

For the near-origin probability in a maximal-angular-momentum hydrogen state, integrate the normalized radial probability density:Throughout this small interval the exponential is $1+O(r_p/a)$ for fixed $l$. HenceUsing the normalization above,The high power of the small radius reflects the near-origin behavior $R\propto r^l$ of the hydrogen atom state.

The baryon-to-photon ratio is extremely small, of order $10^{-9}$. Thus even when the typical photon energy is below $I$, the high-energy tail of the blackbody radiation contains enough photons above the ionization energy to reionize almost every newly formed hydrogen atom. Cosmological recombination becomes effective only when this exponentially small tail is also small compared with the baryon-to-photon ratio. Consequently recombination occurs at a temperature much lower than $I/k$, consistent with $kT\simeq 0.03I$. The exact freeze-out history also depends on reaction rates and expansion; the equilibrium estimate explains the large suppression of the temperature.

Expand the state in the energy eigenstates of $H_0$ asSubstitution into the Time-dependent Schrodinger equation and projection onto $⟨k|$ gives the exact interaction picture equationsIn first-order time-dependent perturbation theory, put $c_j(t)={\delta }_{j0}$ on the right-hand side and integrate from $0$ to $t$:

For a uniform electric field, the electric-dipole interaction is $\Delta (t)=e{\mathcal{E}}_{0}z{e}^{-t/\tau }$ up to an irrelevant overall sign. The supplied hydrogen atom wavefunctions giveFor $t\gg \tau$, the time integral isThe long-time transition fraction is consequentlyBoth $|100⟩$ and $|200⟩$ have even parity, whereas $z$ is odd, so the electric-dipole selection rule gives $⟨200|z|100⟩=0$. Thus the requested fraction in $|200⟩$ is zero.

Write $Q={\dot{N}}_{\mathrm{ion}}$ for the constant hydrogen-ionizing photon production rate. Assume an initially neutral gas of hydrogen atoms, a sharp spherical ionization front, complete ionization inside it, and a fixed gas density. The Electron and proton number densities inside are then both $n$. Take the recombination coefficient $\alpha$ to be constant, with the usual Case B recombination interpretation: photons from recombination directly to the ground state are absorbed locally. Neglect absorption by dust, pressure-driven gas motion, and light-travel delays.

The total rate of Case B recombination within radius $R$ isAt photoionization equilibrium, $Q=\mathcal{R}(R_S)$, giving the Strömgren radiusThe square of the density appears because recombination requires encounters between an Electron and a proton.

During Ionization-front growth of a Strömgren sphere, each photon either replaces an ion lost through Case B recombination or ionizes an additional atom. There are $N_i=(4\pi /3)n{R}^3$ ions inside the ionization front, so photon counting givesThe recombination time is $t_{\mathrm{rec}}=(\alpha n{)}^{-1}$. Solving this first-order differential equation with $R(0)=0$ givesFor an initial radius $R_0$, the corresponding result is $R^3(t)=R_S^3+(R_0^3-R_S^3)e^{-t/t_{\mathrm{rec}}}$. Initially, the idealized photon-counting solution has $R\simeq [3Qt/(4\pi n){]}^{1/3}$; at late times, $R/R_S\simeq 1-\frac{1}{3}{e}^{-t/t_{\mathrm{rec}}}$. The initially divergent front speed in this approximation is a consequence of neglecting light-travel delays; the formula describes the growth once that approximation is applicable.

For a power law density, specify a reference radius $r_0$ so that $n(r)=n_0(r/r_0{)}^{-\beta }$. With an inner cutoff $r_{\mathrm{in}}>0$, the recombination integral for a power-law nebula isEquivalently, a logarithmic interval in radius contributes $d\mathcal{R}/d\ln r=4\pi \alpha n_0^2{r}_0^{2\beta }{r}^{3-2\beta }$. For $\beta <3/2$, the central integral converges as $r_{\mathrm{in}}\to 0$, and the recombination rate grows as $R^{3-2\beta }$. At $\beta =3/2$, each logarithmic interval contributes equally and the centre gives a logarithmic divergence. For $\beta >3/2$, the unmodified profile gives a power divergence at the centre, while its outer recombination integral converges. The change at $\beta =3/2$ is set by the competition between the spherical volume element and the squared density.

To regularize the centre, choose the following explicit cored power-law density profile, taking $r_0=r_c$ and $n_0$ as the core density:For $R\le r_c$, $\mathcal{R}(R)=(4\pi /3)\alpha n_0^2{R}^3$. For $R>r_c$, put $u=R/r_c$ and integrate the core and envelope separately:This rate is continuous and strictly increasing with $R$. If $\beta >3/2$, it has a finite supremum, the critical ionizing photon rate of a cored power-law nebula:For $0<Q<Q_{\mathrm{crit}}$ there is a unique finite radius at photoionization equilibrium. At $Q=Q_{\mathrm{crit}}$ the equality is approached only as $R\to \infty$; for $Q>Q_{\mathrm{crit}}$ even ionizing the whole profile cannot consume all photons through Case B recombination, so no equilibrium radius exists. For $\beta \le 3/2$, the envelope integral grows without bound and there is no finite critical rate in this infinite-extent model.

For a concrete choice, take any $n_0>0$, $r_c>0$, and $\beta =2$. Then $Q_{\mathrm{crit}}=16\pi \alpha n_0^2{r}_c^3/3$. In the envelope, defining $q=Q/(4\pi \alpha n_0^2{r}_c^3)$ gives $R=r_c/(4/3-q)$ for $1/3<q<4/3$, explicitly showing the divergence as the critical rate is approached. A different core shape changes the numerical coefficient of $Q_{\mathrm{crit}}$, while the outer-slope condition $\beta >3/2$ remains the same.

The anti-muon has the same charge as a proton, so the Coulomb potential seen by the electron is the same as in the hydrogen atom. The main difference is the source mass. Separating the centre-of-mass motion replaces $m_e$ by the Coulomb two-body reduced massSince $m_{\mu }\gg m_e$, this is close to $m_e$, though less close than the electron-proton reduced mass. The smaller reduced mass gives a slightly larger Bohr radius and energies of slightly smaller magnitude. Under the assumptions in the question, the Coulomb form, angular eigenfunctions and quantum numbers are otherwise unchanged; anti-muon decay and spin effects are being neglected.

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.