The Radial Equation & the Hydrogen Atom

The angular problem is solved once and forever; what remains is one radial equation per potential. A single substitution, $u = rR$, turns it into a one-dimensional Schrödinger equation — every tool from Course P.5 applies — and for the Coulomb potential the series solution terminates only at special energies: $-13.6\ \text{eV}/n^2$, Bohr's lucky guess, now derived. This is the calculation that convinced physics that wave mechanics was true. Approach it the way a cat approaches a full food bowl: unhurried, but missing nothing.

Learning Objectives

After this lesson you will be able to:

1. Reduce a central-potential problem to a 1D equation for $u(r) = rR(r)$ with the effective potential $V_{\text{eff}}(r)$ and its centrifugal barrier.
2. Apply the correct boundary conditions on $u$ and solve the infinite spherical well for $\ell = 0$.
3. Carry out the full hydrogen solution: asymptotics, series, the recursion relation, and termination.
4. Derive the Bohr energies $E_n = -13.6\ \text{eV}/n^2$, the Bohr radius $a_0$, and the $n^2$ degeneracy.
5. Work with the radial functions $R_{n\ell}$ and the radial probability density $P(r) = r^2|R|^2$, computing most probable and mean radii.
6. Interpret the quantum numbers $(n,\ell,m)$, use spectroscopic notation, and contrast the Bohr and Schrödinger pictures of the atom.

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Intuition

Classically, conservation of angular momentum flattens any central-force problem into one-dimensional radial motion in an effective potential $V(r) + L^2/2mr^2$ — the orbiting piece of the kinetic energy masquerades as a repulsive barrier (P.1.2 Lagrangian Mechanics). Quantum mechanics repeats the maneuver verbatim, with $L^2 \to \hbar^2\ell(\ell+1)$: the electron with angular momentum is pushed away from the origin by the same centrifugal effect that keeps a whirling ball on a string taut.

So the hydrogen atom is, at heart, a Course P.5 problem: a particle in a 1D well (Coulomb attraction plus centrifugal repulsion), with bound states at discrete energies. The magic is in the numbers that come out — energies matching every observed spectral line, and a natural length scale, the Bohr radius, assembled from $\hbar$, $m$, and $e$ with no free parameters.

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Theory

From 3D to effective 1D: the radial equation

Separation in P.6.1 left the radial equation

$$\frac{d}{dr}\!\left(r^2\frac{dR}{dr}\right) - \frac{2mr^2}{\hbar^2}\bigl[V(r) - E\bigr]R = \ell(\ell+1)\,R .$$

Substitute $u(r) \equiv r\,R(r)$. Then $R = u/r$, $\dfrac{dR}{dr} = \dfrac{u'}{r} - \dfrac{u}{r^2}$, so $r^2\dfrac{dR}{dr} = r u' - u$ and $\dfrac{d}{dr}\bigl(r u' - u\bigr) = r u''$. The equation becomes $r u'' - \frac{2mr}{\hbar^2}(V - E)\,u = \ell(\ell+1)\,\frac{u}{r}$, i.e.

$$-\frac{\hbar^2}{2m}\frac{d^2u}{dr^2} + \underbrace{\left[V(r) + \frac{\hbar^2\ell(\ell+1)}{2mr^2}\right]}_{V_{\text{eff}}(r)} u = E\,u .$$

This is exactly a 1D Schrödinger equation, with $V$ replaced by the effective potential: the true potential plus the centrifugal barrier $\hbar^2\ell(\ell+1)/2mr^2$ — the same $L^2/2mr^2$ term as in the classical central-force problem (P.1.2), quantized. Normalization also cooperates: $\int|\psi|^2 d^3r = \int_0^\infty |R|^2 r^2 dr = \int_0^\infty |u|^2 dr = 1$.

Boundary conditions: the 1D toolkit applies

Two conditions replace the free constants of the 1D problems:

• At the origin: $u(0) = 0$. If $u(0)\ne 0$ then $R \sim u(0)/r$, and $\nabla^2(1/r) = -4\pi\delta^3(\mathbf r)$ produces a delta function no finite $V$ can cancel — not a solution. (For $\ell > 0$ the centrifugal barrier enforces even faster vanishing, as we'll see.)
• At infinity: $u \to 0$ for a bound state ($E < V(\infty)$), just as in P.5.2 Step, Well & Barrier.

With $u$ vanishing at both ends, all the machinery of Course P.5 — node counting, matching, shooting-method numerics — transfers wholesale to three dimensions.

Warm-up: the infinite spherical well

Let $V(r) = 0$ for $r < a$ and $\infty$ outside. For $\ell = 0$ the radial equation is just $u'' = -k^2 u$ with $k = \sqrt{2mE}/\hbar$ — the 1D box. The condition $u(0) = 0$ selects $u = A\sin(kr)$, and $u(a) = 0$ quantizes $k = n\pi/a$:

$$E_n = \frac{n^2\pi^2\hbar^2}{2ma^2}, \qquad n = 1, 2, 3, \dots$$

with $R_{n0} = A\sin(n\pi r/a)/r$, perfectly finite at the origin. For $\ell > 0$ the solutions regular at $r=0$ are the spherical Bessel functions, $R \propto j_\ell(kr)$, and the energies come from their zeros — Exercise E5 works out $\ell = 1$.

The hydrogen atom I: setting up

Now the main event. The proton's Coulomb potential is

$$V(r) = -\frac{e^2}{4\pi\epsilon_0}\frac1r ,$$

so the radial equation reads

$$-\frac{\hbar^2}{2m}\frac{d^2u}{dr^2} + \left[-\frac{e^2}{4\pi\epsilon_0 r} + \frac{\hbar^2\ell(\ell+1)}{2mr^2}\right]u = E\,u .$$

Bound states have $E < 0$. Define

$$\kappa \equiv \frac{\sqrt{-2mE}}{\hbar}, \qquad \rho \equiv \kappa r, \qquad \rho_0 \equiv \frac{me^2}{2\pi\epsilon_0\hbar^2\kappa} ,$$

($\kappa$ is real; $\rho$ is dimensionless radius; $\rho_0$ packages the strength of the Coulomb attraction relative to the energy). Dividing the equation by $E$ and tidying up:

$$\frac{d^2u}{d\rho^2} = \left[1 - \frac{\rho_0}{\rho} + \frac{\ell(\ell+1)}{\rho^2}\right]u .$$

The hydrogen atom II: asymptotics and the series

Large $\rho$: the bracket $\to 1$, so $u'' \approx u$ and $u \sim Ae^{-\rho} + Be^{\rho}$; normalizability kills $B$. Small $\rho$: the centrifugal term dominates, $u'' \approx \dfrac{\ell(\ell+1)}{\rho^2}u$, with solutions $\rho^{\ell+1}$ and $\rho^{-\ell}$; the second violates $u(0) = 0$ (and normalizability), so $u \sim C\rho^{\ell+1}$. Peel off both behaviors:

$$u(\rho) = \rho^{\ell+1}\,e^{-\rho}\,v(\rho) .$$

Substituting (two applications of the product rule) turns the equation for $u$ into

$$\rho\,\frac{d^2v}{d\rho^2} + 2(\ell + 1 - \rho)\,\frac{dv}{d\rho} + \bigl[\rho_0 - 2(\ell+1)\bigr]v = 0 .$$

Try a power series $v(\rho) = \sum_{j=0}^\infty c_j\rho^j$. Then $\rho\,v'' = \sum_j j(j+1)c_{j+1}\rho^j$, $2(\ell+1)v' = \sum_j 2(\ell+1)(j+1)c_{j+1}\rho^j$, and $-2\rho v' = -\sum_j 2j\,c_j\rho^j$ (each after shifting the summation index). Collecting the coefficient of $\rho^j$:

$$j(j+1)c_{j+1} + 2(\ell+1)(j+1)c_{j+1} - 2j\,c_j + \bigl[\rho_0 - 2(\ell+1)\bigr]c_j = 0 ,$$

which solves to the recursion relation

$$c_{j+1} = \frac{2(j + \ell + 1) - \rho_0}{(j+1)(j + 2\ell + 2)}\;c_j .$$

One constant $c_0$ (fixed later by normalization) generates the whole series.

The hydrogen atom III: quantization

Suppose the series never terminates. For large $j$ the recursion is $c_{j+1} \approx \dfrac{2}{j+1}c_j$, whose solution is $c_j \approx \dfrac{2^j}{j!}c_0$ — the series of $c_0e^{2\rho}$. Then $u \approx c_0\,\rho^{\ell+1}e^{-\rho}e^{2\rho} = c_0\,\rho^{\ell+1}e^{+\rho}$: the asymptotic disease we amputated grows back. The only escape is termination: some maximal index $j_{\max}$ must give $c_{j_{\max}+1} = 0$, i.e.

$$2(j_{\max} + \ell + 1) = \rho_0 .$$

Define the principal quantum number $n \equiv j_{\max} + \ell + 1 \in \{1, 2, 3, \dots\}$, so $\rho_0 = 2n$. Unpacking $\rho_0$'s definition, $\kappa = \dfrac{me^2}{4\pi\epsilon_0\hbar^2 n}$, and $E = -\dfrac{\hbar^2\kappa^2}{2m}$ gives

$$E_n = -\left[\frac{m}{2\hbar^2}\left(\frac{e^2}{4\pi\epsilon_0}\right)^2\right]\frac{1}{n^2} = -\frac{13.6\ \text{eV}}{n^2}, \qquad n = 1, 2, 3, \dots$$

— the Bohr formula, exactly, but derived from a real dynamical theory rather than postulated. The natural length scale also assembles itself: $\kappa = 1/(a_0 n)$ with

$$a_0 = \frac{4\pi\epsilon_0\hbar^2}{me^2} = 0.529\ \text{Å} = 5.29\times10^{-11}\ \text{m},$$

the Bohr radius. Two structural facts come free. Since $j_{\max} = n - \ell - 1 \ge 0$,

$$\ell \le n - 1 \qquad (\ell = 0, 1, \dots, n-1),$$

and counting all states at energy $E_n$ ($2\ell+1$ values of $m$ for each allowed $\ell$):

$$d(n) = \sum_{\ell=0}^{n-1}(2\ell + 1) = 2\cdot\frac{(n-1)n}{2} + n = n^2 .$$

The $m$-degeneracy is guaranteed by spherical symmetry; the extra $\ell$-degeneracy (2s degenerate with 2p) is an "accident" special to the pure $1/r$ potential.

The wavefunctions

The terminating polynomials $v(\rho)$ are (up to normalization) the associated Laguerre polynomials $L_{n-\ell-1}^{2\ell+1}(2\rho)$, with $\rho = r/na_0$, so $\psi_{n\ell m} = R_{n\ell}(r)\,Y_\ell^m(\theta,\varphi)$ with $R_{n\ell} = \rho^{\ell+1}e^{-\rho}v(\rho)/r$. The first few, normalized:

$$R_{10} = \frac{2}{a_0^{3/2}}\,e^{-r/a_0}, \qquad R_{20} = \frac{1}{\sqrt2\,a_0^{3/2}}\left(1 - \frac{r}{2a_0}\right)e^{-r/2a_0}, \qquad R_{21} = \frac{1}{\sqrt{24}\,a_0^{3/2}}\,\frac{r}{a_0}\,e^{-r/2a_0}.$$

The physically meaningful radial quantity is the radial probability density

$$P(r) = r^2\,|R(r)|^2 ,$$

the probability per unit radius of finding the electron in the spherical shell $[r, r+dr]$ (the $r^2$ is the shell's growing area). For the ground state, $P(r) = \frac{4}{a_0^3}r^2e^{-2r/a_0}$; setting $dP/dr = 0$ gives $2re^{-2r/a_0}(1 - r/a_0) = 0$, so the most probable radius is exactly $a_0$ — Bohr's orbit radius reborn as the peak of a distribution. The mean radius is larger, because the distribution has a long tail:

$$\langle r\rangle = \int_0^\infty r\,P(r)\,dr = \frac{4}{a_0^3}\int_0^\infty r^3 e^{-2r/a_0}dr = \frac{4}{a_0^3}\cdot 3!\left(\frac{a_0}{2}\right)^4 = \frac32\,a_0 .$$

Caution. The electron is not "at the Bohr radius," and it is not on a trajectory at all — $P(r)\,dr$ is the probability of finding it in a shell upon measurement, per the Born rule (P.4.1). And do not confuse $|R|^2$ with $r^2|R|^2$: the ground-state density $|\psi_{100}|^2 \propto e^{-2r/a_0}$ peaks at $r = 0$, while $P(r)$ peaks at $a_0$. Both statements are true; they answer different questions.

Quantum numbers, notation, and Bohr vs Schrödinger

Symbol	Name	Allowed values	Determines
$n$	principal	$1, 2, 3, \dots$	energy $E_n = -13.6\ \text{eV}/n^2$
$\ell$	orbital (azimuthal)	$0, 1, \dots, n-1$	shape; $\lvert\mathbf L\rvert = \hbar\sqrt{\ell(\ell+1)}$
$m$	magnetic	$-\ell, \dots, +\ell$	orientation; $L_z = m\hbar$

Spectroscopic notation labels $\ell = 0,1,2,3,\dots$ as $s, p, d, f,\dots$ prefixed by $n$: the ground state is 1s; the $n=2$ level holds 2s and 2p; and so on.

	Bohr model (P.2.2)	Schrödinger theory
Energies	$-13.6\ \text{eV}/n^2$ (right, luckily)	same — now derived
Angular momentum	$L = n\hbar$; ground state $L = \hbar$	$\lvert\mathbf L\rvert = \hbar\sqrt{\ell(\ell+1)}$; ground state $\ell = 0$, $L = 0$!
Picture	planetary orbits of definite radius	stationary probability clouds ("orbitals, not orbits")
States per level	one orbit	$n^2$ degenerate states $(n,\ell,m)$

The Rydberg formula, fitted to data in P.2.2, is now a theorem: a photon emitted in the transition $n_i \to n_f$ carries $h\nu = E_{n_i} - E_{n_f}$, so

$$\frac{1}{\lambda} = R_H\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right), \qquad R_H = \frac{m}{4\pi c\hbar^3}\left(\frac{e^2}{4\pi\epsilon_0}\right)^2 = 1.097\times10^7\ \text{m}^{-1}.$$

Not every pair of levels actually radiates, though: coupling to the electromagnetic field enforces the selection rule $\Delta\ell = \pm1$ (2p → 1s shines; 2s → 1s is forbidden and that state is metastable). Deriving this requires time-dependent perturbation theory, beyond our scope here.

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Worked Examples

Example 1 — Lyman-α from first principles

The $n = 2 \to 1$ transition emits $\Delta E = E_2 - E_1 = -3.40 - (-13.6) = 10.2\ \text{eV}$. The wavelength: $\lambda = hc/\Delta E = (1240\ \text{eV·nm})/10.2\ \text{eV} = 121.6\ \text{nm}$ — deep ultraviolet, matching the measured Lyman-α line ($121.567\ \text{nm}$) to our precision. The same arithmetic on $3\to2$, $4\to2$, ... reproduces the visible Balmer series a century of spectroscopists catalogued.

Example 2 — Where is the ground-state electron?

What is the probability of finding it beyond the Bohr radius? With $x = 2r/a_0$:

$$P(r > a_0) = \frac{4}{a_0^3}\int_{a_0}^\infty r^2 e^{-2r/a_0}\,dr = \frac12\int_2^\infty x^2 e^{-x}\,dx = \frac12\Bigl[-(x^2 + 2x + 2)e^{-x}\Bigr]_2^\infty = \frac{10\,e^{-2}}{2} = 5e^{-2} \approx 0.677 .$$

Two-thirds of measurements find the electron outside the "orbit" it supposedly occupies. The Bohr radius marks the peak of $P(r)$, not a boundary — the atom has no edge, only an exponential fade.

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Hands-on (Python)

Shooting method on $u(r)$, in Bohr-radius units where the equation becomes $u'' = [\ell(\ell+1)/x^2 - 2/x - \varepsilon]u$ with $\varepsilon = E/(13.6\ \text{eV})$:

import numpy as np

def shoot(eps, l, x_max=40.0, n=8000):
    """Integrate u'' = k(x) u outward; return u(x_max). Eigenvalue <=> u(x_max) ~ 0."""
    x = np.linspace(1e-4, x_max, n); h = x[1] - x[0]
    k = l*(l + 1)/x**2 - 2.0/x - eps
    u = np.zeros(n); u[0], u[1] = x[0]**(l + 1), x[1]**(l + 1)   # u ~ x^{l+1} near 0
    for i in range(1, n - 1):
        u[i+1] = 2*u[i] - u[i-1] + h*h*k[i]*u[i]
    return u[-1]

def eigenvalue(l, lo, hi):
    for _ in range(60):                       # bisect on the sign of u(x_max)
        mid = 0.5*(lo + hi)
        if shoot(lo, l)*shoot(mid, l) < 0: hi = mid
        else: lo = mid
    return 0.5*(lo + hi)

for l, brackets in [(0, [(-1.5, -0.6), (-0.35, -0.15), (-0.14, -0.09)]),
                    (1, [(-0.35, -0.15), (-0.14, -0.09)])]:
    print(l, [round(eigenvalue(l, *b)*13.6057, 2) for b in brackets])
# Expect: l=0 -> [-13.61, -3.40, -1.51];  l=1 -> [-3.40, -1.51]
# E depends only on n -- the accidental Coulomb degeneracy, found numerically.

Radial probability densities $P(r) = r^2R^2$ in units of $a_0$:

import matplotlib.pyplot as plt

r = np.linspace(0, 25, 1000)                                # r in units of a0
R = {(1,0): 2*np.exp(-r),
     (2,0): (1 - r/2)*np.exp(-r/2)/np.sqrt(2),
     (2,1): r*np.exp(-r/2)/np.sqrt(24),
     (3,0): (2/np.sqrt(27))*(1 - 2*r/3 + 2*r**2/27)*np.exp(-r/3)}
for (n, l), Rnl in R.items():
    plt.plot(r, r**2*Rnl**2, label=f"$P_{{{n}{l}}}$")
plt.xlabel("$r/a_0$"); plt.ylabel("$P(r)\\,a_0$"); plt.legend(); plt.show()
# Expect: P_10 peaks at r = a0; P_20 has a node (2 bumps); P_21 one bump peaking
# at 4 a0; P_30 has two nodes. Check each curve integrates to 1 with np.trapezoid.

Monte Carlo orbital clouds — rejection-sample $|\psi|^2$ and scatter the points:

rng = np.random.default_rng(7)

def psi2_210(x, y, z):                       # |psi_210|^2 ∝ e^{-r} z^2   (a0 = 1)
    return np.exp(-np.sqrt(x*x + y*y + z*z))*z**2

def psi2_320(x, y, z):                       # |psi_320|^2 ∝ r^4 e^{-2r/3}(3cos^2θ - 1)^2
    r = np.sqrt(x*x + y*y + z*z)
    return r**4*np.exp(-2*r/3)*(3*(z/np.maximum(r, 1e-12))**2 - 1)**2

def sample(pdf, half_box, f_max, n_keep=4000):
    kept = np.empty((0, 3))
    while len(kept) < n_keep:                # rejection sampling in a cube
        p = rng.uniform(-half_box, half_box, (100_000, 3))
        acc = p[rng.uniform(0, f_max, 100_000) < pdf(*p.T)]
        kept = np.vstack([kept, acc])
    return kept[:n_keep]

fig = plt.figure(figsize=(11, 5))
for i, (pdf, box, fmax, name) in enumerate(
        [(psi2_210, 15, 0.6, "2p$_z$"), (psi2_320, 30, 95.0, "3d$_{z^2}$")]):
    pts = sample(pdf, box, fmax)
    ax = fig.add_subplot(1, 2, i + 1, projection='3d')
    ax.scatter(*pts.T, s=1, alpha=0.3); ax.set_title(name)
plt.show()
# Expect: the 2p_z dumbbell along z; the 3d_z2 dumbbell wearing a donut at the
# waist -- the textbook orbital pictures, generated from |psi|^2 itself.

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Exercises

E1 (easy). Compute the wavelength of the Balmer H-α line ($n = 3 \to 2$). Which part of the spectrum is it in?

Solution

$\Delta E = 13.6\left(\frac14 - \frac19\right) = 13.6\cdot\frac{5}{36} = 1.89\ \text{eV}$; $\lambda = 1240/1.89 = 656\ \text{nm}$ — red, visible; the famous H-α line of astronomy.

E2 (easy). Verify by explicit integration that $R_{10}$ is normalized: $\int_0^\infty R_{10}^2\,r^2\,dr = 1$.

Solution

$\int_0^\infty \frac{4}{a_0^3}e^{-2r/a_0}r^2dr = \frac{4}{a_0^3}\cdot\frac{2!}{(2/a_0)^3} = \frac{4}{a_0^3}\cdot\frac{2a_0^3}{8} = 1$✓ (using$\int_0^\infty r^ne^{-br}dr = n!/b^{n+1}$).

E3 (medium). Find the most probable radius of the 2p state ($R_{21}$) and compare with the Bohr model's prediction for $n = 2$.

Solution

$P(r) \propto r^2R_{21}^2 \propto r^4e^{-r/a_0}$. Then $dP/dr \propto (4r^3 - r^4/a_0)e^{-r/a_0} = 0$ gives $r = 4a_0$ — exactly Bohr's $n^2a_0$ for $n = 2$. The circular Bohr orbit survives as the peak of the maximal-$\ell$ radial distribution; small-$\ell$ states (like 2s) have no classical counterpart at all.

E4 (medium). For the ground state compute $\langle 1/r\rangle$, then $\langle V\rangle$ and $\langle T\rangle$. Verify the virial pattern $\langle V\rangle = 2E_1$, $\langle T\rangle = -E_1$.

Solution

$\langle 1/r\rangle = \frac{4}{a_0^3}\int_0^\infty r\,e^{-2r/a_0}dr = \frac{4}{a_0^3}\cdot\frac{a_0^2}{4} = \frac{1}{a_0}$. So $\langle V\rangle = -\frac{e^2}{4\pi\epsilon_0}\langle 1/r\rangle = -\frac{e^2}{4\pi\epsilon_0 a_0} = -27.2\ \text{eV} = 2E_1$, and $\langle T\rangle = E_1 - \langle V\rangle = -13.6 + 27.2 = +13.6\ \text{eV} = -E_1$ ✓ — the quantum virial theorem for a $1/r$ potential, $2\langle T\rangle = -\langle V\rangle$.

E5 (hard). Solve the infinite spherical well for $\ell = 1$: show that $u(r) = A\bigl[\sin(kr)/(kr) - \cos(kr)\bigr]$ satisfies the radial equation and the origin condition, derive the quantization condition, and compare the lowest $\ell=1$ energy to the $\ell=0$ ground state.

Solution

With $u = \sin(kr)/(kr) - \cos(kr)$ (this is $kr\,j_1(kr)$), differentiating twice gives $u'' = -\frac{k\sin kr}{r} - \frac{2\cos kr}{r^2} + \frac{2\sin kr}{kr^3} + k^2\cos kr$, while $\bigl(k^2 - \frac{2}{r^2}\bigr)u = \frac{k\sin kr}{r} - k^2\cos kr - \frac{2\sin kr}{kr^3} + \frac{2\cos kr}{r^2}$; the sum vanishes term by term, so $u'' + \bigl[k^2 - 2/r^2\bigr]u = 0$ — the $\ell = 1$ radial equation with $V = 0$. As $r\to0$, $\sin(kr)/(kr) \to 1 - (kr)^2/6$ and $\cos(kr)\to 1 - (kr)^2/2$, so $u \to (kr)^2/3 \to 0$ ✓ (indeed $u\sim r^{\ell+1}$). The wall imposes $u(a) = 0$: $\tan(ka) = ka$, whose smallest positive root is $ka \approx 4.4934$. Hence $E_{1,\ell=1} = (4.4934)^2\hbar^2/2ma^2 \approx 20.2\,\hbar^2/2ma^2$, about $2.05\times$ the $\ell = 0$ ground energy $\pi^2\hbar^2/2ma^2 \approx 9.87\,\hbar^2/2ma^2$ — the centrifugal barrier raises every $\ell = 1$ level, and unlike hydrogen there is no accidental degeneracy.

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Checkpoint

1. What is $V_{\text{eff}}(r)$, and where does its extra term come from — classically and quantum mechanically?
2. Why must $u(0) = 0$?
3. Trace the logic: how does demanding a normalizable solution produce $E_n = -13.6\ \text{eV}/n^2$?
4. Why is the degeneracy of level $n$ equal to $n^2$, and which part of it is "accidental"?
5. The ground-state $|\psi|^2$ peaks at $r = 0$ but $P(r)$ peaks at $a_0$. Reconcile.

Answers

1. $V_{\text{eff}} = V(r) + \hbar^2\ell(\ell+1)/2mr^2$. The barrier is the angular kinetic energy: classically $L^2/2mr^2$ from conservation of $\mathbf L$; quantum mechanically the $\hat L^2/2mr^2$ term acting on $Y_\ell^m$.
2. Otherwise $R\sim1/r$ and $\nabla^2\psi$ contains $-4\pi\delta^3(\mathbf r)$, which nothing in the equation can balance; the wavefunction would not solve the Schrödinger equation at the origin.
3. Asymptotics force $u = \rho^{\ell+1}e^{-\rho}v$; the series for $v$ obeys $c_{j+1} = \frac{2(j+\ell+1)-\rho_0}{(j+1)(j+2\ell+2)}c_j$; a non-terminating series grows like $e^{2\rho}$ and ruins normalizability; termination forces $\rho_0 = 2n$, which converts to $E_n \propto -1/n^2$.
4. $d(n) = \sum_{\ell=0}^{n-1}(2\ell+1) = n^2$. The $2\ell+1$ $m$-degeneracy follows from rotational symmetry; the degeneracy across different $\ell$ is accidental, special to the $1/r$ potential.
5. They answer different questions: $|\psi|^2$ is probability per unit volume (largest at the origin), $P(r) = 4\pi r^2|\psi|^2$ is probability per unit radius, and the growing shell area $\propto r^2$ moves its peak out to $a_0$.

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Further Reading

• [Gri] Griffiths & Schroeter, §4.2 — the hydrogen atom; this lesson follows its route and notation ($\kappa$, $\rho$, $\rho_0$).
• [ER] Eisberg & Resnick, Ch. 7 — one-electron atoms with full experimental context and the classic orbital plots.
• [Gold] Goldstein, Poole & Safko, Ch. 3 — the classical central-force problem; compare the effective potentials side by side.
• [Sha] Shankar, Ch. 13 — the hydrogen atom, with a sharper look at the accidental degeneracy and its hidden symmetry.

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