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.
7. Solve the finite spherical well for $\ell = 0$ and explain why, unlike in one dimension, a three-dimensional well needs a minimum depth to bind.
8. Separate the two-body problem into centre-of-mass and relative motion, and correct hydrogen's spectrum with the reduced mass.

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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.1.3 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.1.3), 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 finite spherical well: a minimum depth to bind

Now let the walls be finite: $V(r) = -V_0$ for $r < a$ and $0$ outside, with $V_0 > 0$, and look for $\ell = 0$ bound states, $-V_0 < E < 0$. Inside, $u'' = -k^2u$ with $k = \sqrt{2m(E + V_0)}/\hbar$, and $u(0) = 0$ again selects $u = A\sin kr$. Outside, $u'' = \kappa^2u$ with $\kappa = \sqrt{-2mE}/\hbar$, and normalizability selects $u = Be^{-\kappa r}$. Matching $u'/u$ at $r = a$ gives the quantization condition $k\cot ka = -\kappa$.

That is exactly the condition for the odd states of the 1D finite well of half-width $a$ (P.5.2): with $u(0) = 0$, the $\ell = 0$ problem is a 1D problem on the half-line with a node at the origin. And it changes the physics of binding. In one dimension every attractive well, however shallow, holds at least one bound state, the even ground state; the odd states need a minimum depth. Here only the odd ones exist. Writing $k_0 = \sqrt{2mV_0}/\hbar$ for the largest possible $k$, a solution needs $\cot ka < 0$, so $ka$ must exceed $\pi/2$; the first bound state appears only when $k_0a > \pi/2$, that is when $V_0a^2 > \pi^2\hbar^2/8m$. Each further multiple of $\pi$ in $k_0a$ adds one more state, so the number of $\ell = 0$ bound states is $N = \lfloor k_0a/\pi + \tfrac12\rfloor$. A three-dimensional well has to be deep enough, or wide enough, to bind anything at all; for $\ell > 0$ the centrifugal barrier makes it harder still.

The deuteron, the proton–neutron bound state, is the classic case. Model the nuclear attraction as a square well of radius $a \approx 2$ fm. The threshold depth is then a few tens of MeV, and fitting the measured binding energy of 2.22 MeV puts the well only modestly above it, about 34 MeV: deep enough for one bound state, and nowhere near the depth a second would need, nine times the threshold. That is why the deuteron is so weakly bound and has no excited states (E7).

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 $E < 0$. States with $E > 0$ are not bound: the electron can reach infinity, any positive energy is allowed, and the spectrum there is continuous. This ionization continuum begins exactly at the series limits of P.2.2. We want the discrete part. 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.

Two bodies, not one: the reduced mass

Everything above treated the proton as a fixed centre. It is not: electron and proton both move, about their common centre of mass. The repair is exact. For two particles of masses $m_1, m_2$ interacting through $V(\lvert\mathbf r_1 - \mathbf r_2\rvert)$, switch to the centre-of-mass and relative coordinates

$$\mathbf R = \frac{m_1\mathbf r_1 + m_2\mathbf r_2}{M}, \qquad \mathbf r = \mathbf r_1 - \mathbf r_2, \qquad M = m_1 + m_2 .$$

The chain rule turns the kinetic energy into $-\frac{\hbar^2}{2m_1}\nabla_1^2 - \frac{\hbar^2}{2m_2}\nabla_2^2 = -\frac{\hbar^2}{2M}\nabla_R^2 - \frac{\hbar^2}{2\mu}\nabla_r^2$ (E6), with the reduced mass $\mu = m_1m_2/(m_1 + m_2)$. Since the potential depends on $\mathbf r$ alone, the equation separates: $\Psi = \phi(\mathbf R)\,\psi(\mathbf r)$ with total energy $E_{\text{cm}} + E$, where

$$-\frac{\hbar^2}{2M}\nabla_R^2\,\phi = E_{\text{cm}}\,\phi, \qquad \Big[-\frac{\hbar^2}{2\mu}\nabla_r^2 + V(r)\Big]\psi = E\,\psi .$$

The atom as a whole moves freely, as a plane wave of mass $M$, and the internal motion is that of a single fictitious particle of mass $\mu$ in the potential $V$. Every result above therefore holds with $m \to \mu$. For hydrogen $\mu = m_e/(1 + m_e/m_p) = 0.99946\,m_e$: the energies shrink by 0.054%, to $E_1 = -13.598$ eV, the Bohr radius grows by the same fraction, and the Rydberg constant becomes $R_H = R_\infty/(1 + m_e/m_p)$, the correction met in P.2.2. Spectroscopy resolves it easily: deuterium's heavier nucleus moves its H-α line by 0.18 nm, which is how deuterium was discovered. In positronium, an electron bound to a positron (P.3.1), $\mu = m_e/2$: every energy halves, to $E_1 = -6.8$ eV, and every length doubles.

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.

The Schrödinger hydrogen atom is exact for the Hamiltonian written down, but that Hamiltonian leaves three things out. It has no spin, which P.6.3 adds and which doubles every count to $2n^2$. It is non-relativistic: relativity and the coupling of spin to the orbital motion split each level by a relative amount of order $\alpha^2 \approx 5\times10^{-5}$, the fine structure that Sommerfeld fitted (P.2.3) and Dirac's equation explains. And it treats the electromagnetic field classically: quantizing the field gives the excited states their finite lifetimes (spontaneous emission) and adds the Lamb shift, about a gigahertz in the $n = 2$ level. Each is a small correction; the $-13.6\ \text{eV}/n^2$ skeleton derived here survives them all.

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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.

E6 (medium). (a) With $\mathbf R$ and $\mathbf r$ as defined in the text, show that $\nabla_1 = \frac{m_1}{M}\nabla_R + \nabla_r$ and $\nabla_2 = \frac{m_2}{M}\nabla_R - \nabla_r$, and hence that $\frac{1}{m_1}\nabla_1^2 + \frac{1}{m_2}\nabla_2^2 = \frac1M\nabla_R^2 + \frac1\mu\nabla_r^2$. (b) Find the Lyman-α wavelength ($n = 2 \to 1$) for hydrogen with the reduced mass, and for positronium. Take $13.606$ eV for the infinite-nuclear-mass Rydberg energy and $m_p/m_e = 1836.2$.

Solution

(a) $\mathbf r_1$ enters $\mathbf R$ with weight $m_1/M$ and $\mathbf r$ with weight $+1$, so the chain rule gives $\nabla_1 = \frac{m_1}{M}\nabla_R + \nabla_r$; likewise $\nabla_2 = \frac{m_2}{M}\nabla_R - \nabla_r$. Squaring, $\frac{1}{m_1}\nabla_1^2 = \frac{m_1}{M^2}\nabla_R^2 + \frac2M\nabla_R\cdot\nabla_r + \frac{1}{m_1}\nabla_r^2$ and $\frac{1}{m_2}\nabla_2^2 = \frac{m_2}{M^2}\nabla_R^2 - \frac2M\nabla_R\cdot\nabla_r + \frac{1}{m_2}\nabla_r^2$. The cross terms cancel, and adding gives $\frac{1}{M}\nabla_R^2 + \big(\frac1{m_1} + \frac1{m_2}\big)\nabla_r^2$, with $\frac1{m_1} + \frac1{m_2} = \frac1\mu$ ✓. (b) Lyman-α carries $\tfrac34$ of the Rydberg energy times $\mu/m_e$. Hydrogen: $\tfrac34\times13.606\times(1 + 1/1836.2)^{-1} = 10.199$ eV, so $\lambda = 1239.84/10.199 = 121.57$ nm, the measured value; with a fixed nucleus it would be 121.50 nm. Positronium: $\mu/m_e = \tfrac12$, so 5.102 eV and $\lambda = 243.0$ nm, in the near ultraviolet.

E7 (hard). Model the deuteron as a proton and neutron (reduced mass $\mu c^2 = 469.5$ MeV) in a spherical well of radius $a = 2.1$ fm, with $\hbar c = 197.3$ MeV·fm. (a) Find the minimum depth $V_0$ for any bound state. (b) The deuteron's binding energy is 2.22 MeV. Find the well depth numerically from $k\cot ka = -\kappa$. (c) How deep would the well have to be for an excited $\ell = 0$ state?

Solution

(a) $V_0 > \pi^2(\hbar c)^2/8\mu c^2a^2 = 9.87\times(197.3)^2/(8\times469.5\times4.41) = 23$ MeV. (b) $\kappa = \sqrt{2\mu c^2B}/\hbar c = \sqrt{2\times469.5\times2.22}/197.3 = 0.232\ \mathrm{fm^{-1}}$ and $k = \sqrt{2\mu c^2(V_0 - B)}/\hbar c$. Try $V_0 = 34$ MeV: $k = 0.875\ \mathrm{fm^{-1}}$, $ka = 1.838$, $k\cot ka = -0.240\ \mathrm{fm^{-1}}$, slightly too negative. At $V_0 = 33.7$ MeV: $k = 0.871$, $ka = 1.830$, $k\cot ka = -0.231\ \mathrm{fm^{-1}}$ ✓. So $V_0 \approx 34$ MeV, about 1.5 times the threshold. Note $ka \approx 1.83$, just past $\pi/2 = 1.57$: the wavefunction barely turns over inside the well, and most of the probability lies outside it, in the tail $e^{-\kappa r}$ of range $1/\kappa \approx 4.3$ fm. (c) A second state needs $k_0a > 3\pi/2$, i.e. nine times the threshold depth, about 210 MeV. The real deuteron has no excited state, consistent with a well of about 34 MeV.

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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.
6. Why does a three-dimensional square well need a minimum depth to bind, while any one-dimensional attractive well binds?
7. How does including the proton's motion change hydrogen's spectrum, and what does the same reasoning predict for positronium?

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 $a_0$.
6. For $\ell = 0$, $u = rR$ must vanish at the origin, so the radial problem is the 1D well restricted to odd solutions. The even 1D ground state, which exists for any depth, is not allowed; the first odd state needs $k_0a > \pi/2$, i.e. $V_0a^2 > \pi^2\hbar^2/8m$.
7. The two-body problem separates into free centre-of-mass motion and relative motion of mass $\mu = m_1m_2/(m_1 + m_2)$, so $m_e \to \mu$ everywhere: hydrogen's energies shrink by 0.054% and $R_H = R_\infty/(1 + m_e/m_p)$. Positronium has $\mu = m_e/2$, so its energies are halved ($E_1 = -6.8$ eV) and its lengths doubled.

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Further Reading

• [Gri] Griffiths & Schroeter, §4.2 — the hydrogen atom; this lesson follows its route and notation ($\kappa$, $\rho$, $\rho_0$).
• [Gri] Griffiths & Schroeter, §5.1 — two-particle systems, the centre-of-mass separation and the reduced mass.
• [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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