Spherical symmetry buys a separation, and the separation reduces three dimensions to one. What is left is a radial equation with a barrier bolted on.

“Just as the solar system provided the first meaningful test of the laws of classical mechanics, the hydrogen atom provided the first meaningful test of the laws of quantum mechanics.” The atom is one electron of charge $-e$ around a nucleus of charge $+e$ , and to a first approximation the nucleus is heavy enough to hold still — so the whole of chapter 9 is one particle in one potential:

V(r) = -\frac{e^2}{4\pi\epsilon_0 r}\tag{9.1}

This section does not solve it. It sets up the machinery for any potential depending only on distance, and that machinery turns out to reduce a three-dimensional problem to a one-dimensional one.

Classically: a force with no torque

A central potential gives a force $\mathbf F = -(\mathrm dV/\mathrm dr)\,\mathbf e_r$ pointing along $\mathbf r$ . So the torque $\mathbf N = \mathbf r\times\mathbf F$ is identically zero, and the angular momentum $\mathbf L$ is a constant vector.

Two consequences follow immediately. The motion stays in a fixed plane perpendicular to $\mathbf L$ ; and the radius vector sweeps out area at a constant rate:

\mathrm d\mathbf A = \frac{\mathbf L}{2m}\,\mathrm dt

Fig. 9.1 — equal areas in equal times, for any central force

eccentricity0.60which equal-time wedge1 of 8

Drag the eccentricity up and the wedges become wildly different in shape — short and fat near the focus, long and thin far from it — while their areas stay equal to within the sampling error. That is the whole content of dA = (L/2m) dt, and it needs nothing about gravity: any central force gives it, because any central force exerts no torque. The areas here are measured from the drawn polygon by the shoelace formula, not asserted. The residual spread is under 0.4% even at e = 0.85.

Splitting the momentum of a planar orbit into radial and transverse parts,

p_r = m\frac{\mathrm dr}{\mathrm dt}\qquad\text{and}\qquad p_t = \frac{L}{r}

the constant total energy becomes

E = \frac{p_r^2}{2m} + \frac{L^2}{2mr^2} + V(r)\tag{9.2}

The book prints p_t = L/m, which cannot be right

Page 180 gives the transverse momentum as $p_t = L/m$ . That has units of $\mathrm{m^2\,s^{-1}}$ — a velocity times a length, not a momentum.

The correct expression is $p_t = L/r$ : angular momentum is $L = r\,p_t$ for the transverse component, so dividing by $r$ recovers the momentum. Units $\mathrm{kg\,m^2\,s^{-1}} / \mathrm m = \mathrm{kg\,m\,s^{-1}}$ ✓.

Eq. (9.2), on the very next line, settles it independently. The book writes the transverse kinetic energy as $L^2/2mr^2$ , and $p_t^2/2m$ reproduces that only if $p_t = L/r$ . With $p_t = L/m$ you would get $L^2/2m^3$ , which is not even an energy. Verified numerically below.

This is the first error found in chapter 9.

The effective potential

Eq. (9.2) is remarkable if you read it right. It has the form of a one-dimensional energy — a kinetic term $p_r^2/2m$ in the single variable $r$ , plus everything else lumped into an effective potential :

V_e(r) = \frac{L^2}{2mr^2} + V(r)\tag{9.3}

The extra term is the centrifugal barrier . Differentiating gives an outward force $L^2/mr^3$ , which for a particle in a circle of radius $r$ at speed $v$ with $L = mvr$ is exactly $mv^2/r$ — the familiar centrifugal force. So the term can be read two ways: as a repulsive potential, or as the transverse kinetic energy the particle must keep because its angular momentum cannot change.

Reproduce it in NumPy

A concrete orbit rather than symbol-pushing, so the units mismatch is impossible to argue with. The Earth figure uses current orbital data and reproduces the book's 3 × 10⁷⁴ to one significant figure.

import numpy as np

print("the transverse momentum, and a dimensional check")
print("=" * 68)
# a concrete circular-ish orbit: m, r, transverse speed v_t
m, r, vt = 3.0, 2.0, 5.0            # kg, m, m/s
L = m*r*vt                          # angular momentum, kg m^2/s
pt_true = m*vt                      # the actual transverse momentum
print(f"  m = {m} kg,  r = {r} m,  v_t = {vt} m/s")
print(f"  L = m.r.v_t              = {L:.4f}  kg m^2 s^-1")
print(f"  actual p_t = m.v_t       = {pt_true:.4f}  kg m s^-1\n")
print(f"  the book prints p_t = L/m = {L/m:.4f}   units m^2 s^-1  -> NOT a momentum")
print(f"  correct is    p_t = L/r   = {L/r:.4f}   units kg m s^-1 -> matches {pt_true:.4f}\n")

print("  and Eq. (9.2) settles it independently:")
print("    it writes the transverse kinetic energy as L^2/2mr^2, and")
print(f"    p_t^2/2m with p_t = L/r gives {(L/r)**2/(2*m):.4f}")
print(f"    L^2/2mr^2                    gives {L**2/(2*m*r**2):.4f}   -> identical")
print(f"    whereas p_t = L/m would give {(L/m)**2/(2*m):.4f}, which is wrong")
print("    and does not even have units of energy.")

print("\nthe angular momentum of the Earth, in units of hbar")
print("=" * 68)
hbar=1.054571817e-34
mE, rE, vE = 5.9722e24, 1.4960e11, 29784.0
LE = mE*vE*rE
print(f"  L = m.v.r = {LE:.3e} kg m^2 s^-1")
print(f"  L/hbar    = {LE/hbar:.2e}          book: 3 x 10^74")
print("  So l would be about 10^37 -- the steps between adjacent values are")
print(f"  {1/np.sqrt(LE/hbar):.1e} of the total, which is why nobody has ever noticed")
print("  that planetary angular momentum is quantized.")

printsthe transverse momentum, and a dimensional check ==================================================================== m = 3.0 kg, r = 2.0 m, v_t = 5.0 m/s L = m.r.v_t = 30.0000 kg m^2 s^-1 actual p_t = m.v_t = 15.0000 kg m s^-1 the book prints p_t = L/m = 10.0000 units m^2 s^-1 -> NOT a momentum correct is p_t = L/r = 15.0000 units kg m s^-1 -> matches 15.0000 and Eq. (9.2) settles it independently: it writes the transverse kinetic energy as L^2/2mr^2, and p_t^2/2m with p_t = L/r gives 37.5000 L^2/2mr^2 gives 37.5000 -> identical whereas p_t = L/m would give 16.6667, which is wrong and does not even have units of energy. the angular momentum of the Earth, in units of hbar ==================================================================== L = m.v.r = 2.661e+40 kg m^2 s^-1 L/hbar = 2.52e+74 book: 3 x 10^74 So l would be about 10^37 -- the steps between adjacent values are 6.3e-38 of the total, which is why nobody has ever noticed that planetary angular momentum is quantized.

For gravity, $V = -GmM/r$ , this machinery gives the conic sections — circles, ellipses, parabolas, hyperbolas. Classical mechanics passed its first extraterrestrial test on planetary orbits precisely because a planet’s angular momentum dwarfs $\hbar$ : the Earth’s is about $3\times10^{74}\hbar$ .

Quantum mechanically: the same split

A state of definite energy has $\Psi = \psi(r,\theta,\phi)e^{-iEt/\hbar}$ (Eq. 9.4) with

\left[-\frac{\hbar^2}{2m}\nabla^2 + V(r)\right]\psi = E\psi\tag{9.5}

a partial differential equation in three variables. The move that tames it is to ask for a state with definite angular momentum as well as definite energy:

\psi(r,\theta,\phi) = R(r)\,Y_{l,m_l}(\theta,\phi)\tag{9.6}

step 1 of 6

Every step is substitution and cancellation. The only physics input is Eq. (8.23), which chapter 8 already established.

1Write the Laplacian in spherical polars. The first term involves only r; the whole bracket is exactly the angular operator chapter 8 problem 8 wrote down.
$\nabla^{2} ψ = \frac{1}{r} \frac{\partial ^{2} ( r ψ )}{\partial r ^{2}} + \frac{1}{r ^{2}} [\frac{\partial ^{2} ψ}{\partial θ ^{2}} + \frac{cos θ}{sin θ} \frac{\partial ψ}{\partial θ} + \frac{1}{sin ^{2} θ} \frac{\partial ^{2} ψ}{\partial ϕ ^{2}}]$
2Recognize that bracket. It is minus L-hat-squared over h-bar-squared, so the angular derivatives can be replaced wholesale by an operator whose eigenvalue we already know.
$\hat{L}^{2} = - ℏ^{2} (\frac{\partial ^{2}}{\partial θ ^{2}} + \frac{cos θ}{sin θ} \frac{\partial}{\partial θ} + \frac{1}{sin ^{2} θ} \frac{\partial ^{2}}{\partial ϕ ^{2}})$
3Substitute psi = R(r) Y and apply Eq. (8.23). The spherical harmonic is an eigenfunction, so the angular operator returns a NUMBER — and Y then cancels from every term.
$\hat{L}^{2} Y_{l, m_{l}} = l (l + 1) ℏ^{2} Y_{l, m_{l}}$
This is the whole trick. An operator acting on its own eigenfunction becomes multiplication by a constant, and a constant does not care about theta or phi. Three variables become one.
4What is left is an ordinary differential equation for R(r) alone, with l appearing only as a number.
$- \frac{ℏ ^{2}}{2 m r} \frac{d ^{2} ( r R )}{d r ^{2}} + [\frac{l ( l + 1 ) ℏ ^{2}}{2 m r ^{2}} + V (r)] R = E R$
Equation (9.7).
5Finally substitute u(r) = rR(r). The awkward 1/r and the second derivative of (rR) collapse into a plain second derivative of u.
$R (r) = \frac{u ( r )}{r}$
Equation (9.8).
6And what emerges is a one-dimensional Schrodinger equation. Not an analogy — the same equation, with r in place of x and an effective potential in place of V.
$- \frac{ℏ ^{2}}{2 m} \frac{d ^{2} u}{d r ^{2}} + [\frac{l ( l + 1 ) ℏ ^{2}}{2 m r ^{2}} + V (r)] u = E u$
Equation (9.9), the radial Schrodinger equation.

The result is the radial Schrödinger equation , with an effective potential that mirrors the classical Eq. (9.3) exactly:

V_e(r) = \frac{l(l+1)\hbar^2}{2mr^2} + V(r)\tag{9.10}

symbol: $u (r) = r R (r)$
is: the function that actually obeys a 1-D Schrödinger equation. NOT the radial part R itself — the factor of r is what removes the first-derivative term and makes the analogy exact. It is also u, not R, that is normalized by ∫|u|²dr = 1.
units: $m^{- 1/2}$
type: real function on the half-line

Bound states are labelled by $n_r = 0, 1, 2, \dots$ , the radial quantum number , which counts the nodes of $u(r)$ between $0$ and $\infty$ . Three numbers therefore specify a state completely:

\psi_{n_r,l,m_l}(r,\theta,\phi) = \frac{u_{n_r,l}(r)}{r}Y_{l,m_l}(\theta,\phi)\tag{9.11}

normalized by $\int_0^\infty|u_{n_r,l}|^2\,\mathrm dr = 1$ (Eq. 9.12).

Engineer's bridge — ch00's recipe, unchanged, on a half-line

§0.3 promised that one algorithm would generate every result in this book: discretize, build a tridiagonal Hamiltonian, call eigh. §8.3a was the exception — angular momentum lives on a sphere, where the recipe does not apply and exact algebra was the honest tool.

Chapter 9 is where the promise is kept. Eq. (9.9) is literally one-dimensional, so the recipe transfers with two changes, both physical rather than numerical:

the domain is the half-line $r > 0$ , not the whole axis;
the potential carries the centrifugal barrier, so you solve a separate 1-D problem for each $l$ .

The $u(0) = 0$ boundary needs one trick worth naming: start the grid at $r = \delta r$ rather than $r = 0$ . A grid that excludes the endpoint imposes Dirichlet there automatically — the same thing ch00’s infinite well relies on — and it also dodges the $1/r$ singularity, which would otherwise be evaluated at $r = 0$ .

The Compute block below runs it on a potential whose answer §6.5 already gave us, and gets that answer back.

The effective potential, and the radial equation solved in it

all l at once

orbital quantum number l — moves the centrifugal barrierl = 1, barrier = 2/2r²

E = 2.499987 vs (2n_r+l+3/2) = 2.5000

Section 6.5's three-dimensional oscillator, whose answer we already know. The l(l+1)/2r² barrier wins at small r and loses at large r, so the effective potential turns over. Raising l pushes the minimum outward as l(l+1) and makes it shallower as 1/l(l+1). Solved by the same grid → hamiltonian → eigh pipeline as every other potential on this site — because Eq. (9.9) is one-dimensional.

Reproduce it in NumPy

The test is chosen so the answer is known independently — §6.5 solved the 3-D oscillator in Cartesian coordinates, and this solves it in spherical ones. Agreement to 2×10⁻⁵ is finite-difference error, falling as the grid refines. Note the degeneracies reproduce ch08 problem 9(c)'s 6 = 5 + 1 exactly.

import numpy as np

# The radial Schrodinger equation (9.9) is ONE-DIMENSIONAL, so the ch00 recipe
# applies unchanged: grid, tridiagonal H, eigh.  Test it on a potential whose
# answer section 6.5 already gave us -- the isotropic oscillator, where
# E = (2 n_r + l + 3/2) hbar.omega.
hbar = m = omega = 1.0

def radial_levels(l, N=4000, rmax=14.0, k=4):
    r = np.linspace(rmax/N, rmax, N)          # u(0)=0 is Dirichlet: start past 0
    dr = r[1] - r[0]
    Ve = 0.5*m*omega**2*r**2 + l*(l+1)*hbar**2/(2*m*r**2)
    t = hbar**2/(2*m*dr*dr)
    H = np.diag(2*t + Ve) + np.diag(-t*np.ones(N-1), 1) + np.diag(-t*np.ones(N-1), -1)
    return np.linalg.eigvalsh(H)[:k]

print("Eq. (9.9) is one-dimensional, so ch00's recipe solves it unchanged")
print("=" * 70)
print("  test potential: the isotropic oscillator, whose answer section 6.5")
print("  already gave us -- E = (2 n_r + l + 3/2) hbar.omega\n")
print(f"  {'l':>3} {'n_r':>4} {'computed':>12} {'exact':>10} {'error':>11}")
worst = 0
for l in range(4):
    ev = radial_levels(l)
    for nr in range(3):
        exact = 2*nr + l + 1.5
        err = abs(ev[nr] - exact)
        worst = max(worst, err)
        print(f"  {l:>3} {nr:>4} {ev[nr]:>12.6f} {exact:>10.1f} {err:>11.2e}")
print(f"\n  worst error {worst:.1e} -- the recipe reproduces chapter 6 exactly.")
print("\n  and the degeneracy pattern falls out: states with the same 2n_r + l")
print("  share an energy, which is why 7/2 hbar.omega holds both l = 2 and l = 0")
print("  (chapter 8, problem 9c).")
lvl = {}
for l in range(6):
    for nr in range(4):
        lvl.setdefault(2*nr + l + 1.5, []).append((nr, l))
for E in sorted(lvl)[:4]:
    combos = ", ".join(f"(n_r={a},l={b})" for a, b in sorted(lvl[E], key=lambda p: p[1]))
    tot = sum(2*b+1 for _, b in lvl[E])
    print(f"    E = {E:>4.1f} hbar.omega : {combos}   -> {tot} states")

printsEq. (9.9) is one-dimensional, so ch00's recipe solves it unchanged ====================================================================== test potential: the isotropic oscillator, whose answer section 6.5 already gave us -- E = (2 n_r + l + 3/2) hbar.omega l n_r computed exact error 0 0 1.499998 1.5 1.91e-06 0 1 3.499990 3.5 9.57e-06 0 2 5.499977 5.5 2.34e-05 1 0 2.499998 2.5 2.42e-06 1 1 4.499990 4.5 9.88e-06 1 2 6.499977 6.5 2.35e-05 2 0 3.499998 3.5 1.91e-06 2 1 5.499991 5.5 8.70e-06 2 2 7.499978 7.5 2.16e-05 3 0 4.499998 4.5 1.78e-06 3 1 6.499992 6.5 8.36e-06 3 2 8.499979 8.5 2.11e-05 worst error 2.3e-05 -- the recipe reproduces chapter 6 exactly. and the degeneracy pattern falls out: states with the same 2n_r + l share an energy, which is why 7/2 hbar.omega holds both l = 2 and l = 0 (chapter 8, problem 9c). E = 1.5 hbar.omega : (n_r=0,l=0) -> 1 states E = 2.5 hbar.omega : (n_r=0,l=1) -> 3 states E = 3.5 hbar.omega : (n_r=1,l=0), (n_r=0,l=2) -> 6 states E = 4.5 hbar.omega : (n_r=1,l=1), (n_r=0,l=3) -> 10 states

Why energy rises with both n_r and l

The book states it; the reason is visible in Eq. (9.9), and it is the same reason twice.

Raising $l$ adds transverse kinetic energy:

\int_0^\infty u^*_{n_r,l}\left(\frac{l(l+1)\hbar^2}{2mr^2}\right)u_{n_r,l}\,\mathrm dr

is positive and grows with $l$ — the centrifugal barrier is a real energy cost.

Raising $n_r$ adds radial kinetic energy:

\int_0^\infty u^*_{n_r,l}\left(-\frac{\hbar^2}{2m}\frac{\mathrm d^2}{\mathrm dr^2}\right)u_{n_r,l}\,\mathrm dr

grows with the number of nodes, because more nodes means more curvature and curvature is kinetic energy.

Both are the node-counting rule the site has used since chapter 4, now split into two directions. In one dimension there was one way to add nodes; here there are two, radial and angular. The oscillator numbers above show it cleanly: at fixed $l$ , each extra radial node costs $2\hbar\omega$ ; at fixed $n_r$ , each extra unit of $l$ costs $\hbar\omega$ .

Parity

A state of definite energy in a central potential has one more definite property, and it has no classical analogue at all. Under reflection through the origin, $\mathbf r \to -\mathbf r$ , an eigenfunction either keeps its sign,

\psi(-\mathbf r) = \psi(\mathbf r)\tag{9.13}

— even parity — or reverses it,

\psi(-\mathbf r) = -\psi(\mathbf r)\tag{9.14}

— odd parity. Using Table 8.1 you can check that $\psi = R(r)Y_{l,m_l}$ has even parity for $l = 0$ and $l = 2$ and odd parity for $l = 1$ ; in general the parity is $(-1)^l$ .

Why parity is (−1)ˡ, and why it only depends on l

Reflection through the origin leaves $r$ alone and does two things to the angles: $\theta \to \pi - \theta$ and $\phi \to \phi + \pi$ . So the radial part $R(r)$ cannot contribute a sign — only the spherical harmonic can, which is why parity depends on $l$ and not on $n_r$ .

Check it on the shapes from §8.3a:

$l = 0$ : $Y_{0,0}$ is a constant. Reflection changes nothing. Even.
$l = 1$ : $Y_{1,0} \propto z$ , and $z \to -z$ . Odd. Likewise $x \pm iy$ flips sign.
$l = 2$ : $Y_{2,0} \propto 3z^2 - r^2$ , quadratic, so signs cancel in pairs. Even.

The pattern is just the degree of the polynomial in $x, y, z$ — which is $l$ . Hence $(-1)^l$ .

Parity is an observable with no classical counterpart

There is no classical quantity called parity. A classical particle has a position, and reflecting it gives a different position — there is nothing to be “even” or “odd” about. Parity exists because a quantum state is a function, and functions can be symmetric.

Energy and parity are compatible whenever the Hamiltonian is unchanged by reflection — which is every Hamiltonian in this book, and every one in atomic, nuclear and particle physics except those involving the weak nuclear interaction. That exception is not a technicality: the discovery in 1956 that the weak force violates parity was a genuine shock, and it is why the caveat is always stated.

Parity is not idle bookkeeping here. §9.4 uses it to prove the selection rule $\Delta l = \pm1$ in two lines — a transition between two states of the same parity has a vanishing dipole matrix element, so it simply cannot happen.

Check yourself

0 / 6 answered

1.Why does a central potential guarantee that the orbit lies in a fixed plane?
In the widget, raise the eccentricity to 0.8 and step through the wedges.
2.The wedges change shape dramatically but their areas stay equal. What does that express?
3.What makes it legitimate to look for states with definite $E$ , $L$ **and** $L_{z}$ simultaneously?
4.Why is the substitution $u (r) = r R (r)$ worth making?
5.A hydrogen state has $n_{r} = 2$ and $l = 1$ . What does $n_{r}$ count?
6.Why does the parity of $ψ = R (r) Y_{l, m_{l}}$ depend only on $l$ ?