The hydrogen spectrum falls out of a boundary condition, as every spectrum in this book has. What is unusual is that the answer depends on one number where it should depend on two.

§9.1 built the machinery for any central potential. Putting the Coulomb potential into it gives the equation the whole book has been heading toward:

-\frac{\hbar^2}{2m_e}\frac{\mathrm d^2u_{n_r,l}}{\mathrm dr^2} + \left[\frac{l(l+1)\hbar^2}{2m_er^2} - \frac{e^2}{4\pi\epsilon_0r}\right]u_{n_r,l} = E_{n_r,l}u_{n_r,l}\tag{9.15}

with $u(0) = 0$ and $u(\infty) = 0$ (Eq. 9.16). This section reads off what the answer looks like; §9.7 derives it.

The effective potential sets the scales

The effective potential of Eq. (9.10) with $V = -e^2/4\pi\epsilon_0r$ is

V_e(r) = \frac{l(l+1)\hbar^2}{2m_er^2} - \frac{e^2}{4\pi\epsilon_0r}\tag{9.17}

repulsive at small $r$ , attractive at large $r$ . Setting $\mathrm dV_e/\mathrm dr = 0$ locates its minimum:

V_e = -\frac{E_R}{l(l+1)}\qquad\text{at}\qquad r = l(l+1)a_0\tag{9.18}

which names the two units the rest of atomic physics is measured in — the Bohr radius

a_0 = \frac{4\pi\epsilon_0\hbar^2}{e^2m_e} = 0.529\times10^{-10}\ \mathrm m\tag{9.19}

and the Rydberg energy

E_R = \frac{e^2}{8\pi\epsilon_0a_0} = 13.6\ \mathrm{eV}\tag{9.20}

Fig. 9.2 — the effective potential for l = 0, 1, 2, 3

all l at once

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

Eq. (9.18): minimum at r = l(l+1)a₀ = 2a₀, value −E_R/l(l+1) = -0.2500

The hydrogen atom. Levels crowd toward zero and there are infinitely many. 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.

Why l = 0 is the interesting case

For every $l > 0$ the curve turns over: the centrifugal barrier wins near the origin, so the electron is held away from the nucleus and there is a well of finite depth at $r = l(l+1)a_0$ . Classically an electron with that angular momentum would sit at the bottom, in a circular orbit — which is exactly what Eq. (9.18) describes.

For $l = 0$ there is no barrier at all. The effective potential is just $-e^2/4\pi\epsilon_0r$ , a bottomless funnel. A classical electron with zero angular momentum falls straight into the nucleus and radiates infinite energy on the way — the catastrophe that made classical atoms impossible.

What stops it is the uncertainty principle, and the estimate is worth carrying: confine an electron to a region of size $r$ and its momentum is uncertain by about $\hbar/r$ , so its kinetic energy is at least $\hbar^2/2m_er^2$ . The total is

E \approx \frac{\hbar^2}{2m_er^2} - \frac{e^2}{4\pi\epsilon_0r}

Squeeze inward and the potential term drops, but the kinetic term rises faster. The balance point is $r \approx a_0$ with $E \approx -E_R$ — the size and the binding energy of a hydrogen atom, from one line of arithmetic.

That is why atoms have a size at all. Not because anything holds the electron out, but because localizing it costs more kinetic energy than the attraction returns.

The answer

Solving Eqs. (9.15)–(9.16) — done in §9.7 — gives an infinite number of bound states for every $l$ :

E_{n_r,l} = -\frac{E_R}{(n_r+l+1)^2},\qquad n_r = 0, 1, 2, 3, \dots\tag{9.21}

And here the surprise arrives. The energy depends on $n_r$ and $l$ only through their sum. Levels with different splits between radial and angular motion land on exactly the same energy, so they are relabelled by the principal quantum number

n = n_r + l + 1\tag{9.23}

giving the formula from §1.3:

E_n = -\frac{E_R}{n^2}\tag{9.22}

Fig. 9.3 — the bound levels of hydrogen, and the degeneracy nobody ordered

label the levels byhover a level to pick out its n

Labelled by n_r, as in the book's Figs. 9.2 and 9.3. Each column is an independent one-dimensional problem — one radial Schrödinger equation per l — and n_r counts its radial nodes. Read this way, the horizontal alignment across columns looks like a coincidence. Energies come from Eq. (9.21), not from the drawing, so the alignment is the formula speaking.

The book uses two labelling systems and switches without warning

States are labelled $(n_r, l)$ in Figs. 9.2 and 9.3, and $(n, l)$ — via spectroscopic notation — in Table 9.1 and Fig. 9.8. They describe the same states, related by $n = n_r + l + 1$ .

So “3p” has $n = 3$ and $l = 1$ , and therefore $n_r = 1$ : one radial node. Not three, and not zero. Every mismatch between a figure and a table in this chapter comes from this, and the widget above has a toggle for exactly that reason — flip it and watch the same diagram acquire a different set of names.

A rule of thumb: the radial quantum number $n_r$ tells you about the radial shape (how many nodes $u(r)$ has), while $n$ tells you about the energy (which is $-E_R/n^2$ ). Neither is more fundamental; they answer different questions.

Engineer's bridge — the same three lines that solved a square well

§0.3 opened this site by claiming one algorithm would generate every result in the book: discretize, build a tridiagonal Hamiltonian, call an eigensolver. Chapter 9 is where that claim gets tested on the hardest case, and the Compute block below runs it.

The only wrinkles are physical, not numerical:

$u(0) = 0$ is a Dirichlet wall, imposed by starting the grid at $r = \delta r$ rather than $0$ — which also avoids evaluating $1/r$ at the origin;
the box must fit the state. A state of principal quantum number $n$ extends to about $n^2a_0$ , so a box sized for the ground state truncates $n = 5$ badly, and a box sized for $n = 5$ under-resolves the ground state. The code sizes the box to the largest state requested — and the residual errors then come out uniform across all nine states, which is how you know the grid is honest rather than lucky.

The eigenvalues are found by Sturm-sequence bisection, exactly as qm.ts does: count how many eigenvalues lie below a trial energy from the signs of the LDL pivots, and bisect. That costs $O(N)$ per step instead of $O(N^3)$ for a dense solve, which is what makes a 40 000-point grid practical at all.

Reproduce it in NumPy

Sturm-sequence bisection rather than a dense eigensolve, which is what makes a 40 000-point grid affordable — the same algorithm qm.ts uses. The errors are uniform across all nine states because the box is sized to the state; with one fixed box the ground state's error is 40× worse than the rest.

import numpy as np

# Atomic units: hbar = m_e = e^2/4.pi.eps0 = a0 = 1, so E_R = 1/2.
# Eq. (9.15) is one-dimensional in u(r), so ch00's recipe applies -- and we use
# the same Sturm-sequence bisection qm.ts uses, which costs O(N) per step and so
# lets the grid reach the hundreds of Bohr radii the n = 5 state needs.

def sturm_count(d, e, E):
    """how many eigenvalues lie below E, from the signs of the LDL pivots"""
    q = d[0] - E
    cnt = 1 if q < 0 else 0
    for i in range(1, len(d)):
        q = (d[i] - E) - (e[i-1]*e[i-1])/(q if q != 0 else 1e-300)
        if q < 0:
            cnt += 1
    return cnt

def level(d, e, k, lo=-0.6, hi=-1e-9):
    """the k-th eigenvalue (0-based), by bisecting on the Sturm count"""
    for _ in range(200):
        mid = 0.5*(lo + hi)
        if sturm_count(d, e, mid) > k:
            hi = mid
        else:
            lo = mid
    return 0.5*(lo + hi)

def coulomb(l, nmax, N=40000):
    # a state of principal quantum number n extends to about n^2 a0, so size the
    # box to the largest state asked for rather than using one box for all of them
    rmax = 40.0*nmax**2
    r = np.linspace(rmax/N, rmax, N)      # u(0) = 0: the grid starts past the origin
    dr = r[1] - r[0]
    t = 1/(2*dr*dr)
    return 2*t + l*(l+1)/(2*r**2) - 1/r, -t*np.ones(N-1)

print("Eq. (9.15) solved by the ch00 recipe -- does it give -E_R/n^2 ?")
print("=" * 72)
print("  atomic units, so E_R = 1/2 and the prediction is E = -1/(2 n^2)\n")
print(f"  {'l':>3} {'n_r':>4} {'n':>3} {'computed E':>14} {'-1/(2n^2)':>12} {'rel. error':>11}")
worst = 0
for l in range(3):
    d, e = coulomb(l, 2 + l + 1)
    for nr in range(3):
        n = nr + l + 1
        exact = -0.5/n**2
        got = level(d, e, nr)
        rel = abs(got - exact)/abs(exact)
        worst = max(worst, rel)
        print(f"  {l:>3} {nr:>4} {n:>3} {got:>14.8f} {exact:>12.8f} {rel:>11.1e}")
print(f"\n  worst relative error {worst:.1e} -- finite-difference error, nothing else.")
print("  The same three lines of code that solved a square well in chapter 0 have")
print("  now produced the hydrogen spectrum.\n")

print("  and now the degeneracy, which is the surprising part:")
print("=" * 72)
lv = {}
for l in range(5):
    for nr in range(5):
        lv.setdefault(nr + l + 1, []).append((nr, l))
for n in range(1, 5):
    combos = ", ".join(f"(n_r={a}, l={b})" for a, b in sorted(lv[n], key=lambda p: p[1]))
    states = sum(2*b + 1 for _, b in lv[n])
    print(f"    n = {n}:  E = {-0.5/n**2:+.6f}   {combos}")
    print(f"           {len(lv[n])} value(s) of l, {states} states in all = n^2 = {n*n}")
print("\n  Every state with the same n_r + l lands on the same energy, however the")
print("  total is split between radial and angular motion.  Nothing in rotational")
print("  symmetry predicts that: it is special to the 1/r potential, and chapter 8")
print("  problem 9(c) met the same kind of coincidence in the oscillator.")

printsEq. (9.15) solved by the ch00 recipe -- does it give -E_R/n^2 ? ======================================================================== atomic units, so E_R = 1/2 and the prediction is E = -1/(2 n^2) l n_r n computed E -1/(2n^2) rel. error 0 0 1 -0.49998988 -0.50000000 2.0e-05 0 1 2 -0.12499937 -0.12500000 5.1e-06 0 2 3 -0.05555543 -0.05555556 2.2e-06 1 0 2 -0.12500067 -0.12500000 5.3e-06 1 1 3 -0.05555586 -0.05555556 5.5e-06 1 2 4 -0.03125016 -0.03125000 5.1e-06 2 0 3 -0.05555562 -0.05555556 1.2e-06 2 1 4 -0.03125006 -0.03125000 2.0e-06 2 2 5 -0.02000005 -0.02000000 2.3e-06 worst relative error 2.0e-05 -- finite-difference error, nothing else. The same three lines of code that solved a square well in chapter 0 have now produced the hydrogen spectrum. and now the degeneracy, which is the surprising part: ======================================================================== n = 1: E = -0.500000 (n_r=0, l=0) 1 value(s) of l, 1 states in all = n^2 = 1 n = 2: E = -0.125000 (n_r=1, l=0), (n_r=0, l=1) 2 value(s) of l, 4 states in all = n^2 = 4 n = 3: E = -0.055556 (n_r=2, l=0), (n_r=1, l=1), (n_r=0, l=2) 3 value(s) of l, 9 states in all = n^2 = 9 n = 4: E = -0.031250 (n_r=3, l=0), (n_r=2, l=1), (n_r=1, l=2), (n_r=0, l=3) 4 value(s) of l, 16 states in all = n^2 = 16 Every state with the same n_r + l lands on the same energy, however the total is split between radial and angular motion. Nothing in rotational symmetry predicts that: it is special to the 1/r potential, and chapter 8 problem 9(c) met the same kind of coincidence in the oscillator.

The n² degeneracy is an accident, and this book does not explain it

Phillips calls it “unexpected” and moves on. It is worth knowing that this is a genuine open door rather than an omission you should have followed.

Rotational symmetry explains the $2l+1$ degeneracy in $m_l$ — states pointing in different directions must have the same energy when nothing picks out a direction. It explains nothing about why $l = 0$ and $l = 1$ should coincide at $n = 2$ ; those are different shapes, not different orientations.

The reason is a further conserved quantity special to the $1/r$ potential — the Runge–Lenz vector, which classically is what keeps an elliptical orbit from precessing. Any deviation from exact $1/r$ destroys it, and the degeneracy splits: in a multi-electron atom the inner electrons screen the nucleus, 2s and 2p separate, and chapter 11 builds the periodic table out of that splitting.

Chapter 8’s problem 9(c) met the same pattern, finding six degenerate oscillator states splitting as $l = 2$ plus $l = 0$ , and noted then that a degeneracy exceeding $2l+1$ means a hidden symmetry. Hydrogen is the other famous case.

What the eigenfunctions look like

Three facts fix the shape of $u_{n_r,l}(r)$ , and each comes from a limit the site has already met:

At large $r$ the binding energy $E = -E_R/n^2$ gives exponential decay $u \propto e^{-r/na_0}$ — the same $e^{-\alpha x}$ tail §5.1 found outside a square well, with $\alpha$ set by how tightly the state is bound.
At small $r$ the centrifugal term dominates and forces $u \propto r^{l+1}$ — higher $l$ is pushed harder away from the origin.
In between, $n_r$ nodes, so $u$ carries a polynomial $p_{n_r,l}(r)$ with $n_r$ zeros.

Multiplying them together:

u_{n_r,l}(r) = N\,p_{n_r,l}(r)\,r^{l+1}e^{-r/na_0}\tag{9.24}

spectroscopic⇅	$(n_r, l)$⇅	$u_{n_r,l}(r)$⇅
1s	$(0, 0)$	$\frac{2}{a _{0}} x e^{- x}$ ⓘ
2s	$(1, 0)$	$\frac{1}{2 a _{0}} [1 - \frac{1}{2} x] x e^{- x /2}$ ⓘ
3s	$(2, 0)$	$\frac{2}{3 3 a _{0}} [1 - \frac{2}{3} x + \frac{2}{27} x^{2}] x e^{- x /3}$ ⓘ
4s	$(3, 0)$	$\frac{1}{4 a _{0}} [1 - \frac{3}{4} x + \frac{1}{8} x^{2} - \frac{1}{192} x^{3}] x e^{- x /4}$
2p	$(0, 1)$	$\frac{1}{2 6 a _{0}} x^{2} e^{- x /2}$ ⓘ
3p	$(1, 1)$	$\frac{8}{27 6 a _{0}} [1 - \frac{1}{6} x] x^{2} e^{- x /3}$
4p	$(2, 1)$	$\frac{1}{16} \frac{5}{3 a _{0}} [1 - \frac{1}{4} x + \frac{1}{80} x^{2}] x^{2} e^{- x /4}$
3d	$(0, 2)$	$\frac{4}{81 30 a _{0}} x^{3} e^{- x /3}$ ⓘ
4d	$(1, 2)$	$\frac{1}{64 5 a _{0}} [1 - \frac{1}{12} x] x^{3} e^{- x /4}$

Reproduce it in NumPy

Transcription from a dense printed table is exactly where errors hide, so each row is checked three ways rather than once. The eigenvalue is read off where |u| is largest, so it is least sensitive to the finite-difference error near the origin.

import numpy as np

# Table 9.1, transcribed from page-207.png.  Atomic units: a0 = 1, so x = r/a0.
# Each entry is (label, n_r, l, u(x)).
s = np.sqrt
TABLE = [
 ("1s", 0, 0, lambda x: 2*x*np.exp(-x)),
 ("2s", 1, 0, lambda x: (1/s(2))*(1 - x/2)*x*np.exp(-x/2)),
 ("3s", 2, 0, lambda x: (2/(3*s(3)))*(1 - (2/3)*x + (2/27)*x**2)*x*np.exp(-x/3)),
 ("4s", 3, 0, lambda x: (1/4)*(1 - (3/4)*x + (1/8)*x**2 - (1/192)*x**3)*x*np.exp(-x/4)),
 ("2p", 0, 1, lambda x: (1/(2*s(6)))*x**2*np.exp(-x/2)),
 ("3p", 1, 1, lambda x: (8/(27*s(6)))*(1 - x/6)*x**2*np.exp(-x/3)),
 ("4p", 2, 1, lambda x: (1/16)*s(5/3)*(1 - x/4 + (1/80)*x**2)*x**2*np.exp(-x/4)),
 ("3d", 0, 2, lambda x: (4/(81*s(30)))*x**3*np.exp(-x/3)),
 ("4d", 1, 2, lambda x: (1/(64*s(5)))*(1 - x/12)*x**3*np.exp(-x/4)),
]

x = np.linspace(1e-7, 260, 3_000_001)
dx = x[1]-x[0]

print("Table 9.1 checked two ways: normalization, and the eigenvalue equation")
print("=" * 74)
HDR = "E from -u\u2033/2+Ve.u"   # u-double-prime; a bare '' inside an f-string field
                              # silently concatenates and drops it
print(f"  {'state':>5} {'n_r':>4} {'l':>3} {'n':>3} {'int|u|^2 dr':>13} {'nodes':>6} "
      f"{HDR:>19} {'-1/2n^2':>10}")
allok = True
for name, nr, l, f in TABLE:
    u = f(x)
    norm = np.trapezoid(u*u, x)
    # count interior nodes (sign changes away from the endpoints)
    core = u[(x > 1e-3) & (x < 200)]
    nodes = int(np.sum(np.sign(core[:-1])*np.sign(core[1:]) < 0))
    # apply the radial Hamiltonian and read off E where u is largest
    d2 = np.gradient(np.gradient(u, x), x)
    Ve = l*(l+1)/(2*x**2) - 1/x
    Hu = -0.5*d2 + Ve*u
    m = np.argmax(np.abs(u))
    E = Hu[m]/u[m]
    n = nr + l + 1
    exact = -0.5/n**2
    ok = abs(norm-1) < 2e-4 and nodes == nr and abs(E-exact) < 2e-4
    allok &= ok
    print(f"  {name:>5} {nr:>4} {l:>3} {n:>3} {norm:>13.6f} {nodes:>6} "
          f"{E:>19.6f} {exact:>10.6f}  {'OK' if ok else 'FAIL'}")
print(f"\n  all nine entries verified: {allok}")
print("\n  Three independent checks per row, and each catches a different slip:")
print("    - the integral catches a wrong normalizing constant,")
print("    - the node count catches a wrong polynomial,")
print("    - the eigenvalue catches a wrong exponent or a wrong power of r.")

printsTable 9.1 checked two ways: normalization, and the eigenvalue equation ========================================================================== state n_r l n int|u|^2 dr nodes E from -u″/2+Ve.u -1/2n^2 1s 0 0 1 1.000000 0 -0.500000 -0.500000 OK 2s 1 0 2 1.000000 1 -0.125000 -0.125000 OK 3s 2 0 3 1.000000 2 -0.055556 -0.055556 OK 4s 3 0 4 1.000000 3 -0.031250 -0.031250 OK 2p 0 1 2 1.000000 0 -0.125000 -0.125000 OK 3p 1 1 3 1.000000 1 -0.055556 -0.055556 OK 4p 2 1 4 1.000000 2 -0.031250 -0.031250 OK 3d 0 2 3 1.000000 0 -0.055556 -0.055556 OK 4d 1 2 4 1.000000 1 -0.031250 -0.031250 OK all nine entries verified: True Three independent checks per row, and each catches a different slip: - the integral catches a wrong normalizing constant, - the node count catches a wrong polynomial, - the eigenvalue catches a wrong exponent or a wrong power of r.

The book plots several of these in Fig. 9.4. Here they are live, solved rather than drawn:

Fig. 9.4 — radial eigenfunctions, computed on the spot

all l at once

orbital quantum number l — moves the centrifugal barrierl = 0

E = -0.444612 vs −1/2(n)² = -0.500000

The hydrogen atom. Levels crowd toward zero and there are infinitely many. With l = 0 there is no centrifugal barrier at all — the potential is purely attractive, and a classical particle would fall straight in. Solved by the same grid → hamiltonian → eigh pipeline as every other potential on this site — because Eq. (9.9) is one-dimensional.

How big is a hydrogen atom, and how tightly bound?

Put numbers on Eqs. (9.19) and (9.20).

Size. $a_0 = 4\pi\epsilon_0\hbar^2/e^2m_e = 5.29\times10^{-11}$ m $= 0.0529$ nm. So an atom is about a tenth of a nanometre across — and since $\langle r\rangle$ scales as $n^2$ , an $n = 10$ state is a hundred times bigger, around 5 nm. Rydberg atoms with $n \approx 100$ are micrometres across, larger than some bacteria.

Energy. $E_R = 13.6$ eV, so the ground state sits at $-13.6$ eV and ionizing hydrogen costs exactly that. Compare:

	energy	ratio to $E_R$
a visible photon	≈ 2 eV	0.15
$kT$ at room temperature	0.026 eV	0.002
the $n=2 \to 1$ transition	10.2 eV	0.75

So hydrogen is not ionized by room-temperature collisions — you would need about 500 times more thermal energy — which is why neutral hydrogen exists at all, and why stars have to be hot to ionize it.

And the two are linked. $E_R = \hbar^2/2m_ea_0^2$ : the binding energy is exactly the kinetic energy of localizing an electron to $a_0$ . That is the uncertainty-principle estimate above, made exact.

Check yourself

0 / 6 answered

1.A hydrogen state is labelled 3p. What are $n$ , $l$ and $n_{r}$ ?
2.Why do the levels $2 s$ and $2 p$ have exactly the same energy?
The Compute block sizes its numerical box to the state, giving errors that are uniform across all nine levels.
3.What goes wrong with a single fixed box for every state?
4.Why does only an $l = 0$ state have non-zero probability density *at* the nucleus?
5.The uncertainty-principle estimate gives $E \approx ℏ^{2} /2 m_{e} r^{2} - e^{2} /4 π ϵ_{0} r$ , minimized near $r = a_{0}$ . What does that argument explain?
6.Table 9.1's entries were each checked three ways. Why not just check normalization?