Which lines a spectrum contains is decided by a symmetry, not by an energy. The gaps say what could be emitted; parity says what actually is.

The states of §9.2 are stationary — left alone, an excited hydrogen atom would stay excited forever. Real atoms emit light. This section turns the electromagnetic field on and asks which transitions can happen, then makes one small correction that turns out to have discovered an isotope.

What drives a transition

The most probable radiative transitions are electric dipole transitions, driven by the interaction of the field $\mathbf E$ with the electron–nucleus dipole moment $\hat{\mathbf d} = -e\hat{\mathbf r}$ :

\hat H_I = -\hat{\mathbf d}\cdot\mathbf E\tag{9.26}

and the probability of a transition between states $i$ and $f$ is proportional to

\left|\int\psi^*_{n_f,l_f,m_{l_f}}(\mathbf r)\,\hat H_I\,\psi_{n_i,l_i,m_{l_i}}(\mathbf r)\,\mathrm d^3\mathbf r\right|^2\tag{9.27}

Parity kills most of them

The argument is two lines and needs no integration.

Why a transition between same-parity states is exactly forbidden

Change the integration variable from $\mathbf r$ to $-\mathbf r$ . The integral is over all space, so its value cannot change.

$\hat H_I \propto \mathbf r$ changes sign.
Each wave function keeps its sign if it has even parity and flips it if odd — that is §9.1’s Eqs. (9.13)–(9.14).

If $\psi_i$ and $\psi_f$ have the same parity, their two sign changes cancel, and the integrand as a whole picks up the single minus sign from $\hat H_I$ . So the integral equals minus itself — and the only number equal to its own negative is zero.

Electric dipole transitions therefore always change parity. Since parity is $(-1)^l$ , that immediately forbids $\Delta l = 0$ and $\Delta l = \pm2$ : any even change of $l$ preserves parity.

Working through the spherical-harmonic algebra narrows the survivors further — the integral vanishes unless $l_f - l_i$ is $+1$ or $-1$ exactly:

\Delta l = \pm1\tag{9.28}

Reproduce it in NumPy

Only the ANGULAR factor is computed, because that is where the selection rule lives — the radial integral is generally non-zero and merely scales the rate. The z component is used; x and y give Δm = ±1 instead of 0, which is why a magnetic field splits a line into three.

import numpy as np
pi = np.pi

# The electric dipole operator is d = -e r.  Its z component sandwiched between
# two states factorizes into a radial integral and an ANGULAR one:
#     <n'l'm'| z |nlm>  =  (radial) x  int Y*_{l'm'} cos(theta) Y_{lm} dOmega
# The selection rule lives entirely in the angular factor, so that is what we
# compute -- no radial functions needed.

t = np.linspace(0, pi, 3001)
p = np.linspace(0, 2*pi, 3002)
T, P = np.meshgrid(t, p, indexing='ij')
s = np.sqrt

Y = {
 (0,0):  s(1/(4*pi))*np.ones_like(T),
 (1,0):  s(3/(4*pi))*np.cos(T),
 (1,1): -s(3/(8*pi))*np.sin(T)*np.exp(1j*P),
 (1,-1): s(3/(8*pi))*np.sin(T)*np.exp(-1j*P),
 (2,0):  s(5/(16*pi))*(3*np.cos(T)**2-1),
 (2,1): -s(15/(8*pi))*np.sin(T)*np.cos(T)*np.exp(1j*P),
 (2,-1): s(15/(8*pi))*np.sin(T)*np.cos(T)*np.exp(-1j*P),
 (2,2):  s(15/(32*pi))*np.sin(T)**2*np.exp(2j*P),
 (3,0):  s(7/(16*pi))*(5*np.cos(T)**3-3*np.cos(T)),
}
def ang(a, b):
    """int Y*_a cos(theta) Y_b dOmega"""
    f = np.conj(Y[a])*np.cos(T)*Y[b]*np.sin(T)
    return np.trapezoid(np.trapezoid(f, p, axis=1), t)

print("the angular part of the dipole matrix element, <Y' | cos(theta) | Y>")
print("=" * 74)
print("  (cos(theta) is z/r, so this is the z component of the dipole operator)\n")
print(f"  {'initial':>10} {'final':>10} {'dl':>4} {'dm':>4} {'|integral|':>12}  verdict")
rows = [((0,0),(1,0)), ((1,0),(0,0)), ((1,0),(2,0)), ((2,0),(1,0)),
        ((1,1),(2,1)), ((1,1),(2,2)), ((0,0),(2,0)), ((1,0),(3,0)),
        ((0,0),(0,0)), ((1,0),(1,0)), ((2,0),(2,0)), ((1,1),(2,-1))]
for a, b in rows:
    v = abs(ang(a, b))
    dl, dm = b[0]-a[0], b[1]-a[1]
    allowed = abs(dl) == 1 and dm == 0
    mark = "ALLOWED" if v > 1e-6 else "forbidden"
    consistent = (v > 1e-6) == allowed
    print(f"  {str(a):>10} {str(b):>10} {dl:>+4} {dm:>+4} {v:>12.2e}  {mark}"
          f"{'' if consistent else '   <-- INCONSISTENT'}")
print("\n  Every non-zero entry has dl = +-1 and dm = 0; every other pairing")
print("  vanishes to 1e-16.  That is Eq. (9.28).\n")

print("and the parity half of the argument, which needs no integration at all")
print("=" * 74)
print("  Parity is (-1)^l, and z is odd under r -> -r.  So the integrand")
print("  Y*_l' . z . Y_l has parity (-1)^l' . (-1) . (-1)^l = (-1)^(l+l'+1).")
print("  If l and l' have the SAME parity, that is odd -- and an odd function")
print("  integrated over all space gives exactly zero.\n")
for l, lp in ((0,0),(0,1),(0,2),(1,1),(1,2),(2,2)):
    same = (l % 2) == (lp % 2)
    print(f"    l = {l}, l' = {lp}: parities {'match' if same else 'differ'}"
          f"  -> integrand {'odd -> ZERO' if same else 'even -> may survive'}")
print("\n  So parity alone forbids dl = 0 and dl = +-2.  The full spherical-harmonic")
print("  algebra then narrows the survivors to exactly dl = +-1.")

printsthe angular part of the dipole matrix element, <Y' | cos(theta) | Y> ========================================================================== (cos(theta) is z/r, so this is the z component of the dipole operator) initial final dl dm |integral| verdict (0, 0) (1, 0) +1 +0 5.77e-01 ALLOWED (1, 0) (0, 0) -1 +0 5.77e-01 ALLOWED (1, 0) (2, 0) +1 +0 5.16e-01 ALLOWED (2, 0) (1, 0) -1 +0 5.16e-01 ALLOWED (1, 1) (2, 1) +1 +0 4.47e-01 ALLOWED (1, 1) (2, 2) +1 +1 1.88e-19 forbidden (0, 0) (2, 0) +2 +0 0.00e+00 forbidden (1, 0) (3, 0) +2 +0 0.00e+00 forbidden (0, 0) (0, 0) +0 +0 0.00e+00 forbidden (1, 0) (1, 0) +0 +0 0.00e+00 forbidden (2, 0) (2, 0) +0 +0 0.00e+00 forbidden (1, 1) (2, -1) +1 -2 1.84e-17 forbidden Every non-zero entry has dl = +-1 and dm = 0; every other pairing vanishes to 1e-16. That is Eq. (9.28). and the parity half of the argument, which needs no integration at all ========================================================================== Parity is (-1)^l, and z is odd under r -> -r. So the integrand Y*_l' . z . Y_l has parity (-1)^l' . (-1) . (-1)^l = (-1)^(l+l'+1). If l and l' have the SAME parity, that is odd -- and an odd function integrated over all space gives exactly zero. l = 0, l' = 0: parities match -> integrand odd -> ZERO l = 0, l' = 1: parities differ -> integrand even -> may survive l = 0, l' = 2: parities match -> integrand odd -> ZERO l = 1, l' = 1: parities match -> integrand odd -> ZERO l = 1, l' = 2: parities differ -> integrand even -> may survive l = 2, l' = 2: parities match -> integrand odd -> ZERO So parity alone forbids dl = 0 and dl = +-2. The full spherical-harmonic algebra then narrows the survivors to exactly dl = +-1.

Fig. 9.8 — electric dipole transitions, and the one that cannot happen

label the levels byobey Δl = ±1hover a level to pick out its n

Labelled by n = n_r + l + 1, as in Table 9.1 and Fig. 9.8. Now the aligned levels share a name, and the spectroscopic labels 1s, 2s, 2p, 3d appear. Same diagram, same physics — only the bookkeeping changed, and this is the switch the book makes without announcing it. Energies come from Eq. (9.21), not from the drawing, so the alignment is the formula speaking.

Dotted lines are the allowed electric dipole transitions — every one steps sideways by exactly one column, because Δl = ±1. Notice what is missing: nothing connects 2s to 1s, since both have l = 0. That is why the 2s state is metastable.

Forbidden means forbidden, not unlikely

A selection rule is not a statement about probabilities being small. If the matrix element of Eq. (9.27) is exactly zero, the transition does not occur by this mechanism at all — the rate is not suppressed, it vanishes.

What can happen is that a different, weaker mechanism takes over. Electric dipole radiation is only the leading term; magnetic dipole and electric quadrupole transitions follow different rules and are typically millions of times slower. So a “forbidden” line is usually not absent from a spectrum, just extraordinarily faint and slow — which is exactly what the next box is about.

The 2s state cannot decay, and lives 10⁸ times longer for it

Look at the diagram with the rule switched on. Nothing connects 2s to 1s.

Both have $l = 0$ , so $\Delta l = 0$ and the electric dipole matrix element vanishes. And 2s has nothing else below it — 1s is the only lower level in the atom.

The consequences are dramatic:

transition	mechanism	mean lifetime
2p → 1s	electric dipole, allowed	$1.6\times10^{-9}$ s
2s → 1s	forbidden; proceeds by two-photon emission	$0.14$ s

A factor of about $10^{8}$ . The 2s state is metastable : it eventually decays, but by emitting two photons at once rather than one, which is a far less probable process.

That longevity is useful. Metastable states are where you can accumulate a population — which is the basic requirement for a laser — and the 2s state of hydrogen is a standard target in precision spectroscopy precisely because its long life gives it an extremely narrow linewidth.

The spectrum

Transitions can be induced by an external field oscillating at the resonant frequency $\hbar\omega = |E_{n_f} - E_{n_i}|$ , absorbing energy when $E_{n_f} > E_{n_i}$ and emitting it otherwise. Spontaneous transitions look uncaused but are not: they are driven by the quantized electromagnetic field, which is present even around a perfectly isolated atom.

Either way the emitted photon carries $E_{n_i} - E_{n_f}$ , giving

\frac{hc}{\lambda} = E_R\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)\tag{9.29}

Engineer's bridge — this is the same data file chapter 1 used

§1.3 drew this spectrum eight chapters ago, computing it from the book’s Eq. (1.9) — which at that point was simply asserted, the Balmer formula fitted to observation in 1885.

This page reuses ch01-hydrogen.ts unchanged rather than recomputing the wavelengths. That is deliberate: a page that quietly recomputed them could drift from the earlier one by a rounding choice, and a reader flipping between the two would find two different values for the same line with no way to tell which is right.

The physics content of chapter 9 is not new numbers — it is that the old numbers now follow from something. Eq. (9.29) has the same form as Eq. (1.9) because it is Eq. (1.9), now derived from the Coulomb potential and the Schrödinger equation instead of fitted to a spectrograph.

The reduced mass effect

One assumption has been quietly load-bearing: that the nucleus stays put. It does not — nucleus and electron both orbit their common centre of mass. Writing the classical energy in that frame turns the two-body problem into a one-body one with the reduced mass

\mu = \frac{m_em_N}{m_e + m_N}

and the quantum problem transforms the same way. So every result so far holds with $m_e$ replaced by $\mu$ — the length scale becomes $a_0' = (m_e/\mu)a_0$ , the energy scale $E_R' = (\mu/m_e)E_R$ , and

E_n' = -\frac{E_R'}{n^2} = \frac{\mu}{m_e}\left(-\frac{E_R}{n^2}\right)\tag{9.30}

A correction of one part in two thousand, which found an isotope

For hydrogen $m_N = 1836\,m_e$ , so $\mu/m_e = 1836/1837 = 0.99946$ — a shift of 0.05%. For deuterium the nucleus is twice as heavy, $\mu/m_e = 3671/3672 = 0.99973$ .

The difference between the two is only 0.027%, which sounds hopeless. But spectroscopy does not measure energies to a percentage — it measures positions of lines, and 0.027% of 656 nm is 0.18 nm, which any diffraction grating resolves comfortably.

So each hydrogen line has a faint companion 0.18 nm to the blue, appearing at about 1/6000 the brightness because natural hydrogen is 0.0156% deuterium. Harold Urey found exactly that in 1934 and won a Nobel Prize for it — an isotope discovered because a two-body correction shifted a spectral line by less than a part in a thousand.

Reproduce it in NumPy

The absolute wavelengths sit 1.5 ppm from the book's because it rounds E_R to 13.6 eV; the separation — the quantity Urey actually measured — agrees to four decimal places. Tritium and the infinite-mass limit are included to show the trend: heavier nucleus, shorter wavelength, converging on 656.1123 nm.

import numpy as np
me = 9.1093837015e-31
amu = 1.66053906660e-27
h, c, e = 6.62607015e-34, 299792458.0, 1.602176634e-19
ER = 13.605693          # eV, the Rydberg energy for an infinitely heavy nucleus

def halpha(mN):
    """wavelength of the n = 3 -> 2 line, Eq. (9.29) scaled by mu/m_e"""
    mu = me if mN is None else me*mN/(me + mN)
    return h*c/((mu/me)*ER*e*(1/4 - 1/9))*1e9, mu/me

print("the reduced-mass effect, Eq. (9.30):  E_n' = (mu/m_e) x (-E_R/n^2)")
print("=" * 72)
print(f"  {'nucleus':<24} {'m_N/m_e':>9} {'mu/m_e':>11} {'H-alpha (nm)':>14}")
for name, mN in (("hydrogen  (proton)",   1.007276466*amu),
                 ("deuterium (deuteron)", 2.013553213*amu),
                 ("tritium   (triton)",   3.015500856*amu),
                 ("infinitely heavy",     None)):
    lam, ratio = halpha(mN)
    mr = "-" if mN is None else f"{mN/me:9.1f}"
    print(f"  {name:<24} {mr:>9} {ratio:>11.7f} {lam:>14.4f}")

lamH, _ = halpha(1.007276466*amu)
lamD, _ = halpha(2.013553213*amu)
print(f"\n  hydrogen  H-alpha = {lamH:.4f} nm     book: 656.4686 nm")
print(f"  deuterium H-alpha = {lamD:.4f} nm     book: 656.2899 nm")
print(f"  separation        = {lamH-lamD:.4f} nm     book: {656.4686-656.2899:.4f} nm")
print(f"  relative shift    = {(lamH-lamD)/lamH*1e6:.0f} parts per million")
print("  (the 1.5 ppm offset from the book is its rounded E_R = 13.6 eV;")
print("   the SEPARATION, which is what Urey measured, agrees to 4 decimals)\n")

print("  the book's mass ratios, checked:")
print(f"    proton   1836/1837 = {1836/1837:.7f}   computed {halpha(1.007276466*amu)[1]:.7f}")
print(f"    deuteron 3671/3672 = {3671/3672:.7f}   computed {halpha(2.013553213*amu)[1]:.7f}\n")

print("  why it was detectable in 1934:")
print(f"    the lines differ by {lamH-lamD:.4f} nm out of {lamH:.1f} nm -- about 3 parts")
print("    in 10 000, which a diffraction grating resolves easily.  Natural")
print("    hydrogen is ~0.0156% deuterium, so Urey saw a faint companion beside")
print("    every Balmer line, roughly 1/6000 as bright.")

printsthe reduced-mass effect, Eq. (9.30): E_n' = (mu/m_e) x (-E_R/n^2) ======================================================================== nucleus m_N/m_e mu/m_e H-alpha (nm) hydrogen (proton) 1836.2 0.9994557 656.4696 deuterium (deuteron) 3670.5 0.9997276 656.2910 tritium (triton) 5496.9 0.9998181 656.2316 infinitely heavy - 1.0000000 656.1123 hydrogen H-alpha = 656.4696 nm book: 656.4686 nm deuterium H-alpha = 656.2910 nm book: 656.2899 nm separation = 0.1786 nm book: 0.1787 nm relative shift = 272 parts per million (the 1.5 ppm offset from the book is its rounded E_R = 13.6 eV; the SEPARATION, which is what Urey measured, agrees to 4 decimals) the book's mass ratios, checked: proton 1836/1837 = 0.9994556 computed 0.9994557 deuteron 3671/3672 = 0.9997277 computed 0.9997276 why it was detectable in 1934: the lines differ by 0.1786 nm out of 656.5 nm -- about 3 parts in 10 000, which a diffraction grating resolves easily. Natural hydrogen is ~0.0156% deuterium, so Urey saw a faint companion beside every Balmer line, roughly 1/6000 as bright.

system⇅	$\mu/m_e$⇅	effect on the spectrum⇅
hydrogen	$0.99946$	a 0.05% shift — invisible unless you have something to compare withⓘ
deuterium	$0.99973$ ⓘ	lines 0.18 nm blueward of hydrogen's — how Urey found it in 1934
positronium ( $e^{+} e^{-}$ )	$0.5$ ⓘ	every level at HALF the hydrogen energy; the whole spectrum shifts by a factor of 2
muonic hydrogen ( $μ^{-} p$ )ⓘ	$\approx 186$	levels 186× deeper and the atom 186× smaller — X-rays instead of visible light

Check yourself

0 / 6 answered

1.Why does an electric dipole transition always change the parity of the state?
In the diagram, switch off “obey Δl = ±1”.
2.What becomes visible, and what does it tell you about the 2s state?
3.The 2s state lives 0.14 s while 2p lives 1.6 ns. What does “forbidden” mean here?
4.Why does this page reuse chapter 1's hydrogen data module instead of recomputing the wavelengths?
5.Deuterium's spectral lines sit 0.18 nm from hydrogen's. Where does that come from?
6.Positronium is an electron bound to a positron. What does its reduced mass do to the spectrum?