Three interactions of very different size act on the same electrons, and which you treat first decides which quantum numbers survive to label the answer.

§11.1a left carbon as $(1s)^2(2s)^2(2p)^2$ — one configuration, one energy. That is not what a spectrometer sees. Carbon’s ground configuration contains fifteen states at five distinct energies, and this section recovers them.

Two corrections are responsible. The first is residual electron–electron repulsion : the part of the repulsion that no central potential could absorb. The second is spin–orbit coupling, the same interaction §9.6 applied to hydrogen. Before either can be discussed, the states need better labels than a configuration.

Conventions first, because the notation is dense

Four rules, and the book states them as bullets because there is no deriving them:

quantity⇅	one electron⇅	two or more⇅
orbital angular momentum	l, m_l	L, M_L
spin angular momentum	s, m_s	S, M_S
combined (orbital + spin)	j, m_j	J, M_J
spectroscopic letter	s, p, d, f for l = 0,1,2,3	S, P, D, F for L = 0,1,2,3

Any angular momentum has magnitude $\sqrt{j(j+1)}\hbar$ and $z$ component $m_j\hbar$ , with $j$ any non-negative multiple of $\tfrac12$ and $m_j$ running from $j$ down to $-j$ in integer steps. Two of them combine (Eq. 8.6) to

j = j_1 + j_2,\; j_1 + j_2 - 1,\; \ldots,\; |j_1 - j_2|

Capital S now has FOUR jobs, and the last one is new

ch10 warned about three. This page adds a fourth:

$S$ — the combined spin quantum number, 0 or 1 for two electrons.
$^{(S)}$ — a parenthesised superscript meaning symmetric under exchange.
$S = \sqrt{s(s+1)}\hbar$ — the magnitude of a spin angular momentum (§8.1).
$S$ as a spectroscopic letter, meaning $L = 0$ — new here.

So the term symbol $^1S$ means $L = 0$ and $S = 0$ , with the letter and the superscript describing different quantities. And $^3S$ means $L = 0$ with $S = 1$ — a symbol in which the same letter appears twice with two meanings and two values.

There is no fixing this; it is universal notation. The only defence is position: a capital letter in the middle of a term symbol is always $L$ , the superscript is always $2S+1$ , and the subscript is always $J$ .

Two ways to add up the angular momenta

An atom has several electrons, each with an $l$ and an $s$ . There are two sensible orders in which to combine them — $L$ – $S$ coupling and $j$ – $j$ coupling — and which is right depends on which interaction is stronger.

scheme⇅	how it combines⇅	appropriate when⇅
L–S (Russell–Saunders)	all the l into L, all the s into S, then L + S → J	residual repulsion ≫ spin–orbit — light atoms
j–j	each electron's own l + s → j, then all the j into J	spin–orbit ≫ residual repulsion — heavy atoms

Couple the strong interaction first

The rule behind the rule is worth stating, because it is not arbitrary.

Whichever interaction is stronger defines the good quantum numbers, so you combine along it first and treat the weaker one as a perturbation on the result. In a light atom the electrons repel each other far more than any one electron’s spin talks to its own orbit, so the orbital motions lock together into a definite $L$ and the spins into a definite $S$ . In a heavy atom spin–orbit coupling wins — it grows roughly as $Z^4$ , from §9.6 — and each electron’s own $l$ and $s$ lock to each other first.

This is the same reasoning that made $j$ rather than $m_l$ and $m_s$ the good quantum numbers in hydrogen’s fine structure. A perturbation that is off-diagonal in your basis is telling you to change basis.

Carbon’s fifteen states

Take $(1s)^2(2s)^2(2p)^2$ and ask what is actually available.

The closed subshells contribute nothing. The two 1s electrons have a symmetric spatial wave function (both in the same orbital), so Eq. (10.23) forces an antisymmetric spin state — the singlet, $S = 0$ . Their orbital angular momenta are both zero. Same for 2s. A closed subshell always has $L = S = 0$ and can be ignored entirely.

So everything rests on the two 2p electrons. Each has $l = 1$ , so Eq. (8.6) allows $L = 2, 1, 0$ ; each has $s = \tfrac12$ , so $S = 1$ or $0$ . That is six combinations — and only three survive.

The reason is ch10’s, applied to orbital angular momentum instead of spin. The book states the orbital exchange symmetry:

L = 2:\ \text{symmetric} \qquad L = 1:\ \text{antisymmetric} \qquad L = 0:\ \text{symmetric}

and we already know $S = 1$ is symmetric, $S = 0$ antisymmetric. The total must be antisymmetric, so the two factors must have opposite symmetry:

\text{(i) } S=1,\,L=1 \qquad \text{(ii) } S=0,\,L=2 \qquad \text{(iii) } S=0,\,L=0

which in term-symbol notation are $^3P$ , $^1D$ and $^1S$ .

Which terms a configuration has — counted, not quoted

Equivalent electrons — same n and same l, so the Pauli principle forbids the two electrons from sharing a spin-orbital. That single restriction is what removes terms. Carbon's ground configuration, and §11.1b's worked example. Equivalent electrons — Pauli bites.

M_L ＼ M_S	+1	0	-1
+2	·	1	·
+1	1	2	1
0	1	3	1
-1	1	2	1
-2	·	1	·

15 micro-states. Peeling removes one whole (2L+1)(2S+1) block at a time.

peel:

term	L	S	(2S+1)(2L+1)	J = \|L−S\| … L+S
1D	2	0	1 × 5 = 5	1D2
3P	1	1	3 × 3 = 9	3P0, 3P1, 3P2
1S	0	0	1 × 1 = 1	1S0
total			15	matches the micro-state count ✓

Why the missing terms are missing. With S = 1 the spins are parallel, so the two electrons must differ in m_l — look at the M_S = +1 column, which stops short of the top row. The largest M_L available with parallel spins is 1, not 2, so no triplet with L = 2 can exist. Nothing was forbidden by hand; the table simply has no cell to build it from.

What the peeling is really doing

The table is every way of placing two electrons in the six 2p spin-orbitals, sorted by $M_L$ and $M_S$ . Pauli enters exactly once, as the requirement that the two spin-orbitals differ — which is why the $M_S = \pm1$ columns are shorter than the middle one.

Then: the highest occupied cell must belong to some term, and a term with quantum numbers $L$ and $S$ occupies a full $(2L+1)\times(2S+1)$ block. So take the top cell, remove its whole block, and repeat. The terms come out in descending order and nothing is left over.

Switch to (2p)(3p) and watch what changes. Same two $l$ values, but the electrons now differ in $n$ , so Pauli never engages, the table is the full outer product, and there are six terms instead of three. The entire difference between the two configurations is one line of bookkeeping.

Reproduce it in NumPy

The peeling algorithm is the same one the widget above runs, so the picture and the numbers cannot drift apart. Pauli enters in exactly one place — combinations() for equivalent electrons versus product() for inequivalent — and every difference between (2p)² and (2p)(3p) follows from that single line. The Fig. 11.3 comparison uses NIST's tabulated carbon levels, so it checks the book against measurement rather than against theory.

from itertools import combinations, product
from collections import Counter

LET = 'SPDFGH'

def spin_orbitals(l):
    """every (m_l, 2*m_s) slot in a subshell -- 2(2l+1) of them"""
    return [(ml, ms2) for ml in range(l, -l-1, -1) for ms2 in (1, -1)]

def micro(l1, l2, equivalent):
    """The (M_L, M_S) census. For EQUIVALENT electrons Pauli forces the two
       spin-orbitals to differ and the pair is unordered -- that one line is
       the entire difference between (2p)^2 and (2p)(3p)."""
    if equivalent:
        pairs = combinations(spin_orbitals(l1), 2)
    else:
        pairs = product(spin_orbitals(l1), spin_orbitals(l2))
    return Counter((a[0]+b[0], a[1]+b[1]) for a, b in pairs)

def peel(box):
    """Take the highest occupied (M_L, M_S); it must belong to a term with
       L = M_L and S = M_S. Remove that term's whole block. Repeat."""
    box, out = Counter(box), []
    while box:
        ML = max(m for m, _ in box)
        S2 = max(s for m, s in box if m == ML)
        out.append((ML, S2))
        for ml in range(-ML, ML+1):
            for ms2 in range(-S2, S2+1, 2):
                box[(ml, ms2)] -= 1
                if box[(ml, ms2)] == 0: del box[(ml, ms2)]
    return out

def report(name, l1, l2, equivalent):
    box = micro(l1, l2, equivalent)
    ts  = peel(box)
    n   = sum(box.values())
    tot = sum((s+1)*(2*l+1) for l, s in ts)
    names = ', '.join(f"{s+1}{LET[l]}" for l, s in ts)
    print(f"  {name:<11}{'equivalent' if equivalent else 'different n':<13}"
          f"{n:>7}{tot:>8}   {names}")
    return ts, n

print("Which terms exist, counted from the micro-states rather than quoted")
print("=" * 78)
print(f"  {'config':<11}{'electrons':<13}{'micro':>7}{'sum g':>8}   terms")
ts, n = report("(2p)^2", 1, 1, True)
report("(2p)(3p)", 1, 1, False)
print("""
  Same two l values, and six terms instead of three. The ONLY difference is
  that (2p)^2's electrons share a subshell, so Pauli forbids them the same
  spin-orbital. Everything section 11.1b says about carbon follows from that.""")

print("\nCarbon's (2p)^2 in detail -- the book says 3P, 1D, 1S and 15 states")
print("=" * 78)
print(f"  {'term':<6}{'L':>3}{'S':>4}{'(2S+1)(2L+1)':>15}{'J values':>22}")
tot = 0
for L, S2 in ts:
    S = S2/2
    g = (S2+1)*(2*L+1); tot += g
    Js = [abs(L-S)+i for i in range(int(2*min(L,S))+1)]
    lab = ', '.join(f"{S2+1}{LET[L]}_{j:g}" for j in Js)
    print(f"  {str(S2+1)+LET[L]:<6}{L:>3}{S:>4g}{str(S2+1)+' x '+str(2*L+1)+' = '+str(g):>15}{lab:>22}")
print(f"  {'total':<6}{'':>3}{'':>4}{tot:>15}   matches the {n} micro-states: {tot==n}")

print("""
  Only 3P splits under the spin-orbit interaction, because only 3P has more
  than one J. That is exactly what Fig. 11.3's third column shows.

  Why no 3D? With S = 1 the spins are parallel, so the two electrons must sit in
  different m_l orbitals. The largest M_L you can then build is 1 + 0 = 1, not
  1 + 1 = 2 -- so L = 2 simply has no parallel-spin state to be built from.
  Nothing is forbidden by decree; the table has no cell for it. Problem 7 asks
  the same question about 3S, and the answer is read off the same table.""")

print("\nFig. 11.3's splittings against the measured carbon spectrum (NIST, cm^-1)")
print("=" * 78)
lv = {'3P0':0.0, '3P1':16.417, '3P2':43.413, '1D2':10192.657, '1S0':21648.030}
eV = 1/8065.544
print(f"  {'level':<7}{'cm^-1':>12}{'eV above 3P0':>15}")
for k, v in lv.items():
    print(f"  {k:<7}{v:>12.3f}{v*eV:>15.4f}")
print(f"""
  3P1 - 3P0 = {(lv['3P1']-lv['3P0'])*eV*1000:>5.2f} meV    the book's figure says 2.0 meV
  3P2 - 3P1 = {(lv['3P2']-lv['3P1'])*eV*1000:>5.2f} meV    the book's figure says 3.4 meV
  1D2 - 3P  = {(lv['1D2']-lv['3P0'])*eV:>5.2f} eV     the book's figure says 1.3 eV
  1S0 - 1D2 = {(lv['1S0']-lv['1D2'])*eV:>5.2f} eV     the book's figure says 1.4 eV

  Every number in Fig. 11.3 is right, and the three scales are worth noticing:
  the configuration is fixed to within nothing, residual repulsion splits it by
  ELECTRONVOLTS, and spin-orbit splits it again by MILLI-electronvolts. Three
  levels of approximation, three orders of magnitude apart.""")

print("\nProblems 5, 6 and 7 come off the same routine")
print("=" * 78)
print(f"  {'config':<11}{'electrons':<13}{'micro':>7}{'sum g':>8}   terms")
for name, l1, l2, eq in (("(1s)^2",0,0,True), ("(1s)(2s)",0,0,False), ("(1s)(2p)",0,1,False),
                         ("(1s)(3d)",0,2,False), ("(2p)(3s)",1,0,False), ("(2p)(3p)",1,1,False)):
    report(name, l1, l2, eq)
print("""
  Problem 7 in one line: (2p)(3p) has a 3S and (2p)^2 does not, because only in
  the equivalent case must the two electrons differ in their spin-orbital.""")

printsWhich terms exist, counted from the micro-states rather than quoted ============================================================================== config electrons micro sum g terms (2p)^2 equivalent 15 15 1D, 3P, 1S (2p)(3p) different n 36 36 3D, 1D, 3P, 1P, 3S, 1S Same two l values, and six terms instead of three. The ONLY difference is that (2p)^2's electrons share a subshell, so Pauli forbids them the same spin-orbital. Everything section 11.1b says about carbon follows from that. Carbon's (2p)^2 in detail -- the book says 3P, 1D, 1S and 15 states ============================================================================== term L S (2S+1)(2L+1) J values 1D 2 0 1 x 5 = 5 1D_2 3P 1 1 3 x 3 = 9 3P_0, 3P_1, 3P_2 1S 0 0 1 x 1 = 1 1S_0 total 15 matches the 15 micro-states: True Only 3P splits under the spin-orbit interaction, because only 3P has more than one J. That is exactly what Fig. 11.3's third column shows. Why no 3D? With S = 1 the spins are parallel, so the two electrons must sit in different m_l orbitals. The largest M_L you can then build is 1 + 0 = 1, not 1 + 1 = 2 -- so L = 2 simply has no parallel-spin state to be built from. Nothing is forbidden by decree; the table has no cell for it. Problem 7 asks the same question about 3S, and the answer is read off the same table. Fig. 11.3's splittings against the measured carbon spectrum (NIST, cm^-1) ============================================================================== level cm^-1 eV above 3P0 3P0 0.000 0.0000 3P1 16.417 0.0020 3P2 43.413 0.0054 1D2 10192.657 1.2637 1S0 21648.030 2.6840 3P1 - 3P0 = 2.04 meV the book's figure says 2.0 meV 3P2 - 3P1 = 3.35 meV the book's figure says 3.4 meV 1D2 - 3P = 1.26 eV the book's figure says 1.3 eV 1S0 - 1D2 = 1.42 eV the book's figure says 1.4 eV Every number in Fig. 11.3 is right, and the three scales are worth noticing: the configuration is fixed to within nothing, residual repulsion splits it by ELECTRONVOLTS, and spin-orbit splits it again by MILLI-electronvolts. Three levels of approximation, three orders of magnitude apart. Problems 5, 6 and 7 come off the same routine ============================================================================== config electrons micro sum g terms (1s)^2 equivalent 1 1 1S (1s)(2s) different n 4 4 3S, 1S (1s)(2p) different n 12 12 3P, 1P (1s)(3d) different n 20 20 3D, 1D (2p)(3s) different n 12 12 3P, 1P (2p)(3p) different n 36 36 3D, 1D, 3P, 1P, 3S, 1S Problem 7 in one line: (2p)(3p) has a 3S and (2p)^2 does not, because only in the equivalent case must the two electrons differ in their spin-orbital.

Three approximations, three orders of magnitude

Fig. 11.3 is the chapter’s best picture: the same configuration described three times, each more precisely than the last.

Click a gold level for its energy and degeneracy.

No transitions shown in this view.

The three scales, and why the figure cannot be drawn to scale

Configuration → one level. $(1s)^2(2s)^2(2p)^2$ .
Residual repulsion → three levels, split by electronvolts. $^1D$ lies 1.26 eV above $^3P$ ; $^1S$ lies a further 1.42 eV above $^1D$ .
Spin–orbit → five levels, split by milli-electronvolts. $^3P_1$ is 2.04 meV above $^3P_0$ ; $^3P_2$ a further 3.35 meV.

The ratio between the second and third stages is about 600. The book’s own caption admits the $^3P$ splittings “have not been drawn to scale”, and any honest rendering must do the same — at true scale the three $^3P$ levels are one line.

Every one of these numbers matches the measured carbon spectrum, which the calculation above checks against NIST’s tabulated levels. Unlike §9.6’s fine-structure coefficients, Fig. 11.3 is exactly right.

Selection rules

Most of what is known about atomic levels comes from the light emitted between them, and not every transition is allowed. §9.4 gave $\Delta l = \pm1$ for one electron; for a many-electron atom the rules are richer.

Always, for electric dipole radiation:

\Delta J = 0,\ \pm1, \qquad\text{but}\qquad J_i = 0 \to J_f = 0 \ \text{is strictly forbidden}\tag{11.7}

And when $L$ – $S$ coupling describes the states well:

\Delta L = 0,\ \pm1 \qquad\text{and}\qquad \Delta S = 0\tag{11.8}

plus the parity rule: the parity must change. For a configuration $(n_1l_1)(n_2l_2)\ldots$ the parity is even if $l_1 + l_2 + \ldots$ is even and odd if that sum is odd.

ΔS = 0 is why the ³P levels are metastable, and why 'forbidden' does not mean impossible

$\Delta S = 0$ says a photon cannot flip a spin — the electric field couples to charge, not to spin. So a $^1D \to {}^3P$ transition is forbidden, and carbon’s $^1D_2$ state cannot easily decay to the ground state even though 1.26 eV lies below it.

But $\Delta S = 0$ holds only when $L$ – $S$ coupling is accurate, and it never is exactly. A little spin–orbit mixing gives every state a small admixture of other $S$ , so the transition proceeds — very slowly. These are forbidden lines, and they are not rare: they dominate the spectra of nebulae, where densities are so low that a metastable atom can wait millions of seconds without a collision. Astronomers write them in brackets, like [O III].

$J_i = 0 \to J_f = 0$ is different. That one is strictly forbidden — a photon carries one unit of angular momentum, and there is no way to hand it to a system that has none and ends with none.

Where this is going

Carbon’s ground state is $^3P_0$ : of the fifteen states, the one with the largest $S$ , then the largest $L$ , then the smallest $J$ for a less-than-half-filled subshell. That ordering rule is Hund’s, and its origin is ch10’s exchange integral — parallel spins keep electrons apart, which lowers the Coulomb energy.

§11.2 steps back down to configurations, where the periodic table lives, and none of this notation is needed again until you read a real spectrum.

Check yourself

0 / 6 answered

Two 2p electrons could in principle have $L = 2, 1, 0$ combined with $S = 1, 0$ — six combinations.
1.Why do only three of them exist?
2.In the widget's micro-state table, why are the $M_{S} = \pm 1$ columns shorter than the $M_{S} = 0$ column?
3.Fig. 11.3 splits the configuration by electronvolts and then by milli-electronvolts. Which interaction does which?
4.Why does spin–orbit coupling split $^{3} P$ into three levels but leave $^{1} D$ and $^{1} S$ alone?
5.Configuration $(2 p)^{2}$ gives three terms and $(2 p) (3 p)$ gives six. What accounts for the difference?
6.The selection rule $Δ S = 0$ makes many transitions "forbidden". Does that mean they never happen?