Good Quantum Numbers and Symmetry-Adapted Bases

Degenerate Perturbation Theory ended with a promissory note: build the $g\times g$ matrix $W$ and diagonalize — unless a symmetry does it for you. This article pays the note in full. The claim is startlingly strong: find one hermitian operator $\hat{A}$ that commutes with both $\hat{H}_0$ and $\hat{H}'$ and whose eigenvalues tell the degenerate states apart, and the entire secular apparatus deactivates — the $\hat{A}$ -eigenbasis is the good basis, the first-order shifts are bare diagonal elements, and the non-degenerate formulas apply verbatim inside a degenerate subspace. No determinant, no mixing angle: the symmetry knew the answer before you asked. The label of such an operator is a good quantum number, a phrase carrying two meanings this article keeps separated and then reconnects. The dynamical meaning — $[\hat{A},\hat{H}]=0$ makes the label a constant of the motion — belongs to Constants of the Motion and Conservation Laws and is recalled, not re-proved. The perturbative meaning is this chapter's: a label that survives the perturbation, staying attached to a definite zeroth-order state instead of leaking across the block. The bridge between the meanings is a two-line identity with an outsized career: $\langle a'|[\hat{A},\hat{H}']|a\rangle=(\alpha_{a'}-\alpha_a)W_{a'a}$ . Everything is prosecuted on live examples — the previous serial's $\lambda xy$ laboratory re-solved by symmetry in one line, a no-symmetry counterexample where the shortcut honestly fails, an oscillator pair checked against an exact normal-mode solution — and the chapter's payoff, which operators bless the fine-structure and Zeeman calculations, is set up as a decision table with every energy formula left to its owner serial. Each commutator and number below is machine-verified — including one sign the base article printed backwards.

Key takeaways

The theorem: if $[\hat{A},\hat{H}_0]=[\hat{A},\hat{H}']=0$ and the degenerate states carry distinct $\hat{A}$ -eigenvalues, $W$ is diagonal in the $\hat{A}$ -basis: those states are the good states, $E^{(1)}=W_{aa}$ , non-degenerate machinery legal. Proof in one line: $0=\langle a'|[\hat{A},\hat{H}']|a\rangle=(\alpha_{a'}-\alpha_a)W_{a'a}$ .
"Good" has a dynamics sense ( $[\hat{A},\hat{H}]=0$ : label frozen in time — Constants of the Motion and Conservation Laws) and a perturbation sense (label survives $\hat{H}'$ , no leaking within the block). Same commutator, two readings.
Decision table: operator exists and distinguishes → shifts off the diagonal; none exists → diagonalize $W$ (Degenerate Perturbation Theory); commutes with $\hat{H}_0$ but not $\hat{H}'$ → label bad, change basis.
Live demo: exchange $x\leftrightarrow y$ picks $(\psi_{12}\pm\psi_{21})/\sqrt2$ with $P=\pm1$ — the previous serial's diagonalization answer recovered without diagonalizing (its numbers quoted, verified equal).
Honest failure: $\hat{H}'=\lambda x^2y$ commutes with nothing useful — diagonals 14.13/16.03 meV, coupling 3.24 meV, θ = 53.2°: the full $W$ machinery genuinely needed (verified vs exact diagonalization).
Fine-structure credentials only: $\hat{L}^2,\hat{S}^2,\hat{J}^2,\hat{J}_z$ commute with $\hat{L}\!\cdot\!\hat{S}$ ; $\hat{L}_z,\hat{S}_z$ do not — $|n\,\ell\,j\,m_j\rangle$ is the good frame (energies owned by Spin-Orbit Coupling and Fine Structure of Hydrogen).
Key idea: a good quantum number is a label the perturbation cannot change — and every operator that commutes with both $\hat{H}_0$ and $\hat{H}'$ pre-diagonalizes $W$ , converting a degenerate problem back into a non-degenerate one.

The core insight

Find a hermitian $\hat{A}$ with $[\hat{A},\hat{H}_0]=[\hat{A},\hat{H}']=0$ whose eigenvalues distinguish the degenerate states: its eigenbasis is automatically the good basis — $W$ arrives diagonal, the first-order shifts are plain diagonal elements, and the degenerate problem dissolves back into the non-degenerate one.

Where this lives today

The theorem is production infrastructure wherever big Hamiltonians meet symmetry. Quantum-chemistry and nuclear-structure codes block-diagonalize by point-group or angular-momentum labels before any numerics — "symmetry-adapted basis" is a literal input flag, and the saving is $g^3$ versus $\sum_i g_i^3$ across blocks. The topological-materials industry runs on exhaustive tables of exactly this article's question — which labels stay good under which distortions and fields. And quantum computing engineers with the contrapositive: decoherence-free subspaces and symmetry-protected qubits are degenerate blocks whose labels the noise operators cannot change, because they commute with the wrong things — a $W$ made to vanish by design.

The physical picture

A degenerate level is a flat plateau: $\hat{H}_0$ charges every direction the same energy and cannot name its own basis. Degenerate Perturbation Theory handled this by brute honesty — ask the tilt, diagonalize $W$ . This article's observation is that the question often has a chaperone: a symmetry shared by the bowl and the tilt. If exchanging $x\leftrightarrow y$ leaves both the square well and $\lambda xy$ unchanged, no amount of perturbing can mix an exchange-even state with an exchange-odd one — they live in rooms the dynamics cannot connect. Labels of such shared symmetries are good: they survive.

Figure

Two frames for one subspace: the shared symmetry pre-sorts what the perturbation may touch

The deepest version of the sorting is group-theoretic — states organized by irreducible representations of the perturbation's symmetry group cannot be mixed by it (Schur's lemma wearing lab clothes) — and the mechanism is one physicists already trust in a different costume: a selection rule. "Which matrix elements vanish by symmetry" is precisely the question answered by parity rules in spectroscopy, and the portrait band tells how that idea, born in atomic line catalogues, became the working grammar here. A good quantum number is a selection rule running inside a degenerate block: the forbidden elements are exactly the off-diagonals of $W$ . One warning shapes the whole subject: goodness is contextual — the label must commute with the perturbation actually acting, not just with $\hat{H}_0$ . Hydrogen's $m_\ell$ is perfectly conserved by the bare Coulomb problem and bad for spin-orbit coupling, because $\hat{L}_z$ fails to commute with $\hat{L}\!\cdot\!\hat{S}$ (matrix-verified below, with the base's printed sign corrected). Good is decided perturbation by perturbation — the decision table below is this article's real deliverable.

The theorem

First pin the two meanings of "good." The dynamical sense: if $\hat{A}$ commutes with the full Hamiltonian, its expectation and its entire distribution are frozen —

i\hbar\,\frac{\mathrm{d}}{\mathrm{d}t}\langle\hat{A}\rangle=\langle[\hat{A},\hat{H}]\rangle=0 \qquad (1)

where: this is the conservation statement of Constants of the Motion and Conservation Laws — recalled, not re-proved. A label with this property survives time. This chapter needs the label to survive the perturbation: to stay attached to a definite zeroth-order state instead of leaking across the degenerate block. Both senses meet in the commutator; the perturbative one is settled by a single identity.

Derivation AThe good-states theorem

We prove the shortcut that retires the secular equation: if a hermitian $\hat{A}$ commutes with both $\hat{H}_0$ and $\hat{H}'$ , and the members of a degenerate block carry distinct $\hat{A}$ -eigenvalues, then the perturbation matrix $W$ is diagonal in the $\hat{A}$ -adapted frame — so those states are already the good states and the first-order shifts are bare diagonal elements. The engine is the invariant-eigenspace mechanism of The Commutator and Compatible Observables, run inside one degenerate block; the punchline is that the theorem needs three hypotheses, each of which fails somewhere in this chapter, so each earns a checkpoint.

Step 1. Set the frame. $[\hat{A},\hat{H}_0]=0$ lets us choose the degenerate basis $\{|n_a^{(0)}\rangle\}$ to be simultaneous eigenstates: $\hat{A}|n_a^{(0)}\rangle=\alpha_a|n_a^{(0)}\rangle$ (compatibility theorem, consumed). This is hypothesis one: without it, "the $\hat{A}$ -basis" need not consist of $\hat{H}_0$ -eigenstates at all, and $W$ is not even defined on it.

Step 2. Sandwich the second commutator. For any two members of the block,

0=\langle n_{a'}^{(0)}|[\hat{A},\hat{H}']|n_a^{(0)}\rangle=(\alpha_{a'}-\alpha_a)\,\langle n_{a'}^{(0)}|\hat{H}'|n_a^{(0)}\rangle=(\alpha_{a'}-\alpha_a)\,W_{a'a}

the left equality is hypothesis two, $[\hat{A},\hat{H}']=0$ ; the middle one lets $\hat{A}$ act left on the bra and right on the ket.

Step 3. Invoke hypothesis three — distinctness. Whenever $\alpha_{a'}\neq\alpha_a$ , the prefactor is nonzero, so $W_{a'a}=0$ : every off-diagonal element between different labels is forced to vanish. If all $g$ labels differ pairwise, $W$ is fully diagonal; the $\hat{A}$ -eigenstates are eigenvectors of $W$ , i.e. the good states of the previous serial, and $E^{(1)}_a=W_{aa}=\langle n_a^{(0)}|\hat{H}'|n_a^{(0)}\rangle$ — the non-degenerate formula, now legal inside the block. Why distinctness is not decorative: if two states share an $\hat{A}$ -eigenvalue, Step 3 says nothing about their mutual element, and $W$ is only block-diagonal — smaller secular problems inside each repeated-label block. That is still progress (the fine-structure calculations live on exactly this partial sorting), but the residual blocks must be diagonalized honestly, or a second commuting operator found to split the tie (the CSCO instinct of Simultaneous Diagonalization and Complete Sets of Commuting Observables, one level down).

Step 4. Close the loop on the divergence. In this frame the dangerous first-order mixing terms of the naive expansion have numerator $W_{a'a}=0$ wherever the denominator vanishes — the same cancellation Degenerate Perturbation Theory engineered by diagonalization now holds by symmetry, to all such pairs at once. The label $\alpha$ rides through the perturbation unchanged: that is, operationally, what "good quantum number" means. Verified: the identity and the forced zeros are checked symbolically and on random commuting pairs (off-diagonals $<10^{-12}$ ); the generic non-commuting contrast leaves finite couplings; hypothesis-failure cases exhibited in the worked examples (article verifier, 35 checks).

Go deeper · Derivation B · 1 minGoodness is exact, not perturbative

Derivation BGoodness is exact, not perturbative

The theorem above spoke first-order language — $W$ diagonal, shifts free. We now prove the stronger statement hiding underneath: when $\hat{A}$ commutes with both pieces, its label survives to all orders — indeed exactly. The strategy is to forget the expansion entirely, show the full Hamiltonian preserves each $\hat{A}$ -eigenspace, and then read the perturbative corollary off that exact fact. This is what elevates "good quantum number" from a first-order bookkeeping trick to a structural property of the problem.

Step 1. $[\hat{A},\hat{H}_0]=[\hat{A},\hat{H}']=0$ implies $[\hat{A},\hat{H}(\lambda)]=0$ for the full $\hat{H}(\lambda)=\hat{H}_0+\lambda\hat{H}'$ — at every coupling strength, since the commutator is linear in $\hat{H}$ . The symmetry is not approximately shared; it is shared, period.

Step 2. Apply the invariant-eigenspace lemma of The Commutator and Compatible Observables to the exact pair: $\hat{H}(\lambda)$ maps each $\hat{A}$ -eigenspace into itself, so the exact problem splits into independent blocks, one per label $\alpha$ , at every $\lambda$ . Exact eigenstates can be chosen as $\hat{A}$ -eigenstates throughout — and levels of different labels may cross freely as $\lambda$ grows: the loophole in the no-crossing rule.

Step 3. Extract the perturbative corollary. Every order- $k$ correction $|n^{(k)}\rangle$ is built by applying $\hat{H}'$ and resolvent factors to $|n^{(0)}\rangle$ — all of which commute with $\hat{A}$ or act within blocks — so the entire tower stays inside the starting $\alpha$ -room. The label never leaks at any order: first-order goodness was the visible tip of an exact conservation. Why this matters practically: it licenses computing ANY downstream quantity — matrix elements, higher-order shifts, responses to further perturbations — entirely within one symmetry block, the workhorse economy of every symmetry-adapted computation in the applications below.

Step 4. Certify on the laboratory. For $\hat{H}_0+\lambda\,xy$ the full commutator $[\hat{P},\hat{H}(\lambda)]$ vanishes identically, and the exact eigenvectors of the 196-state matrix carry pure $P=\pm1$ to machine precision — not approximately, not to first order: exactly. Verified: $[\hat{P},\hat{H}_{\text{full}}]=0$ and exact-eigenvector parity purity $<10^{-10}$ on the 196-state laboratory (article verifier).

The theorem's contrapositive is the practical decision engine:

[\hat{A},\hat{H}_0]=0\ \ \text{but}\ \ [\hat{A},\hat{H}']\neq0\quad\Longrightarrow\quad \alpha\ \text{is a bad label for this perturbation} \qquad (2)

where: "bad" is precise: $\hat{H}'$ has matrix elements between different $\alpha$ -rooms, so amplitude leaks, the label does not survive, and shifts computed as diagonal elements in that basis are wrong. Note what is not claimed: badness of one candidate says nothing about the problem being hopeless — it says look for a different operator. The full search protocol:

situation	verdict	route
$\hat{A}$ commutes with both; labels distinct in block	$W$ diagonal — shifts free	this article's theorem
$\hat{A}$ commutes with both; labels partially repeat	$W$ block-diagonal	theorem + small secular blocks (or find a second $\hat{A}$ )
every candidate fails $[\hat{A},\hat{H}']=0$	no shortcut exists	full $W$ diagonalization — Degenerate Perturbation Theory
$\hat{A}$ commutes with $\hat{H}'$ but not $\hat{H}_0$	frame isn't made of $\hat{H}_0$ -eigenstates	useless as-is; symmetrize differently
two perturbations compete	goodness depends on which dominates	regime taxonomy — Zeeman Effect's decision table

The laboratory, re-solved by symmetry

The previous serial split the 2D square well pair $\psi_{12},\psi_{21}$ under $\hat{H}'=\lambda xy$ by diagonalizing $W$ . Watch the theorem do it without touching a determinant. The exchange operator $\hat{P}:(x,y)\mapsto(y,x)$ satisfies all three hypotheses — it commutes with $\hat{H}_0$ and with $xy$ — and the combinations

\frac{\psi_{12}\pm\psi_{21}}{\sqrt2}\quad\text{carry}\quad P=\pm1 \qquad (3)

where: the labels are distinct — so equation (3) states the good basis by inspection and the shifts are the two diagonal elements. They evaluate to exactly the eigenvalues the previous serial obtained the hard way — $\lambda(L^2/4\pm256L^2/81\pi^4)$ , i.e. $28.24$ and $21.76$ meV at its lab settings — quoted from Degenerate Perturbation Theory, never recomputed; the machine check confirms the $\hat{P}$ -frame off-diagonal is zero to $10^{-14}$ . One line of symmetry replaced the whole secular ritual.

Figure

The exchange shortcut on the λxy laboratory: symmetry labels replace the secular equation

The same pair under $\hat{H}'=\lambda(x^2+y^2)$ shows the theorem's other face: exchange still commutes with everything, but the diagonal elements come out equal ( $\langle x^2\rangle_1+\langle x^2\rangle_2$ both times, verified) and the off-diagonal vanishes by orthogonality. $W\propto\mathbb{1}$ : sorted, unsplit. "Good basis found" and "degeneracy lifted" are independent verdicts — a distinction the fine-structure story leans on.

Limiting cases

case	result	meaning
$\hat{A}$ exists, labels all distinct	$W$ diagonal: $E^{(1)}_a=W_{aa}$	secular equation retired; non-degenerate formulas legal
labels repeat within block	$W$ block-diagonal	smaller secular problems; hunt a second operator
$[\hat{A},\hat{H}']\neq0$	$\alpha$ bad for this perturbation	change basis or diagonalize $W$ honestly
$[\hat{A},\hat{H}]=0$ exactly	$\alpha$ good to all orders, frozen in time	dynamics sense and PT sense coincide
$W\propto\mathbb{1}$ in the good frame	sorted but unsplit	symmetry too strong to lift at first order
competing perturbations	goodness regime-dependent	dominant term picks the frame; intermediate regime has no good frame

Common trapA label conserved by H₀ is not thereby good for the perturbation

The classic trap in this chapter: hydrogen's $m_\ell$ commutes beautifully with the Coulomb Hamiltonian, so students treat $|n\,\ell\,m_\ell\,m_s\rangle$ as sacred and evaluate spin-orbit shifts as its diagonal elements. But goodness is a property of the label and the perturbation jointly: $\hat{L}_z$ does not commute with $\hat{L}\!\cdot\!\hat{S}$ (the ladder part $\hat{L}_x\hat{S}_x+\hat{L}_y\hat{S}_y$ trades $m_\ell$ against $m_s$ ), so the uncoupled frame leaves $W$ riddled with off-diagonals and its diagonal elements are not shifts of anything. Always test the commutator against $\hat{H}'$ — the operator actually doing the perturbing — before trusting any label. When two perturbations compete, test against each: the answer may differ, and that difference is physics (the Zeeman regimes), not bookkeeping.

Worked examples

Example 1 — three perturbations, one pair: the full decision tree on live numbers. The 2D well pair $\psi_{12},\psi_{21}$ ( $L=1$ nm, level $5E_1=1.8801$ eV) under three different tilts.

Step 1. $\hat{H}'=\lambda xy$ , $\lambda=0.1$ eV/nm². Exchange commutes with both operators, labels $P=\pm1$ distinct: theorem applies. Good states $(\psi_{12}\pm\psi_{21})/\sqrt2$ , shifts = diagonal elements = $28.24$ and $21.76$ meV — Degenerate Perturbation Theory's answer recovered with zero algebra (its lab, its numbers, quoted).

Step 2. $\hat{H}'=\lambda(x^2+y^2)$ . Same symmetry, same good frame — but $W_{++}=W_{--}$ (each equals $\lambda[\langle x^2\rangle_1+\langle x^2\rangle_2]$ , computed symbolically) and $W_{+-}=0$ . Verdict: sorted, unsplit. The pair rides up rigidly; its internal ambiguity waits for a less even-handed perturbation.

Step 3. $\hat{H}'=\lambda x^2y$ , $\lambda=0.1$ eV/nm³. Now $\hat{P}\hat{H}'\hat{P}=\lambda y^2x\neq\hat{H}'$ — hypothesis two fails and the full machinery is compulsory: $W_{aa}=14.13$ meV, $W_{bb}=16.03$ meV, $W_{ab}=\lambda\langle1|x^2|2\rangle\langle2|y|1\rangle=+3.24$ meV (with $\langle1|x^2|2\rangle=-16L^2/9\pi^2$ exactly — two negatives cancel). Diagonalizing: shifts $11.70$ and $18.46$ meV, splitting $6.76$ meV, $\theta=53.2°$ — neither $0°$ (frame not good) nor $45°$ (diagonals unequal); the 196-state exact diagonalization confirms within 3%. Answer: symmetry route where symmetry exists, secular route where it does not — one subspace demonstrating both, plus the sorted-but-unsplit middle case.

Example 2 — the 2D oscillator pair, checked against an exact solution. The isotropic 2D oscillator's $N=1$ level is the pair $|10\rangle,|01\rangle$ at $2\hbar\omega$ ; perturb with $\hat{H}'=\lambda xy$ .

Step 1. Symmetry route. Exchange commutes with $\hat{H}_0$ and $xy$ ; the combinations $(|10\rangle\pm|01\rangle)/\sqrt2$ carry $P=\pm1$ . Theorem: they are the good states, and the shifts are diagonal elements. Using ladder-operator matrix elements (Ladder Operators and the Algebraic Oscillator, consumed): $\langle0|x|1\rangle=\sqrt{\hbar/2m\omega}$ , so $E^{(1)}_\pm=\pm\lambda\langle1|x|0\rangle\langle0|y|1\rangle=\pm\lambda\hbar/2m\omega$ . Splitting $\lambda\hbar/m\omega$ .

Step 2. Exact check — this problem secretly solves in closed form. Rotate to normal modes $u=(x+y)/\sqrt2$ , $v=(x-y)/\sqrt2$ : the coupled Hamiltonian becomes two independent oscillators with $\omega_\pm=\omega\sqrt{1\pm\lambda/m\omega^2}$ . The $N=1$ pair sits at $E_{00}+\hbar\omega_\pm$ , so the exact splitting is $\hbar(\omega_+-\omega_-)=\lambda\hbar/m\omega+O(\lambda^3)$ — the sympy series confirms first order lands exactly on the perturbative answer, and the good states are exactly the $u$ / $v$ excitations, i.e. the $P=\pm1$ combinations. Answer: $E^{(1)}=\pm\lambda\hbar/2m\omega$ , certified twice — once by the theorem, once by an exact diagonalization the theorem never needed to know existed.

Which labels bless the chapter ahead — credentials only

The chapter's application run is one long exercise in this article's theorem, and the credentials — which commutators vanish — can be laid out now, with every energy left to its owner. For the spin-orbit perturbation $\propto\hat{L}\!\cdot\!\hat{S}$ , matrix verification (in the $\ell=1\otimes s=\tfrac12$ representation) gives

[\hat{L}_z,\hat{L}\!\cdot\!\hat{S}]=i\hbar\,(\hat{L}_y\hat{S}_x-\hat{L}_x\hat{S}_y)\neq0,\qquad [\hat{J}_z,\hat{J}^2,\hat{L}^2,\hat{S}^2\,;\ \hat{L}\!\cdot\!\hat{S}]=0 \qquad (4)

where: the first commutator convicts $m_\ell$ (and symmetrically $m_s$ ) as bad labels — the ladder part of $\hat{L}\!\cdot\!\hat{S}$ trades one against the other — while the four vanishing commutators certify the coupled labels $(\ell,s,j,m_j)$ as good: $\hat{L}\!\cdot\!\hat{S}$ is a scalar under joint rotations. (The base article printed the nonzero commutator with its sign transposed, $i\hbar(\hat{L}_x\hat{S}_y-\hat{L}_y\hat{S}_x)$ ; the machine check fixed it — the doctrine is unchanged, the sign matters if you ever chain such identities.) The evaluation of $\langle\hat{L}\!\cdot\!\hat{S}\rangle$ and everything it splits belongs to Spin-Orbit Coupling; the assembled $j$ -only spectrum to Fine Structure of Hydrogen. For a magnetic field along $z$ , the perturbation carries $\hat{L}_z+2\hat{S}_z$ : it commutes with $\hat{L}_z$ and $\hat{S}_z$ (uncoupled labels good) but not with $\hat{J}^2$ (verified) — when the field dominates, $j$ itself dies. Weak field: $(j,m_j)$ good — Weak-Field (Anomalous) Zeeman Effect. Strong field: $(m_\ell,m_s)$ good — Strong-Field Paschen-Back Effect. In between, no frame is good — the regime map with numbers is Zeeman Effect's opening move. The Stark perturbation $e\mathcal{E}z$ keeps $m_\ell$ good while butchering parity — the block structure Linear Stark Effect in Hydrogen exploits. Each is the same two-line theorem pointed at a different operator.

Origins · 1 minSelection Rules and how this was discovered

Selection Rules

1924–1931 · Ann Arbor, Berlin, Princeton

The idea that symmetry forbids matrix elements — the engine of this article's theorem — entered physics through a spectroscopist's ledger, not a theorem. In 1924 Otto Laporte, a 22-year-old German cataloguing the ferociously complex iron arc spectrum, noticed that its thousands of levels split cleanly into two families, and that observed lines only ever connected one family to the other — never even to even, never odd to odd. The Laporte rule, stated years before anyone could say why, is parity selection: the dipole operator is odd, so its matrix elements between same-parity states vanish identically. The why arrived with Eugene Wigner's 1927 papers and 1931 Gruppentheorie monograph, which recast Laporte's bookkeeping as representation theory — an operator transforming one way cannot connect states whose symmetry labels differ the wrong way, and the vanishing is exact, not approximate. That is precisely the mechanism running inside every degenerate block above: $(\alpha_{a'}-\alpha_a)W_{a'a}=0$ is a selection rule applied to the perturbation itself, with "forbidden transition" translated to "forced zero of $W$ ." Laporte himself emigrated to Michigan, worked on fluid dynamics and diplomatic science policy, and lived to see his empirical rule become a two-line corollary of symmetry — the fate, he is said to have remarked, of all the best spectroscopy.

History

1924

— Laporte's analysis of the iron arc spectrum: even and odd terms never intercombine

1925

— Russell and Saunders systematize term labels, the raw material of atomic selection rules

1927

— Wigner derives spectroscopic selection rules from group representation theory

1929

— Weyl's Gruppentheorie und Quantenmechanik makes symmetry labels standard equipment

1931

— Wigner's monograph fixes the modern grammar: forbidden = forced zero by symmetry

Going further

Griffiths & Schroeter, Introduction to Quantum Mechanics, 3rd ed., §7.2-7.4 — the "good states" theorem at source register and its fine-structure payoff.
Sakurai & Napolitano, Modern Quantum Mechanics, 2nd ed., §5.1-5.2 — degenerate theory with projector language.
Cohen-Tannoudji, Diu & Laloë, Vol. II, Ch. XI — rigor on the $\lambda\to0$ limit and symmetry-adapted frames.
Wigner, Group Theory and its Application to Quantum Mechanics — the representation-theoretic deep version.

Provenance note: Conceptual framework and 'good states' theorem from Griffiths & Schroeter §7.2–7.4; spin-orbit and fine-structure results from Griffiths §7.3 and Sakurai §5.3; Zeeman regimes from Griffiths §7.4 and Cohen-Tannoudji Vol. II. A verifier should recompute: (1) the commutators $[L_z,\mathbf{L}\cdot\mathbf{S}]\neq0$ and $[J_z,\mathbf{L}\cdot\mathbf{S}]=0$ ; (2) the eigenvalue $\langle\mathbf{L}\cdot\mathbf{S}\rangle=\tfrac{\hbar^2}{2}[j(j+1)-\ell(\ell+1)-s(s+1)]$ ; (3) the Landé factor formula; (4) the off-diagonal-vanishing identity $\langle a_i|[A,H']|a_j\rangle=(a_i-a_j)\langle a_i|H'|a_j\rangle$ . All numbers quoted (Bohr magneton appears symbolically as $\mu_B$ ) are standard. — Items (1) and (4) are executed in this article's verify script (35 checks); ⚠ the base's printed result line for $[L_z,\mathbf{L}\cdot\mathbf{S}]$ carried a transposed sign, $i\hbar(L_xS_y-L_yS_x)$ — machine-verified to be the negative of the correct $i\hbar(L_yS_x-L_xS_y)$ , corrected in equation (4). Edition note: by chapter ownership the base's ENERGY evaluations are relocated to their owner serials with the commutator credentials retained here: the $\langle\mathbf{L}\cdot\mathbf{S}\rangle$ eigenvalue and fine-structure shift (base derivation 2, verifier item 2) → Spin-Orbit Coupling; the Landé $g_J$ formula and weak-field shift (base eq. 8, verifier item 3) → Weak-Field (Anomalous) Zeeman Effect (whose derivation is owned by The Projection Theorem for Vector Operators); the strong-field $\mu_BB(m_\ell+2m_s)$ evaluation (base derivation 3) → Strong-Field Paschen-Back Effect.

Is a good quantum number the same thing as a conserved quantity?

They share the commutator but answer different questions. Conserved ( $[\hat{A},\hat{H}]=0$ , the full Hamiltonian): the label survives time evolution — Constants of the Motion and Conservation Laws' territory. Good in this chapter's sense ( $[\hat{A},\hat{H}_0]=[\hat{A},\hat{H}']=0$ ): the label survives the perturbation — it stays attached to a definite zeroth-order state. When both commutators hold the senses coincide, which is the usual case; but a label can be conserved by $\hat{H}_0$ alone and still be bad, which is exactly the $m_\ell$ -under-spin-orbit trap.

What if my symmetry operator's eigenvalues fail to distinguish some of the degenerate states?

Then the theorem delivers a partial sort: $W$ is block-diagonal, with one block per repeated eigenvalue, and you diagonalize the surviving blocks — still a huge saving. The professional instinct is to hunt a second operator commuting with $\hat{H}_0$ , $\hat{H}'$ , and the first operator, and keep going until the joint labels are unique — the CSCO program of Simultaneous Diagonalization and Complete Sets of Commuting Observables, running inside a single degenerate level.

Does the good basis depend on the perturbation's strength?

No — only on its operator structure. The theorem's hypotheses are commutation relations, blind to the coefficient in front of $\hat{H}'$ . What strength decides is which of two competing perturbations dominates, and hence which one you diagonalize first when they demand incompatible frames: that is the weak-versus-strong Zeeman story, where the good labels flip from $(j,m_j)$ to $(m_\ell,m_s)$ as the field grows past the fine structure. Within one perturbation, though, weak or strong, the good frame is the same frame.

If the symmetry route and the W route disagree, which one is wrong?

Neither — a disagreement means a hypothesis was violated, almost always $[\hat{A},\hat{H}']=0$ . The two routes are provably identical when the theorem's conditions hold (the forced zeros make the $\hat{A}$ -basis an eigenbasis of $W$ ), so treat any mismatch as a unit test failing: recheck the commutator with the perturbation explicitly. The $\lambda x^2y$ contrast is the canonical example — exchange "feels" right for the pair, but the commutator says no, and $W$ 's 53.2° angle is the truth.

Can a quantum number be good at first order but fail at higher orders?

If $\hat{A}$ genuinely commutes with both $\hat{H}_0$ and $\hat{H}'$ , its label is exact to all orders — the full Hamiltonian conserves it, and perturbation order never enters. The approximate versions arise when $[\hat{A},\hat{H}']$ is nonzero but small: then the label is approximately good, leaks slowly, and higher orders (or long times) expose the leak. Spectroscopists' "nearly good quantum numbers" — $\ell$ in alkali atoms, $j$ in light-atom intermediate coupling — are exactly this, and quantifying the leak is what the intermediate-regime matrix treatments do.

How do I actually hunt for the operator A in an unfamiliar problem?

Interrogate the perturbation's geometry: what transformations leave $\hat{H}'$ invariant — reflections, rotations about the field axis, exchange of identical coordinates, parity? Each invariance is a candidate; check it against $\hat{H}_0$ (usually automatic), then check its labels distinguish your block. Fields along an axis keep rotations about that axis ( $m$ good); scalar couplings of two angular momenta keep joint rotations ( $j$ good). Five minutes of geometry routinely deletes a $g\times g$ diagonalization.