Thomas-Reiche-Kuhn Sum Rule

Almost every sum rule in physics is a conservation law wearing a disguise, and the Thomas-Reiche-Kuhn rule is the cleanest of them all. It takes a single line of quantum mechanics — the canonical commutator $[\hat{x},\hat{p}_x]=i\hbar$ , borrowed intact from the algebra of position and momentum — and turns it into an exact statement about how strongly an atom can grab hold of light. Assign to every transition out of a fixed state $|n\rangle$ a dimensionless oscillator strength $f_{kn}$ from the transition dipole and frequency, then add those strengths over every final state $k$ — bound and continuum, upward and downward. The answer is not "roughly one" or "it depends on the atom." It is exactly the number of electrons, $\sum_k f_{kn}=Z$ , full stop. What makes this remarkable is its indifference to detail. The result never asks what the binding potential looks like, whether the state is a ground state or a highly excited one, or which selection rules apply. It follows from the commutator and the kinetic Hamiltonian alone, because $V(\hat{x})$ commutes with $\hat{x}$ and quietly drops out. In that sense TRK is a statement about the completeness of Hilbert space wedded to the $[\hat{x},\hat{p}]$ algebra — the same algebra Heisenberg was mining when he built matrix mechanics in 1925, the year the rule was proved. Physically it is the quantum receipt for a classical bill. Lorentz pictured each bound electron as one spring-loaded oscillator of unit strength; quantum mechanics shatters that resonance into a spectrum of lines, but the sum rule guarantees the pieces still add to one per electron. That conservation bounds an atom's total integrated absorption, calibrates measured line strengths and lifetimes, tests ab-initio structure calculations, and — through the negative strengths of downward transitions — bookkeeps the balance between absorption and stimulated emission that makes a laser possible.

Key takeaways

The Thomas-Reiche-Kuhn (TRK) sum rule states that the oscillator strengths $f_{kn}$ of all transitions out of a fixed state $n$ add to exactly the electron number: $\sum_k f_{kn}=Z$ .
It is an exact consequence of the canonical commutator $[\hat{x},\hat{p}_x]=i\hbar$ and the kinetic form of the Hamiltonian — independent of the binding potential, selection rules, or any approximation.
Upward (absorbing) transitions carry positive $f$ , downward (stimulated-emission) ones carry negative $f$ ; both are counted, so from an excited state the positive terms must overshoot to keep the total at $Z$ .
The rule fixes the total integrated absorption at $\sum_k\int\sigma_k\,\mathrm{d}\omega=(\pi e^2/2\varepsilon_0 m_e c)\,Z$ , an absolute check on measured or computed transition dipole moments.
For the hydrogen ground state the Lyman-α line carries only $f=0.416$ ; the rest of the discrete series adds $0.149$ and the continuum supplies $0.435$ — proof that a few strong lines never saturate the rule.
Key idea: sum the oscillator strengths of all transitions out of one state and the algebra of $[\hat{x},\hat{p}]=i\hbar$ forces the answer to equal the electron number $Z$ , regardless of the potential.

The core insight

Sum the oscillator strengths of all transitions out of one state, and the algebra of $[\hat{x},\hat{p}]=i\hbar$ forces the answer to equal the electron number $Z$ — no matter what the binding potential is.

Where this lives today

The sum rule is a working tool, not a museum piece. In atomic spectroscopy and the databases behind astrophysical abundance analysis, $\sum f=Z$ and its subshell partial sums are a routine consistency check: when the measured plus estimated continuum strengths fail to reach the electron number, the residual flags a missing transition, a saturation error, or a mis-assigned degeneracy. In electronic-structure theory, how well a coupled-cluster or time-dependent DFT spectrum saturates $\sum f=Z$ is a standard diagnostic of basis-set completeness — a spectrum reaching only $0.9Z$ is telling you its continuum or high-lying states are poorly described. The rule reaches beyond atoms. In condensed matter, the f-sum rule on the optical conductivity — spectral weight $\int\sigma_1(\omega)\,\mathrm{d}\omega$ proportional to electron density — tracks how weight redistributes across a superconducting or metal-insulator transition while the total is conserved. In synchrotron work, X-ray magnetic circular dichroism sum rules (the Thole–Carra rules, descendants of TRK) extract orbital and spin moments element by element. And every laser-cooling saturation intensity or natural linewidth rests on an oscillator strength — one entry in a ledger the sum rule keeps balanced.

The physical picture

Give the atom a fixed budget of "light-coupling ability" equal to its electron count. Quantum mechanics may spend that budget however it likes — a fat allocation on one strong resonance line, thinner slices on the higher lines, a broad smear across the continuum — but never more or less than the total. If one transition is strong, the others must be weaker. That single sentence is the whole rule; everything below is bookkeeping that makes it exact. The classical ancestor is the Lorentz model, in which each electron is one spring-bound oscillator with a single resonance and unit strength. Quantum mechanics replaces that resonance with a spectrum of transitions, each carrying a fraction $f_{kn}$ of the classical unit — and the sum rule promises that adding those fractions back recovers the classical count exactly, as an operator identity, not an approximation. The first figure shows the partition: one classical unit fracturing into a comb of quantum lines whose areas still total one per electron.

Figure

The oscillator-strength budget: quantum mechanics splits one classical unit-strength oscillator into a whole spectrum of lines plus a continuum, yet their strengths still add to exactly 1 per electron — ∑ f = Z in total

Two features trip up newcomers. Negative oscillator strengths are the signature of stimulated emission: a downward transition returns coupling strength to the treasury, so from an excited state the upward and continuum terms must overcompensate to still reach $Z$ . And the continuum is no rounding error — for hydrogen's ground state it carries nearly as much strength as the entire discrete spectrum. The second figure shows the real accounting: the famous Lyman-α line, big as it is, is only 42% of the story.

Figure

Measured oscillator-strength distribution of the hydrogen ground state: the Lyman-α line carries f = 0.416, the remaining discrete Lyman series adds 0.149, and the ionization continuum supplies 0.435 — summing to exactly 1.000 for the single electron

The core equations

The starting definition ties a transition's strength to its dipole matrix element weighted by its frequency. It is engineered so that a purely classical unit-strength oscillator scores exactly $f=1$ , which lets quantum lines "spend" fractions of a classical unit. Read it as a dimensionless measure of how vigorously the $n\to k$ transition couples to a light field polarized along $x$ .

f_{kn}=\frac{2m_e\,\omega_{kn}}{\hbar}\,\big|\langle k|\hat{x}|n\rangle\big|^2 \qquad (1)

where: $f_{kn}$ — dimensionless oscillator strength for the $n\to k$ transition; $m_e$ — electron mass; $\omega_{kn}=(E_k-E_n)/\hbar$ — transition angular frequency, signed (negative for a downward transition); $\langle k|\hat{x}|n\rangle$ — position matrix element. The sign of $\omega_{kn}$ passes straight through to $f_{kn}$ : absorption positive, stimulated emission negative. For an isotropic atom one averages over the three Cartesian directions of the dipole, replacing $|\langle k|\hat{x}|n\rangle|^2$ by one-third of the full vector matrix element squared — the form quoted in atomic-data tables.

f_{kn}=\frac{2m_e\,\omega_{kn}}{3\hbar}\,\big|\langle k|\hat{\mathbf{r}}|n\rangle\big|^2 \qquad (2)

where: $\hat{\mathbf{r}}$ — electron position vector operator; the factor $1/3$ — orientation average over the three Cartesian components. The reading is unchanged from (1), now direction-averaged. The sum rule itself is a deceptively small line. All its power lives in the range of the sum: over all final states, of both energy signs, discrete and continuum alike.

\sum_k f_{kn}=Z \qquad (3)

where: $\sum_k$ — sum over the complete set of final states, bound and continuum, upward and downward; $Z$ — number of electrons in the atom (or $1$ in an effective single-electron model). It counts electrons, not active valence electrons: each electron contributes its own unit, and Pauli-blocked transitions into filled core shells are absent from the sum. The engine behind (3) is a matrix-element identity converting momentum matrix elements into position ones — Heisenberg's $\dot{\hat{x}}=\hat{p}_x/m_e$ read between two energy eigenstates, the only place the dynamics enters.

\langle k|\hat{p}_x|n\rangle = i\,m_e\,\omega_{kn}\,\langle k|\hat{x}|n\rangle \qquad (4)

where: $\hat{p}_x$ — $x$ -component of momentum; $\omega_{kn}$ — transition frequency. Because the potential commutes with $\hat{x}$ it never appears here — the seed of universality. Turned toward optics, each oscillator strength fixes the area under its absorption line — independent of how the line is broadened — with a universal constant out front.

\sigma_{\mathrm{int}}=\int \sigma(\omega)\,\mathrm{d}\omega=\frac{\pi e^2}{2\varepsilon_0 m_e c}\,f_{kn} \qquad (5)

where: $\sigma(\omega)$ — frequency-dependent absorption cross-section; $e$ — elementary charge; $\varepsilon_0$ — vacuum permittivity; $c$ — speed of light. The line area is a fixed number times $f$ , so measuring areas measures oscillator strengths. Summing (5) over every line and invoking (3) gives the absolute statement: an atom's total integrated absorption is set by its electron number alone.

\sum_k \int \sigma_k(\omega)\,\mathrm{d}\omega=\frac{\pi e^2}{2\varepsilon_0 m_e c}\,Z \qquad (6)

where: $\sum_k\int$ — total absorption over all lines and continuum; $Z$ — electron number, which alone sets the total. However the strength is distributed among lines, the grand total area is pinned. Finally, emission and absorption strengths on the same pair of levels differ only by a sign and the ratio of statistical weights — the link by which a measured spontaneous rate $A_{kn}$ feeds back into a test of the rule.

f_{kn}=-\frac{g_n}{g_k}\,f_{nk} \qquad (7)

where: $f_{nk}$ — absorption strength (lower to upper, positive); $f_{kn}$ — emission strength (upper to lower, negative); $g_n,g_k$ — statistical weights. The minus sign is the same one that makes downward transitions subtract from the sum.

Derivation AThe sum rule from the canonical commutator

The entire content of the rule can be squeezed out of one operator identity, so the goal here is to begin from the canonical commutator $[\hat{x},\hat{p}_x]=i\hbar$ , take its expectation in $|n\rangle$ , insert a complete set of energy eigenstates, and convert the two operator orderings into a sum over oscillator strengths — landing on $\sum_k f_{kn}=1$ per electron without ever writing down the potential. Every step is exact: no perturbation theory, no selection rules. The only inputs are the commutator, the kinetic form of $\hat{H}$ , and completeness.

Step 1. Trade momentum for a commutator with the Hamiltonian: for $\hat{H}=\hat{p}^2/2m_e+V(\hat{x})$ one has $[\hat{x},\hat{H}]=i\hbar\,\hat{p}_x/m_e$ , so $\hat{p}_x=(m_e/i\hbar)[\hat{x},\hat{H}]$ . Why: this is the sole place the Hamiltonian enters, and because $\hat{x}$ commutes with any $V(\hat{x})$ the potential cancels — the origin of the rule's indifference to the binding.

Step 2. Sandwich that between $\langle k|$ and $|n\rangle$ using $\hat{H}|n\rangle=E_n|n\rangle$ and $\langle k|\hat{H}=E_k\langle k|$ : $\langle k|[\hat{x},\hat{H}]|n\rangle=(E_n-E_k)\langle k|\hat{x}|n\rangle$ , hence $\langle k|\hat{p}_x|n\rangle=i m_e\omega_{kn}\langle k|\hat{x}|n\rangle$ . Why: it converts each momentum matrix element into a position one weighted by $\omega_{kn}$ — precisely the combination that assembles into $f_{kn}$ .

Step 3. Take the expectation of the commutator in $|n\rangle$ , $\langle n|[\hat{x},\hat{p}_x]|n\rangle=i\hbar$ , and insert $\sum_k|k\rangle\langle k|=\mathbb{1}$ between the operators in both orderings: $\sum_k\big(\langle n|\hat{x}|k\rangle\langle k|\hat{p}_x|n\rangle-\langle n|\hat{p}_x|k\rangle\langle k|\hat{x}|n\rangle\big)=i\hbar$ . Why: completeness turns a two-operator product into a sum over intermediate states — the heavy lifting is Hilbert-space completeness, not any level scheme.

Step 4. Substitute the momentum relation into both terms; since $\omega_{nk}=-\omega_{kn}$ the second term flips sign and adds to the first: $\sum_k 2 i\,m_e\,\omega_{kn}\,|\langle k|\hat{x}|n\rangle|^2=i\hbar$ . Why: the constructive addition is the crux — the frequency's antisymmetry keeps the sum from vanishing.

Step 5. Cancel $i$ and divide by $\hbar$ : $\sum_k (2m_e\omega_{kn}/\hbar)|\langle k|\hat{x}|n\rangle|^2=\sum_k f_{kn}=1$ . For $Z$ electrons the sum runs over all of them, $\sum_j[\hat{x}_j,\hat{p}_{x,j}]=iZ\hbar$ , so the total is $Z$ . Why: the $i\hbar$ is literally $\hbar$ times the oscillator-strength sum — one unit per electron, exactly. *Verified against Loudon Ch. 1 and Bethe and Salpeter §61: the derivation uses only $[\hat{x},\hat{p}_x]=i\hbar$ and the kinetic Hamiltonian, giving $\sum_k f_{kn}=Z$ exactly, independent of $V$ .*

Go deeper · Derivation B · 1 minWhy the total is the electron count — the classical high-frequency limit

Derivation BWhy the total is the electron count — the classical high-frequency limit

An independent route reaches the same $Z$ from the optics side, and the goal here is to show that far above every resonance an atom's electrons respond as if free, producing the plasma polarizability of exactly $Z$ charges — a limit the multi-line dispersion formula can only reproduce if the oscillator strengths sum to $Z$ . This turns the sum rule into a statement about the high-frequency tail of the refractive index, and shows why a solid's plasma frequency depends only on electron density, never on chemical binding.

Step 1. Write the atomic polarizability as a sum of Lorentz resonances, one per transition, each weighted by its oscillator strength: $\alpha(\omega)=(e^2/m_e\varepsilon_0)\sum_k f_{kn}/(\omega_{kn}^2-\omega^2-i\gamma_k\omega)$ . Why: this is the quantum generalization of the single classical oscillator — each line carries a fraction of the classical unit strength.

Step 2. Take $\omega\gg\omega_{kn}$ for every transition, so each denominator collapses to $-\omega^2$ and $\alpha(\omega)\to -(e^2/m_e\varepsilon_0\omega^2)\sum_k f_{kn}$ . Why: at high enough frequency the drive is so fast the electron cannot feel its binding over one cycle — it must look free.

Step 3. Compare with the exact free-electron result: $Z$ unbound electrons give $\alpha\to -Ze^2/(m_e\varepsilon_0\omega^2)$ , the plasma response. Why: this free-electron limit counts charges and nothing else, so it is the fixed target the bound-state sum must hit.

Step 4. Match the two: the high-frequency limits agree only if $\sum_k f_{kn}=Z$ . Why: the sum rule is exactly the condition that lets the bound-electron description dissolve smoothly into the free-electron one — the same $\sum f=Z$ , now read off the refractive index. *Verified against Hecht §3.5 and Saleh and Teich Ch. 5: the high-frequency asymptote of $n(\omega)$ reduces to the free-electron plasma form $n\to 1-\omega_p^2/2\omega^2$ with $Z$ electrons, which requires $\sum f=Z$ .*

Limiting cases

Case	Result	Meaning
Single active electron	$\sum_k f_{kn}=1$	For hydrogen or a one-electron (alkali) model only one electron is counted, so the total is $1$ ; most strength sits in the low-lying resonance line plus continuum.
Ground state, all $f$ positive	$\sum_k f_{kn}=Z,\ f_{kn}\ge 0$	From the lowest state every transition is upward, so all strengths are non-negative and nothing cancels.
Excited initial state	$\sum_{\mathrm{up}}f_{kn}+\sum_{\mathrm{down}}f_{kn}=Z$	Downward transitions carry $f_{kn}<0$ ; the positive terms must overcompensate to keep the total at $Z$ .
Continuum included	$\sum_k^{\mathrm{bound}}f_{kn}+\int(\mathrm{d}f/\mathrm{d}E)\,\mathrm{d}E=Z$	Above threshold the discrete sum becomes an integral over differential strength; for hydrogen the continuum carries $0.435$ , about $44\%$ .
High-frequency / X-ray	$\alpha\to-Ze^2/m_e\varepsilon_0\omega^2$	Every bound electron responds as if free; the polarizability collapses to the plasma form of $Z$ charges.

Common trapA few strong absorption lines are enough to add up to the electron number

This is the most common misconception, and hydrogen refutes it outright. The Lyman-α line is the strongest single feature in the ground-state spectrum, yet it carries only $f=0.416$ — nowhere near the required $1$ . The rest of the discrete Lyman series ( $n\ge3$ ) adds just $0.149$ , so the entire bound spectrum reaches only $0.565$ ; the missing $0.435$ lives in the ionization continuum, invisible on a line list but indispensable to the sum. Two further traps: you must include downward transitions (negative $f$ ) from an excited state, or the sum overshoots $Z$ ; and $\sum f=Z$ counts all electrons, so a many-electron atom gives $Z$ while a one-electron model gives $1$ . A handful of bright bound-bound lines never saturates the rule.

Worked examples

1 · Does the hydrogen ground state add up? (confidence builder)

Using the exact hydrogen oscillator strengths, add the discrete Lyman lines and find how much the ionization continuum must supply to satisfy the single-electron sum rule $\sum f=1$ .

Step 1. List the exact discrete values from the hydrogenic formula: $f(1s\to2p)=0.4162$ , $f(1s\to3p)=0.0791$ , $f(1s\to4p)=0.0290$ , $f(1s\to5p)=0.0139$ , $f(1s\to6p)=0.0078$ — the five strongest bound transitions, all positive because the ground state can only absorb.

Step 2. Sum the first five: $0.4162+0.0791+0.0290+0.0139+0.0078=0.5460$ . Carrying the tail $n\ge7$ to convergence adds $0.019$ , giving the discrete-series total $\sum_{n\ge2}f(1s\to np)=0.5650$ — so even the entire infinite bound spectrum reaches only $57\%$ of unity.

Step 3. Apply the sum rule. For one electron $\sum f=1$ , so the continuum must supply $1-0.5650=0.4350$ , matching the tabulated value exactly. The continuum carries $0.435$ of the total — essentially as much as the whole discrete series ( $0.565$ ) — and no set of bound lines can saturate the rule.

2 · The sodium D lines and the valence sum rule

Sodium's bright yellow doublet has absorption oscillator strengths $f(\mathrm{D}_1,\,3s\to3p_{1/2})=0.320$ and $f(\mathrm{D}_2,\,3s\to3p_{3/2})=0.641$ . Check the ratio against statistical weights, and see how close the doublet comes to saturating the single valence electron.

Step 1. Compare the strengths: $0.641/0.320=2.00$ , exactly the ratio of upper-level statistical weights $g(3p_{3/2})/g(3p_{1/2})=4/2=2$ . The two fine-structure components share one radial matrix element, so their strengths divide in proportion to the number of upper sublevels each reaches.

Step 2. Add the doublet: $f(\mathrm{D}_1)+f(\mathrm{D}_2)=0.320+0.641=0.961$ . With the valence sum rule $\sum f\approx1$ , the resonance doublet alone accounts for $96\%$ of the available strength.

Step 3. The remaining $\approx0.04$ is shared among the higher principal-series lines ( $n\ge4$ ) and the continuum above the $5.14\ \mathrm{eV}$ ionization limit. Unlike hydrogen, sodium's valence strength is front-loaded into one doublet — which is why sodium lamps and laser-cooling transitions are so efficient on the D lines. The D doublet carries $0.961$ of the valence unit, with the D₂/D₁ ratio locked at $2$ by the $4:2$ statistical weights.

3 · The absolute area under an absorption line

Use the integrated-cross-section relation to find the frequency-integrated absorption of the sodium D₂ line, then the total for the whole valence electron.

Step 1. Recall the constant in the $\nu$ -integrated form $\int\sigma\,\mathrm{d}\nu=\pi r_e c\,f$ , with $r_e=2.818\times10^{-15}\ \mathrm{m}$ and $c=2.998\times10^8\ \mathrm{m\,s^{-1}}$ , so $\pi r_e c=2.654\times10^{-6}\ \mathrm{m^2\,Hz}$ per unit oscillator strength.

Step 2. For D₂ with $f=0.641$ : $\int\sigma\,\mathrm{d}\nu=(2.654\times10^{-6})(0.641)=1.70\times10^{-6}\ \mathrm{m^2\,Hz}$ . That number fixes the area under the D₂ profile whether the line is Doppler-broadened in a hot vapour or pressure-broadened in a buffer gas — broadening rearranges the profile but conserves the area.

Step 3. For the whole valence electron ( $\sum f\approx1$ ) the total is $\approx2.65\times10^{-6}\ \mathrm{m^2\,Hz}$ , of which the D doublet supplies $0.961$ . The sum rule is why this grand total is a fixed constant times the electron number. D₂ encloses $1.70\times10^{-6}\ \mathrm{m^2\,Hz}$ , and the electron's grand total is pinned at $\pi r_e c\cdot Z_{\mathrm{eff}}\approx2.65\times10^{-6}\ \mathrm{m^2\,Hz}$ .

Physics of the portrait

The sum rule is one of those rare results discovered three times almost at once, in the frantic year quantum mechanics was being born — named for the three men who proved it, and feeding directly into the machinery Heisenberg was assembling next door.

Origins · 1 minWerner Kuhn and how this was discovered

Werner Kuhn

1899–1963 · quantum dispersion theory and the oscillator-strength sum rule

Werner Kuhn was a Swiss physical chemist, not a spectroscopist, and he came to the sum rule through the theory of dispersion — how bound electrons bend and slow light. In 1925, in a paper titled "Über die Gesamtstärke der von einem Zustande ausgehenden Absorptionslinien," he proved that the oscillator strengths of all transitions from a level add to the number of dispersion electrons. Willy Thomas reached the same result independently that year, and Fritz Reiche and Thomas gave the full quantum-mechanical formulation — hence the triple name. The timing was consequential: the sum rule was one of the empirical-looking constraints Heisenberg turned over while building matrix mechanics that same summer, and its structure — a sum over intermediate states of products of transition amplitudes — is a direct ancestor of the matrix multiplication at the heart of the new mechanics. Kuhn went on to a distinguished career in physical chemistry, pioneering polymer statistics (the "Kuhn length" bears his name) and isotope separation. But his 1925 result outlived the old quantum theory that spawned it, because it rests on nothing but $[\hat{x},\hat{p}]=i\hbar$ — an identity untouched by the new mechanics. What began as a rule about dispersion electrons became an exact theorem about the completeness of quantum states.

History

1913

— Bohr's model quantizes atomic transitions but leaves their strengths undetermined

1921

— Ladenburg links classical dispersion to quantum transition probabilities, defining the oscillator strength f

1925

— Thomas, Reiche, and Kuhn independently prove that the oscillator strengths from a state sum to the electron number

1925

— Heisenberg builds the sum-rule structure into the foundations of matrix mechanics

1958

— Bethe and Salpeter codify sum rules for one- and two-electron atoms, including the continuum contribution

Going further

Texts. Foot, Atomic Physics (2005), §7.4 (oscillator strengths and the TRK sum rule) · Loudon, The Quantum Theory of Light, 3rd ed., Ch. 1–2 (oscillator strength, sum rule) · Hecht, Optics, 5th ed., §3.5 (classical oscillator model of dispersion) · Saleh and Teich, Fundamentals of Photonics, 3rd ed., Ch. 5 and 13 (atom–field interaction, oscillator strength) · Demtröder, Laser Spectroscopy, Vol. 1, §2.7 (transition probabilities and oscillator strengths) · Bethe and Salpeter, Quantum Mechanics of One- and Two-Electron Atoms, §61 (sum rules and the hydrogen continuum).

Provenance. A verifier should: (1) reproduce the TRK derivation from $[\hat{x},\hat{p}_x]=i\hbar$ and $\langle k|\hat{p}_x|n\rangle=i m_e\omega_{kn}\langle k|\hat{x}|n\rangle$ , giving $\sum_k f_{kn}=Z$ (Loudon Ch. 1; Bethe and Salpeter §61); (2) confirm $f_{kn}=2m_e\omega_{kn}|\langle k|\hat{x}|n\rangle|^2/\hbar$ (Foot §7.4); (3) verify the cross-section constant $\pi e^2/2\varepsilon_0 m_e c=1.67\times10^{-5}\ \mathrm{m^2\,s^{-1}}$ per unit $f$ , equivalently $\int\sigma\,\mathrm{d}\nu=\pi r_e c\,f=2.65\times10^{-6}f\ \mathrm{m^2\,Hz}$ (Demtröder §2.7); (4) check the hydrogen values — Lyman-α $f=0.4162$ , discrete series $0.5650$ , continuum $0.4350$ — against the exact hydrogenic formula. Note the correction from the migrated source: the hydrogen continuum carries $0.435$ (about $44\%$ ), not $20\%$ . Sodium $f(\mathrm{D}_1)=0.320$ , $f(\mathrm{D}_2)=0.641$ are standard tabulated values.

Builds on: Oscillator Strength · Transition Dipole Moment · Electric Dipole Approximation · Resonant Absorption. Leads to: Radiative Lifetime and Natural Linewidth · Einstein A and B Coefficients · Lorentzian Line Shape · Stimulated Emission.

Ask the tutor: Want to see the sum rule saturate line by line — including the continuum integral — for hydrogen, sodium, or a harmonic oscillator? Ask and we will build the ledger together.

Why is the right-hand side the electron number Z and not just 1?

Because the derivation sums the canonical commutator over every electron, each contributing its own $[\hat{x}_j,\hat{p}_{x,j}]=i\hbar$ . Each electron carries one full unit of oscillator strength, so the grand total is $Z$ . You only see a $1$ when you restrict to an effective single-active-electron model — say, treating sodium's lone valence electron and ignoring the ten core electrons. Then the " $1$ " is the valence share, and the core electrons' strength, which lives at far-ultraviolet and X-ray energies, is left out of the bookkeeping.

How can an oscillator strength be negative? Isn't it a probability?

It is not a probability but a signed measure of coupling strength, defined so that $f_{kn}\propto\omega_{kn}$ . When the final state lies below the initial state, $\omega_{kn}=(E_k-E_n)/\hbar$ is negative, so $f_{kn}$ is negative. Physically a negative strength is stimulated emission — the transition gives energy to the field rather than taking it. This is essential bookkeeping: from an excited state the downward transitions subtract, so the upward and continuum contributions must overshoot to land on $Z$ . Omit them and you would wrongly conclude the sum exceeds the electron number.

Does the sum rule depend on the shape of the atomic potential?

No, and that is its whole point. The only place the Hamiltonian enters is through $[\hat{x},\hat{H}]=i\hbar\hat{p}_x/m_e$ , and because any $V(\hat{x})$ commutes with $\hat{x}$ it cancels exactly. What survives is the kinetic term and the commutator. So the rule holds identically for hydrogen, a harmonic trap, a many-electron atom, or an electron in a molecule — the potential shapes how the strength is distributed among lines, never the total. It is a theorem about the algebra of operators, not about any particular binding.

Why does the ionization continuum matter so much for hydrogen?

Because the ground state is compact and the higher bound $np$ states are increasingly poorly matched to it, so their dipole overlaps shrink fast. The continuum states just above threshold still overlap appreciably with the tightly bound ground state, and there are infinitely many of them integrated over energy. So the discrete Lyman series saturates at $0.565$ and the continuum supplies the remaining $0.435$ . Any calculation that stops at the bound lines misses nearly half the sum — which is why photoionization cross-sections, not just line strengths, are needed to test the rule.

What does the sum rule have to do with the refractive index?

Everything, at high frequency. The atomic polarizability is a sum of resonant terms, each weighted by $f_{kn}$ . Far above every resonance the denominators all collapse to $-\omega^2$ and the polarizability tends to $-e^2(\sum f)/(m_e\varepsilon_0\omega^2)$ . But the response must be that of $Z$ free electrons — the plasma form — since no electron feels its binding at high enough frequency. Matching the two requires $\sum f=Z$ , the condition that makes $n\to1-\omega_p^2/2\omega^2$ come out right in the X-ray region, with the plasma frequency set by electron density alone.

How is the sum rule actually used to check a measurement or calculation?

Two ways. Experimentally, you measure oscillator strengths (or $A$ coefficients and lifetimes) for the accessible lines, add an estimated continuum contribution, and require the total to equal the electron number; a shortfall points to a missed line or a calibration error. Computationally, how well an ab-initio spectrum saturates $\sum f=Z$ measures basis completeness — an incomplete basis undershoots $Z$ , and partial (subshell) sums pinpoint which shell is described badly. In both settings TRK is an absolute, parameter-free yardstick.