Absorption and Stimulated Emission

Shine a beam on a cloud of atoms, half in the ground state and half excited. The ground-state atoms absorb photons and climb. What do the excited ones do? Intuition offers "nothing, until they decay." The quantum answer is stranger: the same beam that pushes ground-state atoms up pushes excited atoms down, forcing them to emit a photon that is a phase-perfect copy of the drive. Absorption and stimulated emission are the two resonant halves of one coupling, and this article proves they proceed at exactly the same intrinsic rate. The framework article built the coupling and split the drive into its $e^{-i\omega t}$ and $e^{+i\omega t}$ halves; here we do the rate bookkeeping. We derive the absorption rate for an atom bathed in broadband radiation of spectral energy density $u(\omega)$ , defining Einstein's $B$ coefficient, and we prove the symmetry theorem $B_{12}=B_{21}$ — first in three lines from Hermiticity, then in its full degenerate form $g_1B_{12}=g_2B_{21}$ , machine-verified on hydrogen's $m$ -sublevels. We also settle a debt: the Thomas–Reiche–Kuhn sum rule, stated upstream, is proved here from a double commutator, with the harmonic oscillator as its cleanest exact exhibit and hydrogen's ledger — $0.565$ discrete, $0.435$ continuum — as its honest one. Master this and lasers stop being magic: a two-level medium can never lase precisely because up and down doors are equally wide, and every optical amplifier ever built is an exercise in cheating that symmetry with a third level.

Key takeaways

A monochromatic drive splits into $e^{-i\omega t}$ (absorption) and $e^{+i\omega t}$ (stimulated emission) halves with the same matrix element — so the induced rates are identical.
In broadband radiation the first-order $\mathrm{sinc}^2$ lineshape integrates to a rate that grows linearly in time: $R=B\,u(\omega_0)$ with $B=\pi|\langle b|\mathbf{d}|a\rangle|^2/3\epsilon_0\hbar^2$ .
The $1/3$ comes from averaging $|\hat{\boldsymbol\epsilon}\cdot\mathbf{d}|^2$ over isotropic, unpolarized light — a solid-angle integral, done once and verified.
Symmetry theorem: $B_{12}=B_{21}$ (nondegenerate) and $g_1B_{12}=g_2B_{21}$ (degenerate) — equilibrium bookkeeping later forces exactly this.
Beware conventions: $B$ defined against $u(\omega)$ differs from $B$ against $u(\nu)$ by a factor $2\pi$ — tabulated values are convention-bound.
The Thomas–Reiche–Kuhn sum rule $\sum_f f_{if}=1$ follows from $[\hat x,[\hat H,\hat x]]=\hbar^2/m$ : one electron's worth of response, conserved and merely distributed.
KEY IDEA: one coupling, two resonant doors — light drives upward and downward transitions with identical strength, and only population differences decide which door wins.

The core insight

Within the dipole approximation, absorption ( $a\to b$ ) and stimulated emission ( $b\to a$ ) share one and the same coupling matrix element up to complex conjugation, so their induced rates are identical — $B_{12}=B_{21}$ .

Where this lives today

Every erbium-doped fibre amplifier in the internet's backbone is equation (3) run in reverse: pump erbium ions into inversion and each data photon stimulates phase-perfect copies — all-optical gain. The symmetry theorem also sets the noise floor: the same inversion emits spontaneously, and that amplified-spontaneous-emission noise limits long-haul capacity. In the sky, natural OH and water masers amplify microwave lines by $10^{12}$ wherever shocks invert populations, and opacity projects still use TRK-style sum rules as consistency audits. Even saturation is now a tool: saturated-absorption spectroscopy exploits the two-level bleaching ceiling to erase Doppler broadening — the standard trick behind laser frequency stabilization.

The physical picture

Picture the atom as a swing and the light as a hand pushing at the swing's own frequency. A swing at rest gains energy from resonant pushing — absorption. A swing at full amplitude pushed in antiphase is drained by the same periodic contact — it does work on the hand. Stimulated emission is the antiphase case: the atom hands its $\hbar\omega_0$ to the field, and the emitted photon joins the driving mode with the same frequency, direction, polarization and phase. The drive does not merely trigger the release; it dictates the copy.

Figure

Two doors, one hinge: the same coupling matrix element drives climbing and descending transitions; only populations break the tie

Three consequences precede any algebra. A thermal gas has $N_2<N_1$ , so absorption wins on population alone — cold matter dims light. Engineer $N_2>N_1$ (an inversion) and the identical door widths force the medium to amplify: laser gain is a one-line corollary of the symmetry theorem. And pump a strict two-level system as hard as you like — the best you reach is $g_2N_1=g_1N_2$ , balanced flows, saturation to transparency. Two-level lasers are forbidden by the very symmetry that makes amplifiers possible; real lasers cheat with a third or fourth level.

From the two halves to a rate

Semiclassical Theory of Atom–Radiation Interaction built the coupling and split a monochromatic drive into an $e^{-i\omega t}$ half (the atom climbs) and an $e^{+i\omega t}$ half (it descends) — one operator, two doors. The rate bookkeeping it deliberately left here. The first-order machinery lives in First-Order Transition Amplitudes and Probabilities: under a near-resonant harmonic drive the jump probability carries the famous $\mathrm{sinc}^2$ profile,

P_{a\to b}(t)=\frac{|V_{ba}|^2}{\hbar^2}\,\frac{\sin^2\!\big[(\omega_0-\omega)t/2\big]}{(\omega_0-\omega)^2/4}, \qquad (1)

where: $\omega_0=(E_b-E_a)/\hbar$ is the Bohr frequency and $V_{ba}=-\tfrac{1}{2}E_0\,\langle b|\hat{\mathbf d}\cdot\hat{\boldsymbol\epsilon}|a\rangle$ the half-amplitude coupling. On resonance the probability grows as $t^2$ ; off resonance it oscillates. Neither is a "rate" — for a strictly monochromatic drive no constant rate exists; that coherent story, Rabi flopping included, is told exactly in Transition Probability for a Periodic Two-Level System. A rate emerges the moment the light is broadband: a continuum of incoherent frequencies whose $\mathrm{sinc}^2$ contributions add as probabilities, not amplitudes. That single integration step is the first derivation below; everything else falls out of it.

Derivation AThe Broadband Rate and the Einstein B Coefficient

We want the transition rate for a two-level atom in isotropic, unpolarized, broadband radiation of spectral energy density $u(\omega)$ (energy per volume per unit angular frequency). Strategy: take the monochromatic probability (1), replace the field amplitude by each spectral slice's energy density, average the dipole projection over directions, then integrate over frequency. Two watch-points: the $\mathrm{sinc}^2$ factor must act as a delta at large $t$ (that is what births a rate), and the direction average must be done honestly — it is the origin of the ubiquitous $1/3$ .

Step 1. Convert amplitude to energy density. A monochromatic wave of amplitude $E_0$ carries time-averaged energy density $\bar u=\tfrac{1}{2}\epsilon_0E_0^2$ , so $|V_{ba}|^2=\tfrac{1}{4}E_0^2\,|\langle b|\hat{\boldsymbol\epsilon}\cdot\hat{\mathbf d}|a\rangle|^2=\frac{\bar u}{2\epsilon_0}|\langle b|\hat{\boldsymbol\epsilon}\cdot\hat{\mathbf d}|a\rangle|^2$ .

Step 2. Average over directions and polarizations. For isotropic unpolarized light the projection $|\hat{\boldsymbol\epsilon}\cdot\mathbf d_{ba}|^2$ averages to $|\mathbf d_{ba}|^2\,\langle\cos^2\theta\rangle_{\rm sphere}$ , and the average of $\cos^2\theta$ over the sphere is

\frac{1}{4\pi}\int_0^{2\pi}\!\!\int_0^{\pi}\cos^2\theta\,\sin\theta\,d\theta\,d\phi=\frac{1}{3}. \qquad (2)

Why $1/3$ and not $1/2$ ? Because the average runs over the three-dimensional sphere of relative orientations: each Cartesian component of $\mathbf d$ carries an equal share of $|\mathbf d|^2$ , and the field picks out one axis.

Step 3. Add the spectral slices. Each frequency slice $u(\omega)\,d\omega$ contributes its own copy of (1). Summing probabilities (incoherent light — random relative phases kill the cross terms):

P_{a\to b}(t)=\frac{|\mathbf d_{ba}|^2}{6\epsilon_0\hbar^2}\int u(\omega)\,\frac{\sin^2[(\omega_0-\omega)t/2]}{(\omega_0-\omega)^2/4}\,d\omega .

Step 4. Let the lineshape become a delta. For $t$ long compared with the inverse spectral width of $u$ , the $\mathrm{sinc}^2$ factor is sharply peaked at $\omega=\omega_0$ and $u(\omega)$ slides out of the integral at $u(\omega_0)$ . The remaining integral is elementary: with $x=\omega_0-\omega$ , $\int_{-\infty}^{\infty}\sin^2(xt/2)/x^2\,dx=\pi t/2$ . The probability now grows linearly in time — that is the birth of a rate:

R_{a\to b}=\frac{P_{a\to b}(t)}{t}=\frac{\pi\,|\mathbf d_{ba}|^2}{3\epsilon_0\hbar^2}\,u(\omega_0)\;\equiv\;B_{ab}\,u(\omega_0). \qquad (3)

Why does linear growth require broadband light? The $t^2$ coherent pile-up of a single frequency is spread by the integral over detunings into a steady drip: the width of the $\mathrm{sinc}^2$ peak shrinks as $1/t$ while its height grows as $t^2$ , and the product is exactly linear.

Step 5. Read off the coefficient and its twin. $B_{ab}=\pi|\mathbf d_{ba}|^2/3\epsilon_0\hbar^2$ is Einstein's absorption coefficient in the $\omega$ -convention. Running the identical steps on the $e^{+i\omega t}$ half — same operator, matrix element conjugated — gives the stimulated-emission rate $R_{b\to a}=B_{ba}u(\omega_0)$ with $|\mathbf d_{ab}|^2=|\mathbf d_{ba}|^2$ , hence $B_{ab}=B_{ba}$ . Verified: the sphere average (2) equals exactly $1/3$ and the lineshape integral equals exactly $\pi t/2$ , both symbolically in sympy; the assembled $B(1s\to2p)=1.27\times10^{21}\ \mathrm{m^3\,J^{-1}\,s^{-2}}$ is reproduced numerically in this article's verification script.

where in equation (3): $u(\omega_0)$ is evaluated at the Bohr frequency only — broadband light interrogates the atom exclusively at its resonance; $|\mathbf d_{ba}|^2=e^2|\langle b|\mathbf r|a\rangle|^2$ is the squared transition dipole; and the $3$ in the denominator is the orientation average, not a degeneracy factor. Reach for (3) whenever the light is spectrally wide and weak; reach for the Rabi solution when it is narrow and strong. The golden-rule grammar — rate $\propto$ coupling squared $\times$ density of what's available at resonance — is the same one Fermi's Golden Rule runs for a continuum of final states; here the continuum is in the field, there in the atom.

The symmetry theorem, twice

The equality $B_{12}=B_{21}$ looks innocent — one Hermitian conjugate. Its degenerate generalization is where real spectroscopy lives, because real levels carry $m$ -sublevels.

Go deeper · Derivation B · 1 minUp Equals Down: the B-Symmetry Theorem with Degeneracy

Derivation BUp Equals Down: the B-Symmetry Theorem with Degeneracy

We prove that the total induced rates between two levels (not just two states) obey $g_1B_{12}=g_2B_{21}$ , where $g_1,g_2$ count the degenerate sublevels. The strategy: first the three-line nondegenerate proof from Hermiticity, then the honest sublevel bookkeeping, machine-verified on hydrogen's $1s\to2p$ manifold. The watch-point: $B$ between levels means averaging over initial sublevels and summing over final ones — get that order wrong and the theorem seems to fail.

Step 1. Nondegenerate case. The dipole operator is Hermitian: $\langle b|\hat{\mathbf d}|a\rangle=\langle a|\hat{\mathbf d}|b\rangle^*$ . Therefore $|\mathbf d_{ba}|^2=|\mathbf d_{ab}|^2$ , and since (3) depends on the matrix element only through this modulus squared, $B_{ab}=B_{ba}$ . Absorption and stimulated emission are the same door, read in opposite directions.

Step 2. Degenerate case — set up the sums. Let level 1 have sublevels $\{a_i\}$ , $i=1..g_1$ , and level 2 sublevels $\{b_j\}$ , $j=1..g_2$ . Define the line strength $S=\sum_{ij}|\langle b_j|\mathbf d|a_i\rangle|^2$ , which is symmetric in the two levels by Step 1 applied termwise.

Step 3. Rates between levels. An atom sitting in a particular lower sublevel $a_i$ absorbs at a rate proportional to $\sum_j|\langle b_j|\mathbf d|a_i\rangle|^2$ ; averaging over the $g_1$ equally-populated sublevels gives $B_{12}\propto S/g_1$ . Likewise $B_{21}\propto S/g_2$ . Dividing,

g_1B_{12}=g_2B_{21}. \qquad (4)

Step 4. Machine check on hydrogen $1s\to2p$ : $g_1=1$ , $g_2=3$ . Explicit Gaunt integrals give every sublevel the same strength, $\sum_{q}|\langle 2p\,m|r_q|1s\rangle|^2=\mathcal R^2/3$ with $\mathcal R$ the radial integral. So $S=\mathcal R^2$ , $B_{12}\propto\mathcal R^2$ , $B_{21}\propto\mathcal R^2/3$ : the per-atom door down is three times narrower than the door up, exactly balancing the three ways to sit upstairs — $1\cdot B_{12}=3\cdot B_{21}$ . Why must equilibrium enforce exactly this? Run time backwards: a photon bath at temperature $T$ holds $N_2/N_1=(g_2/g_1)e^{-\hbar\omega_0/kT}$ ; balance of up and down flows at all $T$ is only consistent with Planck's spectrum if (4) holds — Einstein's 1917 argument, executed in full in Spontaneous Emission and the Einstein A Coefficient, forces the symmetry we just derived microscopically. Two independent routes, one identity. Verified: the per-sublevel angular strengths equal $1/3$ each (symbolic Gaunt integrals over $Y_{00},Y_{1m}$ ), so $S=\mathcal R^2$ and $g_1B_{12}=g_2B_{21}$ checks exactly; hermiticity equality verified numerically on the same elements.

The convention trap: which u is your B measured against

There is no such thing as "the" $B$ coefficient — only $B$ paired with a spectral convention. Against $u(\omega)$ , $B^{(\omega)}=\pi|\mathbf d|^2/3\epsilon_0\hbar^2$ ; against $u(\nu)$ the same physical rate requires $B^{(\nu)}=2\pi B^{(\omega)}$ , since $u(\nu)d\nu=u(\omega)d\omega$ squeezes the same energy into a $2\pi$ -narrower unit. The Einstein ratio transforms too: $A/B^{(\omega)}=\hbar\omega_0^3/\pi^2c^3$ versus $A/B^{(\nu)}=8\pi h\nu_0^3/c^3$ — both correct, both common, and mixing them silently costs a factor of $6.28$ . When in doubt, reconstruct $B$ from the convention-free $A$ or the oscillator strength. Speaking of which: the cross-section is the third, laboratory-facing currency. Integrating the absorption profile over frequency gives a result independent of any lineshape details,

\int\sigma(\nu)\,d\nu=\frac{e^2}{4\epsilon_0 m_ec}\,f=2.654\times10^{-6}\ \mathrm{m^2\,Hz}\times f, \qquad (5)

where: $f$ is the oscillator strength defined in the framework article ( $f(1s\to2p)=0.4162$ , its number). Equation (5) is why $f$ values are the astronomer's currency: an absorption line's integrated depth counts atoms, once you know $f$ . And $f$ values obey an exact conservation law — stated upstream, owed a proof, delivered now.

Go deeper · Derivation C · 2 minThe Thomas-Reiche-Kuhn Sum Rule from a Double Commutator

Derivation CThe Thomas-Reiche-Kuhn Sum Rule from a Double Commutator

We prove that the oscillator strengths out of any state $|i\rangle$ sum to exactly one (per electron): $\sum_f f_{i\to f}=1$ , where $f_{i\to f}=(2m_e\omega_{fi}/3\hbar)\sum_q|\langle f|r_q|i\rangle|^2$ and the sum runs over all final states, discrete and continuum. The strategy is the classic double-commutator evaluation: compute $[\hat x,[\hat H,\hat x]]$ two ways — once as an operator identity, once resolved over the energy eigenbasis. The watch-point: the rule needs only $[\hat H,\hat x]$ 's kinetic part, so it holds for any local potential — that generality is its power.

Step 1. Operator route. With $\hat H=\hat p^2/2m_e+V(\hat{\mathbf r})$ , the potential commutes with $\hat x$ , so $[\hat H,\hat x]=-i\hbar\,\hat p_x/m_e$ , and

[\hat x,[\hat H,\hat x]]=\big[\hat x,\,-i\hbar\hat p_x/m_e\big]=\frac{\hbar^2}{m_e}.

A pure number — no wavefunctions involved.

Step 2. Spectral route. Sandwich the same double commutator in $|i\rangle$ and insert a complete set of energy eigenstates $|f\rangle$ between every product. Each term picks up energy denominators-turned-numerators, and collecting the two orderings gives

\langle i|[\hat x,[\hat H,\hat x]]|i\rangle=2\sum_f (E_f-E_i)\,|\langle f|\hat x|i\rangle|^2 .

Step 3. Equate and dress. Setting the two routes equal: $\sum_f(E_f-E_i)|\langle f|\hat x|i\rangle|^2=\hbar^2/2m_e$ . Repeat for $\hat y,\hat z$ and add: $\sum_f(E_f-E_i)\sum_q|\langle f|r_q|i\rangle|^2=3\hbar^2/2m_e$ . Multiplying by $2m_e/3\hbar^2$ turns the left side into exactly $\sum_f f_{i\to f}$ :

\sum_f f_{i\to f}=1. \qquad (6)

Why "one electron's worth"? For an $N$ -electron atom the same algebra on $\sum_k\hat x_k$ gives $\sum f=N$ : the total capacity of matter to respond to light counts electrons, exactly — the 1925 result that predates matrix mechanics by weeks.

Step 4. The exact exhibit. For the harmonic oscillator, $\langle n|\hat x|0\rangle$ vanishes except $n=1$ , with $|\langle1|\hat x|0\rangle|^2=\hbar/2m_e\omega$ ; then $f_{0\to1}=(2m_e\omega/\hbar)(\hbar/2m_e\omega)=1$ exactly, all other $f$ 's zero. One transition carries the entire sum rule — the cleanest possible machine check, and the reason the classical Lorentz-oscillator model of dispersion worked as well as it did.

Step 5. The honest exhibit. For hydrogen's ground state the discrete Lyman series carries $f(1s\to2p)=0.4162$ , $f(1s\to3p)=0.0791$ , $f(1s\to4p)=0.0290$ , … — summing all discrete lines reaches only $\approx0.565$ . The missing $0.435$ sits above the ionization edge: bound-free transitions into the continuum. The sum rule is exact; the ledger just has a page most tables omit. Verified: the oscillator $f_{0\to1}=1$ symbolically; hydrogen $f$ ladder computed from explicit radial integrals for $n=2..7$ ( $0.4162/0.0791/0.0290/\ldots$ ), discrete total $\approx0.565$ with the $n^{-3}$ tail, continuum share $\approx0.435$ ; operator identity $[\hat x,[\hat H,\hat x]]=\hbar^2/m_e$ dimension-checked.

Figure

The TRK ledger for hydrogen 1s: every transition takes a slice of exactly one electron — and the biggest slice is not a line at all

Limiting cases

case	result	meaning
monochromatic, strong drive	no constant rate; Rabi oscillations at $\Omega_R$	coherent regime — owner: Transition Probability for a Periodic Two-Level System
broadband, weak drive	$R=B\,u(\omega_0)$ , linear in $t$	this article's regime: incoherent rate bookkeeping
$u\to$ Planck at $T$ , $\hbar\omega_0\gg kT$	stimulated $\ll$ spontaneous ( $\bar n\ll1$ )	optical transitions at room temperature ignore stimulated emission
$u\to$ Planck, $T>\hbar\omega_0/k\ln2$	stimulated exceeds spontaneous ( $\bar n>1$ )	for a 2 eV line this needs $T\approx3.3\times10^4$ K — hotter than the Sun's surface
$g_1=g_2$	$B_{12}=B_{21}$ exactly	the textbook symmetry as a special case of (4)
two-level pumping, $u\to\infty$	populations lock at $g_2N_1=g_1N_2$ ; net absorption $\to0$	saturation: a two-level medium can bleach but never lase

Worked examples

Example 1 — how fast does the Sun pump Lyman-α? A hydrogen atom floats at the solar surface, $T=5777$ K. Enumerate what is needed: the matrix element (upstream number), the $B$ coefficient (equation 3), the Planck energy density at the line (evaluate), the rate (multiply).

Step 1. Matrix element to $B$ . The framework article's radial integral gives $|\langle 2p\,m|\mathbf r|1s\rangle|^2=\mathcal R^2/3$ per sublevel with $\mathcal R=1.2903\,a_0$ ; summing the three final sublevels, the level-to-level absorption coefficient is $B_{12}=\pi e^2\mathcal R^2 a_0^2/3\epsilon_0\hbar^2=1.27\times10^{21}\ \mathrm{m^3\,J^{-1}\,s^{-2}}$ ( $\omega$ -convention).

Step 2. Planck density at the line. $\hbar\omega_0=10.20$ eV gives $x=\hbar\omega_0/kT=20.5$ , deep in the Wien tail: $u(\omega_0)=\dfrac{\hbar\omega_0^3}{\pi^2c^3}\,\dfrac{1}{e^{x}-1}=1.9\times10^{-21}\ \mathrm{J\,m^{-3}\,s}$ .

Step 3. Multiply. $R=B_{12}\,u(\omega_0)=\mathbf{2.4\ s^{-1}}$ — a ground-state atom waits on average $0.42$ s to absorb a Lyman-α photon. Cross-check by the occupation route: $R=3A\bar n$ with $A=6.27\times10^8\ \mathrm{s^{-1}}$ (the next article's flagship) and $\bar n=1/(e^{20.5}-1)\approx1.3\times10^{-9}$ gives the same $2.4\ \mathrm{s^{-1}}$ — two routes, one number.

Step 4. Read it. Even at a stellar surface the Wien factor $e^{-20.5}$ throttles a nanosecond-class transition to seconds: rates are matrix elements times photon supply, and the supply usually wins.

Example 2 — balancing the TRK books for hydrogen. Claim: the discrete lines cannot account for hydrogen's response to light. Enumerate the ledger, then check the total.

Step 1. Discrete entries. From explicit radial integrals: $f(1s\to2p)=0.4162$ , $f(1s\to3p)=0.0791$ , $f(1s\to4p)=0.0290$ , $f(1s\to5p)=0.0139$ , with the higher series falling off as $\sim1.6/n^3$ .

Step 2. Sum the series. $n=2..7$ gives $0.546$ ; the $n^{-3}$ tail to $n\to\infty$ adds $\approx0.019$ . Discrete total: $\mathbf{0.565}$ .

Step 3. Invoke the rule. Equation (6) says the total is exactly $1$ . The balance, $\mathbf{0.435}$ , must be bound-free absorption — photoionization continuum above $13.6$ eV. No discrete table, however long, closes the books.

Step 4. Read it. Nearly half of hydrogen's light-response is not spectral lines at all — opacity codes that skimp on photoionization misplace 43% of the coupling, which is why the sum rule serves as the standard bookkeeping audit.

Origins · 1 minStimulated Emission and how this was discovered

Stimulated Emission

1917–1960 · Berlin, Columbia, Malibu

Stimulated emission was born as a bookkeeping necessity and grew into a civilization-scale technology. Einstein introduced the downward induced process in his 1916–17 quantum theory of radiation because equilibrium with Planck's spectrum was impossible without it — thermodynamics forcing a term a decade before wave mechanics could compute it. Ladenburg and Kopfermann's negative-dispersion measurements in excited neon (1928) showed the term was physically real. The engineering took a generation more: Townes's Columbia group ran the ammonia maser in 1954 on an inverted population, and in May 1960 Maiman at Hughes Research Laboratories in Malibu flashed a ruby rod into the first laser. Every laser photon since is a stimulated copy of a copy — Einstein's balancing entry, amplified.

History

1917

— Einstein's "Zur Quantentheorie der Strahlung" introduces the A and B coefficients and the stimulated-emission term

1925

— Thomas, Reiche and Kuhn publish the oscillator-strength sum rule, weeks before Heisenberg's matrix mechanics

1928

— Ladenburg and Kopfermann observe negative dispersion in excited neon: stimulated emission is real, not bookkeeping

1954

— Townes, Gordon and Zeiger operate the ammonia maser: the first device run on population inversion

1960

— Maiman's ruby laser at Hughes Research Laboratories: optical-frequency stimulated emission, coherent and bright

Going further

References. Griffiths & Schroeter, Introduction to Quantum Mechanics 3rd ed., §11.2–11.3 · Sakurai & Napolitano, Modern Quantum Mechanics §5.6–5.7 · Cohen-Tannoudji, Diu & Laloë Vol. II, Ch. XIII and Complement A_XIII · Shankar Ch. 18 · Hilborn, "Einstein coefficients, cross sections, f values…" (Am. J. Phys. 50, 982) for the convention zoo.

Provenance. Equations and numbers are standard textbook results. A verifier should recompute: (1) the first-order amplitude and resonance lineshape (Griffiths 3rd ed. §11.2; Sakurai §5.6); (2) the sinc-squared integral $\int \sin^2(xt/2)/x^2\,\mathrm{d}x=\pi t/2$ and the resulting $B_{ab}=\pi|\langle b|\mathbf{d}|a\rangle|^2/(3\epsilon_0\hbar^2)$ (Griffiths §11.2.3); (3) the Einstein relations $B_{ab}=B_{ba}$ and $A/B=\hbar\omega_0^3/\pi^2c^3$ via Planck-law matching (Sakurai §5.7, Griffiths §11.3); (4) the Hermiticity equality of matrix elements. Cross-check against Cohen-Tannoudji Vol. II Complement A_XIII and Shankar Ch. 18. (Lane note: item (1)'s lineshape derivation is owned upstream by First-Order Transition Amplitudes and Probabilities and Harmonic (Sinusoidal) Perturbations and Resonance; items (2)–(4) are executed and machine-verified here; the Planck-matching half of (3) is completed by Spontaneous Emission and the Einstein A Coefficient per the band's ownership map.)

Prerequisites: Harmonic (Sinusoidal) Perturbations and Resonance · First-Order Transition Amplitudes and Probabilities · Fermi's Golden Rule · The Dipole Approximation · Semiclassical Theory of Atom–Radiation Interaction

Leads to: Spontaneous Emission and the Einstein A Coefficient · Transition Probability for a Periodic Two-Level System · Density of Final States

If B₁₂ = B₂₁, why doesn't a driven gas just freeze with nothing happening?

Because rates multiply populations: net absorption is $B\,u\,(N_1-N_2)$ . A cold gas has $N_1\gg N_2$ and absorbs strongly; as the drive pumps population upstairs the net rate shrinks, vanishing when populations equalize — the medium bleaches transparent, with enormous two-way photon traffic and zero net flow.

Is the stimulated photon really identical to the driving one, or just similar?

Identical in every mode label: frequency, direction, polarization, phase. In field-quantized language the emission amplitude into an $n$ -photon mode scales as $\sqrt{n+1}$ : the " $n$ " piece is stimulated emission into the occupied mode — coherent gain, beams staying beams — while the " $+1$ " piece, into empty modes with random direction and phase, is spontaneous emission, deliberately outside this article's semiclassical reach.

Where exactly does the derivation stop being valid if I crank up the intensity?

When the first-order probability stops being small over the dynamics' coherence time — i.e. when the Rabi frequency $\Omega_R=|d\,E_0|/\hbar$ rivals the linewidth or detuning spread. Then population cycles instead of flowing, and you must swap rate equations for the exact two-level solution: the rate picture is the incoherent, weak-field face of the same physics.

Does the TRK sum rule survive in real multi-electron atoms with all their interactions?

Yes — the proof used only $[\hat H,\hat x]=-i\hbar\hat p_x/m_e$ , which holds for any local, velocity-independent potential including full electron–electron repulsion: for $N$ electrons the sum is exactly $N$ . Correlations only shift the allocation between lines and continuum. A TRK violation in a calculation is a bug (or relativity/nonlocality), which is why opacity projects run it as an audit.

My laser is monochromatic, not broadband — does B u(ω) even apply to laser experiments?

Often yes, with a role swap: if the atom's line is much broader than the laser's spectrum, the atom supplies the smooth function and the laser the delta — the same integral runs exchanged, giving $R=\sigma(\omega_L)\Phi$ with $\Phi$ the photon flux. Narrow laser and narrow line and strong drive is different: coherent Rabi territory, owned by the periodic two-level article.

Why did Einstein need the stimulated term at all — couldn't absorption plus spontaneous decay balance?

Try it: balance $N_1Bu=N_2A$ with Boltzmann populations and you get Wien's law $u\propto e^{-\hbar\omega/kT}$ , not Planck's. The $-1$ in Planck's denominator arises precisely from the stimulated term in $N_1B_{12}u=N_2(A+B_{21}u)$ . Thermodynamics demanded induced emission in 1917; dynamics caught up a decade later. The full argument opens the next article.