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The following original model isolates amplification by inverted electric dipoles. Let $N=N_u-N_l$ be the excess number in the upper member of one equally degenerate resonant pair. Let $a$ be a slowly varying cavity amplitude with photon number $n=|a{|}^2$, and let $p$ describe the collective molecular dipole component in the amplifying phase convention. On exact resonance, stipulate the linear equationsThe dipole relaxation rate is $\Gamma$, and $g_0=|d_{ul}{\mathcal{E}}_{\mathrm{vac}}|/\hslash$ is a single-molecule coupling in the chosen mode normalization. Here $d_{ul}$ is an electric dipole moment matrix element, ${\mathcal{E}}_{\mathrm{vac}}$ is the field's zero-point amplitude, and $\hslash =h/(2\pi )$. Field overlap and dipole orientation enter this coupling. The characteristic polynomial for growth rate $\lambda$ isFor $N>0$, one root becomes positive precisely when $g_0^{2}N>\kappa \Gamma /2$. If dipole relaxation is fast enough to eliminate $p$, then $p\approx g_{0}Na/\Gamma$ andThus positive population inversion alone is insufficient: the gain rate $GN$ must exceed the passive decay rate. This reduction describes the onset near resonance; it does not predict output phase noise, actual transit dynamics or an exact historical threshold. Nikolay Basov and Aleksandr Prokhorov's molecular-generator theory likewise treats self-excitation as competition between molecular gain and resonator-plus-load losses.^[ref]

To include replenishment, suppose upper-state molecules enter at rate $F$ and both levels leave with residence rate $\gamma$. A deliberately simple ordinary differential equation model isOne downward transition removes two units of the difference $N_u-N_l$, so $\dot{N}=F-\gamma N-2GNn$. Below onset the unsaturated inversion is $N_0=F/\gamma$, givingFor fixed mode volume $V$, the equivalent number density condition is $N_0/V>\kappa /(GV)$. Raising selected beam intensity raises gain; increasing quality factor lowers the loss rate at fixed frequency. These statements hold with $G$, residence time and mode overlap fixed.

Above onset, the nonzero steady state hasThe inversion is clamped while excess supply supports output. A finite initial seed is understood: the noiseless equations leave $n=0$ exactly unchanged even above onset. Actual startup has spontaneous emission and field fluctuations.

A set of resonance lines arising from resolved spin coupling. Weak-coupling patterns can be assigned through coupling paths and line intensities; line count alone can be ambiguous.

A resonance with no resolved scalar splitting under the acquisition conditions. This does not prove that every coupling is exactly zero.

Guglielmo Marconi and Karl Ferdinand Braun shared the award equally for developing wireless telegraphy.^[ref] Braun's receiver coupling experiment connects that history to resonance and the selectivity of a receiver.

For a tank $0<x<L$, $-h<z<0$, let the left wall move by $\mathrm{Re}(c{e}^{-i\omega t})$ with small amplitude, while the right wall and bottom remain fixed. A periodic velocity potential $\mathrm{Re}(f{e}^{-i\omega t})$ has $f_x(0,z)=-i\omega c$. Add the harmonic functionto obtain $G=f+Q$ with homogeneous wall and bottom Neumann boundary conditions. Thus $G=a_0/2+{\sum }_{n\ge 1}{a}_n\cos (k_{n}x)\cosh (k_n(z+h))$, where $k_n=n\pi /L$. The linearized free-surface boundary conditions give $G_z-\lambda G=H$ at $z=0$, with $\lambda ={\omega }^2/g$ andIts Fourier cosine coefficients $H_n=(2/L){\int }_0^{L}H(x)\cos (k_{n}x)dx$ give the title expression for $n\ge 1$ and $a_0=-H_0/\lambda$. In particular $H_n=2i\omega c\lambda /(L{k}_n^2)$ for $n\ge 1$. The denominator vanishes at the corresponding normal modes of surface gravity waves in a rectangular tank, giving resonance. Exact undamped resonance produces secular growth instead of a bounded periodic response. Free transients do not generally decay in this inviscid model.

An ideal inductor and capacitor in parallel have electrical admittance $1/Z=1/(j\omega L)+j\omega C$ for phasors with time dependence $e^{j\omega t}$. The branch currents cancel at angular frequency ${\omega }_0=1/\sqrt{LC}$, giving zero total electrical admittance and an infinite limiting electrical impedance. Below this resonance its electrical reactance is positive; above it the electrical reactance is negative. Losses make the peak finite.

An unstable particle state associated with a resonance in quantum scattering.

Choose the relative phases of the two energy eigenstates so that $d=⟨{\psi }_2|{\hat{D}}_0|{\psi }_1⟩$ is real and nonnegative. Define the resonant Rabi angular frequency $\Omega =d/\hslash$, the Rabi detuning $\delta ={\omega }_0-\omega$, and $\Sigma ={\omega }_0+\omega$. Substitution into the Schrödinger equation, followed by projection onto the two eigenstates, gives the exact interaction picture equationsKeep both probability amplitudes dynamical: replacing $c_1$ by one would incorrectly impose weak excitation and miss complete population transfer.

For $|\delta |,\Omega \ll {\omega }_0$, apply the rotating-wave approximation to discard the rapidly oscillating terms at $\Sigma$. Differentiating the second remaining equation and using the first givesThe initial conditions are $c_2(0)=0$ and ${\dot{c}}_2(0)=-i\Omega /2$. With ${\Omega }_R=\sqrt{{\Omega }^2+{\delta }^2}$, their solution isConsequently,This result permits order-one excitation, including full population transfer on resonance, even though the drive is small compared with the optical angular frequency.

For an average from switch-on to time $\tau$, integrate the squared sine:On resonance this limiting average is $1/2$, although the instantaneous probability repeatedly reaches one. The zero-coupling, zero-Rabi detuning case is understood separately as $P_2=0$.

Near resonance alone does not make this an exact result for arbitrarily strong driving. The stated sinusoidal Hamiltonian also needs the rotating-wave approximation. A concrete counterexample is $\omega ={\omega }_0$ and $\Omega ={\omega }_0$: the exact equations give $P_2(t)={\Omega }^2{t}^2+O(t^4)$ at switch-on, while the claimed rotating-wave sinusoid gives ${\Omega }^2{t}^2/4+O(t^4)$. Thus “without assuming weak radiation” can mean without assuming small excitation, but cannot remove the drive-to-carrier scale separation. The infinite-time average above belongs to the rotating-wave model; omitted corrections can accumulate over sufficiently long times.

Use first-order time-dependent perturbation theory, setting $c_1=1$ in the equation for $c_2$ and retaining the generally complex matrix element $d=⟨{\psi }_2|{\hat{D}}_0|{\psi }_1⟩$. With $\delta ={\omega }_0-\omega$ and $\Sigma ={\omega }_0+\omega$, the probability amplitude isFor $t>0$, split the integral at the pulse maximum. For either $x=\delta$ or $x=\Sigma$,Thus the full first-order answer, without discarding the counter-rotating contribution, isAt long times it givesThis full expression is not exactly one squared Lorentzian line shape. To obtain the intended near-resonance spectral line profile, additionally assume $b,|\delta |\ll {\omega }_0$, so the pulse has a narrow positive-frequency lobe and $F(\Sigma ,t)$ is negligible. ThenThe weak-excitation condition near resonance is $|d|/(\hslash b)\ll 1$. The squared Lorentzian line shape occurs because the symmetric exponential pulse has a Lorentzian line shape amplitude Fourier transform, which is then squared to obtain probability.

At half the maximum, $[1+(\delta /b{)}^2{]}^2=2$, soThe full width at half maximum refers to varying the drive frequency through the positive-frequency resonance. For a concrete failure of the unqualified squared-Lorentzian line shape claim, set $b=\omega ={\omega }_0$ while keeping $|d|/(\hslash b)\ll 1$. The full first-order resonant probability is $36/25$ times the one-lobe prediction. Weak radiation alone therefore does not justify dropping the second Fourier transform lobe.

For a closed absorbing two-level system, write $N=N_1+N_2$, let $A=1/\tau$, and use ${\sigma }_a={\sigma }_{12}$ for the absorption cross section. This differs from the stimulated-emission cross section convention in part (a). With $r=g_2/g_1$, the rate equations areAt steady state, $N_2=W_{a}N/[A+(1+1/r)W_a]$ for $W_a={\sigma }_{a}I/(\hslash \omega )$. Hence the positive absorption coefficient isFor the given sodium fine-structure transition, the statistical weights are $g_1=2$, $g_2=4$, so $r=2$ and $1+1/r=3/2$. The small-signal transmission gives the optical depthTo account for the finite decrease of intensity through the sample, integrate $dI/dz=-\kappa I$ rather than treating the stronger beam as exactly uniform. If $T=I_{\mathrm{out}}/I_{\mathrm{in}}$, this givesTaking $270\mathrm{W}{\mathrm{m}}^{-2}$ as the incident intensity and $T=0.967$,Using the photon energy $hc/\lambda$ then givesIgnoring the small variation of the stronger intensity would instead give about $1.11\times 10^{-13}{\mathrm{m}}^2$; the integrated value uses the stated transmissions more consistently. The length is not needed to eliminate $N$, but gives the check $N\simeq 9.3\times 10^{14}{\mathrm{m}}^{-3}$.

The resonant absorption cross section is of optical-wavelength-squared size, much larger than a typical atomic geometric collision area. Pressure broadening cross sections involve phase-changing interatomic encounters and need not equal hard-sphere collision areas; their numerical value depends on collision partners and speed. No precise ratio follows from the supplied data. A useful distinction is that resonant absorption involves coherent response on the optical scale, so its cross section need not resemble an chemical atom's geometric area.

For homogeneous spectral broadening, all participating chemical atoms must share the same resonance and single-atom response width. In a dilute sample dominated by natural linewidth, the weak-beam requirement isUsing the sodium atomic mass, equality occurs near $T=0.017\mathrm{K}$, so $T\ll 17\mathrm{mK}$ is a sufficient thermal condition. Spatial Zeeman energy shifts, light shifts and transit effects should also be small, the laser should address the same selected component, and the interaction should last long enough for steady populations. Uniform impact pressure broadening can also be homogeneous, but would change the width and effective cross section; it is not included in the lifetime-only numerical model.

Real sodium also has hyperfine structure and optical pumping among magnetic sublevels. The inferred ${\sigma }_a$ is the effective cross section of the stated $g_J=2J+1$ two-level system model, assuming statistical sublevel populations and no loss into unaddressed states. A measurement on a resolved, polarized closed hyperfine structure component requires its actual transition dipole moment and selection rules; a universal polarization-independent sodium cross section cannot be inferred from this simplified model. The dependence of resonant cross sections and saturation intensities on sublevel preparation is tabulated explicitly for the real sodium D lines.^[ref]

C gives spectrum 3. For allyl acetate, $\mathrm{C}{\mathrm{H}}_3\mathrm{C}(=\mathrm{O})\mathrm{OC}{\mathrm{H}}_2\mathrm{CH}=\mathrm{C}{\mathrm{H}}_2$, assignThe 65.1-ppm resonance is the oxygen-bound methylene. Its attached proton test phase matches the terminal alkene methylene and the nonprotonated carbonyl, while the alkene CH and acetyl methyl have the opposite phase. The carbonyl is not conjugated with the alkene because oxygen and a saturated methylene interrupt the carbon–carbon conjugated path. This explains its higher carbonyl shift relative to B without requiring the two esters to have identical electronic environments.

B gives spectrum 4. Number the unsaturated acyl chain from its carbonyl carbon. For $\mathrm{MeO}-\mathrm{C}(=\mathrm{O})-\mathrm{CH}=\mathrm{CH}-\mathrm{C}{\mathrm{H}}_3$, the assignments areThe strongly deshielded beta carbon and relatively shielded alpha carbon reflect polarization of the conjugated ester. All four protonated carbons here are CH or methyl environments and share one attached proton test phase, whereas the carbonyl has the opposite phase. The methoxy resonance is decisive: B has an O-bound methyl, not an O-bound methylene.

On carbon protonation, the C–N bond length increases and the N–N bond length decreases. Forming the fourth C sigma bond changes the carbon from approximately planar $s{p}^2$ to tetrahedral $s{p}^3$, removing its effective C–N pi bonding. The product is described by ${\mathrm{H}}_3\mathrm{C}-{\mathrm{N}}^{+}\equiv \mathrm{N}$: C–N becomes predominantly single and N–N predominantly triple. The resonance of the unprotonated species makes its actual bond orders intermediate rather than exact integers.

There are $\boxed{4+2(1)+2(5)=16}$ valence electrons, all paired, so diazomethane exhibits diamagnetism. A localized orbital occupancy consistent with the double-bond construction in 2(c)(i) isHere $c,t$ mean central and terminal. Each listed occupied spatial orbital contains two electrons of opposite spin. The central N's two p atomic orbitals are perpendicular, so they can support the two perpendicular pi bonds. No central-N lone pair is left over: its positive formal charge accompanies four bonding pairs in this Lewis structure.

Allowing resonance mixes a terminal-N lone pair into the C–N–N $\pi$ orbitals. In the alternative terminal-$sp$ basis, the same electron count is four occupied $\sigma$ orbitals ($8$ electrons), one terminal-$sp$ lone pair ($2$), one perpendicular N–N $\pi$ orbital ($2$), and two occupied orbitals of a three-centre C–N–N $\pi$ system ($4$). The $16$-electron count and the paired ground state survive this change of description; the localized picture does not assert two perfectly isolated double bonds.

One consistent localized orbital hybridization description uses the Lewis structuresFor methyldiazonium, take C as $s{p}^3$ and both N atoms as $sp$. For diazomethane, the double-bond resonance contributor suggests C as $s{p}^2$, central N as $sp$, and terminal N as $s{p}^2$. The central N uses two opposite $sp$ hybrids for the linear C–N–N sigma bond framework. In the localized double-bond picture, terminal N uses one $s{p}^2$ hybrid for its sigma bond, two for lone pairs, and its remaining p atomic orbital for a pi bond.

These labels specify an approximate bonding construction, not uniquely measurable atomic properties. An alternative representation takes terminal N as $sp$, with one $sp$ lone pair and one p lone pair; the latter can participate in the delocalized C–N–N $\pi$ system. It describes the same resonance more explicitly. Neither construction requires an unpaired electron or a bent central N.

The primary chloride undergoes an SN2 reaction; the triphenylmethyl chloride undergoes an SN1 reaction. The primary site is accessible to back-side attack, whereas three phenyl groups make back-side attack at the triphenylmethyl site extremely hindered and stabilize its carbocation through resonance.

For the SN2 reaction, a suitable transition state isThe entering C and leaving Cl lie on opposite sides of the reacting carbon, with partial C–C and C–Cl bonds. The total charge is $-1$. This transition state is not a discrete reaction intermediate.

For the SN1 reaction, the approximately planar $\mathrm{P}{\mathrm{h}}_3{\mathrm{C}}^{+}$ carbocation is a genuine reaction intermediate. Assuming its formation by C–Cl heterolysis is rate determining and trapping is rapid,Thus the limiting rate law is independent of ethanol concentration. The successive attack and proton transfer occur after the slow ionization; a cation-stabilizing environment and efficient trapping underpin this idealized SN1 description.

For the original Lagrangian of solution, differentiation with respect to the generalized velocities givesIn contrast, using $L^{\prime }$ in the nonrotating case gives $p_{x_1}^{\prime }=m{\dot{x}}_1$, $p_{y_1}^{\prime }=m{\dot{y}}_1$; this is the expected momentum shift under adding a total derivative.

The Euler-Lagrange equations in the general cup frame areThese include the Coriolis and centrifugal terms when the cup axes really rotate. Set $z_q=x_1+i{y}_1$ to obtainFor ${\omega }^2\ne {\Omega }^2$, its complete solution iswhere the two complex constants provide the four real initial data. The driving denominator is independent of cup-axis rotation: rotation changes the coordinate description, not the physical trap frequency.

In the requested special case $\nu =0$, an equivalent real form isFor example $a_x=x_1(0)-D$, $b_x={\dot{x}}_1(0)/\omega$, $a_y=y_1(0)$ and $b_y=[{\dot{y}}_1(0)-D\Omega ]/\omega$ determine the constants.

At resonance ${\omega }^2={\Omega }^2$ with nonzero $\Omega$, replace the particular term byIn the nonrotating cup frame this has components $(R\Omega t\sin \Omega t/2,-R\Omega t\cos \Omega t/2)$, plus the homogeneous solution. Secular growth eventually invalidates a finite cup or the neglected vertical kinetic energy. If the ride does not rotate, its forcing vanishes and ordinary free oscillations remain.

Let $\mathcal{R}=|r{|}^2$ be the single-mirror power reflectivity and ${\theta }_g$ the propagation angle inside the gas. Define $\lambda$ as the vacuum wavelength. Successive transmitted complex amplitudes differ by a round-trip factor $\mathcal{R}{e}^{i\delta }$, where, for real dielectric reflection coefficients whose product has positive sign,The cosine arises by comparing successive emerging rays on the same outgoing wavefront; it is not the literal zigzag distance $2d/\cos {\theta }_g$. Equivalently, the round-trip phase is $2k_{z}d$. If the source's $\theta$ is the angle in external air, Snell's law gives $n\sin {\theta }_g=\sin \theta$ andIf $\theta$ denotes the internal angle, simply set ${\theta }_g=\theta$.

For identical lossless mirrors, their power transmission is $1-\mathcal{R}$. In an energy-normalized amplitude convention this is $|t{|}^2=1-\mathcal{R}$; more general interface amplitudes require the corresponding wave impedance factors. The geometric series of transmitted amplitudes givesHenceThis is Fabry–Pérot transmission from multiple reflections.^[ref]

At normal incidence, for nondispersive $n$ and zero net reflection phase modulo $2\pi$, the resonance condition is $\delta =2\pi m$. The resonance frequencies and free spectral range areFor finite $\mathcal{R}<1$ the resonances have finite width, and transmission away from them is nonzero. Thus “only discrete frequencies” is an ideal high-reflectivity description of the selected resonances, not an exact statement about the support of the transmitted spectrum. For example, $\mathcal{R}=0.9$ at $\delta =\pi$ still gives $I_t/I_0=(0.1/1.9{)}^2\simeq 0.00277$.

At half the unit peak, $Q{\sin }^2(\delta /2)=1$. Measuring the phase displacement from a resonance yieldsThe last expression is the high-reflectivity approximation, with finesse $\pi \sqrt{\mathcal{R}}/(1-\mathcal{R})$. At $\mathcal{R}=0.9$, the full phase width is about $0.2109$ radians and $\Delta {\nu }_{\mathrm{FWHM}}/\Delta {\nu }_{\mathrm{FSR}}\simeq 0.03357$. At $\mathcal{R}=0.1$, the minimum transmission is $0.6694$, so the trace never reaches half the peak and this full width at half maximum is not defined. The exact half-maximum condition requires $\mathcal{R}\ge 3-2\sqrt{2}$.

For the absorbing Fabry–Pérot mirror phase and transmission, write $r=\sqrt{\mathcal{R}}{e}^{i{\varphi }_r}$ and let $\mathcal{T}$ be the power transmission of one mirror. Optical absorption makes $\mathcal{R}+\mathcal{T}<1$. The transmitted intensity becomesHere the phase formula is at normal incidence. The peak is reduced below unity, while a change in $\mathcal{R}$ changes the width. The complex reflection phase also moves the resonances: if ${\varphi }_r$ is approximately constant,If either $n$ or ${\varphi }_r$ depends on frequency, this is an implicit condition and the free spectral range need not be constant. No numerical shift is determined without the mirrors' optical response.

Seek the periodic response of a surface gravity wave forced by a traveling bottom, with the same wavenumber $k$ and angular frequency $kc$ as the bottom. The linear equations have constant coefficients in $x,t$, so they preserve each such harmonic. There is no dissipative phase lag in this model; away from resonance, the surface response is in phase or in opposition with the bottom. Thus write $\eta =b\sin \theta$, $\theta =k(x-ct)$, with a signed coefficient $b$. Its physical amplitude is $|b|$. Arbitrary additional free surface gravity waves depend on initial data; the paper's expression selects the forced traveling component.

For $k>0$, $c\ne 0$, use $\varphi =F(y)\cos \theta$. The Laplace equation gives $F^{\prime \prime }-k^{2}F=0$. At $y=0$, the two linearized free-surface boundary conditions implyThese determine both constants in the vertical solution:Its vertical derivative at the bottom isMatching it to $-akc\cos \theta$ and multiplying by $c$ givesand thereforeAn equivalent expression for the velocity potential, which makes the bottom condition immediate, isLet ${\omega }_f=kc$ be the forcing angular frequency, and let ${\omega }_0^2=gk\tanh (kd)$ be the natural dispersion relation from Airy wave theory. ThenThe denominator vanishes at $c^2=(g/k)\tanh (kd)$, when the forcing travels at the free-wave phase velocity. For nonzero forcing there is no bounded periodic forced solution at resonance; the undamped linear response develops secular growth. Below that speed $b<0$, giving a phase reversal, while above it $b>0$. At $c=0$ the bottom is stationary: the rest solution has $\varphi =0$ and $\eta =0$, consistent with the zero limit of $b$, though the formula containing $1/c$ must be interpreted through that limit.

Linearize about fluid at rest, with mean free surface $y=0$ and mean bottom $y=-d$. Regard $a$, $\eta$ and the perturbation velocity potential as first-order quantities. A Taylor expansion from $y=\eta$ to $y=0$ produces products of first-order quantities, as do ${\varphi }_x{\eta }_x$ and $|\nabla \varphi {|}^2$. Discarding these second-order terms gives the linearized free-surface boundary conditionsEquivalently, differentiate the second condition in time and substitute the first to obtain ${\varphi }_{tt}+g{\varphi }_y=0$ at $y=0$.

The bottom is also impermeable and moving. Its exact kinematic boundary condition follows from $D(y-h)/Dt=0$:Here $h_t=-akc\cos (k(x-ct))$, while ${\varphi }_x{h}_x$ and the change from evaluating ${\varphi }_y$ at $h$ to evaluating it at $-d$ are second order. HenceThe interior Laplace equation remains ${\varphi }_{xx}+{\varphi }_{yy}=0$ on $-d<y<0$. This linearization requires small slopes and small resulting excursions as well as $a\ll d$; close to resonance a small prescribed bottom displacement can produce a response too large for it.