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Wave interference is the reinforcement or cancellation that occurs when overlapping wave amplitudes add by the superposition principle. For two monochromatic waves of the same frequency, mutual optical coherence means that their relative wave phase is stable during the observation. Equal wave phases give constructive interference; a half-cycle phase difference gives the destructive interference intensity minimum. Wave amplitudes add before their squares determine wave intensity. ^[ref].

Consider two overlapping beams in a homogeneous medium, with the same polarization of an electromagnetic wave and individual local time-averaged light intensities $I_1,I_2$. Their phase difference $\delta$ is measured in radians. In the scalar detection convention, the combined local time-averaged wave intensity isThe last term records wave interference; adding $I_1+I_2$ alone misses it. The formula describes irradiance when both fields propagate along the same direction, or when the same detector geometry factor is included in all three intensities. For crossing beams it describes the squared electric field response, proportional to scalar wave intensity; a specified component of the Poynting vector also requires the propagation geometry. The ideal crossing plane waves below make that distinction explicit.

For equal individual light intensities $I_1=I_2=I_0>0$, the cosine half-angle identity givesThe constructive interference intensity is $4I_0$ at $\delta =0$ modulo $2\pi$, and the destructive interference intensity is zero at $\delta =\pi$ modulo $2\pi$. At phase quadrature, $\delta =\pi /2$, it is $2I_0$. With $I_1=1$ and $I_2=0.25$ in the same arbitrary wave intensity unit, the maximum is $2.25$, the minimum is $0.25$, and the phase quadrature value is $1.25$: unequal wave amplitudes cannot cancel completely.

Both curves use the same wave intensity unit and a horizontal angle in radians. The dashed levels are the averages over a full phase difference cycle. Equal beams have interferometric visibility one; the unequal example has interferometric visibility $0.8$.

A spatially varying phase difference turns the same law into interference fringes. Take two ideal plane electromagnetic waves with common wavelength $\lambda$, equal individual light intensities $I_0$, and propagation directions at angles $+\theta$ and $-\theta$ from the $z$ axis in the $xz$ plane, with $0<\theta <\pi /2$. Both have linear polarization along $y$, perpendicular to both propagation directions. Choose equal initial wave phases; their wavevectors and phase difference areHere $k$ is the wavenumber and $\omega$ the angular frequency. The common $z$ and time terms cancel, so the interference fringes stay at fixed $x$ for every $z$. Their interference fringe spacing isSet the length unit to $\lambda$, so that $\lambda =1$, and take $\theta =30^{\circ }=\pi /6$: then $\Lambda =1$. The constructive interference intensity bands lie at integer $x/\lambda$; the destructive interference intensity bands lie at half-integer $x/\lambda$.

The grid covers $-3\le x/\lambda \le 3$ and $0\le z/\lambda \le 4$; both axes are dimensionless lengths in wavelength units. The colour scale is $I/I_0$ from zero to four, using the same single-beam normalization as the equal-beam curve above. This is an ideal plane wave superposition, with no slits or point sources and no near-field approximation. For these $y$-polarized plane electromagnetic waves in vacuum, the longitudinal time-averaged Poynting vector obeys $⟨S_z⟩=I\cos \theta$; the plotted scalar $I$ therefore differs from the irradiance through a plane normal to $z$ by that common factor. This relation follows by adding the magnetic fields as well as the electric fields.

The bright peaks do not create energy. Averaging over one interference fringe spacing, or equivalently over one full phase difference cycle, givesbecause the integral of the cosine is zero, with sine as its antiderivative. Equal beams thus average to $2I_0$, despite bright bands at $4I_0$. For the crossing-wave geometry, the average longitudinal energy flux is correspondingly $2I_0\cos \theta$. Bright and dark regions redistribute the incoming energy; conservation of energy concerns the full energy flux, rather than the height of one peak.

Wave interference is useful because phase differences convert small changes in optical path length into measurable changes of wave intensity, as in a Mach-Zehnder interferometer. Its beauty is that the same superposition principle produces both complete cancellation and fourfold brightening; adding more coherent contributions also produces the narrow peaks of a diffraction grating. Feynman's treatment connects this many-source pattern directly to addition of complex amplitudes. ^[ref].

A short proof outline uses local real fields $E_j(t)=A_j\cos (\omega t+{\varphi }_j)$ with parallel linear polarizations. Write $I=C⟨(E_1+E_2{)}^2{⟩}_t$ and $I_j=C{A}_j^2/2$ for the same positive conversion factor $C$. Averaging over many wave periods while the phase difference stays fixed uses $⟨{\cos }^2(\omega t+{\varphi }_j){⟩}_t=1/2$ and $⟨\cos (\omega t+{\varphi }_1)\cos (\omega t+{\varphi }_2){⟩}_t=\cos \delta /2$, giving the stated law. Equivalently, add the complex amplitudes and take the squared complex modulus. ^[ref].

For different fixed linear polarizations in a common transverse plane, the cross term acquires the dot product of their real unit vectors, equal to $\cos \alpha$ for relative angle $\alpha$. Orthogonal linear polarizations remove the cross term in polarization-insensitive scalar detection; different orientations generally only reduce it. A common polarizer can select overlapping components and recover wave interference. Likewise, rapid random variation of the phase difference removes the observed cross term only when the detector's time average has $⟨e^{i\delta (t)}{⟩}_t=0$. Slowly varying wave phases can instead move resolved interference fringes, and biased or incompletely sampled wave phases can leave residual contrast. Finite optical coherence therefore depends on the observation time as well as the fields.

Across a horizontal material interface with continuous background mass density but a velocity jump, an internal gravity wave preserves laboratory frequency and horizontal wavenumber. The matching conditions are continuity of displacement $w/\hat{\omega }$ and pressure, not continuity of $w$. For wave amplitudes and upward phase-line angles measured from the vertical ${\theta }_1,{\theta }_2$, define $r={\hat{\omega }}_2/{\hat{\omega }}_1$. Then the transmission coefficient isThe displacement condition is $A_i+A_r=A_t\cos {\theta }_2/(r\cos {\theta }_1)$, while pressure continuity gives $A_i-A_r=r{A}_t\sin {\theta }_2/\sin {\theta }_1$. Adding eliminates the reflected amplitude and gives the coefficient. It assumes nonzero intrinsic frequencies and propagating outgoing branches.

Zhu, Miao, Xu and Cao's 2007 experiment drove a triangular metal frame electromagnetically with a sinusoidal generator. In a roughly $3\mathrm{cm}$-deep, temperature-regulated pool, it produced nearly plane capillary waves. A $632.8\mathrm{nm}$ laser illuminated the surface obliquely; the moving ripples acted as a reflection diffraction grating. One split beam monitored laser stability, while a CCD recorded diffraction orders on a screen $6.76\mathrm{m}$ away. An air-supported optical table reduced ambient vibration. ^[ref]

The frequency sweep was $100$–$290\mathrm{Hz}$ in $10\mathrm{Hz}$ steps, using distilled water and kerosene at $20{}^{\circ }\mathrm{C}$. For each setting they averaged at least three measurements of first-order spot separation. Known optical geometry and the grating equation yielded $k=2\pi /\lambda$; generator readings yielded $\omega =2\pi f$. Agreement with the predicted log–log slope $3/2$ tests the capillary law, and its intercept gives surface tension if mass density is known. The authors identify surface contamination as a major error source; noncontact optical detection reduces additional disturbance. ^[ref]

The optical measurement can be understood from the grating equation. Let ${\lambda }_L$ be the laser wavelength, $\alpha$ the incident grazing angle above the mean surface, and ${\beta }_m$ the outgoing grazing angle of diffraction order $m$, with the sign of $m$ chosen so thatThus calibrated spot positions determine outgoing angles, which determine the ripple wavelength $\lambda$, independently of the mechanical dispersion relation. This optical geometry treats the surface as a reflection diffraction grating at the laser wavelength. ^[ref]

The inferred surface tension in the full deep-water model isThis formula makes useful experimental controls explicit. Measure mass density at the sample's temperature; check depth with $kh$; keep the driven wave amplitude small enough that $ka\ll 1$; and distinguish $f$ from angular frequency $\omega$. For the stipulated parameters, the $100$–$290\mathrm{Hz}$ window corresponds to calculated wavelengths about $3.62$–$1.76\mathrm{mm}$ and $kh\simeq 52$–$107$ at $h=30\mathrm{mm}$. Depth corrections are negligible here, but the gravity contribution is about $4.5\%$–$1.1\%$ of the capillary contribution. These are ideal predictions, not digitized results from the experiment.

Omitting gravity would infer ${\gamma }_{\mathrm{cap}}=\rho {\omega }^2/k^3=\gamma +\rho g/k^2$, biasing surface tension upward. In the capillary limit, the differential estimate $d\gamma /\gamma \simeq d\rho /\rho +2d\omega /\omega -3dk/k$ also shows why careful spot-spacing calibration matters. A clean surface and weak viscosity are modeling assumptions: appreciable damping, surfactant stresses or large wave amplitudes call for additional physics. The experiment tests the short-wave regime; its drive window does not establish the crossover or the long-wave surface gravity wave limit.

Here light intensity means the time-averaged energy flux per perpendicular area for light. It is the propagation-direction component of the averaged Poynting vector. For plane electromagnetic waves in the same medium it is proportional to squared electric field wave amplitude, giving the projection-squared factor in Malus's law.

At normal incidence, let $z$ measure distance from the reflector into a uniform transparent layer with refractive index $n$. For a monochromatic wave of vacuum wavelength ${\lambda }_0$, the layer wavenumber is $k=2\pi n/{\lambda }_0$. Represent the incident and returning electric fields by complex wave amplitudes with time factor $e^{-i\omega t}$ suppressed:The time-averaged squared electric field, normalized to the incident contribution, isThis electric exposure pattern has maxima separated byThe wave phase of the reflector shifts the maxima's positions but leaves their spacing unchanged in this uniform-layer model. The electric pattern, rather than the zero net energy flux of an ideal standing wave, is the relevant exposure proxy. An imperfect mirror reduces its contrast: $(I_{\max }-I_{\min })/(I_{\max }+I_{\min })=2|r|/(1+|r{|}^2)$, where $I$ denotes this exposure proxy.

Choose ${\lambda }_0=600\mathrm{nm}$ and $n=1.5$. Then the medium wavelength is $400\mathrm{nm}$ and $d=200\mathrm{nm}$. For an ideal phase-reversing reflector, $r=-1$ and the normalized pattern is $4{\sin }^2(kz)$: its first two maxima are at $z=100$ and $300\mathrm{nm}$. These are chosen calculations, not measured properties of a historical photographic emulsion or the optical response of Mercury.

For multiple mutually incoherent wavelengths, time averaging adds exposure contributions rather than their instantaneous electric fields. In a linearized recording model,Here $S$ is incident spectral exposure density, $q$ a wavelength-dependent recording sensitivity and $D$ a model exposure response. An optical spectrum therefore writes overlapping depth oscillations, not necessarily visibly separated sheets. This is a useful connection to the Fourier transform: approximately, depth is paired with optical spatial frequency $2n/{\lambda }_0$. Finite recording depth, varying sensitivity and nonlinear photographic processing prevent treating this expression as an exact inversion rule for an actual plate.

The localization length $\xi$ measures the decay scale of a localized eigenstate's wave amplitude envelope. Here $a$ is the site spacing and $n_0$ its centre. With this amplitude convention, probability decays on half that scale:Individual random tails fluctuate; this notation describes their typical envelope, not exact equality at every site. A localization length much greater than a finite sample's length makes a state look extended inside that sample. The participation ratio below is an effective site count, not an independent fitted value of $\xi$.

For incident light with linear polarization, an ideal linear polarizer transmitswhere $I$ is light intensity and $\theta$ is the angle between the incident electric field's line of oscillation and the polarizer's pass axis. The input and output are compared in the same medium. The classical explanation is an orthogonal projection: for a unit vector $\mathbf{u}$ along the pass axis,The dot product reduces the transmitted wave amplitude by $\cos \theta$. For a plane electromagnetic wave in the same medium, light intensity is proportional to squared electric field wave amplitude, giving Malus's law. For example, $\theta =60^{\circ }$ transmits one quarter of the incident light intensity, although the transmitted wave amplitude is one half. ^[ref].

Start with light already in linear polarization along the horizontal $0^{\circ }$ axis and let its light intensity be $I_0$. Two ideal absorbing linear polarizers with pass axes $0^{\circ }$ and $90^{\circ }$ give $I_0{\cos }^2{0}^{\circ }{\cos }^{2}90^{\circ }=0$. Insert a $45^{\circ }$ polarizer between them: the first leaves $I_0$ unchanged, the middle transmits $I_0/2$ with linear polarization along $45^{\circ }$, and the last transmits $I_0/4$ with linear polarization along $90^{\circ }$. Each angle in Malus's law is relative to the preceding output, so the final step uses $90^{\circ }-45^{\circ }=45^{\circ }$:

For these ideal absorbing polarizers, the rejected electric field component transfers energy to the filter. In the three-filter arrangement the middle absorbs the fraction $I_0/2$, and the last absorbs $I_0/4$, leaving $I_0/4$ transmitted. Conservation of energy holds at each step. Inserting the middle polarizer changes the linear polarization arriving at the last one; it does not create light. The effect illustrates why successive orthogonal projections cannot generally be replaced by a single comparison of the first and last axes.

The initially horizontal assumption matters. For unpolarized light of light intensity $I_0$, the first ideal linear polarizer transmits $I_0/2$: averaging ${\cos }^2\theta$ over uniformly distributed incident orientations gives $1/2$. The same $0^{\circ },45^{\circ },90^{\circ }$ sequence then transmits $I_0/8$. This extra first-stage factor does not apply to the already horizontally polarized input in the figure.

Mechanical impedance is the ratio of a driving force complex amplitude to the resulting velocity complex amplitude at a specified point and in a specified direction. Its units are $\mathrm{N}\mathrm{s}{\mathrm{m}}^{-1}$. The direction of the force, the time convention and whether one is describing a travelling component or a total field must be stated. A positive real travelling-wave mechanical impedance transfers mean power $⟨P⟩=Z|v{|}^2/2$ when $|v|$ is the peak velocity wave amplitude.

For a real wavefront error $h={\sum }_k{a}_k{s}_k$, its squared spatial root mean square is $⟨h^2⟩={\sum }_{k,l}{a}_k{a}_l⟨s_k{s}_l⟩$. Orthogonal unit-peak sine modes have mean square $1/2$, so $h_{\mathrm{rms}}^2=\frac{1}{2}{\sum }_k{a}_k^2$, whereas the coefficient norm is $\sqrt{{\sum }_k{a}_k^2}$. On a finite optical pupil orthogonality needs checking; independent uniform random wave phases give the same $1/2$ diagonal after wave phase averaging. This normalization matters when a speckle contrast of a sinusoidal wavefront error is specified by peak wave amplitude.

A monochromatic wave has a single angular frequency. Its sinusoidal components can have different wave amplitudes and wave phases while sharing that angular frequency, as in the polarization of waves examples.

Optical coherence describes correlations between the wave amplitudes and wave phases of light sampled at different places or times. Its most direct classical test is wave interference: combine two samples of a field and see how much the detected light intensity changes when their relative wave phase is scanned. First-order optical coherence makes this statement quantitative; temporal optical coherence compares delayed samples, while spatial optical coherence compares different positions. These properties are relative to the samples and averaging procedure, rather than a single yes-or-no label for a source. ^[ref].

A useful surprise is that two copies of a fluctuating source can interfere strongly when they have matched delays: the copies share the fluctuations. Conversely, two individually steady-looking beams need not have correlated wave phases. The ideal diffraction grating calculation requires spatial optical coherence across its illuminated elements; unequal-arm measurements with a Michelson interferometer test temporal optical coherence.

Optical fiber attenuation is the reduction of transmitted optical power along an optical fiber. Define a positive loss in decibels by comparing optical power $P(0)$ already launched into the guide with surviving optical power $P(z)$:The factor ten applies to a power ratio; the factor twenty applies to a wave amplitude ratio only when power is proportional to its square under the same conditions.^[ref]

For a uniform illustrative model with constant loss coefficient ${\alpha }_{\mathrm{dB}}$ in $dB/km$ and distance $z$ in kilometres,For ${\alpha }_{\mathrm{dB}}=0.2dB/km$ over $10\mathrm{km}$, the loss is $2\mathrm{dB}$ andThus about $63.1\%$ of the launched optical power survives and $36.9\%$ is lost. This is an illustrative coefficient, not a property of all optical fibers, and is independent of the earlier illustrative refractive indices.

The exponential decay follows from assuming each equal distance interval transmits the same fraction. For two such intervals, power ratios multiply; their logarithms add, so the losses in decibels add. Two $5\mathrm{km}$ sections each transmit $10^{-0.1}\simeq 0.794328$; together they transmit $(0.794328{)}^2\simeq 0.630957$. The loss curve is straight in decibels and curved on a linear power scale.

Wave absorption and scattering of light can remove guided optical power even in a straight fibre, while bends can let it escape. Entrance losses, joints and connectors contribute additional losses; they belong separately from a uniform distance coefficient. This is why a measured end-to-end power ratio need not equal the straight-fibre model.^[ref]

Optical gain is amplification of light by a medium. Here the small-signal intensity gain coefficient $g$ has units ${\mathrm{m}}^{-1}$: for fixed populations, a weak beam obeysin the nondegenerate resonant two-level model, with common transition cross-section $\sigma$. Thus $g$ includes the competition between stimulated emission and photon absorption on that transition. Separate passive intensity loss $\alpha$ gives $I(z)=I(0)e^{(g-\alpha )z}$. Both coefficients describe light intensity, not wave amplitude; the corresponding field-amplitude exponent is half as large. ^[ref].

For $g=0.30{\mathrm{m}}^{-1}$, $\alpha =0.10{\mathrm{m}}^{-1}$ and $z=0.50\mathrm{m}$, the single-pass intensity multiplier is $e^{0.10}\approx 1.10517$. This is amplification during one passage, before mirror losses. Population inversion can therefore yield positive optical gain while the complete oscillator remains below laser threshold.

Optics studies light, its propagation, and instruments for forming images and measuring optical spectra.

The central question is how light carries information from a source to an observer. A useful description must explain both where the light goes and how its distribution changes. In geometrical optics, light rays locate paths and images; in physical optics, an electromagnetic wave has a wave amplitude and a wave phase, so overlapping contributions can produce wave interference and diffraction. These descriptions address different levels of detail within the same subject. ^[ref].

An image is especially interesting because a single object point emits light in many directions. An optical lens redirects part of that spreading bundle toward a corresponding image point; repeating this construction over the object preserves spatial information. Simply gathering more light is insufficient: light rays from distinct object points must arrive at distinct locations. This is why an optical camera needs both an optical lens and a correctly positioned receiving surface. The thin lens example below makes that correspondence quantitative. ^[ref].

Substitution of the proposed normal mode into the wave equation gives the ordinary differential equationsA nonzero exterior displacement field has exponential decay at infinity only when $c<c_2$. Withits wave amplitude is $f_2(y)=D{e}^{-\ell (y-h)}$. At $c=c_2$ the exterior wave amplitude is affine and cannot decay; at $c>c_2$ it is oscillatory and cannot decay. Setting the exterior wave amplitude identically zero would impose both $f_1(h)=0$ and $f_1^{\prime }(h)=0$, forcing the entire normal mode to vanish.

We also need $c>c_1$. Indeed, if $c<c_1$, the rigid boundary condition gives $f_1(y)=A\sinh (qy)$ with $q=k\sqrt{1-c^2/c_1^2}>0$. At the interface,which contradicts continuity of traction because both values of the shear modulus are positive. At $c=c_1$, the layer wave amplitude is $Ay$ and the same contradiction follows from $f_1^{\prime }(h)/f_1(h)=1/h>0$.

Therefore $c_1<c<c_2$. Define the positive transverse wavenumberThe rigid boundary condition now gives $f_1(y)=A\sin (my)$. The interface boundary conditions giveNeither $\sin (mh)$ nor $\cos (mh)$ can vanish for a nontrivial normal mode with $m,\ell >0$. Dividing therefore gives the rigid-backed shear-wave dispersion relation:Conversely, positive $m,\ell$ satisfying these relations generate a nonzero normal mode through the wave amplitudes above. Its physical displacement field is the real part of the complex expression.

The annihilation operator eigenvalue property givesThis is a classical monochromatic electric field: the modulus of $\alpha$ fixes the wave amplitude, while its argument fixes the wave phase relative to the chosen mode convention. Direct normal ordering also gives $(\Delta E{)}_{\alpha }^2=|\mathcal{E}{|}^2$, independent of $\alpha$. Hence the fluctuation divided by the peak mean-field amplitude is $1/(2|\alpha |)$, small in the bright coherent state limit. A relative fluctuation evaluated exactly at a field zero is not a useful classicality test.

Put $s=z-z_0$. In the source's complex beam parameter convention,Substitution separates the transverse wave amplitude and phase:Thus the Gaussian beam is circularly symmetric, has a Gaussian function transverse light intensity, and narrows to a beam waist at $z_0$ before spreading by diffraction. Its wavefront is planar at the beam waist, converging before it and diverging after it. The quadratic transverse phase identifies the signed radius of curvature:Define spot size as the wave amplitude $1/e$ radius, equivalently the light intensity $1/e^2$ radius. ThenThe Rayleigh range is where $w=\sqrt{2}{w}_0$; the far-field beam divergence half-angle is $\lambda /(\pi w_0)$. The factor $1/q$ also produces a longitudinal phase shift relative to a plane wave, changing by $\pi$ across the beam waist; its sign here follows the given $e^{i(kz-\omega t)}$ convention. The on-axis light intensity decreases as $1/(s^2+z_R^2)$ while transverse integration leaves the power constant. These are scalar paraxial approximation results.

A thin lens has ray transfer matrixThe complex beam parameter transformation gives $q_2=q_1/(1-q_1/f)$, henceThe sign is also obtained by multiplying the incident wave amplitude by the thin lens phase $\exp [-ik{\rho }^2/(2f)]$.

Choose the thin lens plane as $z=0$ and put $a=\pi w_0^2/\lambda$, the incident Rayleigh range. Since the incident beam waist lies there, $q_1=-ia$. Immediately after the thin lens,Subsequent free propagation adds $z$, so write $q(z)=z-z_f-i{z}_R^{\prime }$. Comparing real and imaginary parts givesFor a positive focal length, the new beam waist lies downstream but closer than the geometrical focus. At fixed $f,\lambda$, as $w_0\to \infty$,This is the formal unbounded-aperture scalar limit; eventually finite aperture and the paraxial approximation limit the extrapolation.

A chirp is variation of the instantaneous frequency through a signal. Here the group delay relative to the pulse centre at the optical lens isConsequently $\eta >0$ orders lower frequencies before higher frequencies; $\eta <0$ reverses that ordering. A representative strongly chirped optical pulse iswith a smooth localized envelope $A(t)$. The carrier oscillations become closer together towards later times for positive $\eta$ and farther apart for negative $\eta$.

The relation between the spectral phase and a locally assigned instantaneous frequency is a stationary phase method, not an exact reconstruction from spectral phase alone: the spectral wave amplitude also matters. For example, if the spectral envelope is $e^{-(\omega -{\omega }_0{)}^2/(2{\sigma }^2)}$, a Fourier transform of a Gaussian gives the exact real-field phase ${\omega }_{0}t+\eta t^2/(4|a{|}^2)$ up to a constant, where $a=1/(2{\sigma }^2)-i\eta$. The slope retains the sign of $\eta$ and approaches $1/(2\eta )$ for $|\eta |{\sigma }^2\gg 1$.

Quantize a single electromagnetic normal mode of angular frequency ${\omega }_c$ and represent one isolated atomic transition as a two-level system with raising operator ${\sigma }_{+}=|e⟩⟨g|$. The bare Hamiltonian operator isThe first main approximation is the electric-dipole approximation: the field varies negligibly across an atom, $k{a}_{\mathrm{atom}}\ll 1$, and the leading coupling is $H_d=-\mathbf{d}\cdot \mathbf{E}({\mathbf{r}}_0)$. For states of definite opposite parity, the relevant dipole part is off-diagonal. After choosing operator phases, this gives $H_d=\hslash g(a+a^{†})({\sigma }_{+}+{\sigma }_{-})$, with $g$ fixed by the transition dipole and the one-photon field wave amplitude.

In the interaction picture, $a{\sigma }_{+}$ and $a^{†}{\sigma }_{-}$ oscillate at the difference ${\omega }_a-{\omega }_c$, whereas $a{\sigma }_{-}$ and $a^{†}{\sigma }_{+}$ oscillate at the sum ${\omega }_a+{\omega }_c$. The second main approximation is the rotating-wave approximation: near resonance and for $g\sqrt{n+1}\ll {\omega }_a+{\omega }_c$, the latter terms average away on the interaction time scale. The resulting Jaynes-Cummings model hasThe two-level system and single-normal mode truncations are additional assumptions: other atomic transitions and optical normal modes must be sufficiently far off resonance or weakly coupled. Dissipation is neglected over the interaction time. The retained terms exchange one atomic excitation for one photon, conserving $a^{†}a+{\sigma }_{+}{\sigma }_{-}$.