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The Airy disk is the central bright region, up to the first dark ring, in the scalar diffraction pattern of a uniformly illuminated, unobscured circular aperture. The whole pattern includes the weaker surrounding rings; its radius is distinct from its full width at half maximum.

For a circular optical pupil of diameter $D$ under normal, uniform monochromatic plane wave illumination, Fraunhofer diffraction givesHere $I_0$ is the on-axis light intensity, $\theta$ is the angle from the optical axis, $\lambda$ is the wavelength in the propagation medium, and $J_1$ is the order-one Bessel function of the first kind. The paraxial approximation replaces $\sin \theta$ by $\theta$, measured in radians. The first positive zero $j_{1,1}=3.831705970\dots$ givesThis is the angular radius of the Airy disk; the angular diameter is twice as large. The Airy disk derivation explains the origin of the Bessel function of the first kind and of the numerical coefficient.^[ref]

The image and both radial curves show the same calculated light intensity, divided by its peak. The image's logarithmic scale makes the weak rings visible; its colour is a display choice. The linear curve shows how small the ring peaks are compared with the centre. A dark ring marks a zero of wave amplitude, produced by cancellation across the optical pupil; it is not the edge of all transmitted light. The first Airy disk contains about $83.78\%$ of the total radiant flux, as calculated in Airy disk normalization and encircled energy.

For an original design example, take an ideal optical lens in air with focal length $f=100\mathrm{mm}$, circular optical pupil diameter $D=5.00\mathrm{mm}$ and wavelength $\lambda =550\mathrm{nm}$. In the focal plane, the radial coordinate $r\simeq f\theta$ givesThe first-zero diameter is $26.8327\mu \mathrm{m}$. Solving $P(u_{1/2})=1/2$ gives $u_{1/2}=1.616339948\dots$, so the diameter measured by full width at half maximum is onlyThus a quoted “spot diameter” is incomplete without its definition. Doubling $D$ halves both widths at fixed wavelength and focal length; increasing magnification alone changes the image scale without improving object-space discrimination.

The Airy disk supplies a calculable reference for a point spread function, allowing diffraction to be distinguished from defocus and optical aberrations. Its beauty is that one circular boundary determines an entire sequence of bright and dark rings through a Fourier transform. The ideal formula assumes an unobscured pupil, uniform illumination, ideal optical focus, a single wavelength, and a scalar description with small output angles. A blocked centre, nonuniform pupil transmission or a significant pupil wave phase error changes the response. High numerical aperture may require polarization of an electromagnetic wave and a vector field description. The circular-aperture diffraction laboratory experiment tests the far-field pattern directly; the Rayleigh criterion uses its first zero to define a two-source convention.

Let $P(t)$ be the pulse's power integrated over its transverse section, and $U$ its energy. Its peak power is ${\max }_{t}P(t)$. For a chosen Gaussian power envelope with full width at half maximum $\tau$,The numerical shape factor is approximately $0.9394$. It matters when comparing quoted peak powers; the estimate $U/\tau$ assumes a shape and a width convention.

If an otherwise unchanged pulse is stretched by a factor $S$ without loss, its energy stays fixed and its peak power falls by $S$. For energy gain $G$ and compressor energy transmission $\eta$, ideal restoration of the original shape givesThese relations require a specified shape and preserved spectrum; they do not model amplifier saturation. For the diagram's chosen $1\mathrm{nJ}$, $30\mathrm{fs}$ seed, $S=9.2959$, $G=1000$ and $\eta =0.8$, the successive peak powers are $31.31\mathrm{kW}$, $3.369\mathrm{kW}$, $3.369\mathrm{MW}$ and $25.05\mathrm{MW}$. The last pulse contains $0.8\mu \mathrm{J}$: the compressor increases peak power while losing some energy.

Fluence is time-integrated light intensity, $F(x,y)=\int I(x,y,t)dt$, measured in $\mathrm{J}{\mathrm{m}}^{-2}$. For a uniform beam of area $\mathcal{A}$, $F=U/\mathcal{A}$ and $I_{\mathrm{pk}}=P_{\mathrm{pk}}/\mathcal{A}$. At fixed transverse profile, lossless temporal stretching lowers peak light intensity but leaves fluence unchanged. Absorbed energy, heat removal, beam size, coatings and pulse duration still constrain optical components. For nonoverlapping identical pulses at repetition frequency $f_r$, the average power is $U{f}_r$; it is a different quantity from peak power.

Choose $N$ equal-amplitude, monochromatic odd components with orders $q_j=q_0+2j$, $j=0,\dots ,N-1$, and odd $q_0$. Fix the complex-field conventionHere $I$ is a normalized light intensity after averaging the fast carrier oscillations. Setting every ${\varphi }_j=0$ and summing a geometric series givesAt the removable singularities take the continuous limit. Maxima repeat every $T/2=\pi /\omega$; the first zeros on either side of a maximum lie at $t=\pm T/(2N)$. The zero-to-zero width is $T/N$, which is not the full width at half maximum.

Choose orders $13,15,17,19,21$ and driving wavelength $\lambda =800\mathrm{nm}$. Using $T=\lambda /c$ with the speed of light $c$ gives $T=2.66851\mathrm{fs}$ and burst spacing $1.33426\mathrm{fs}$. Solving the exact half-maximum equation gives a full width at half maximum of approximately $240.6$ attoseconds. These are values of the specified five-component model, not measurements from the 2001 experiment.

A deliberately chosen spectral phase ${\varphi }_q=q\omega t_0$ translates the whole attosecond pulse train by $t_0$ without changing its shape. Adding $\kappa (q-17{)}^2/2$ makes neighboring phase increments unequal: the same component intensities then produce a different temporal profile. For the choice $t_0=0.20\mathrm{fs}$ and $\kappa =0.08$ radians per squared harmonic order, the adjacent-component delay estimates arenamely $0.0981$, $0.1660$, $0.2340$ and $0.3019\mathrm{fs}$ for $q=13,15,17,19$. This finite difference estimates the group delay at the pair's midpoint. It is not a single common arrival time.

The infinite monochromatic sum is periodic forever and has infinite total energy. A real attosecond pulse train has a finite envelope and finite component linewidths. This calculation isolates coherent addition; it cannot describe an isolated burst or establish its contrast against satellites. As a further check, averaging $I(t)$ over one half-period gives $N|a{|}^2$ for any constant component spectral phases, because the cross terms integrate to zero by Fourier orthogonality. Spectral phase redistributes the light intensity in time while leaving that average fixed.

If phase-randomizing collisions form a Poisson process of mean interval ${\tau }_c$, field correlation survives only when no such collision occurs during the delay. Thus its magnitude decays as $e^{-|\tau |/{\tau }_c}$. The Wiener-Khinchin theorem gives the normalized Lorentzian line shapeIts full width at half maximum is $2/{\tau }_c$ in angular frequency or $1/(\pi {\tau }_c)$ in ordinary frequency. A population lifetime is a different time constant: lifetime-limited amplitude decay at half the population rate introduces the familiar factor of two. The definition of the supplied lifetime must therefore accompany any numerical linewidth.

For an isolated fluorescent molecule at displacement $\mathbf{r}$ from the scanning optical focus, write $E(\mathbf{r})$ for relative excitation light intensity and $D_a(\mathbf{r})$ for the fraction of its collected emission passed by a centred detection pinhole of radius $a$. The expected response, apart from a common brightness and collection factor, isHere $\mathbf{u}$ is position in the detection-pinhole plane and $p_{\mathrm{em}}$ is the emitted light intensity distribution there conditional on collection, with units of inverse area. This point spread function normalization retains throughput: $D_a$ is a fraction, not a curve rescaled to have peak one. The product expresses excitation followed by detection; it is an intensity/probability product, not multiplication of coherent wave amplitudes from different molecules. ^[ref]

Assume linear, unsaturated fluorescence, independent emitters whose light intensities add, a stationary specimen, fixed collection efficiency and quantum efficiency, and negligible changes from scattering, absorption or bleaching during a scan. Different excitation and emission wavelengths generally give different excitation and detection responses. For a sufficiently small pinhole, $D_a$ is approximately its area times $p_{\mathrm{em}}$ at the centre; this is the familiar excitation-times-detection point spread function. It is not universally the square of one point spread function. A spatially invariant response further allows the image to be expressed as a convolution with the emitter distribution; depth-dependent optical aberrations can invalidate that approximation.

For an explicitly synthetic axial example, put $t=z/z_0$ and choose an on-axis excitation $E(t)=1/(1+t^2)$, inspired by a Gaussian beam. Stipulate a circular Gaussian function for the detection-plane light intensity, with standard deviation $s(t)=s_0\sqrt{1+t^2}$. In dimensionless coordinates $\mathbf{v}=\mathbf{u}/s_0$, put $\alpha =a/s_0$. ThenThe expression for $D_{\alpha }$ follows by integrating $p$ over a disk using polar coordinates. The assumptions specify a toy model with matched axial scales; its parameters are not a calibration for a real objective optical lens. It omits Airy disk rings, vector diffraction, and unequal excitation and emission wavelengths.

For a vanishingly small pinhole, the peak-normalized synthetic response tends to $(1+t^2{)}^{-2}$, whose full width at half maximum is $2\sqrt{\sqrt{2}-1}{z}_0\simeq 1.2872z_0$. With all collected emission accepted it tends to $(1+t^2{)}^{-1}$, of width $2z_0$. Meanwhile $H_{\alpha }(0)=1-e^{-{\alpha }^2/2}\to 0$ as the pinhole closes. The narrower normalized curve therefore accompanies a vanishing detected signal; peak normalization must not conceal that loss.

At one depth, scan the illumination spot across $x$ and $y$ and assign the detector reading during each dwell interval to a pixel. Move the specimen or objective optical lens axially to acquire the next depth. A stack of these optical sections samples three-dimensional structure. With beam scanning, the returning fluorescence commonly retraces the scanning optics: this descanning holds the image of the currently illuminated point at the detection pinhole. With stage scanning, the fixed spot and pinhole remain aligned while the specimen moves. ^[ref]

An optical section is a depth-weighted image of an intact specimen, with a finite axial response and tails. A physical cut actually separates a slice of material; its thickness and cutting damage are properties of the preparation. Changing optical focus selects another optical section without removing a slice, although illumination can still damage a specimen. Neither a thin physical slice nor a sharply displayed optical section makes the imaging point spread function infinitely narrow. ^[ref]

Point response and section thickness answer different questions. For a spatially uniform, very thin fluorescent sheet, its response versus axial displacement is proportional to the lateral integral of the three-dimensional confocal response,Consequently a trace through a single bead, $H_a(0,0,z)$, does not by itself give the sheet's optical-section thickness. State which object and which full width at half maximum are being measured. A smaller axial sampling step gives more samples of the same response, rather than automatically improving diffraction-limited resolution. ^[ref]

Construct an original illustrative spectrum using ordinary frequency detuning $x={\nu }_L-{\nu }_0$. Write the longitudinal Doppler shift in frequency units as $u=v/{\lambda }_0$, where ${\lambda }_0=c/{\nu }_0$. Choose a normal distribution with standard deviation ${\sigma }_D$ and a weak-probe Lorentzian line shape with half width at half maximum ${\gamma }_{\nu }$:Thus $B(0)=1$. Choose ${\sigma }_D=120\mathrm{MHz}$ and natural linewidth $\Gamma /(2\pi )=6\mathrm{MHz}$, so ${\gamma }_{\nu }=3\mathrm{MHz}$. The Gaussian-only full width at half maximum is $2\sqrt{2\ln 2}{\sigma }_D\simeq 282.58\mathrm{MHz}$; the averaged profile also includes the finite Lorentzian line shape width.

Add a deliberately phenomenological Lamb dip:Here $A_s$ is the absorption coefficient normalized to the pump-off line-centre value. The factor $0.24$ is chosen, not inferred from populations or a fit. The dip's assigned full width at half maximum $w_s$ borrows the single-atom power broadening scale purely to illustrate contrast versus width. It is not the preceding overlap integral or a fit to historical spectra. There are no crossover features in this isolated-transition toy spectrum.

For $s=1$, the subtracted dip has depth $0.12$ and width $8.49\mathrm{MHz}$; for $s=9$, depth $0.216$ and width $18.97\mathrm{MHz}$. Its half-depth points are $x=\pm w_s/2$. Measure that width relative to the curved background $B(x)$, not a horizontal line through the total absorption minimum.

For a chosen cell optical depth ${\tau }_s(x)=0.4A_s(x)$, the line-centre pump-off transmission is $e^{-0.4}\simeq 0.6703$. At $s=1$ it is $e^{-0.352}\simeq 0.7033$. The transmitted probe therefore increases by about $4.92$ percent relative to pump off. Absorption contrast, transmission contrast and spectral width are three different quantities.

For an ideal air-gap Fabry–Pérot etalon at normal incidence, the free spectral range is ${\nu }_{\mathrm{FSR}}=c/(2d)$ and the peak full width at half maximum is $\delta \nu ={\nu }_{\mathrm{FSR}}/\mathcal{F}$. To separate a pair of lines spaced by $\Delta \nu$ by at least $M$ peak widths, require $d\mathcal{F}\ge Mc/(2\Delta \nu )$. Putting both lines inside one unambiguous free spectral range additionally requires $d<c/(2\Delta \nu )$, so $\mathcal{F}>M$. In terms of their vacuum wavelengths, $c/\Delta \nu ={\lambda }_1{\lambda }_2/|{\lambda }_2-{\lambda }_1|$.

The strict order condition needs a margin if tails from adjacent periodic peaks are to be negligible. If every neighbouring peak in a scanned two-line periodic spectrum must be separated by $M$ widths, the stronger condition is $\min (\Delta \nu ,{\nu }_{\mathrm{FSR}}-\Delta \nu )\ge M{\nu }_{\mathrm{FSR}}/\mathcal{F}$. Ordinary resolving power alone does not prevent ambiguous repeated orders. A fixed etalon's transmission as a function of wavelength is a periodic-in-frequency filter; it is not automatically a plot of the incident source's spectral lines.

The finesse of a Fabry–Pérot interferometer is its free spectral range divided by a transmission peak's full width at half maximum. Large finesse means a narrow accepted frequency interval compared with the spacing between repeated orders. Here half maximum means $T=1/2$ of the ideal unit peak, with zero detector background subtracted; it does not mean halfway between the transmission minimum and maximum.

Write $A=4R/(1-R{)}^2$ for the coefficient multiplying ${\sin }^2(\delta /2)$ in Fabry–Pérot transmission from multiple reflections. This coefficient is sometimes called the coefficient of finesse; it is not the finesse $\mathcal{F}$. About a resonance $\delta =2\pi m$, put $ϵ=\delta -2\pi m$. Solving $T=1/2$ givesand, since one free spectral range advances the wave phase by $2\pi$,These are exact for the ideal nondispersive model with constant $R$ and linear frequency dependence of $\delta$. The inverse sine requires $(1-R)/(2\sqrt{R})\le 1$, or $R\ge 3-2\sqrt{2}\simeq 0.171573$. At equality, neighbouring peaks merely touch half maximum at their shared minima: the nominal width is one full free spectral range and $\mathcal{F}=1$. Isolated half-maximum peaks require the strict inequality. Below this threshold the transmission never falls to $1/2$, so this peak-width definition supplies no finesse; at $R=0$ the transmission is flat.

For large reflectivity, the small-angle approximation $\arcsin x\simeq x$ givesThis is a high-reflectivity approximation, not an identity for every $R$. The original dimensionless examples below have exact full width at half maximum values obtained from the inverse sine formula.

The beauty of the narrowing is visible directly in the geometric series: large $R$ preserves contributions from many round trips, so a small phase difference accumulates across many terms and spoils their addition. For the stipulated $10\mathrm{mm}$ vacuum gap, $R=0.95$ gives an ideal full width at half maximum of about $244.792\mathrm{MHz}$. This instrumental width is not a measured spectral line width. A real scan also depends on the source profile, accepted angles, mirror shape, alignment, and scan stability. The University of Washington handout separately discusses mirror-reflectivity, mirror-flatness, and detector-pinhole contributions; increasing $R$ alone cannot remove the others.^[ref]

Use the Driven two-level Rabi model in the ordered basis $(|e⟩,|g⟩)$, with Pauli matrices satisfying ${\sigma }_z=\mathrm{diag}(1,-1)$. Let the Rabi detuning be $\delta ={\omega }_0-\omega =2\pi ({\nu }_0-\nu )$, in angular frequency, while ${\nu }_0$ and $\nu$ are ordinary frequencies in hertz. Assume two identical square pulses, equal drive oscillation phases, constant positive Rabi angular frequency $\Omega$, constant $\delta$ during pulses and flight, and no quantum decoherence or loss. The rotating-wave approximation and isolation of the chosen two-level system are required, just as for the Driven two-level Rabi model. Free flight means no transverse drive; the reference oscillator continues running.

The pulse and free-flight Hamiltonian operators in the rotating frame areThe Pauli matrix multiplication law gives $(\delta {\sigma }_z+\Omega {\sigma }_x{)}^2=q^{2}I$. Thus the exact special unitary group $\mathrm{SU}(2)$ pulse matrix for this effective Hamiltonian operator, and the flight matrix, areThe sequence is $U_p{U}_f{U}_p$. Starting in $|g⟩$, matrix multiplication and the Born rule giveThis retains the tilted rotation axis and quantum phase accumulated during each finite pulse. At resonance it gives $P_{g\to e}(0)={\sin }^2(\Omega \tau )$; with no flight it reduces to the single square-pulse response for duration $2\tau$. Both checks would fail if arbitrary ideal $\pi /2$ rotations were silently substituted at nonzero Rabi detuning.

For resonant pulse area $\Omega \tau =\pi /2$, $|\delta |/\Omega \ll 1$ and $\tau \ll T$, the central fringe approachesIts half-height points are $|\nu -{\nu }_0|=1/(4T)$, so the central peak's full width at half maximum is $1/(2T)$ in hertz, or $\pi /T$ in angular frequency. This width assumes an effectively flat pulse envelope across that central peak, one flight time and unit contrast; it is neither a universal finite-pulse width nor a frequency uncertainty. Longer flight lets a smaller frequency difference accumulate the same observable quantum phase.

The two pulses here have the same sign and the plotted outcome is transfer to $|e⟩$. The earlier Ramsey interferometry example uses an inverse analysis pulse and reports return to its initial label. Specifying the measured outcome and pulse oscillation phases reconciles the two conventions. In a beam the lengths and speed determine $\tau$ and $T$; in the Quantronium Ramsey interference experiment electronics set the pulse times for a stationary circuit.

Choose an initially transform-limited pulse with Gaussian power and complex envelopeHere ${\sigma }_0$ is the standard deviation of the power-weighted time distribution; ${\tau }_0$ is its full width at half maximum. Neither is a field-amplitude width. The spectral amplitude is proportional to $e^{-{\sigma }_0^2{\Omega }^2}$, so the spectral power has standard deviation ${\sigma }_{\Omega }=1/(2{\sigma }_0)$. The Gaussian time-bandwidth product in ordinary frequency is ${\tau }_0\Delta \nu =2\ln 2/\pi \simeq 0.4413$, where both widths are power FWHMs.

Multiplication by $e^{i{\varphi }_2{\Omega }^2/2}$ preserves spectral power and energy. Substituting ${\varphi }^{\prime }={\varphi }_2\Omega$ into the moment identity givesThis is an exact result for a Gaussian spectrum with quadratic spectral phase. It also follows by completing the square in its inverse Fourier transform.

For these chosen inputs, ${\sigma }_0=12.7398\mathrm{fs}$. The stretched pulse has $\tau =278.88\mathrm{fs}$ and root mean square width $118.4282\mathrm{fs}$. Adding $-3000\mathrm{f}{\mathrm{s}}^2$ recovers $30\mathrm{fs}$; leaving $300\mathrm{f}{\mathrm{s}}^2$ gives $40.85\mathrm{fs}$. All phase-only cases retain $1\mathrm{nJ}$.

If GDD is cancelled but ${\varphi }_3=20000\mathrm{f}{\mathrm{s}}^3$ remains, the pulse is asymmetric. Gaussian spectral moments giveThus the centroid is $15.4033\mathrm{fs}$ and the root mean square duration is $25.2354\mathrm{fs}$. Converting this last value to a FWHM with the Gaussian factor would be wrong: the power envelope is no longer Gaussian. The calculation's integrated energy agrees in time and frequency through Parseval's identity, while its centroid and variance agree with these independent moment formulas.

A real amplifier also changes spectral amplitude. For a simple small-signal example, stipulate an intensity gain profileThis specifies finite gain bandwidth, with no saturation or extra amplifier phase. Multiplying two Gaussian spectra givesFor chosen ${\sigma }_g={\sigma }_{\Omega }$, the amplified spectrum is narrower by $\sqrt{2}$, the shortest Gaussian power FWHM becomes $30\sqrt{2}=42.43\mathrm{fs}$, and the total energy is $G_{0}U/\sqrt{2}$. Perfect spectral phase correction cannot recreate the suppressed spectral amplitude. Real gain narrowing and saturation require a material-dependent amplifier model; a constant multiplier is only a limited idealization. Gérard Mourou's 2018 lecture discusses gain narrowing as a limit on the shortest amplified pulses.^[ref]

A Gaussian pulse has a field or intensity envelope specified by a Gaussian function; the distinction must be stated. For the displayed intensity, the temporal full width at half maximum is $2\sqrt{2\ln 2}{\sigma }_t$. A transform-limited pulse with this shape has intensity spectral and temporal widths satisfying $\Delta \nu \Delta t=2\ln 2/\pi$. A quadratic spectral phase can lengthen the pulse without changing its spectral intensity.

A spectral doublet comprises two neighbouring spectral lines. Resolving it requires information about the line response, relative strengths and noise, not only precision in reading an angle. Choose two equal angular responses shaped as Gaussian functions with common standard deviation $\sigma$ and separation $\Delta$. If $x$ is angular displacement from their midpoint, their summed signal isThese are assumed response shapes, not a measured Siegbahn frequency spectrum. By symmetry $f^{\prime }(0)=0$, andThe midpoint changes from a maximum to a minimum at $\Delta =2\sigma$. For this particular equal-width, equal-height model, two distinct maxima occur when $\Delta >2\sigma$. Each isolated Gaussian function has full width at half maximum $2\sqrt{2\ln 2}\sigma \simeq 2.355\sigma$; the two-maxima threshold is therefore about $0.849$ times that width.

In this chosen model, the two Bragg angles are $30.000000^{\circ }$ and $30.033085^{\circ }$. Their separation is $0.827$ times $\sigma$ for $\sigma =0.040^{\circ }$, giving one maximum, and $4.136$ times $\sigma$ for $\sigma =0.008^{\circ }$, giving two. Sharpening the response can reveal the doublet without changing either emitted wavelength.

The Rayleigh criterion uses a first-zero convention for specified diffraction patterns. A Gaussian function has no finite zero, so its two-maxima threshold is not the Rayleigh criterion. Neither convention is a universal inference limit: known response shapes and adequate signal-to-noise ratio can support estimating a separation before two visible maxima appear. Unequal strengths, background, finite sampling and uncertainty in the response can make that inference much harder. Measurement resolution, the precision of an isolated line's centre and absolute wavelength accuracy should therefore be reported separately.

Consider an original illustrative resonance model, with all numerical settings chosen here. On one isolated branch approximate the transition frequency by $\nu (B)={\nu }_0+aB$, with known slope $a=20MHz/mT$ and drive frequency ${\nu }_d=1300\mathrm{MHz}$. Choose ${\nu }_0=1100\mathrm{MHz}$, hence resonance field $B_r=10\mathrm{mT}$. This local linear model omits hyperfine structure and nonlinear level mixing; it is not the historical hydrogen Zeeman splitting calculation.

Use a Lorentzian line shape for a weak-drive current dip,Choose baseline $C=1$, depth $A=0.25$ in arbitrary detector signal units, and full width at half maximum $w_{\nu }=100\mathrm{MHz}$, so $w_B=5\mathrm{mT}$. The half-depth points are $B_r\pm w_B/2$; this checks the factor of four. If an exponential decay with lifetime $\tau$ alone set this width, natural linewidth would give $w_{\nu }=1/(2\pi \tau )$ and $\tau \simeq 1.59\mathrm{ns}$. The chosen width is an explanatory input, not a measurement of a historical lifetime.

Generate 65 equally spaced fields from $2$ to $18\mathrm{mT}$ and add independent Gaussian noise with known standard deviation ${\sigma }_I=0.004$ signal units. A simultaneous nonlinear regression fit of $C,A,B_r,w_B$ gives $B_r\simeq 10.0007\mathrm{mT}$, $w_B\simeq 4.9305\mathrm{mT}$ and ${\nu }_0\simeq 1099.986\mathrm{MHz}$. With residual derivatives $J_{ik}=\partial I(B_i)/\partial {\theta }_k$ and $\theta =(C,A,B_r,w_B)$, the local covariance matrix estimate isIt gives centre standard deviation $0.01427\mathrm{mT}$, or $0.285\mathrm{MHz}$ in the inferred intercept. This uses the stipulated noise variance and a local linear approximation, rather than treating the width as the centre error. Repeated noisy samples can locate a broad line's centre much more closely than its width.

Include a common magnetic calibration offset with standard deviation $u_b=0.020\mathrm{mT}$, a fractional gain error $u_g=0.001$, and independent drive-frequency error $u_d=0.100\mathrm{MHz}$. To first order the inferred variance isHere $u$ denotes a standard uncertainty, not a frequency. The four contributions in $\mathrm{MHz}$ are $0.285$, $0.400$, $0.200$, and $0.100$; adding their squares gives $u({\nu }_0)\simeq 0.540\mathrm{MHz}$. The calibration offset shifts all sampled fields together, so increasing the sample count cannot average it away. Uncertainty in $a$, unresolved components, asymmetric line distortion, and errors in the assumed field dependence would require additional terms or a richer model. The historical inference requires the appropriate Zeeman splitting branches, rather than simply drawing a straight line through one resonance.

For a modern chosen example, the National Institute of Standards and Technology lists neutral magnesium atomic masses $23.985041697\mathrm{u}$, $24.985836976\mathrm{u}$, and $25.982592968\mathrm{u}$, with reference isotopic abundances $0.7899$, $0.1000$, and $0.1101$. They describe the isotopes with mass numbers 24, 25, and 26, respectively.^[ref] These modern inputs are not a reconstruction of historical observations.

Use the dimensionless mass-to-charge ratio coordinate $x=(m/\mathrm{u})/z$. For the illustration only, approximate ion masses by the neutral atomic masses. The singly charged positions are then near $23.98504$, $24.98584$, and $25.98259$; the doubly charged companions are near $11.99252$, $12.49292$, and $12.99130$. The neutral-to-ion correction already given above is essential at higher accuracy. A peak near $x=12$ alone need not identify a chemical element: another ion with a different electric charge can occupy the same mass-to-charge ratio interval.

For the narrow singly charged peak at $x\approx 24.98584$, the chosen full width at half maximum $\Delta x=0.12$ gives mass resolving power $R\approx 208.2$; widening to $\Delta x=1.20$ gives $R\approx 20.82$. Adjacent isotope spacings here are approximately one unit in $x$. Peaks are visibly separated in the first chosen profile and overlap strongly in the second. A small uncertainty in a fitted peak centre would not by itself make the overlapping components resolved.

Even integrated peak areas need a response model. For isotope fraction $f_i$, let ${\eta }_{i,z}$ collect the probability of ionization into electric charge state $z$, transmission through the analyzer, and the detection response. In a linear instrument with overall exposure factor $C$, the component area isEqual ${\eta }_{i,z}$ at a fixed $z$ makes normalized areas equal the fractions; otherwise it does not. In the chosen example, doubling only the mass-25 response changes the normalized areas to $0.71809$, $0.18182$, and $0.10009$, although the neutral sample is unchanged. Photographic darkening introduces another possible nonlinearity: neither line darkness nor peak height is a universal measure of neutral isotopic abundance. Calibration of position and calibration of intensity answer different questions. Matched standards, a known electric charge assignment, and a response correction make isotope analysis useful; a visually impressive set of peaks alone does not supply those controls.

Mass resolving power compares a peak coordinate $x$ with a stated width or separation criterion $\Delta x$. With $\Delta x$ taken as full width at half maximum, a Gaussian function of standard deviation $\sigma$ hasAt the neon-20 coordinate in the synthetic mass spectrum, $\sigma =0.04$ gives $\Delta x\approx 0.0942$ and $R\approx 212$, while $\sigma =0.70$ gives $\Delta x\approx 1.648$ and $R\approx 12.1$. The width criterion and relative peak areas both matter when judging whether a small neighboring isotope can be distinguished. Mass resolution also differs from the accuracy of a calibrated peak position.

Let $v$ be the source velocity towards the absorber, positive for approach along the photon direction. A first-order Lorentz transformation of the photon energy gives $E^{\prime }=E+vp$; hence, for $|v|\ll c$,Moving the source shifts its emission line through the stationary absorption line. A minimum in transmitted photon counts marks their strongest overlap. The absorber is the narrow energy discriminator; the downstream particle detector only needs to count the transmitted gamma rays.

For a thin absorber with unsplit Lorentzian line shapes, the overlap is a convolution of the source and absorber profiles. The combined full width at half maximum is $W={\Gamma }_s+{\Gamma }_a$. A simple expected transmission model isHere $I_0$ is the off-resonance intensity, $v_0$ encodes a relative shift of the two transition energies, and $A$ includes the resonant absorber strength and $f_s{f}_a$. The convolution result can be checked by multiplying the Fourier transforms of two Cauchy distributions: their exponential decay constants add. The width of the absorption dip in velocity is $cW/E_{\gamma }$, rather than the width of one individual spectral line converted to velocity.

For the teaching system with ${\Gamma }_s={\Gamma }_a={\Gamma }_E$ and $v_0=0$, the dip's full width at half maximum is $2.9979\mathrm{mm}{\mathrm{s}}^{-1}$. Stipulate $A=0.25$ at zero temperature and $10,000$ expected off-resonance counts per equal-duration bin: the central bin then has $7,500$ expected counts, and the half-depth bins have $8,750$. These are model expectations, without sampled counting noise. Holding all other parameters fixed while heating both lattices to $k_{B}T=2\hslash \omega$ multiplies $A$ by $[f(T)/f(0){]}^2$. This isolates how a falling recoil-free fraction weakens the dip.

The thin-absorber model is deliberately limited. A thick absorber can saturate and broaden the dip; unresolved hyperfine structure can create multiple components. Differences in chemical surroundings can shift the centers, and uncontrolled motion can smear them. These effects must be modeled before interpreting an observed width as a natural linewidth. Their sensitivity also makes the Mössbauer effect useful: a small displacement of a narrow dip can reveal a small change in the local nuclear environment.

A simple isolated particle resonance model starts with a survival amplitude proportional toIts survival probability is $|a(t){|}^2=e^{-\Gamma t/\hslash }$, giving mean rest-frame lifetime $\tau =\hslash /\Gamma$. The Fourier transform of this one-sided amplitude is proportional to $(\Gamma /2-i(E-E_0){)}^{-1}$; its squared modulus gives a Lorentzian line shape. Write its normalized density asAt $E-E_0=\pm \Gamma /2$ the density is half its peak, so its intrinsic full width at half maximum is $\Gamma$. This is a local isolated-line approximation, extending the offset coordinate over the real line; physical production thresholds, interference and energy-dependent decay widths can require a richer quantum scattering model.

Now stipulate a centered normal distribution for the measurement response, with standard deviation $\sigma$. A measured coordinate $x$ has densityThis convolution describes the smearing of an underlying line, with constant acceptance and no background. For a pair spectrometer the response summarizes invariant-mass resolution; for an idealized collider scan it summarizes incoming center-of-mass energy spread. These are different experimental sources of uncertainty even when the same mathematical response is useful.

Choose an arbitrary scale $E_{\ast }$, intrinsic decay width $\Gamma =0.20E_{\ast }$ and response standard deviations $0.40E_{\ast }$ and $1.20E_{\ast }$. Numerical integration gives apparent full width at half maximum values of approximately $1.05E_{\ast }$ and $2.93E_{\ast }$, respectively. The state and its rest-frame lifetime are identical in both cases. The normal distribution's width alone is $2\sqrt{2\ln 2}\sigma$; adding that width to $\Gamma$ in quadrature is not an exact rule for this convolution.

For a purely illustrative $E_{\ast }=1\mathrm{MeV}$, taking $\hslash \simeq 6.5821\times 10^{-22}\mathrm{MeV}\mathrm{s}$ gives $\tau \simeq 3.2911\times 10^{-21}\mathrm{s}$. These are invented line parameters, not J/psi meson measurements. Substituting the larger apparent width into $\hslash /\Gamma$ would incorrectly shorten the inferred lifetime by nearly a factor of fifteen. An unresolved width can support an upper bound on intrinsic decay width, and consequently a lower bound on lifetime, only after the instrumental response and model assumptions have been assessed.

The discovery peak and its later interpretation answer different questions. The J/psi meson is understood as charmonium, a bound charm quark–charm antiquark system. Richter's 1976 lecture distinguishes the initial discovery from subsequent tests of competing explanations, including the related level structure and later evidence for particles carrying uncancelled charm. A narrow peak alone did not establish all these properties in 1974.^[ref]

For a closed two-level system with spontaneous emission population decay $P_e(t)=P_e(0)e^{-\Gamma t}$ and no additional dephasing, the natural linewidth has full width at half maximum $\Gamma$ in angular frequency, or $\Gamma /(2\pi )$ in hertz. The excited-state lifetime is $1/\Gamma$.

For weak drive in this homogeneous model, half maximum occurs at $|\Delta |=1/T_2$. The absorption full width at half maximum is thereforeThe example gives $2.653\mathrm{Hz}$ or $0.06230\mu \mathrm{T}$. At finite drive the model width is multiplied by $\sqrt{1+s}$. Field inhomogeneity and unresolved components can also broaden an actual line, so its width alone does not determine pure dephasing.

The same width appears by an original Fourier transform calculation. For $a=1/T_2$ and a free homogeneous signal demodulated exactly at resonance,This connects exponential memory to a Lorentzian absorptive line. It is a mathematical connection, not evidence that the 1946 teams performed pulsed Fourier transform spectroscopy. Instrumental signal bandwidth is a separate choice: narrowing a receiver band can reject thermal electrical noise, but too narrow a band distorts a changing signal. A sample linewidth and a receiver bandwidth are not interchangeable.