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For Lebesgue integrable functions $f,g$ on $\mathbb{R}$, use the Fourier transform convention $\hat{f}(\xi )={\int }_{\mathbb{R}}f(x)e^{-2\pi ix\xi }dx$. ThenThus the integral combining all translated products becomes an ordinary multiplication at each frequency. Different Fourier transform normalizations can place a constant in this identity; fixing the convention prevents that ambiguity. Its use in filtering is illustrated by ^[ref].

For the indicator function $q=1_{[0,1]}$,with $\mathrm{sinc}(0)=1$. Therefore the triangular function $q\ast q$ has Fourier transform $e^{-2\pi i\xi }{\mathrm{sinc}}^2(\xi )$. The extra factor of the sinc function improves the large-frequency bound from a constant times $1/|\xi |$ to one times $1/|\xi {|}^2$, matching the smoother output in the overlap picture. Fourier decay of repeated interval convolutions extends this calculation to further repetitions.

For a proof outline, Fubini's theorem permits exchanging the two integrals because the integral of $|f(y)g(x-y)|$ over $(x,y)$ is $\Vert f{\Vert }_1\Vert g{\Vert }_1<\infty$. Set $u=x-y$ in the inner integral:The factorization of the complex exponential is the mechanism behind the simplification. The discrete convolution version similarly turns coefficient sums into products of generating functions.

A diffraction grating is a periodic optical structure whose repeated slits or grooves direct light into diffraction orders through wave interference. Its beauty is that many individually broad contributions cooperate to select narrow directions: the spacing determines where the principal peaks occur, while the number of illuminated elements determines how sharply they are selected. This makes a diffraction grating useful in a spectrometer, where separating wavelengths spatially lets us examine an optical spectrum. ^[ref].

Consider a transmission diffraction grating in air, illuminated at normal incidence by a plane wave. Let $d$ be the centre-to-centre slit spacing, $\lambda$ the wavelength in the surrounding medium, and $\theta$ the signed outgoing angle from the forward normal. In Fraunhofer diffraction, where observation directions specify the far-field pattern, adjacent contributions have path difference $d\sin \theta$ and phase difference $\delta =2\pi d\sin \theta /\lambda$. Their wave amplitudes reinforce whenever $\delta =2\pi m$, giving $d\sin \theta =m\lambda$. This normal-incidence geometry should be kept distinct from the incident and reflected angle convention in the grating equation below. ^[ref].

Angular colour separation follows immediately. At fixed nonzero diffraction order, increasing the wavelength moves the peak farther from the normal. With $d=2\mu \mathrm{m}$, first-order light at $450\mathrm{nm}$ leaves at $13.003^{\circ }$, whereas $650\mathrm{nm}$ leaves at $18.966^{\circ }$: a separation of $5.963^{\circ }$. The zero diffraction order stays at $0^{\circ }$ for both wavelengths. Thus grating dispersion comes from an orderly dependence of wave interference on wavelength, rather than an arbitrary assignment of colours to directions.

For $N$ identical slits with uniform coherent illumination, meaning stable relative wave phases across the array, the normalized wave interference factor isThe apparent $0/0$ at a principal peak has limiting value one. A short explanation uses the geometric series: equal wave phases add $N$ equal complex amplitudes; a small directional change distributes those contributions around the complex plane and allows cancellation. With each slit's contribution held fixed, peak scalar wave intensity scales as $N^2$. The normalized plots remove that height change to make the narrowing visible. Stable relative wave phases, or optical coherence, are essential; a sum of independent intensities would lose this array structure. ^[ref].

For finite slit width $a$, the ideal rectangular-slit model multiplies $F_N$ by a broad single-slit diffraction envelope:The sinc function factor, extended continuously at $u=0$, controls the brightness available to the narrow peaks. For example, $a=0.5\mu \mathrm{m}$ gives envelope values about $0.811$, $0.405$, and $0.090$ at diffraction orders $1$, $2$, and $3$ in the $600\mathrm{nm}$ example. If an envelope zero coincides with an allowed diffraction order, that peak disappears. A permitted direction therefore need not produce a visible bright spot. ^[ref].

Historically, Joseph von Fraunhofer used fine parallel wires stretched between threaded rods to make diffraction gratings. Kenyon's historical apparatus account describes both this construction and a later reconstruction with visible light. The example makes the governing idea tangible: regular spacing can turn a simple array of obstacles into an instrument for distinguishing wavelengths. ^[ref].

The characteristic function of a Dyadic Rademacher series is $\sin t/t$, interpreted as one at zero. The characteristic function of a sum of independent variables gives the finite cosine products, and weak convergence of probability measures gives their displayed limit for every real $t$. This argument includes zeros of the sinc function without taking logarithms of the factors.

An observation interval of finite duration is not an infinitely sharp frequency selector. Combine the lock-in quadratures intoFor unwanted electric potential difference $s(t)=B\cos ((\omega +\delta )t+\psi )$, where $\delta$ is an angular frequency offset and $\omega +\delta >0$, expansion into complex exponentials gives the exact responseThe sinc function formula follows by integrating the exponential function and factoring $e^{\mathrm{i}\nu T}-1=2\mathrm{i}{e}^{\mathrm{i}\nu T/2}\sin (\nu T/2)$; its value at zero follows from a limit. The first term changes slowly after multiplication; the second oscillates at the sum frequency. It vanishes at exact matching when $T$ is an integer number of reference cycles, each lasting one period of a function, but cannot generally be discarded in a short-window calculation.

For $|\delta |\ll \omega$ and many reference cycles, the first term usually dominates: its modulus is $B|\mathrm{sinc}(\delta T/2)|$ and its complex argument includes $\delta T/2$. Thus $|\delta |T\ll 1$ gives almost no rejection. A longer observation separates closer frequencies, with the first zeros of this term at $|\delta |=2\pi /T$. Those zeros are not a brick-wall cutoff: the sinc function has nonzero lobes beyond them. At $\delta =0$ a synchronized contribution remains however large $T$ becomes.

For a numerical check, use $f_0=1\mathrm{Hz}$, $T=4\mathrm{s}$, $\psi =0$ and an unwanted frequency $f=1.125\mathrm{Hz}$. Then $(f-f_0)T=1/2$ and the exact expression giveswhereas the difference-frequency term alone gives $2/\pi \approx 0.636620$. An interference amplitude $B=0.9\mathrm{V}$ therefore contributes $0.539254\mathrm{V}$ to $|Z_T|$ in this constructed example. A nearby frequency can survive strongly even though the three-times-frequency component in lock-in harmonic rejection cancels exactly.

A practical lock-in amplifier's low-pass filter weights past products differently from this rectangular observation interval. Its time constant and roll-off determine near-reference rejection and settling time; narrowing detection slows the response.^[ref]

For $N$ equally spaced samples at sampling rate $f_s$, the discrete Fourier transform has frequency bins spaced by $f_s/N=1/T_{\mathrm{obs}}$. A rectangular observation window broadens a sinusoid into a sinc function profile whose nearest zeros are $1/T_{\mathrm{obs}}$ from its central frequency. This sets a conventional scale for distinguishing nearby spectral components. It is not a universal lower bound on the uncertainty of frequency estimation for an isolated tone: a specified signal model and noise level can allow precision finer than the bin spacing. Padding with zeros interpolates the spectral display without increasing the observation duration.

An even train of rectangular pulses of width $w<T$, period $T$ and unit area per pulse has Fourier coefficients $a_n=(2/T)\mathrm{sinc}(\pi nw/T)$ for $n\ge 0$. Here the sinc function is normalized as $\mathrm{sinc}z=\sin z/z$, and the constant contribution is $a_0/2=1/T$. Integrating the cosine over the pulse proves the formula; evenness eliminates the sine coefficients. At a pulse edge, pointwise convergence of a piecewise smooth Fourier series gives the mean of the one-sided values rather than a prescribed endpoint value.

Consider a uniform plane wave of unit complex amplitude transmitted by an ideal optical slit of full width $D$. Treat the field as invariant along the optical slit height: this is one-dimensional transverse Fresnel propagation, so the prefactor is $1/\sqrt{i\lambda z}$, rather than the two-dimensional prefactor $1/(i\lambda z)$. Removing the carrier $e^{ikz}$ givesHere $I_{\mathrm{inc}}$ is incident irradiance. This infinitely long slit model describes a transverse cut away from the ends of a sufficiently long physical slit; its finite conserved quantity is transmitted power per unit optical slit height.

Put $t=\xi /D$ and $q=xD/(\lambda z)$. Substitution reduces the integral toThis form separates the pattern's shape from its physical scale. Reflection $t\mapsto -t$ proves $A(-q,N_D)=A(q,N_D)$, so the light intensity is an even function of $x$. At $N_D=0$, direct integration gives $A(q,0)=\sin (\pi q)/(\pi q)$, with limit one at $q=0$. Hence the Fraunhofer diffraction comparison is the squared sinc function. In this paragraph the unnormalized sinc function takes argument $\pi q$.

For an original example, choose wavelength $\lambda =633\mathrm{nm}$ and full width $D=0.400\mathrm{mm}$. At $z=0.0200\mathrm{m}$, $N_D\simeq 12.6382$, $N_a\simeq 3.15956$ and the edge-to-centre wave phase change is $9.92604$ radians. At $z=2.00\mathrm{m}$, $N_D\simeq 0.126382$, $N_a\simeq 0.0315956$ and that wave phase change is only $0.0992604$ radians. Even the nearer screen is fifty slit widths away; that geometric comparison alone misses its strong Fresnel diffraction.

These are numerical predictions from the displayed scalar integral, using the composite trapezoidal rule, rather than measured profiles. Each curve is divided by its own peak light intensity; the two horizontal scales differ. At 2 cm the computed central irradiance is about $1.14812I_{\mathrm{inc}}$, with brighter off-axis maxima. At 2 m it is about $0.126272I_{\mathrm{inc}}$ and is the peak. The dashed Fraunhofer diffraction curve has its own central normalization; its first zeros are at $x=\pm \lambda z/D=\pm 3.165\mathrm{mm}$. The finite-distance Fresnel diffraction curve remains slightly nonzero there.

Normalization hides the brightness change, but not the change in shape. The Fourier transform propagation multiplier has unit complex modulus, so Plancherel theorem givesThis is conservation of transmitted power per unit height in the ideal scalar model. An integral over only the displayed screen interval misses the tails; a peak-normalized profile is not a unit-integral probability density. Spreading and decreasing brightness together preserve the transmitted power.

The Nyquist–Shannon sampling theorem says that a bandlimited function can be recovered exactly from equally spaced samples, provided the sampling rate is high enough. It is the bridge between a continuous waveform and a sequence of numbers: the missing values between samples are determined by the restriction on its frequencies.

Here is a precise finite-energy version. Let $f\in L^2(\mathbb{R})$, meaning ${\int }_{\mathbb{R}}|f(t){|}^{2}dt<\infty$, and suppose its Fourier transform, defined through the Plancherel theorem, is zero outside $[-2\pi B,2\pi B]$. Thus $B>0$ is the largest allowed ordinary frequency, in cycles per second, rather than an angular frequency. Set $T=1/f_s$, where $f_s$ is the number of samples per second. If $f_s\ge 2B$, the continuous representative of $f$ is recovered byThe quotient is the normalized sinc function, with value one when its argument is zero. Each shifted sinc function is one at its own sampling time and zero at every other sampling time. Its coefficient is therefore exactly the required sample value. For this L2 space formulation, the reconstruction converges in the L2 space norm and uniformly in $t$. The critical rate $f_s=2B$ is allowed for this class; choosing $f_s>2B$ leaves a gap between the occupied frequencies and the sampling boundary.

For a numerical example, a bandlimited function with no frequencies above $3$ cycles per second can be reconstructed from its samples every $0.1$ seconds: $f_s=10>6$. Exact reconstruction uses the whole sequence indexed by $\mathbb{Z}$. A finite recording and a finite sum of sinc functions introduce truncation error; joining adjacent samples by straight lines is generally a different reconstruction.

The obstruction is aliasing. At sampling rate $f_s=10$, the two distinct waveformshave identical samples at $t=n/10$, becauseThe samples alone cannot tell whether the waveform oscillates twice or eight times per second. If we know in advance that all frequencies are below $5$, the $8$-cycle waveform is excluded and the ambiguity disappears. These pure cosines are an illustrative aliasing example, rather than members of the finite-energy L2 space class in the stated theorem.

Exactly at a sampling boundary, the function class matters. At rate $f_s$, the sine $\sin (2\pi (f_s/2)t)$ has zero values at every $n/f_s$. It is a nonzero waveform indistinguishable from zero on that grid, but has infinite energy and is excluded by the finite-energy hypothesis above. This explains why a rule intended to cover arbitrary pure sinusoids should demand a strict inequality. For spatial data, replace seconds by distance: the Nyquist spatial frequency is $1/(2p)$ for sample pitch $p$.

The proof idea is short: the samples are Fourier coefficients of the Fourier transform restricted to a sampling-sized frequency interval. Because a bandlimited function has no contribution outside that interval, its Fourier series determines the whole Fourier transform; Fourier inversion then gives the sinc function expansion. See sampling expansion by periodic Fourier projection for this connection.

Claude E. Shannon's Communication in the Presence of Noise (1949), Section II, Theorem 1 and equation (7), gives the sampling statement and reconstruction using shifted sinc functions in a communication setting. Shannon explicitly notes earlier mathematical forms and credits Nyquist with emphasizing the sampling interval in telegraphy.^[ref]

Assume an initially unexcited chemical atom, two nonoverlapping square pulses starting at $0,T$, and a common coherent field phase. In the weak-excitation limit $|\Omega |{\tau }_p\ll 1$, substitute $c_1=1+O({\Omega }^2)$ into the second probability amplitude equation. After the second pulse,where the sinc function means $\mathrm{sinc}u=\sin u/u$, with value one at zero. Squaring givesThe printed prefactor is a factor of four too small for the supplied equations and pulse durations. A direct counterexample is $\delta =0$: the exact resonant Rabi oscillation gives $P_2={\sin }^2(\Omega {\tau }_p)$ for two pulses of duration ${\tau }_p$, because the intervening resonant rotating-frame evolution is trivial. Its small-area limit is $(\Omega {\tau }_p{)}^2$, rather than $(\Omega {\tau }_p/2{)}^2$. The latter is the single-pulse result before coherent addition.

The condition $\omega {\tau }_p\ll 1$ printed in the source also does not justify the derivation from a physical carrier field. A consistent rotating-wave approximation pulse regime isIf the supplied equations are treated as the mathematical model regardless, their weak-excitation integral remains valid for any ${\tau }_p$; subcycle pulses of the original real field require the discarded counter-rotating term and depend on carrier phase. $T$ here is the separation of pulse starts, as specified, rather than the dark interval $T-{\tau }_p$.

The ground electron configuration of caesium has a single $6s$ electron, with $J=1/2$. Addition of angular momentum giving $F=3,4$ fixes the nuclear spin to $I=7/2$. With dimensionless angular momentum operators, write the Fermi contact interaction asThe two hyperfine levels therefore differ by $4A$. In the convention ${\mathbf{\mu }}_I=g_I{\mu }_B\mathbf{I}$ and $g_s\simeq 2$,This exposes a missing ingredient in a prediction from the supplied data: $Z$ and $g_I$ alone do not specify the valence electron contact density. Screening effect and penetration of the closed core strongly affect it. If $g_I$ instead denotes the total nuclear moment in units of ${\mu }_B$, it must first be divided by $I$ in the displayed convention.

An explicitly crude unscreened hydrogen-like ion model uses $|{\psi }_{6s}(0){|}^2=Z^3/(\pi 6^3{a}_0^3)$ and gives ${\nu }_0\simeq 5.76\times 10^{11}\mathrm{Hz}$, or ${\omega }_0\simeq 3.62\times 10^{12}{\mathrm{s}}^{-1}$. It describes neither the shielding nor the radial function of neutral caesium. Using a pure unit-charge $6s$ hydrogen orbital instead gives only $3.46\mathrm{MHz}$, illustrating why a unique useful estimate cannot follow from an unspecified effective charge.

For physical neutral ${}^{133}\mathrm{Cs}$, the independently known zero-magnetic field clock transition isThis measured frequency, used to define the second, is supplementary physical information, not a consequence of $Z=55$ and the rounded $g_I$ alone.^[ref] The direct electronic transition is in the microwave range and is a magnetic dipole transition, rather than a direct optical electric dipole transition. An optical implementation needs a two-field Raman transition.

Using the corrected weak-pulse Ramsey spectrum of part (b), the peak at $\delta =0$ has height $(\Omega {\tau }_p{)}^2$. The quantum interference fringes have spacing $2\pi /T$ in angular frequency, with zeros at $(2k+1)\pi /T$, and the sinc function envelope has its first zeros at $\delta =\pm 2\pi /{\tau }_p$. For $T\gg {\tau }_p$, the central peak has approximately full width at half maximum $\pi /T$ in angular frequency. The clock transition between the two $M_F=0$ states has no first-order Zeeman energy, although a quadratic one remains.

The first-order excitation amplitude is the Fourier transform of the temporal pulse envelope evaluated at the Rabi detuning. Two rectangular time windows therefore give exactly the mathematical structure of double-slit interference: each window gives a sinc function amplitude, and their separation gives a relative quantum phase $\delta T$. The transition probability is a sinc function squared envelope multiplied by ${\cos }^2(\delta T/2)$.

Pulse duration corresponds to slit width, pulse-start separation to slit-center separation, and Rabi detuning to the transverse wavevector sampled on the screen. The narrow fringes measure the long separation, while the broad envelope measures the short interaction duration. As in spatial interference, adding probabilities instead of coherent probability amplitudes would lose the fringes. Randomizing the relative pulse phase averages the cosine squared to $1/2$.

The result depends on the component amplitudes and wave phases; adding many frequencies does not by itself guarantee one localized wave packet. For a definite sketch, take many equally spaced wavenumbers $k_j=k_0+j\Delta k$ in the same fixed band, with equal, aligned complex amplitudes $B/N$. Linearize the dispersion relation as $\omega (k_j)\simeq {\omega }_c+v_g(k_j-k_0)$, where ${\omega }_c=k_0^2/m$. Then with $\xi =x-v_{g}t$ the wave envelope isfor odd $N$. This is a normalized Dirichlet kernel: constructive interference occurs near $\xi =0$, while most other positions undergo partial cancellation. At finite $N$ there are repeated groups separated by $2\pi /\Delta k$.

As $N$ grows with the half-bandwidth $K$ fixed, $\Delta k$ decreases and the repetition spacing grows. The continuous flat-band limit isThus a sinc function envelope has a central group between its first zeros at $\xi =\pm \pi /K$, smaller side lobes and rapidly oscillating crests. At fixed bandwidth the limiting central width is $2\pi /K$; it does not shrink indefinitely just because more components are added. Smoother spectral weights can reduce the side lobes, and random wave phases would produce a different shape.

For the exact quadratic dispersion relation,The additional quadratic wave phase makes the wave packet spread and change shape. The translating sinc function approximation is valid while $|t|K^2/|m|\ll 1$; the preceding figure is the exact $t=0$ comparison, where that approximation introduces no error.

For the finite probability distribution, put ${\xi }_j=(X_j+1)/2$. These are independent fair zero-one variables, andThe binary integer ${\sum }_{j=1}^n{2}^{n-j}{\xi }_j$ is uniform on $\{0,\dots ,2^n-1\}$. HenceThese are the midpoints of $2^n$ equal subintervals of $[-1,1]$. For every bounded continuous $f$, the Riemann sums giveThus $\boxed{Y_n\Rightarrow \mathrm{Unif}[-1,1]}$ in the sense of weak convergence of probability measures.

For the almost-sure and $L^2$ assertions, the Dyadic Rademacher series $Y={\sum }_{j\ge 1}{2}^{-j}{X}_j$ converges absolutely on every sign sequence, since $|X_j|=1$. Its tail satisfies $|Y-Y_n|\le 2^{-n}$, proving almost sure convergence and convergence in L2. More precisely, zero means and independence eliminate the cross terms:This identity follows first for finite tails and then by dominated convergence. Bounded continuous test functions and bounded convergence theorem identify the probability distribution of $Y$ with the weak limit above, namely the uniform distribution on $[-1,1]$.

For the requested product identity, each Rademacher random variable has characteristic function $\mathbb{E}{e}^{it{X}_j}=\cos t$. The characteristic function of a sum of independent variables therefore yieldsBy part (a), or directly by bounded convergence theorem, this converges to the characteristic function of a uniform distribution, $\frac{1}{2}{\int }_{-1}^1{e}^{itx}dx=\sin t/t$. Consequently the dyadic cosine product isBoth sides are one at zero. At zeros of the sinc function the identity is understood as the limit of the finite products, so no logarithm or division by a vanishing factor is needed.

The half-open translates $[-\pi ,\pi )+2\pi k$ partition the real line. Exactly one summand is therefore nonzero in the periodization of $|f{|}^2$, giving ${\sum }_k|f(t+2\pi k){|}^2=1$.

Choose the Shannon scaling mask to be the $2\pi$-periodic extension of $1_{[-\pi /2,\pi /2)}$. On the support of $f$, its product with $f$ is $1_{[-\pi /2,\pi /2)}(t)=f(2t)$; outside that support both sides vanish. Thus the scaling refinement equation holds in frequency. The Fourier coefficients give its precise filter normalization:These coefficients are square summable, so the interpretation in part (b) applies.

The Fourier inversion theorem gives the Shannon scaling functionThis is the normalized sinc function. Its Fourier transform is taken in the Plancherel theorem sense: the Shannon scaling function belongs to $L^2$ but not $L^1$, so its Fourier integral is not absolutely convergent. The inverse integral above is an ordinary integral because the rectangular transform belongs to $L^1$. The resulting refinement spaces are exactly the $L^2$ functions with Fourier transform supported in $[-2^j\pi ,2^j\pi ]$; their union is dense and their intersection is zero. Thus this example also gives a full multiresolution analysis.

Use the transform convention $F(\lambda )={\int }_{\mathbb{R}}f(t)e^{-i\lambda t}dt$. The given Fourier inversion theorem givesHere $F$ is continuous, vanishes at both endpoints and belongs to $L^2[-\pi ,\pi ]$. It therefore defines a continuous periodic function. Its Fourier coefficient at index $-n$ is $f(n)$. By part (i),Pair this convergence with $e^{it\lambda }$. Since that function has normalized $L^2$ norm one, the Cauchy-Schwarz inequality yields, uniformly in $t$,The elementary integral is the sinc function,Thus the sampling expansion by periodic Fourier projection isThere is no pointwise interchange with an unproved Fourier series: the calculation first uses finite sums and then an $L^2$ limit.

The convergence can also be made absolute. Parseval's identity gives ${\sum }_n|f(n){|}^2=\Vert F{\Vert }_2^2$, and Bessel's inequality applied to $e^{it\lambda }$ gives ${\sum }_n|D(t-n){|}^2\le 1$. Henceuniformly in $t$. At an integer argument, $D$ is one at zero and zero at the other integers, so the expansion interpolates the samples exactly.

The half-open intervals $[-\pi +2\pi k,\pi +2\pi k)$ tile the line, so exactly one term contributes to ${\sum }_k|f(t+2\pi k){|}^2$. It equals one almost everywhere.

Choose the Shannon scaling mask, a $2\pi$-periodic low-pass filter of a multiresolution analysis which equals one on $[-\pi /2,\pi /2)$ and zero on the rest of $[-\pi ,\pi )$. On the support of $f$ its product with $f(t)$ equals $f(2t)$; outside that support both sides vanish. Its Fourier coefficients giveso it also has the required symbol representation. The inverse Fourier transform gives the Shannon scaling functionThis sinc function has $L^2$ norm one. It illustrates why the Fourier transform convention in part B must allow $L^2$ transforms: the Shannon scaling function is not absolutely integrable on the line.

Laplace identities. The one-sided Laplace transform isfor $\mathrm{Re}p$ sufficiently large that the integral converges. For a piecewise smooth function of exponential order, integration by parts givesDifferentiating under the integral givesIteration also yields $\mathcal{L}[f^{\prime \prime }]=p^{2}F-pf(0)-f^{\prime }(0)$. For real $k$ and $p>0$, taking the imaginary part of $(p-ik{)}^{-1}$ gives

Transformed differential equation. Let $G=\mathcal{L}[g]$, using the initial values $g(0)=1,g^{\prime }(0)=0$. ThenSumming the three contributions givesFor $\alpha \ne 0$, the condition $G(p)\to 0$ as $p\to \infty$ fixes the integration constant:It is simpler to invert the derivative: $\mathcal{L}[tg]=-G^{\prime }=1/(p^2+{\alpha }^2)=\mathcal{L}[\sin (\alpha t)/\alpha ]$. ThusThis is the sinc function. Its Taylor series begins $1-{\alpha }^2{t}^2/6+O(t^4)$, verifying both initial values. For a direct differential check set $h=tg=\sin (\alpha t)/\alpha$. Then $h^{\prime \prime }+{\alpha }^{2}h=t{g}^{\prime \prime }+2g^{\prime }+{\alpha }^{2}tg=0$. For $\alpha =0$, take the continuous limit $\boxed{g(t)=1}$, which also follows from $G=1/p$.

For $T>0$ the difference of Heaviside step functions is an interval indicator, and $v=1/2$ on $(-T,T)$ and zero elsewhere. Its Fourier transform isThe apparent singularity at $\omega =0$ is removable. The formula remains valid for real negative $T$, because the original step-function difference and the oriented integral both reverse sign; for $T=0$ both vanish. This is the Fourier transform of an interval indicator and produces a scaled sinc function.

By Fourier inversion, the sinc function $\sin (At)/(\pi t)$ has Fourier transform equal to one for $|\omega |<A$ and zero for $|\omega |>A$. Thus the two bandwidths add to giveThe requested squared-magnitude spectrum therefore has heights four and one:Values at the jump points depend on the chosen Fourier convergence convention and do not affect any integral. The graph is shown below.

The Parseval identity with this Fourier transform normalization gives

The polarization generated by the relevant third-order nonlinear optical susceptibility is proportional to $E_1{E}_2{E}_3^{\ast }$. Multiplication of its temporal and spatial wave phase factors givesThe two opposing pump arrows in the original diagram mean ${\mathbf{k}}_2=-{\mathbf{k}}_1$. Thus the mixing source points along $-{\mathbf{k}}_3$, and its conjugate signal amplitude drives a backward field. This is counterpropagating-pump phase matching in four-wave mixing.

There is a phase matching qualification to the initial frequency statement. In the stipulated nondispersive isotropic medium, $|{\mathbf{k}}_j|=n{\omega }_j/c$. Exact equality of the backward free-wave vector to $-{\mathbf{k}}_3$ therefore requires ${\omega }_4={\omega }_3$, hence ${\omega }_3={\omega }_P$. For unequal frequencies the backward direction remains the geometry suggested by the source, but the longitudinal phase mismatch has magnitude $2n|{\omega }_P-{\omega }_3|/c$. Efficient generation over a finite length requires that mismatch times $L$ be small; a sinc function factor suppresses conversion otherwise. Nondispersion does not remove the magnitude mismatch. The remainder of the calculation is explicitly degenerate four-wave mixing, so this difficulty disappears.

At the pump frequency, for the strong field $E_1$, self-phase modulation and cross-phase modulation by $E_2$ dominate the weak signal's influence. They change its propagation wave phase through its own intensity and the other pump's intensity. With continuous uniform plane waves, these are nonlinear wave phase shifts; finite pulses acquire temporal chirp, and finite transverse beams can acquire a lens. No universal factor of two should be assigned between the two shifts without specifying their polarization contraction and the field-amplitude convention.

Write $u={\tilde{A}}_3$ and $v={\tilde{A}}_4$. Differentiating the exponential substitutions cancels the terms proportional to $2\kappa$:Consequently $u^{\prime }=i\kappa v^{\ast }$ and $v^{\prime }=-i\kappa u^{\ast }$. Conjugate the latter before eliminating a field: $(v^{\ast }{)}^{\prime }=i\kappa u$, so $u^{\prime \prime }=-{\kappa }^{2}u$. The backward four-wave-mixing boundary problem requires opposite-end boundary conditions: $u(0)=a$, where $a=A_3(0)$, and $v(L)=0$. The transfer matrix isAt $L$, $0=ia\sin \kappa L+b\cos \kappa L$, so $b=-ia\tan \kappa L$. Substitution givesThe printed target expression for $A_4$ has the opposite overall sign. This is present in the original PDF, not just the transcription. With its stated first-order equations, the derivative at the far face must be $v^{\prime }(L)=-i\kappa a^{\ast }/\cos \kappa L$. The boxed result has that derivative; the printed expression has its negative. Reversing the convention for the mixing coefficient could reverse the generated phase, but would also change the coupled equations. The result above satisfies the equations and both boundary conditions together.

At the entrance, $A_4(0)=i{a}^{\ast }\tan \kappa L$, so the returned amplitude is proportional to the complex conjugate of the signal. An ordinary ideal mirror gives $ra$, with a fixed reflection coefficient $r$, rather than $a^{\ast }$. The mixing pumps supply both the extra energy and a wave phase reference. The ideal conjugate reflectivity and signal transmission areThe constancy of $|u{|}^2+|v{|}^2$ along the medium is another direct check. The relation between the incoming signal and outgoing waves allows net output energy to come from the pumps. As $\kappa L\to \pi /2$, the undepleted-pump approximation fails: the predicted fields would extract an appreciable pump fraction. This is the ideal four-wave-mixing oscillation threshold; saturation, pump depletion and noise seeding replace the divergent linearized solution. It is not an unlimited source of reflected power. Amplified phase-conjugated propagation and parametric oscillation are also the effects identified in the primary analysis by Amnon Yariv and David M. Pepper.^[ref]

For the aberrating linear medium, assume its permittivity is real and stationary at this frequency. Taking the complex conjugate of the forward-envelope equation givesThis is the backward-envelope equation with carrier $e^{-i{k}_{3}z}$. Launching the conjugate boundary field therefore permits retracing of the distorted wave. In a transverse lossless phase screen model, let transmission be $t(x)=e^{i\psi (x)}$. A return proportional to $(tU{)}^{\ast }$ acquires $t$ on its second passage and becomes $t(tU{)}^{\ast }=U^{\ast }$. Thus the aberration phase cancels. A mirror instead gives $r{t}^{2}U$, adding the aberration twice; merely reversing the central propagation direction does not reverse the wavefront phase.

The one-dimensional printed equation describes longitudinal variation; genuine transverse aberrations require the corresponding transverse Helmholtz equation or paraxial operator, for which the same real-coefficient conjugation argument applies. Exact reconstruction requires adequate aperture, the correct conjugate amplitude and polarization, uniform pump-product phase, and an unchanged reciprocal lossless medium. Absorption, missing scattered modes or a changed aberrator prevent a universal reconstruction. This is phase conjugation and thin phase-plate cancellation by phase conjugation under specified conditions, rather than reversal of every possible propagation loss.