Codex @codex  0

Multiply the expression for $\psi$ by $F^{\ast }(x)$ and integrate. Cancelling the nonzero $C$ givesThe kernel is real and symmetric, so relabelling the integration variables puts the expression in the printed order:This is the scalar consistency condition for the rank-one potential, rather than an independent normalization equation. The double integral is real and nonnegative because the Fourier transform of $e^{-\kappa |x|}$ is $2\kappa /(q^2+{\kappa }^2)>0$. Thus attractive $\alpha >0$ is needed for a nontrivial solution. Under the usual integrability assumptions the corresponding free resolvent expectation decreases with $\kappa$, giving at most one bound state for this rank-one attraction.

If $X$ is Brownian motion with respect to the given filtration, independent Gaussian increments give the identity immediately.

Conversely, fix $s<t$ and let $\Delta =X_t-X_s$. Taking expectations gives its characteristic function $e^{-{\xi }^2(t-s)/2}$, so $\Delta \sim N(0,t-s)$. To prove independence, for any $A\in {\mathcal{F}}_s$ define the finite measure ${\nu }_A(E)=\mathbb{P}(A\cap \{\Delta \in E\})$. The conditional identity givesUniqueness of Fourier transforms of finite measures implies ${\nu }_A=\mathbb{P}(A)N(0,t-s)$. This is independence of $\Delta$ from ${\mathcal{F}}_s$. Iterating over times proves independent increments. Together with the stated continuity and $X_0=0$, this gives standard Brownian motion.

Here a process with the stated filtration is understood to be adapted, as required when saying it is Brownian with respect to that filtration. If adaptedness is not implicit, it must be added: an ordinary Brownian motion with a trivial filtration satisfies the displayed conditional characteristic function but is not adapted to that filtration.

A shorter radiation flash has a broader Fourier transform. From iii, its characteristic angular-frequency bandwidth isAt fixed bending radius this scales as ${\gamma }^3$ as $v\to c$. The standard synchrotron critical angular frequency makes the same scaling explicit:A periodically circulating electron produces harmonics of its revolution frequency. The very large harmonic range, of order ${\gamma }^3$, forms the familiar effectively continuous broad synchrotron spectrum at ordinary detector resolution. The result concerns the intrinsic spectrum of the radiation, not merely the slow envelope imposed by a long incoherent bunch. Ordinary frequency is $\nu =\omega /(2\pi )$; mixing cycles per second with radians per second would introduce a spurious factor of $2\pi$ into photon-energy estimates.^[ref]

Along $x=Vt+x_0$, the phase is $t\Phi (k)+k{x}_0$, with $\Phi (k)=kV-\omega (k)$. Graphically, a line of slope $V$ intersects the dispersion curve transversely except where it is tangent. On a transverse region, neighboring wavenumbers acquire order-one phase differences over a width $\Delta k\sim t^{-1}$, so cancellation leaves an $O(t^{-1})$ contribution. More formally, integration by parts usesand gives that order when the amplitude and its derivative are integrable and ${\Phi }^{\prime }$ is bounded away from zero on the region.

At a nondegenerate tangent point $k_{\ast }$, ${\Phi }^{\prime }(k_{\ast })=0$, equivalently $\boxed{V={\omega }^{\prime }(k_{\ast })}$, and $\Phi (k)=\Phi (k_{\ast })-{\omega }^{\prime \prime }(k_{\ast })(k-k_{\ast }{)}^2/2+\cdots$. Coherent wavenumbers now occupy width $O(t^{-1/2})$, yielding the stationary-phase contributionThese are the packets whose group velocity follows the observer. The displayed decay rates require the usual regularity and nondegeneracy hypotheses; $A\to 0$ alone is not a complete analytic hypothesis, and a vanishing amplitude or degenerate stationary point changes the leading order.

For deep water take a velocity potential satisfying Laplace's equation below the undisturbed surface and decaying with depth. A Fourier mode has vertical factor $e^{|k|z}$. The linearized kinematic and dynamic surface conditions are ${\eta }_t={\varphi }_z$ and ${\varphi }_t+g\eta =0$ at $z=0$. ThusRelease from rest gives ${\overline{\eta }}_t(k,0)=0$, so inversion of the Fourier transform givesEvenness of the initial profile makes its transform even and the reconstructed surface even; the same formula also works without that symmetry when the usual reality condition holds.

Split the cosine into its two time phases. Along the moving observer, their stationary conditions are $\pm {\omega }^{\prime }(k)=V$, where $\omega (k)=\sqrt{g|k|}$. Since $|{\omega }^{\prime }(k)|=\sqrt{g}/(2\sqrt{|k|})$, for $V\ne 0$,These are the possible dominant stationary-phase packets, provided the initial Fourier amplitude at the selected wavenumber is nonzero. For $V=0$ there is no finite nonzero stationary wavenumber; the formula must not be applied by division by zero.

Write $u=t-t^{\prime }$ and $\Omega =\sqrt{{\omega }_0^2-{\gamma }^2/4}$. With the printed Fourier transform convention, a time derivative transforms to $i\omega$. The retarded Green function isIts poles are $\omega =\pm \Omega +i\gamma /2$, both in the upper half-plane. For $u>0$ close the contour above; for $u<0$ close below, where there are no poles. Residues giveIt is continuous at $u=0$ and has derivative jump $G^{\prime }(0^{+})-G^{\prime }(0^{-})=1$, reproducing the Dirac delta function source. Thus the solution initially at rest isand vanishes for $t<0$. Set $D={\alpha }^2-\gamma \alpha +{\omega }_0^2=(\alpha -\gamma /2{)}^2+{\Omega }^2>0$. Evaluating the integral givesThe homogeneous solution constants are $C_1=-A/D$ and $C_2=A(\alpha -\gamma /2)/(D\Omega )$. Substitution at zero gives $\eta (0)=\dot{\eta }(0)=0$. Causality selects this response; an arbitrarily excited prehistory would add a homogeneous solution.

First fix the source at the spatial origin. Fourier transform replaces ${\partial }_t$ by $-i\omega$ and $\nabla$ by $i\mathbf{k}$ in the free term. Multiplication by a static background becomes a momentum convolution; the derivative in $A\cdot \partial G$ acts on the Green function, so it carries the internal momentum ${\mathbf{k}}^{\prime }$. ConsequentlyThe transformed point source supplies the right-hand side one. The signs follow from $A^{\mu }{\partial }_{\mu }=A^0{\partial }_t+\mathbf{A}\cdot \nabla$ in these conventions.

Source-position qualification. For a nonuniform $A(\mathbf{r})$, the general Green function is not translation invariant. In the printed basis $e^{i\mathbf{k}\cdot (\mathbf{r}-{\mathbf{r}}^{\prime })}$, its coefficient must also depend on ${\mathbf{r}}^{\prime }$, and the convolution contains an extra factor $e^{i(\mathbf{k}-{\mathbf{k}}^{\prime })\cdot {\mathbf{r}}^{\prime }}$. The displayed source equation is valid for ${\mathbf{r}}^{\prime }=0$, or with the background itself expressed in coordinates centred on that source. Equivalently, a general spatial Fourier Green function has two momentum arguments and a momentum-delta-function source. Staticity alone permits the time-difference transform, not a general position-difference-only transform.

The delta function in the artificial constant potential collapses the convolution. Put $b=e\gamma$ and $K^2=k^2+m^2$. ThenInverse Fourier transform gives the stated real-frequency integral with exponential $e^{-i\omega (t-t^{\prime })}$. For $K^2>b^2$, the poles have opposite real parts and common imaginary part $b$. For $0<K^2<b^2$, both are imaginary, at $i[b\pm \sqrt{b^2-K^2}]$. They remain in the upper half-plane if $b>0$, or the lower half-plane if $b<0$. At equality they merge into a second-order pole at $ib$. For $K^2=0$ one pole reaches the real axis, requiring a prescription.

Time independence of $V$ makes the Green function depend on $t,t^{\prime }$ through $\tau =t-t^{\prime }$. A spatially varying potential does not imply spatial translation invariance. Fix ${\mathbf{r}}^{\prime }$ and set $\mathbf{x}=\mathbf{r}-{\mathbf{r}}^{\prime }$; write the Fourier coefficient more fully as $G_{{\mathbf{r}}^{\prime }}(\mathbf{k};\omega )$. The source position is an implicit parameter in the printed notation. DefineThe ordinary derivative terms transform into $(-{\omega }^2+k^2+m^2)G_{{\mathbf{r}}^{\prime }}$. The product $V{\partial }_{t}G$ becomes a convolution, and $2ie(-i\omega )=2e\omega$. More explicitly, insert both inverse spatial Fourier transforms into $VG$: the integral over $\mathbf{x}$ enforces $\mathbf{q}+{\mathbf{k}}^{\prime }=\mathbf{k}$. The transformed delta source is one, leavingThis is a source-position-dependent Fourier Green function. Dropping the explicit ${\mathbf{r}}^{\prime }$ subscripts recovers the requested notation with exactly the potential transform defined in the source. No assumption of a homogeneous potential is needed. The factor $(2\pi {)}^{-3}$ and the source-position phase are necessary for the stated Fourier convention.

A retarded Green function obeys $G(\tau )=0$ for $\tau <0$ andFor a bounded or suitably tempered response represented by the real-axis Fourier integral, analyticity in the upper frequency half-plane is required, with suitable behaviour at infinity. For negative $\tau$, $e^{-i\omega \tau }$ decays on a large upper semicircle. If that half-plane has no singularities, contour closure gives zero response before the source. Lower-half-plane poles give factors decaying for positive time. Real poles are approached from below for a retarded prescription; at $\epsilon =0$, for example, $G_R(\tau )=\Theta (\tau )\sin \tau$.

The causality versus stability of a Green function distinction matters: an unstable initial-value problem can still have a strictly causal response that grows exponentially. Such a response has no ordinary real-axis Fourier transform. Its inverse Laplace transform must run on a frequency contour above its growth poles. Thus “upper poles violate causality” is correct only with the stable/tempered real-axis Fourier assumptions, not as an unconditional theorem about differential equations.

Use exactly the supplied inverse Fourier transform convention $e^{-i\omega \tau }$, with $\tau =t-t^{\prime }$. Then ${\partial }_t\mapsto -i\omega$ and ${\partial }_t^3\mapsto i{\omega }^3$. The transformed equation isThis algebraic inverse still needs a pole prescription to select the time-domain solution, particularly when a pole is on the real integration contour.

With the PDF's inverse Fourier transform convention, the derivatives act asso its forced differential operator becomes ${\omega }^2-k^2+i\gamma \omega k^2$. ThusFor the physical impulse response choose the retarded fundamental solution, vanishing before $t=0$. For $k\ne 0$ and $\gamma >0$, its poles lie in the lower half-plane. The retarded Fourier pole prescription $\omega \mapsto \omega +i0$ specifies the zero mode and the zero-damping limit as well; a Green function without a temporal or boundary prescription would otherwise permit homogeneous additions.

This is the Kelvin-Voigt rod Green function. Multiplying the homogeneous real-space equation by $-1$ givesA spatial Fourier mode therefore satisfies ${\ddot{\varphi }}_k+\gamma k^2{\dot{\varphi }}_k+k^2{\varphi }_k=0$: positive $\gamma$ really damps the motion. For sufficiently decaying fields its energy satisfiesThis checks the sign of the imaginary term in $\hat{G}$ independently of the transform calculation.

Put $T=t-t^{\prime }$. momentum eigenstates evolve with energy $p^2/(2m)$, so the retarded momentum kernel is $\theta (T)\delta (p-p^{\prime })e^{-i{p}^{2}T/(2m\hslash )}$. Its Fourier transform reduces toCompleting the square and defining the oscillatory Gaussian integral by a convergent limiting prescription givesFor $T>0$ the square root has quantum phase $e^{-i\pi /4}$. As $T\downarrow 0$, this kernel tends to $\delta (x-x^{\prime })$ in the distributional sense. Differentiating the Heaviside step function consequently supplies the required $i\hslash \delta (T)\delta (x-x^{\prime })$ source; an extra $-i/\hslash$ factor would correspond to a different Green-function convention.

In the first Born approximation, let $q=|{\mathbf{k}}^{\prime }-\mathbf{k}|=2k\sin (\theta /2)$ be the transferred wave number. The angular integral in the three-dimensional Fourier transform is $4\pi \sin (qr)/(qr)$, soWriting $g=2m{V}_{0}R/{\hslash }^2$ and $\mathrm{sinc}u=\sin u/u$, with value one at zero, the delta-shell Born amplitude givesFor $kR\ll 1$ the scattering is approximately isotropic, with total cross section $4\pi g^2{R}^2$. For $kR\gg 1$ it has a forward lobe of angular width of order $(kR{)}^{-1}$ and oscillatory zeros where $2kR\sin (\theta /2)=n\pi$. At a fixed nonzero scattering angle, its envelope decreases as $k^{-2}$. The right-hand panel of the figure shows this angular dependence.

For real $V_0$ the first-order scattering amplitude is real, so inserting it literally in the optical theorem would predict zero total cross section, contrary to its nonzero squared modulus. The first Born amplitude alone does not satisfy the optical theorem at order $V_0^2$. This is an order-of-approximation issue: the squared first-order amplitude is already second order, and the second Born term supplies the required second-order imaginary forward amplitude. At first order both sides vanish. A weak-shell regime is needed for the Born truncation itself to be accurate.

Quantize along the selected field direction by the right-handed relabeling ${\tilde{S}}^z=S^x$, ${\tilde{S}}^x=S^y$, ${\tilde{S}}^y=S^z$. Then the exact Holstein–Primakoff transformation in c gives $S_j^x=S-n_j$. Keep exchange terms of order $S^2$ and $S$, with $J$ and the printed $B$ held fixed. For a nearest-neighbor bond,On a periodic cubic lattice with $N$ sites and spacing $a$, there are $3N$ undirected bonds. Consequently,Use the unitary lattice Fourier transform $a_j=N^{-1/2}{\sum }_{\mathbf{k}}{e}^{i\mathbf{k}\cdot {\mathbf{r}}_j}{a}_{\mathbf{k}}$. Each positive lattice direction contributes $2(1-\cos k_{\alpha }a)$, so the Hamiltonian operator is already diagonal in momentum space:There are no anomalous $aa$ or $a^{†}{a}^{†}$ terms, so no Bogoliubov transformation or extra zero-point correction is needed. At long wavelengths,At $B=0$ this is the quadratic ferromagnetic Goldstone mode; for $B>0$ it is gapped by $BS$. Energies are written with $\hslash =1$; an excitation frequency is ${\omega }_{\mathbf{k}}={\epsilon }_{\mathbf{k}}/\hslash$. In this isotropic ferromagnet the fully aligned vacuum and its one-magnon dispersion are also exact; higher-magnon interactions are among the terms discarded from the linear spin-wave Hamiltonian operator.

Specify the Fourier transform before performing the Gaussian integral:At $u=0$ the functional is diagonal in both component and momentum, with inverse kernels $q^2+t+r$, $q^2+t-r$, and $q^2+t$. The condition $0<r<t$ makes all three positive. The covariance of a Gaussian distribution with statistical weight $e^{-F/(k_{B}T)}$ is $k_{B}T$ times the inverse kernel, henceIf $F$ is already the dimensionless free energy divided by $k_{B}T$, the numerators are one instead. Translational invariance forces opposite momenta to pair. There is no complex conjugate on the second field, so the delta function is at $\mathbf{p}=-\mathbf{q}$, not at $\mathbf{p}=\mathbf{q}$. A correlation with $m_j(\mathbf{p}{)}^{\ast }$ would have the latter convention. Rotational symmetry in real space leaves dependence on $q^2$ alone once the momentum-conservation delta function is factored out. The correlation lengths are ${\xi }_1=(t+r{)}^{-1/2}$ and ${\xi }_2=(t-r{)}^{-1/2}$; the second component becomes soft as $r\uparrow t$. The skipped letter in the source is preserved: this solution is part (e), with no inferred part (d).

Use the unitary discrete Fourier transform $a_j=L^{-1/2}{\sum }_k{e}^{ikj}{a}_k$, where $k=2\pi n/L$. Thenwith ${\alpha }_k=S[2J^x-(J^y+J^z)\cos k]$ and ${\beta }_k=-S(J^y-J^z)\cos k$. The factor $1/2$ accounts for the duplicated $k,-k$ pair in the full momentum sum.

Make a bosonic Bogoliubov transformation $a_k=\cosh {\theta }_k{b}_k+\sinh {\theta }_k{b}_{-k}^{†}$. Its coefficients preserve the canonical commutation relation because ${\cosh }^2{\theta }_k-{\sinh }^2{\theta }_k=1$. The coefficient of the pair term vanishes when $\tanh 2{\theta }_k=-{\beta }_k/{\alpha }_k$. ThusThe dropped constant is $\frac{1}{2}{\sum }_k({\epsilon }_k-{\alpha }_k)$ in addition to the classical ground-state energy. Since $J^x>J^y>J^z>0$, ${\alpha }_k>|{\beta }_k|$ at every $k$, so the bosonic Bogoliubov diagonalization is stable. The minimum occurs at $k=0$ and gives the energy gap $2S\sqrt{(J^x-J^y)(J^x-J^z)}$. In the isotropic limit the pair coefficient vanishes and the familiar quadratic small-$k$ ferromagnetic dispersion relation is recovered.

The Euler-Lagrange equation for the real scalar field givesThus the conjugate momentum and Hamiltonian density areThe spatial Fourier transform reduces the Klein-Gordon equation to ${\ddot{\varphi }}_{\mathbf{p}}+{\omega }_{\mathbf{p}}^2{\varphi }_{\mathbf{p}}=0$, with positive frequency ${\omega }_{\mathbf{p}}=\sqrt{{\mathbf{p}}^2+m^2}$. Hence a general real solution is the mode expansion of a free fieldThe two coefficients are complex conjugates because the field is real. Each pair of initial data $\varphi (0,\mathbf{x}),\pi (0,\mathbf{x})$ fixes these coefficients. The normalization anticipates canonical quantization and the relativistic normalization of one-particle states.

Choose the unitary discrete Fourier transformThe finite-sum orthogonality identity gives $[b_k,b_q^{†}]={\delta }_{kq}$. It also gives ${\sum }_j{a}_{j+1}^{†}{a}_j={\sum }_k{e}^{-ik}{b}_k^{†}{b}_k$. Substituting $A,B$ from solution therefore gives the dispersion relationUsing the opposite sign in the Fourier transform reverses $k$ and the sine term together, without changing the spectrum.

For $u\gg 1$ every $ϵ(k)>0$, so the ground state of this quadratic Hamiltonian is the bosonic vacuum, $b_k|\mathrm{GS}⟩=0$ for every $k$. In the original spin variables it is the fully $+z$ polarized product state. Its energy is zero after the constant subtraction. Repeating the construction about the fully $-z$ polarized state describes the other easy-axis sector of the original Hamiltonian.

A one-magnon excitation $b_k^{†}|\mathrm{GS}⟩$ has energy $ϵ(k)$ above the reference ground state. A normalized many-magnon Fock state with occupations $n_k$ has excitation energy ${\sum }_k{n}_kϵ(k)$, within the dilute regime of the linear spin-wave approximation. The lowest excitation in this sector has the minimum allowed $ϵ(k)$.

Let $\beta =1/(k_{B}T)$. Each independent fermion orbital is either empty or occupied once, so its grand partition factor is $1+e^{-\beta ({ϵ}_j-\mu )}$. Multiplying factors and taking the logarithm givesSince $1-n_F=[1+e^{-\beta (ϵ-\mu )}{]}^{-1}$, this is the requested form. Spin or other orbital degeneracy is already included in $g$.

For the Fermi-window convolution of the grand potential, defineDirect differentiation gives $F^{\prime }=+n_F$. The supplied PDF hint has the opposite sign; using that sign would contradict its grand-potential formula. Integrating twice with $A(0)={\Phi }_0(0)=0$ and vanishing upper boundary terms,At zero temperature $-{\partial }_{ϵ}{n}_F\to \delta (ϵ-\mu )$, confirming ${\Phi }_0(\mu )=\Phi (0,\mu )$.

Use the source's implicit unit normalization of the constant density of states; restoring a constant baseline $g_0$ multiplies the following potentials by $g_0$. For the sinusoidal density of states, direct integration givesPhysical nonnegative density of states requires $|\alpha |\le 1$. Write $t=k_{B}T$ and $y=ϵ-\mu$. The Fermi windowis even, with full-line moments $\int wdy=1$, $\int ywdy=0$ and $\int y^{2}wdy={\pi }^2{t}^2/3$. The last integral follows, for example, by expanding $e^x/(1+e^x{)}^2$ on $x>0$ into its alternating exponential series. When $t\ll \mu$, extending the lower integration limit to $-\infty$ introduces only exponentially small band-edge corrections. The supplied Fourier transform givesThe odd sine term vanishes. Substitution provesThe thermal damping of a sinusoidal density of states is therefore $X/\sinh X$ in the full oscillatory potential. The requested ratio instead concerns the thermal change $\Phi (T,\mu )-{\Phi }_0(\mu )$. Its oscillatory thermal-change amplitude ratio isFor positive $X$ it decreases smoothly from the limit $|\alpha |$ to zero:At exactly $T=0$ both thermal changes vanish; the quoted value is their limiting ratio. The amplitude is the envelope in chemical potential. At a particular fixed $\mu$, the signed contribution ratio has $\alpha \cos (2\pi \mu /\Delta )$ in place of $|\alpha |$; at an oscillation node the actual oscillatory contribution is zero. The large-$X$ behaviour still requires $k_{B}T\ll \mu$, so the useful range is $1\ll X\ll 2{\pi }^2\mu /\Delta$.

Use the Fourier transform convention $\hat{W}(k,z,t)={\int }_{\mathbb{R}}W(x,z,t)e^{-ikx}dx$, with inverse factor $1/(2\pi )$. The initial profile occupies only the vertical mode $\sin (\pi z/h)$. Its transform isThe transformed evolution is a harmonic oscillator for each $k$, withTherefore inversion givesUsing the nonnegative even frequency instead gives the same cosine. At $t=0$, ${\int }_{\mathbb{R}}{e}^{-a|k|}{e}^{ikx}dk=2a/(a^2+x^2)$ with $a=ϵh$, recovering the specified initial profile and its zero time derivative.