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An antiderivative of the initial velocity is $-2c{e}^{-s^2/2}$. Thus D'Alembert's formula givesThis is a positive Gaussian function moving right and a negative Gaussian function moving left. Their cancellation at $t=0$ gives the zero initial displacement, while direct differentiation gives ${\psi }_t(x,0)=2cx{e}^{-x^2/2}$.

For every $t>0$, the profile is an odd function of $x$, vanishes at $x=0$, is positive for $x>0$ and negative for $x<0$, and tends to zero at both spatial infinities. The signs follow from $(x+ct{)}^2-(x-ct{)}^2=4xct$. The positive and negative lobes move farther apart as $t$ increases; for large separation they approach the individual unit-height Gaussian functions centred at $x=\pm ct$. At finite separation the actual extrema lie slightly beyond these centres: writing $a=ct$, the positive global maximum occurs at the unique $x>a$ satisfying $(x-a)/(x+a)=e^{-2ax}$.

For the Gaussian function, direct differentiation gives $f^{\prime }=-tf$. Applying the Fourier transform identities from part (a),Solving this first-order ordinary differential equation givesIn fact, the normalization can also be fixed: the Gaussian integral gives $F(0)={\int }_{\mathbb{R}}{e}^{-t^2/2}dt=\sqrt{2\pi }$, so $C=\sqrt{2\pi }$ with the convention used here. The transform has the same Gaussian shape.

Take $\lambda >0$ and $t>0$, the positive diffusivity requirement for the forward heat kernel. The PDF does not explicitly specify the sign of $\lambda$; the stated real Gaussian function formula requires this assumption. For a direct calculation, initially assume sufficiently smooth and rapidly decreasing data, so that the Fourier transform, its inverse and the following integrations are justified. The resulting heat-kernel convolution extends to wider classes of initial data.

Using the stated Fourier transform convention and integration by parts,The transformed ordinary differential equation has solution $\tilde{u}(k,t)=e^{-\lambda k^{2}t}{\tilde{u}}_0(k)$. Fourier inversion then givesComplete the square:Part (a), after scaling by $\sqrt{\lambda t}$, evaluates the remaining Gaussian integral even though its shift is imaginary. ThereforeThe heat kernel has integral one and is an approximate identity as $t\downarrow 0$, which recovers the initial condition. Mere decay at spatial infinity does not by itself imply every Fourier integral is absolutely convergent; the regularity assumption above supplies a rigorous setting for the derivation.

Put $s=\lambda t>0$. In the heat-kernel convolution, complete the real square asThe Gaussian integral therefore givesAt $t=0$ the same formula gives $\beta =\alpha$. Equivalently, ${\beta }^{-2}={\alpha }^{-2}+4\lambda t$: the variance of this normalized Gaussian function grows from $1/(2{\alpha }^2)$ to $1/(2{\alpha }^2)+2\lambda t$.

The initial profiles have unit integral and concentrate at zero as $\alpha \to \infty$. Precisely, for every smooth compactly supported test function $\phi$,Thus the limit is the Dirac delta distribution, rather than an ordinary function: pointwise the profiles tend to zero away from the origin and diverge at the origin. At every fixed $t>0$,The limiting evolved profile is the heat kernel for a unit point source.

The GRW model supports robust macroscopic locations, but not universal exact point locations supplied by a list of particle coordinates. It modifies the wave function rather than adding the actual particle configuration of Bohmian mechanics. A claim that objects are located must therefore specify both the collapse dynamics and the proposed ontology. “Localized” is not synonymous with “has compact support” or “is a position eigenstate”.

Using one standard convention, a collapse on constituent $i$ with centre $X$ acts by the GRW localization operatorIts centre has probability density $\Vert L_i(X)\Psi {\Vert }^2$. Independent constituent collapse rates $\lambda$ give a body with $N$ relevant constituents an approximately $N\lambda$ localization rate. Correlations within a macroscopic superposition allow one constituent's localization to suppress an entire alternative configuration: the GRW amplification mechanism.^[ref]

For two narrow packets separated by $d\gg r_C$, a hit near one multiplies the other's relative amplitude approximately by $e^{-d^2/(2r_C^2)}$. This explains a sharply concentrated pointer reading. Yet the Gaussian function is never zero at finite separation. A nonzero distant tail therefore remains, and a finite-width packet is not an exact position eigenstate. This is the tails problem in collapse theories.^[ref] The assertion that each hit produces exact bounded support is false; an initially everywhere nonzero Gaussian wave packet is a concrete counterexample.

A GRW matter-density ontology specifies matter in ordinary SPACE byA macroscopic body's density is concentrated near one region, making its approximate location physically meaningful. The formula is the proposed matter field itself, not a probability distribution over a hidden set of exact point positions. Residual tails correspond to very low density elsewhere, motivating the further question of why such tails do not constitute ordinary rival objects. Concentration provides a practical account of records; a full ontology must also address that structured-tail objection.

A GRW flash ontology instead identifies the primitive events with collapse centres in spacetime. Individual flashes have definite locations and times, while a persisting macroscopic object is a dense pattern of flashes. There is no continuous particle trajectory between them. An isolated microscopic system may flash rarely, so the model must not be read as assigning it an ordinary point location at every instant. Matter-density and flash accounts express distinct ontologies with related empirical predictions.^[ref]

An approximate-location rule, such as requiring almost all matter density inside a suitably sized region, is meaningful if its tolerance and scale are stated. It does not turn the small tail into zero, and arbitrary sharp property thresholds can create artificial counting puzzles. The defensible result is emergent, highly stable macroscopic localization, or exact locations of flashes, depending on the ontology. Exact Bohmian positions are a different theoretical commitment and cannot be attributed to the collapse equation alone.

Write $G(r)=⟨{\psi }^{†}(0)\psi (\mathbf{r})⟩$, $r=|\mathbf{r}|$, and ${\ell }^2=\hslash /(m\omega )$. At zero temperature all $N$ particles occupy ${\varphi }_0(\mathbf{r})=(\ell \sqrt{\pi }{)}^{-1}{e}^{-r^2/(2{\ell }^2)}$. Expanding the bosonic annihilation operator in single-particle orbitals givesThis one-body density matrix is a Gaussian function of width $\ell$; it contains one wavefunction evaluated at $r$, so its exponent is half that of the spatial number density $N|{\varphi }_0(r){|}^2$. In particular,the total number density at the trap centre, including condensed and noncondensed particles.

At positive temperature, with occupations ${\overline{n}}_a$ of the trap energy eigenstates, the exact ideal-gas decomposition isFor $\hslash \omega \ll k_{B}T<k_B{T}_c$, the sketch has a broad Bose-Einstein condensate component on scale $\ell$ and a narrower thermal peak near the origin. On approaching $T_c$ the broad component loses its weight $N_0$, while the thermal part remains. Above $T_c$ there is no macroscopic broad Bose-Einstein condensate component; close to $T_c$, low energy levels can still give a shoulder, so a single narrow Gaussian function is only a high-temperature description. At $k_{B}T\lesssim \hslash \omega$ individual excited energy levels matter and the two-scale description need not apply.

For $T\gg T_c$ the Maxwell-Boltzmann distribution applies. The leading off-diagonal one-body density matrix isHere ${\sigma }_r$ is the Gaussian function width defined by $e^{-r^2/(2{\sigma }_r^2)}$, and ${\lambda }_T$ is the thermal de Broglie wavelength. The thermal cloud itself extends over the much larger spatial scale $\sqrt{k_{B}T/\kappa }$; this is a correlation width, not the cloud radius.

The issue is the square-integrable condition on the whole real line, not just decay at one end. Because the quantum harmonic oscillator potential is even, the even and odd parts of a square-integrable wavefunction are themselves square integrable and satisfy the same equation. It is therefore enough to examine each parity sequence separately.

For a nonterminating sequence, put $b_j=a_{2j+p}$, where $p=0$ or $1$. Its recurrence relation givesSince $ϵ$ is real for an energy eigenvalue, the ratio is eventually positive. After multiplying by one constant phase, the tail coefficients are all positive. Choose $0<\delta <1/2$. For sufficiently large $j$, the ratio is at least $(1-\delta )/(j+1)$, so for some $K>0$ the tail coefficients obey $b_j\ge K(1-\delta {)}^j/j!$. Consequently, for $\xi >0$, the magnitude of $f$ eventually exceeds a positive multiple of ${\xi }^p{e}^{(1-\delta ){\xi }^2}$ after subtraction of a fixed polynomial. Multiplication by $e^{-{\xi }^2/2}$ leaves exponential growth like $e^{(1/2-\delta ){\xi }^2}$, incompatible with the square-integrable condition. Thus each nonzero parity series must terminate. The parity argument also excludes cancellation between the two nonterminating sequences at both infinities. This proves Normalization forces Hermite-series termination.

If the terminating polynomial has degree $n$, its top nonzero coefficient requires $2n+1-ϵ=0$. Conversely, when $ϵ=2n+1$, choosing the seed of parity $n$ nonzero makes the recurrence relation terminate exactly at degree $n$. The opposite seed must be zero, since that sequence does not terminate. ThereforeThe resulting polynomials are proportional to the physicists' Hermite polynomials, and ${\psi }_n\propto H_n(\xi )e^{-{\xi }^2/2}$ is normalizable because any polynomial times this Gaussian function is square integrable. The two parity seeds describe the full solution space of the second-order differential equation; the preceding argument therefore exhausts the normalizable stationary states. Each eigenspace has dimension one. General time-dependent quantum states can of course be superpositions of these energy eigenstates.

Write $E_0=\hslash \omega /2$. Repeated application of the annihilation operator lowers the energy eigenvalue in steps of $\hslash \omega$. The lower bound from part (ii) forces this process to stop. To see why it also quantizes the starting energy, writeThe first $n$ lowering steps have positive squared norms. If $\delta >0$, the next step is also nonzero by part (ii), and its energy would be $E_0+(\delta -1)\hslash \omega <E_0$, contradicting the bound. Thus $\delta =0$, andExistence of the ground state can also be checked directly in the position representation. The equation $a_{-}{\psi }_0=0$ iswhose solution is proportional to $e^{-m\omega x^2/(2\hslash )}$. This Gaussian function is square-integrable, so it gives an actual ground state. Its normalization can be chosen using the Gaussian integral.

All the displayed levels occur: for a nonzero state at $E_n$,Thus repeated raising creates a nonzero energy eigenstate at every $E_n$.

At zero, the Fourier transform is $\hat{f}(0)=\int f=1$, so $\lambda f(0)=1$. In particular $\lambda \ne 0$ and $f(0)=1/\lambda$. The Fourier inversion theorem at zero also givesEquating the two values yields $\boxed{{\lambda }^2=2\pi }$. Since the explicit normalized Gaussian function in part (i) has $f(0)>0$, here $\lambda =\sqrt{2\pi }$.

$\boxed{\text{(a): two-photon excitation; (b): one-photon/confocal excitation.}}$ The compact emitting region in (a) indicates that excitation itself is strongly localized. In (b), the illuminated path generates out-of-focus fluorescence as well as fluorescence near the focus. A confocal detection pinhole can reject much of that unwanted emission from the recorded image; it does not prevent the dye solution from emitting it.

In ordinary one-photon excitation, the local fluorescence-generation rate is approximately proportional to excitation light intensity $I$. In unsaturated two-photon excitation microscopy, it is proportional to $I^2$. Simultaneous absorption is appreciable only in the high-intensity focal region, generally produced using a pulsed laser. Two lower-energy photons jointly excite the fluorophore; the excitation wavelength is therefore typically substantially longer than in its one-photon experiment. This reduces out-of-focus excitation and permits detection without a confocal pinhole.^[ref]

For a small detection pinhole, the effective confocal point spread function is approximatelyHere each factor denotes an intensity response. Excitation and detection jointly select the focal region. Enlarging the pinhole increases collected signal but weakens axial rejection and changes the effective point spread function. In two-photon excitation microscopy with broad collection, the spatial response is dominated byThe pinhole product and excitation square are different physical mechanisms even when simplified mathematical profiles resemble one another.^[ref]

Compare both lateral and axial widths using the actual wavelengths and objective. For a diffraction-limited system, their characteristic scalings arewhere $\mathrm{NA}$ is numerical aperture and $n_{\mathrm{med}}$ is the medium's refractive index. Numerical prefactors depend on the width criterion and high-aperture optical model. Axial resolution is generally worse than lateral resolution. A small confocal pinhole improves axial sectioning and can narrow the lateral response, at the cost of photon throughput.

As a simple Gaussian function example, $h(x)=\exp [-x^2/(2{\sigma }^2)]$ has squared profile $h(x{)}^2=\exp [-x^2/{\sigma }^2]$, so its width is reduced by $1/\sqrt{2}$. However, the two-photon illumination spot itself is broader at its longer excitation wavelength. Two-photon excitation is not automatically higher-resolution than confocal microscopy, and ordinary two-photon sectioning is not intrinsically super-resolution. Compare measured point spread functions, pinhole settings, numerical apertures, scattering and optical aberrations, not just the fact that one profile is squared.

The original model figure below separates focal confinement from wavelength-dependent width. It is an illustrative one-dimensional Gaussian function calculation, not a reproduction or quantitative fit of the supplied photograph.

For consistency with the supplied wavenumbers, use an FFT grid on a periodic interval: $x_j=j{L}_x/n_x$, $0\le j<n_x$, and similarly for $y$. In MATLAB, replace the endpoint-inclusive coordinate definitions by x = (0:nx-1)*Lx/nx; and y = (0:ny-1)*Ly/ny;. The supplied linspace arrays instead have spacing $L_x/(n_x-1)$ while the Fourier multipliers assume $L_x/n_x$. Including both copies of the periodic endpoint creates a grid inconsistency, separate from high-frequency aliasing.

The line references below use the original listing's line numbers.

For this linear homogeneous problem, the exponential update is exact for the retained modes, apart from roundoff error. It is not a forward Euler method step and does not require the explicit finite-difference diffusion stability restriction. Sampling, periodic modelling and spatial resolution still matter.

Aliasing occurs because a sampled grid cannot distinguish certain different continuous Fourier modes. With spacing $h$, modes $k$ and $k+2\pi m/h$ agree at every sample point. Frequencies above the Nyquist frequency can therefore be misidentified as lower frequencies, changing the resolved spectrum and its subsequent diffusion rate.

For the corrected grid, $h=10/128\mathrm{cm}$ and the maximum resolved angular wavenumber is approximately $\pi /h\simeq 40.2\mathrm{c}{\mathrm{m}}^{-1}$. The initial Gaussian function has spectral envelope proportional to $e^{-a|k{|}^2/4}$. At that one-coordinate cutoff the factor is about $e^{-40.4}$, so the omitted high-frequency tail is tiny. Its continuation mismatch at the distant edges is also tiny. Significant aliasing is not expected for this resolved, smooth initial condition.

The linear constant-coefficient heat equation only damps existing modes; it does not generate higher frequencies through nonlinear products. An unresolved narrow peak, a sharp edge or a nonlinear source would require a separate resolution assessment. The duplicated-endpoint grid issue described above should not be diagnosed as high-frequency folding.

Choose two hypothetical environments A and B of one chemical element, with line centers ${\mu }_A=284\mathrm{eV}$ and ${\mu }_B=286\mathrm{eV}$. Model each line by a normalized Gaussian function with common standard deviation $\sigma$, and stipulate integrated count weights $w_A=3/4$, $w_B=1/4$. On the binding energy axis $E$, the normalized count density isAll energies in this expression use electronvolts. Each component's integrated area equals its count weight, and $\int I(E)dE=1$. There is no background, statistical counting fluctuation, or additional emission feature in this stipulated model. These choices deliberately isolate photoelectron peak overlap and resolution.

Direct integration, or the law of total expectation applied to the component label, gives the centroidThe centroid lies between the chemical environments and is not either component's binding energy. Expanding $(E-\bar{E}{)}^2$ within each component, the odd cross term integrates to zero. The resulting variance isFor $\sigma =0.25\mathrm{eV}$ the variance is $0.8125{\mathrm{eV}}^2$ and the two maxima are separated. For $\sigma =1.00\mathrm{eV}$ it is $1.7500{\mathrm{eV}}^2$ and there is one maximum. Each Gaussian's full width at half maximum is $2\sqrt{2\ln 2}\sigma$, approximately $0.589$ and $2.355\mathrm{eV}$ respectively. The chemical separation remains $2\mathrm{eV}$ in both cases. The number of visible maxima therefore does not equal the number of chemical components.

One interpretation of this change is additional independent Gaussian instrument blur. The convolution of probability densities adds the variances, so broadening $\sigma =0.25$ to $1$ requires response standard deviation $\sqrt{1-0.25^2}=\sqrt{15}/4\simeq 0.968246\mathrm{eV}$. A normalized response with zero mean preserves the centroid and integrated counts. Real line shapes and backgrounds can depart from this model, and separating hidden components is an inverse problem; a plausible fit does not alone establish uniqueness.

Finally, count weights describe detection. In a simple response model, let $n_i$ denote the number of emitting sites sampled in environment $i$, $s_i$ its effective photoionization cross section, and $T_i$ the probability that its emitted electron contributes to the selected measured line. ThenThe common factor $C$ includes exposure. Here $T_i$ includes escape from the sample, analyzer transmission and detection. Count weights equal site fractions only if the relevant response factors are equal or corrected. They are not automatically molecular mole fractions; different molecules can supply different numbers of the selected emitting sites. In an original example, equal site numbers with $s_A{T}_A=3s_B{T}_B$ give the plotted $3:1$ count ratio despite a $1:1$ site ratio.

Let $\Lambda =B{\mathbb{Z}}^d$ be a full-rank Euclidean lattice, with covolume $m(\Lambda )=|\det B|$ and dual lattice ${\Lambda }^{\vee }=B^{-\mathsf{T}}{\mathbb{Z}}^d$. For a Schwartz function $f\in \mathcal{S}({\mathbb{R}}^d)$ use the Fourier transform conventionThen the Poisson summation formula becomesBoth series have absolute convergence. To see the normalization, apply integer-lattice Poisson summation to $g(u)=f(Bu)$. The change of variables formula gives $\hat{g}(k)=|\det B{|}^{-1}\hat{f}(B^{-\mathsf{T}}k)$, yielding both the reciprocal covolume factor and the dual lattice argument. The factor is the physical lattice-point density. In particular, for spacing $h>0$,For the Gaussian function $f(x)=e^{-\pi a\Vert x{\Vert }^2}$ with $a>0$, the product of the one-dimensional Gaussian Fourier transforms givesHere $\Vert \cdot \Vert$ is the Euclidean norm. As a normalization check, take $\Lambda =\mathrm{diag}(2,3){\mathbb{Z}}^2$. Its covolume is $6$, and its dual lattice is $\mathrm{diag}(1/2,1/3){\mathbb{Z}}^2$, soThis also follows by multiplying two Gaussian Poisson summation identities with parameters $4a$ and $9a$. If one instead uses an angular Fourier transform kernel $e^{-i⟨x,\omega ⟩}$, the reciprocal frequencies are $2\pi {\Lambda }^{\vee }$, while the prefactor remains $m(\Lambda {)}^{-1}$.

Put $I=I(1)$. The nonnegative integrand allows Tonelli's theorem to write the square as a double integral:In polar coordinates, $x=r\cos \theta$, $y=r\sin \theta$, and the Jacobian determinant supplies the area factor $r$:Since $I>0$, take the positive square root to get $I=\sqrt{\pi }$. The computation proves finiteness as well as the value. For an explicit limiting argument, first integrate over disks of radius $R$; their integrals are $\pi (1-e^{-R^2})$, which increase to $\pi$ as $R\to \infty$. ^[ref]

The geometric surprise is that radial shells get longer as $r$ increases, even while the Gaussian function gets smaller. Its central height is largest at zero, but the radial contribution $2\pi r{e}^{-r^2}$ vanishes there and peaks at $r=1/\sqrt{2}$. Differentiating gives $2\pi e^{-r^2}(1-2r^2)$, which changes sign at that radius. Keeping the Jacobian determinant is what makes the one-dimensional integral and the plane geometry agree.

The real Gaussian function heat kernel $(4\pi \epsilon z{)}^{-1/2}\exp [-(\varphi -\theta {)}^2/(4\epsilon z)]$ is integrable only when $\epsilon z>0$. For forward time $z>0$, this requires positive diffusivity. A Cole-Hopf transformation is an algebraic substitution for any nonzero coefficient, but that does not make the real forward convolution valid for negative diffusivity. Constant initial heat data already produce a divergent integral with the wrong sign.

The second-order, angle-averaged evolution of a galactic distribution function caused by weak fluctuating gravitational perturbations. In action-angle variables, the first-order response to a Fourier series harmonic feeds a second-order flux proportional to $\mathbf{k}(\mathbf{k}\cdot {\partial }_{\mathbf{J}}{f}_0)$ and to the perturbation's power. Orbital resonances select where transport is strongest; a finite pulse produces Gaussian resonance broadening when its time envelope is a Gaussian function. The general flux retains time memory until a further short-memory or pulse-integrated approximation is made.

For a one-dimensional free particle with $m>0$ and $p>0$, take the positive total energy from the relativistic energy-momentum relation:A fixed plane wave quantum phase obeys $kx-\omega t=\text{constant}$. Its phase velocity is thereforeFor a narrow wave packet centered at $p_0$, expand its dispersion relation as $E(p)\simeq E(p_0)+E^{\prime }(p_0)(p-p_0)$. In the Fourier inversion integral building the wave packet, the envelope then depends on $x-E^{\prime }(p_0)t$. Thus its group velocity isThe superluminal phase velocity follows a repeating phase marker; it does not describe a localized signal or superluminal transport of energy. Higher derivatives of the dispersion relation produce wave packet spreading, so the translated-envelope argument is a narrow-band approximation. For $v=0.6c$, the Lorentz factor is $1.25$, $p=0.75mc$ and $E=1.25m{c}^2$: the group velocity is $0.6c$ while the phase velocity is $5c/3$.

The total-energy convention matters. Define $\psi =e^{im{c}^{2}t/\hslash }\Psi$ to remove the common rest energy from a wavefunction $\Psi$ with total-energy time dependence. ThenThis is exactly the free-particle momentum convention used by the Schrödinger equation here. Removing a common time-dependent global phase changes the reported phase velocity, while leaving the de Broglie wavelength, relative spatial quantum phases and group velocity unchanged. Adding $m{c}^2$ again to a total energy would count the rest energy twice.

A signed spatial filter model of a simple cell assigns a weight $K_{ij}$ to each sampled location and takes its dot product with the visual contrast image $C_{ij}$. Positive weights favor bright features; negative weights favor dark features. The signed driveis a linear function of the image, while the rectified linear unit output $R$ is nonlinear. $D$ and $R$ below are arbitrary model units. To obtain a predicted firing rate one would need a calibrated gain, background response and possibly a different output function. The model does not generate a spike train; a linear-nonlinear-Poisson cascade model adds a stochastic event-generation assumption.

For the first diagram, every one of five rows has weightsA bright central column, with $C_{i3}=1$ and zero elsewhere, gives $D=5/5=1$. A bright middle row gives $D=-1/20-1/20+1/5-1/20-1/20=0$; a uniformly bright patch also gives $D=0$, since every row sums to zero. Move the bright column one site right: $D=5(-1/20)=-1/4$, so $R=0$. Make that shifted column dark instead: $D=1/4$ and $R=1/4$. Thus an OFF-like weight can increase output for darkness. A negative weight is a descriptive sensitivity, not evidence for a particular inhibitory synapse.

For a smooth example use the real, even Gabor filterThe Gaussian function localizes the weights; the cosine supplies alternating polarity. Choose ${\sigma }_x=1/2$, ${\sigma }_y=1$ and $q=\pi$ in arbitrary spatial units. The stripes' spatial frequency is $q/(2\pi )=1/2$ cycles per unit. Their axis makes angle $\theta$ with horizontal, while brightness changes across the perpendicular axis. $\phi$ is their spatial phase. Normalize the full-plane integral of $K{C}_{\theta ,\phi }$ by its value at $\theta =\pi /2$, $\phi =0$. The Gaussian integral and the product identity for cosines givewhereAt $\phi =0$, horizontal stripes give $R\approx 0.004159$, $60^{\circ }$ stripes give $R\approx 0.286741$, and vertical stripes give $R=1$. At the same vertical orientation, $\phi =\pi /2$ gives $D=R=0$, while reversing the polarity with $\phi =\pi$ gives $D=-1$ and $R=0$. The preferred axis persists while the response changes dramatically with position. Shifting the vertical stripes by a quarter of their wavelength changes phase by $\pi /2$ up to sign, without rotating them.

These are two different toy filters, not discretizations of the same fitted receptive field. The discrete filter exactly rejects a uniform contrast patch; this Gabor filter need not reject uniform contrast because its weight integral need not vanish. Neither toy includes stimulus history, recurrent interactions, saturation or a measured response baseline. Their value is a transparent prediction: orientation selectivity can emerge from spatial summation, and it does not by itself imply tolerance to phase.