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The printed integration labels need correction. The parent's four-momentum is fixed, so the two Lorentz-invariant phase-space measures must integrate $p_2$ and $p_3$, rather than the printed $p_1$ and $p_2$. Write $P=p_1=(M,0)$ and $p=p_2$, $k=p_3$. The physical decay width isThe spatial Dirac delta function sets $\mathbf{k}=-\mathbf{p}$. Performing the angular integral in spherical coordinates, with $E=\sqrt{p^2+m^2}$, gives the integrated two-body decay phase spaceThe positive root is $p_{\ast }=\sqrt{M^2/4-m^2}$. Since $d(2E)/dp=2p/E$, the Dirac delta function integration yields ${\Phi }_2=p_{\ast }/(4\pi M)$. ThusThe two final particles have different species, so there is no identical final-state symmetry factor. The result tends to $g^{2}M/(16\pi )$ as $m\to 0$ and vanishes at the two-particle decay threshold.

Put $Q=M-3m$. For $Q<0$ the decay width is zero by conservation of energy. Just above the three-particle decay threshold, all three daughters are nonrelativistic. The decay amplitude of iv remains finite: the internal scalar propagator denominator approaches $(2m{)}^2-m^2=3m^2$, and the Dirac scalar bilinear does not vanish at rest. Thus, for nonzero fixed $g$ and $\lambda$, the leading suppression comes entirely from three-body threshold phase space.

Use the spatial Dirac delta function to eliminate one daughter momentum. The remaining integral has six independent momentum coordinates and one conservation of energy constraint,Rescaling both momenta by $\sqrt{mQ}$ gives $Q^3$ from the measure and $Q^{-1}$ from the Dirac delta function. The smooth factors $1/(2E)$ do not change the power. Consequentlywith a finite positive leading coefficient when $g\lambda \ne 0$; $\Gamma =0$ for $m\ge M/3$. The $1/2!$ for the identical bosons changes only this coefficient. Equivalently, at fixed $M$ the opening is quadratic in $M/3-m$. If $g\lambda =0$, this leading tree-level Feynman diagram vanishes, so the stated leading coefficient is zero.

The quantized field expansion in the complete orthonormal basis of orbitals isThe complex conjugation in the creation operator expansion is essential for complex orbitals. Indeed, ${\hat{a}}_n^{†}=\int d^{3}r{\varphi }_n(\mathbf{r}){\hat{\psi }}^{†}(\mathbf{r})$ creates the wavefunction ${\varphi }_n$. The resolution of the identity in an orthonormal basis givesrecovering the canonical commutation relation and the spatial Dirac delta function.

Use the total density of states $D^{\ast }=A{m}^{\ast }/(2\pi {\hslash }^2)$ and $C=A/(2\pi \hslash {)}^2$ from part (b). In the paper's double-sum convention, $f(\varphi )$ has units of energy and scales as $A^{-1}$. Define its first angular moment byThis convention matters: the Landau quasiparticle energy functional written with an explicit $1/A$ instead uses the intensive kernel $\tilde{f}=Af$.

At zero temperature, the displayed functional is the change in $E-\mu N$; a true Galilean boost preserves $N$, so $\Delta \mathcal{F}=\Delta E$. For a radial displacement $\nu (\theta )$, the first term in the Landau quasiparticle energy functional becomesThe oriented radial integral also works for negative $\nu$: removing occupied levels below the chemical potential costs positive energy. Substituting $\nu =p_0\cos \theta +O(p_0^2/p_F)$ and using $N=C\pi p_F^2$ givesThe exact second-order correction to the boundary is unnecessary here because the unperturbed excitation energy vanishes on the Fermi surface; it changes this term only beyond quadratic order. It must nevertheless be understood when identifying $\Delta \mathcal{F}$ with a fixed-particle number energy change.

For the quadratic interaction term, the first-order occupation change alone suffices. The radial Dirac delta function integrals givePut $\varphi =\theta -{\theta }^{\prime }$. The remaining angular integral usesso only the first cosine component of the interaction survives. Since $D^{\ast }=2\pi C{m}^{\ast }$,Adding both contributions and matching the bare-mass result of part (a) yields the two-dimensional Landau effective-mass relation:Equivalently, using the density of states per unit area $g^{\ast }=D^{\ast }/A$,If dimensionless Landau parameters are defined through the Fourier series $D^{\ast }f(\varphi )=F_0+{\sum }_{\ell \ge 1}{F}_{\ell }\cos (\ell \varphi )$, then $D^{\ast }{\overline{f}}_1=F_1/2$ andDefining the cosine expansion with an extra factor of two would instead absorb that factor into $F_1$; the angular-integral formula above fixes the convention unambiguously. The physical content is that interaction changes accompanying a moving Fermi sea compensate the renormalized quasiparticle velocity, so the total boost energy is governed by the bare mass. For an interaction independent of angle, ${\overline{f}}_1=0$ and this consistency condition requires $m^{\ast }=m$.

At a finite jump in potential energy, the wavefunction and its first derivative must be continuous. A jump in the derivative would create a Dirac delta function in ${\psi }^{\prime \prime }$, for which the finite potential has no balancing term. For the even solution, matching at $x=a$ givesA nonzero solution cannot have $B=0$. It also cannot have $\cos \beta =0$: the first condition would give $D=0$, while the second would force $B\kappa \sin \beta =0$, a contradiction. Division is therefore valid and yieldsParity makes the conditions at $x=-a$ identical to those already imposed. Finally the definitions of the decay and oscillation constants giveThe circle relation fixes the well strength; the tangent relation selects its even bound states.

By definition, the relaxation modulus is the stress generated by a unit step of strain from an initially relaxed state. Set $ϵ(t)=H(t)$, where $H$ is the Heaviside step function, and write $\sigma (t)=G(t)$. Integrating the linear ordinary differential equation across the step shows that neither $G$ nor $G^{\prime }$ can jump: a jump would produce a Dirac delta function or its derivative, absent from the forcing. Hence $G(0^{+})=G^{\prime }(0^{+})=0$.

For $t>0$, a particular solution is $1$. The characteristic equation factors asso $G=1+A{e}^{-t/5}+B{e}^{-t/3}$. The initial conditions give $A+B=-1$ and $A/5+B/3=0$, whenceThe time and modulus units are those implicit in the printed coefficients. A Laplace transform check gives $\hat{G}(s)=1/[s(5s+1)(3s+1)]$.

The original PDF really does put undifferentiated strain on the right-hand side. The resulting response rises from zero to one, with $G^{\prime }(t)=\frac{1}{2}(e^{-t/5}-e^{-t/3})>0$, rather than exhibiting ordinary stress relaxation. There is a stronger physical problem if this is intended as a passive material law: with the convention $e^{i\omega t}$ the stress-to-strain amplitude ratio isThe loss modulus would give negative cycle-averaged supplied power $⟨\sigma \dot{ϵ}⟩=\omega G^{\prime \prime }|{ϵ}_0{|}^2/2$, contrary to passivity in linear viscoelasticity. The boxed expression is therefore the formal relaxation modulus of the equation as printed, not a claim that this equation describes an ordinary passive viscoelasticity model. No derivative of the input has been silently inserted.

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.

Use the Dirac delta function to set $p^{\prime }=p+\alpha T$, and write $X=x-x^{\prime }$. The remaining momentum-dependent quantum phase isShifting the integration variable by $\alpha T/2$ reduces the integral to the free Gaussian integral. ThereforeThe force is $-\alpha$, not $+\alpha$. Both the linear endpoint term and the cubic-time term disappear in the zero-force limit. The prefactor remains the free one.

The quantum propagator is the position-space kernel of the time-evolution operator: $\Psi (x,t)=\int K(x,t|x^{\prime },t^{\prime })\Psi (x^{\prime },t^{\prime })d{x}^{\prime }$. For a time-independent Quantum Hamiltonian, $U=e^{-iH(t-t^{\prime })/\hslash }$. Inserting a complete orthonormal energy eigenbasis givesA continuous spectrum requires the corresponding integral. At equal times, completeness makes this kernel a Dirac delta function.

The free Euclidean action has inverse propagator $D=-m{\partial }_{\tau }^2+\kappa$. Therefore the thermal Green's function obeyswhere ${\delta }_{\beta }$ is the periodic Dirac delta function. It is continuous, has continuous first derivative away from its source, and satisfies $G^{\prime }(0^{+})-G^{\prime }(0^{-})=-1/m$. These conditions follow by integrating the differential equation across the source and using the bosonic thermal trace.

For $\Omega =\sqrt{\kappa /m}$, solving the homogeneous equation on $0<\tau <\beta$ and imposing these conditions givesIn particular $g_0\equiv G^{(0)}(0)=\coth (\beta \Omega /2)/(2m\Omega )$. Equivalently $G^{(0)}={\beta }^{-1}{\sum }_n{e}^{-i{\omega }_n\tau }/(m{\omega }_n^2+\kappa )$ with bosonic Matsubara frequencies ${\omega }_n=2\pi n/\beta$.

Evaluate the two-body Lorentz-invariant phase space in the parent's rest frame, $p=(m,0)$. The spatial Dirac delta function sets ${\mathbf{q}}_2=-{\mathbf{q}}_1$. For massless daughters $E_1=E_2=q$, soThe factor $1/2$ in the last step is the derivative of the delta-function argument. Substituting into the decay width gives, for labelled or distinct daughters,The angular domain is one sphere: $0\le \phi <2\pi$ and $-1\le \cos \theta \le 1$, equivalently $0\le \theta \le \pi$. The printed negative-to-positive polar-angle interval is not a second independent coverage of the sphere.

For identical daughters, integration over the full labelled two-body phase space counts each physical final state twice. One must therefore include an additional factor $1/2!$, or equivalently integrate over an unordered half-domain. This distinction is necessary for the two identical photons in solution.

Write $d{\Pi }_p=d^{3}p/[(2\pi {)}^{3}2{\omega }_p]$, $p^{\mu }=({\omega }_p,\mathbf{p})$ and ${\omega }_p=\sqrt{{\mathbf{p}}^2+m^2}$. The Fourier expansion in the source is compatible with the oscillator canonical commutation relationsFor the free real scalar field, $\pi =\dot{\varphi }$ andSubstitution of the Fourier expansion produces spatial Dirac delta functions. Terms with two annihilation operators or two creation operators cancel: in $P_0$ their coefficient is $-{\omega }_p^2+{\mathbf{p}}^2+m^2=0$, and in $P_i$ their integral is odd under $\mathbf{p}\mapsto -\mathbf{p}$. The two mixed terms combine, with normal ordering removing the vacuum constant, to giveIn particular, $P_i$ contains the lower component $p_i=-{\mathbf{p}}_i$ in this Minkowski metric convention.

To check commutativity without manipulating a square of a Dirac delta function, define $N_f=\int d{\Pi }_{p}f(p)a_p^{†}{a}_p$. The oscillator algebra gives $[N_f,a_q]=-f(q)a_q$ and $[N_f,a_q^{†}]=f(q)a_q^{†}$. Thus $[N_f,a_q^{†}{a}_q]=0$ and $[N_f,N_g]=0$. Taking $f=p_{\mu }$, $g=p_{\nu }$ provesThese identities hold on the usual dense domain of finite-particle states, or can first be proved in a finite box and then continued to the continuum.

An ideal concentrated force. In a one-dimensional distributed-load equation it is represented by its magnitude times a Dirac delta function.

A point mass idealizes a localized body's mass density as $\rho (\mathbf{x},t)=m{\delta }^{(3)}(\mathbf{x}-\mathbf{z}(t))$, using the Dirac delta function. Its second mass moment tensor is $I_{ij}=m{z}_i{z}_j$. The approximation ignores its internal size and structure.

A point spread function is an imaging system's response to an optical point source. Here it means the distribution of image light intensity: an ideal point object becomes a finite spot, often with faint rings, because of diffraction. An unresolved star and a sufficiently small bead emitting fluorescence are practical approximations to an optical point source. Their image tells us how nearby object details will overlap, and reveals blurring from optical aberrations, defocus, and, for an optical telescope, astronomical seeing.

A useful convention separates shape from throughput. At a fixed image plane, let $p(\mathbf{x})\ge 0$ be the point spread function with ${\int }_{{\mathbb{R}}^2}p(\mathbf{x})d^2\mathbf{x}=1$. Its units are inverse area. A point delivering detected radiant flux $F$ at image position ${\mathbf{x}}_0$ produces light intensity $Fp(\mathbf{x}-{\mathbf{x}}_0)$; this point spread function normalization makes the integral over the image equal to $F$. A profile divided by its own peak is convenient for comparing widths, but is a different point spread function normalization and does not by itself compare total radiant flux.

In incoherent imaging, different object points have negligible mutual spatial optical coherence over the measurement, so their image light intensities add. If the response also obeys translation invariance over the chosen field of view, the expected continuous image is the convolutionHere $O$ is the object brightness distribution expressed in image-scaled coordinates, with magnification already included, and $\tau$ is constant transmission. For $O(\mathbf{u})=F_0\delta (\mathbf{u}-{\mathbf{u}}_0)$, where $\delta$ is the two-dimensional Dirac delta function, this gives $I=\tau F_{0}p(\mathbf{x}-{\mathbf{u}}_0)$. A changing response across the field of view instead requires a position-dependent integral kernel $p(\mathbf{x};\mathbf{u})$; one measured point spread function is then only a local calibration.

In coherent imaging, the superposition principle applies to the complex wave amplitudes. With amplitude spread function $h$, the corresponding ideal model isThus $|h\ast E{|}^2$ generally differs from $|h{|}^2\ast |E{|}^2$: coherent contributions produce wave interference terms that depend on their relative wave phases. Their time average vanishes for mutually incoherent emitters, which explains the intensity convolution. Partial optical coherence requires the correlations between object points as well.^[ref]

The figure is an original diffraction model, not measured data. All panels have the same total radiant flux: each member of a pair contributes half the single-source radiant flux. Dashed curves show the separate light intensity contributions; the solid curve is their sum. The point spread functions have tails, so even sources satisfying the Rayleigh criterion overlap. As separation decreases, the image changes continuously; there is no abrupt disappearance of all information at one separation.

A motivating measured example is Measurement of point-spread function (PSF) for confocal fluorescence microscopy by InCheon Song and others (2005). A $543.8\mathrm{nm}$ laser, delivered through an optical fiber, excited fluorescent beads on a slide through a $100\times$, numerical aperture $1.3$ oil-immersion objective. A moving stage scanned the bead; emitted light returned through the objective, an emission filter and a pinhole to a photodiode. The readout was fluorescence light intensity against scanned position. Under the same setup, nominal bead diameters $0.17$, $0.5$, and $1\mu \mathrm{m}$ produced progressively broader measured spots. The size comparison controls for the finite object: the measured response is the convolution of bead emission with the effective point spread function, rather than a pure point response. Inverting this convolution model gave a response compared with the theoretical one. The smaller bead gave a weaker signal relative to noise; uncertainty in actual bead diameter also affected the restored width. These are limits of the calibration. In confocal microscopy, excitation and pinhole detection jointly determine the effective point spread function; this experiment does not measure the ideal widefield Airy disk plotted above.^[ref]

For a new calibration, useful additional controls are to measure detector background, avoid saturating the photodetector, repeat the scan to check stability, and compare isolated beads at several positions in the field of view. Record wavelength, numerical aperture, pinhole setting, sampling interval and optical focus. These are practical recommendations, not extra measurements claimed for the published experiment.

For $E>0$ and the positive-energy branch, the root Jacobian of the Dirac delta function gives this identity. In a massless collinear parton approximation, it produces $\delta (x-\xi )/(P\cdot q)$.

For a free massive Proca field in the mostly-plus Minkowski metric, the covariant inverse of the quadratic action and the canonical time-ordered product of vector potentials differ locally. With Fourier transform $e^{iq\cdot (x-y)}$, the covariant Proca propagator is $-i({\eta }_{ab}+q_a{q}_b/m^2)/(q^2+m^2-i0)$, while the literal canonical two-point correlation function subtracts $i{\delta }_a^0{\delta }_b^0/m^2$ in momentum space. The time component is nondynamical; its physical-mode numerator is $|\mathbf{q}{|}^2/m^2$, rather than $-1+(q^0{)}^2/m^2$. The difference is an instantaneous Dirac delta function, not another propagating state. Both expressions agree away from coincidence. This distinction matters when identifying a covariant Green function with a canonical correlator in a constrained field theory.

A complete orthonormal basis in a Hilbert space resolves the identity as ${\sum }_j|j⟩⟨j|=I$. In infinite dimension the partial sums converge strongly: their action on each fixed vector converges in norm. Generalized position and momentum eigenstates instead give distributional integral resolutions, such as $\int dx|x⟩⟨x|=I$, with their normalization interpreted through the Dirac delta function.

The Schwarz integral formula reconstructs a holomorphic function from boundary values of its real part. For a real boundary function $f$ on the complex upper half-plane satisfying ${\int }_{\mathbb{R}}|f(s)|/(1+s^2)ds<\infty$, the regularized form isIts real part is the Poisson integral of $f$. The subtraction makes the integral converge and changes only the imaginary normalization. Integration by parts, when the boundary terms vanish, gives $W^{\prime }(w)=(\pi i{)}^{-1}\int f^{\prime }(s)/(s-w)ds$; jumps of $f$ are included as Dirac delta function terms in its distributional derivative.

Write an ideal point-transfer model using the Dirac delta function:For this arrival stream with independent increments, distinct arrivals contribute the product of the means, which cancels when centering. The same-arrival term leaves the autocovarianceIts Fourier transform gives the two-sided power spectral density, while folding negative frequencies onto positive ones gives the one-sided convention:These spectra describe $\delta i$, excluding the spectral line from the mean direct current. The amplitude spectral density is $\sqrt{2qI}$ in $\mathrm{A}/\sqrt{\mathrm{Hz}}$; it is not a root mean square electric current until a measurement signal bandwidth is specified. Approximate white noise requires frequencies low enough that the single-transfer pulse shape and any correlations can be neglected. Neither the ideal impulses nor a flat spectrum extending to infinity describe a real detector at every timescale.

All spectra above are per hertz, with $f$ in cycles per second. One possible angular frequency convention defines a density ${\mathcal{S}}_I^{(2)}(\omega )$ integrated using $d\omega$ alone. Since $\omega =2\pi f$, it hasAnother common convention keeps the numerical density $qI$ but integrates it with $d\omega /(2\pi )$. State the integration measure as well as whether a spectrum is one-sided: a change of variable and a folding of signs are separate operations.