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Normalize the incident plane wave to unit amplitude. The scattering asymptotics are $\psi (\mathbf{r})\sim e^{ikz}+f(\theta )e^{ikr}/r$; the outgoing coefficient $f$ is the scattering amplitude. The general axisymmetric free solution for $r>0$ isusing Spherical Bessel functions, spherical Neumann functions and Legendre polynomials.

For a real short-range central potential, the exterior radial solution in each partial wave is proportional to $\cos {\delta }_{\ell }{j}_{\ell }-\sin {\delta }_{\ell }{n}_{\ell }$, whose large-$r$ form is $\sin (kr-\ell \pi /2+{\delta }_{\ell })/(kr)$. This defines the scattering phase shift. Multiplying this combination by $(2\ell +1)i^{\ell }{e}^{i{\delta }_{\ell }}$ matches the incoming spherical component of the incident plane wave: the factor $e^{i{\delta }_{\ell }}$ cancels the shifted incoming phase. The outgoing coefficient instead acquires $e^{2i{\delta }_{\ell }}$. Subtracting the plane wave's outgoing component therefore givesFor the barrier's $s$-wave, the regular reduced radial function inside is $u(r)=A\sinh (\kappa r)$, with $\kappa =\sqrt{{\gamma }^2-k^2}$. Outside take $u(r)=B\sin (kr+{\delta }_0)$. Continuity of $u$ and its derivative at $a$ equates logarithmic derivatives:Use the branch with ${\delta }_0\to 0$ as $k\to 0$. The matching equation gives ${\delta }_0=k(\tanh (\gamma a)/\gamma -a)+o(k)$, so the $s$-wave scattering amplitude tends toIts negative is the repulsive spherical barrier scattering length, positive for this repulsive barrier.

The tree-level Feynman diagram contains one weak charged current vertex. Fermion arrows point along fermion-number flow, so the outgoing antiquark arrow points toward the vertex.

For one fixed matching color, the scattering amplitude isAn overall phase from the vertex has no effect on the width. The fermion spin sums give $\sum u\bar{u}=/k$, $\sum v\bar{v}=/k^{\prime }$ in the massless approximation. For the three initial polarizations, use a spin average of $1/3$. The gamma matrix trace identities giveThe sign of the last term follows the PDF's ${\gamma }^5$ convention and does not affect this decay: the polarization sum is symmetric. Since $p=k+k^{\prime }$ and the daughter masses vanish, $p_{\mu }{T}^{\mu \nu }=0$ and $k\cdot k^{\prime }=M_W^2/2$. ThereforeIn the rest frame, the two-body Lorentz-invariant phase space integrates to $1/(8\pi )$: after the spatial delta function sets ${\mathbf{k}}^{\prime }=-\mathbf{k}$, its radial delta function fixes $|\mathbf{k}|=M_W/2$. Explicitly,The decay formula consequently givesThis is the printed expression, interpreted for one color. For a physical quark-flavor channel there are three orthogonal final color states. Their probabilities add; no initial color average is present for a colorless W boson. ThusThe PDF's formula omits this color multiplicity if read as the ordinary inclusive flavor width. It is also the familiar normalization for a colorless lepton channel.

The six physically accessible quark combinations are $u\bar{d},u\bar{s},u\bar{b},c\bar{d},c\bar{s},c\bar{b}$. A real on-shell W boson cannot produce a top quark; the massless approximation is applied to the accessible daughters and does not open a physically forbidden top channel. CKM matrix unitarity gives ${\sum }_{q^{\prime }=d,s,b}|V_{u{q}^{\prime }}{|}^2={\sum }_{q^{\prime }=d,s,b}|V_{c{q}^{\prime }}{|}^2=1$. Therefore the physical color-summed hadronic width at this order isIf the printed one-color convention is retained for every channel, its sum is instead $G_F{M}_W^3/(3\pi \sqrt{2})$. In the artificial theory where all three up-type flavors, including the top, are kinematically massless, there would be nine channels and the physical sum would be $9G_F{M}_W^3/(6\pi \sqrt{2})$; that is not the on-shell Standard Model channel list.

The authoritative PDF has $m_Z^2/v$, not the dimensionally incorrect $m_Z^2/v^2$ in the TeX conversion. The two identical Z bosons give the vertex factor two, so the scattering amplitude for the decay isup to an irrelevant overall phase. Apply the massive vector polarization sum, identifying the hint's $M_Z$ with $m_Z$:Since $k_1\cdot k_2=(m_H^2-2m_Z^2)/2$, putting $x=m_Z^2/m_H^2$ givesNo initial-spin average is needed for a Higgs boson. In its rest frame the two-body momentum is $|\mathbf{k}|=(m_H/2)\sqrt{1-4x}$. Integrating the energy and momentum delta functions in the Lorentz-invariant phase-space measure gives $d{\Phi }_2=\sqrt{1-4x}d\Omega /(32{\pi }^2)$, hence $\int d{\Phi }_2=\sqrt{1-4x}/(8\pi )$.

The printed generic width formula treats the daughters as labelled. Here the identical final-state symmetry factor is $1/2!$, without which the same physical configuration is counted twice. Thus the Higgs decay to two Z bosons hasEquivalently, $1/v^2=\sqrt{2}{G}_F$ gives the prefactor $G_F{m}_H^3/(16\sqrt{2}\pi )$. The threshold limit is zero and the large-mass limit is $m_H^3/(32\pi v^2)$. The additional factor for identical daughters is a required specialization of the PDF's generic phase-space formula, not an extra factor in the vertex.

For fixed $(P,Q)\ne (0,0)$ the effective Fermi interaction predicts $d\sigma /d\Omega \propto G_F^{2}s$. The angular odd term integrates to zero, while $\int d\Omega (1+{\cos }^2\theta )=16\pi /3$, soThe corresponding fixed-spin scattering amplitudes grow as $G_{F}s$, eventually conflicting with partial-wave unitarity. This is a failure of extrapolating the effective field theory, whose contact approximation requires $s\ll M_W^2$, not a failure of the Standard Model.

Restoring the virtual W boson Feynman propagator replaces its constant low-energy denominator by $M_W^2/(M_W^2-s)$, away from the resonance. The massless external Dirac equations give zero contraction of each current with the exchanged momentum, so the longitudinal numerator does not contribute. Its squared modulus multiplies the effective cross-section. For $s\gg M_W^2$ the resulting tree-level rate decreases as $1/s$ rather than growing as $s$. Near the pole one must include the W boson decay width. These statements assume the same massless conserved external currents; the contact expression alone is not valid there.

Take the quantum electrodynamics interaction density to be ${\mathcal{L}}_{\mathrm{int}}=-e\bar{\psi }{\gamma }^{\mu }\psi A_{\mu }$. The momentum-space Feynman-gauge photon propagator, with Lorentz indices $\mu ,\nu$ and momentum $k$, isThe electron-photon vertex from the QED Feynman rules iswith its spinor indices connecting the incident and outgoing fermion lines. The muon-photon vertex has the same coupling. The Feynman i-epsilon prescription specifies the pole boundary condition; using a different overall charge-sign convention changes both vertices consistently and leaves the squared amplitude unchanged.

For the electron-positron annihilation into a muon pair, the exchanged photon has $k=p+q$ and $k^2=s$. Up to the conventional overall phase, the scattering amplitude is

Both the D meson and kaon are spin-zero pseudoscalars. The strong interaction states obey parity symmetry in quantum field theory. Consequently an axial current matrix element between them would have to be a pseudovector formed from only $p$ and $k$, which is impossible: an totally antisymmetric tensor would require more linearly independent vectors. ThusLorentz covariance then leaves two linearly independent vectors for the vector current matrix element, giving the pseudoscalar-to-pseudoscalar form factorsThe vanishing axial current here follows from strong interaction parity symmetry in quantum field theory, not parity symmetry in quantum field theory of the weak interaction.

To obtain the requested decay formula, use naive factorization of a nonleptonic meson decay: approximate the four-quark matrix element by the product of the ${\bar{D}}^0\to K^{+}$ current matrix element and the vacuum-to-pion matrix element. This is an additional hadronic approximation; tree-level weak vertices alone do not establish it, and nonfactorizable Quantum chromodynamics effects can change the result. The vacuum-to-pion vector current matrix element vanishes by parity symmetry in quantum field theory, while the specified pion decay constant normalization givesThe $\sqrt{2}$ cancels the $1/\sqrt{2}$ in the four-fermion interaction. Up to an irrelevant overall phase, the scattering amplitude is thereforeFor $m_{\pi }=0$, $q^2=0$, so only $f_{+}(0)$ survives. Integrating the two-body decay phase space in the D meson rest frame givesHenceThis coefficient uses exactly the stated pion decay constant convention and the stated massless-pion approximation.

The deep inelastic scattering process contains a virtual photon exchanged between the Electron and the hadron:

Use an electromagnetic vector current $J^{\mu }$ containing the dimensionless quark charges, with the coupling $e$ factored out. The scattering amplitude, up to an irrelevant phase, isTo match the printed prefactor, define the leptonic tensor with a spin sum over both Electron spins and keep the initial spin average $1/2$ outside it. The gamma-matrix trace givesFor a stationary target and massless Electron, the invariant flux factor is $4ME$. The inclusive final-state Lorentz-invariant phase-space measure and target spin average are contained in $4\pi W_H^{\mu \nu }$. Thus the differential scattering cross-section isHere $q^4$ means $(q^2{)}^2$. If the initial spin average is instead built into the leptonic tensor, its normalization is $2(\cdots )$ and the displayed cross-section prefactor must be doubled. The two conventions give the same observable.

The closed string already contains the particle required for quantum gravity. For bosonic string theory in its critical dimension of string theory, the mass-shell condition and closed-string level matching areAt $N=\tilde{N}=1$, the states ${ϵ}_{mn}{\alpha }_{-1}^m{\tilde{\alpha }}_{-1}^n|0;k⟩$ are massless. After imposing the Virasoro constraints and quotienting null string states, the transverse polarization tensor decomposes into symmetric trace-free, antisymmetric and scalar pieces under $SO(D-2)$. They describe a graviton, the Kalb–Ramond field and the dilaton. In particular, the symmetric trace-free sector is a massless spin-two graviton. Its number of degrees of freedom is $D(D-3)/2$.

The state–operator correspondence associates each physical string state with a string vertex operator. For example, the matter part of a massless closed-string vertex operator isIts conformal weights are $(1,1)$, so its integrated form $\int d^{2}z{V}_{ϵ,k}$ is invariant under changes of string worldsheet coordinates. At fixed insertion positions it is accompanied by $c\bar{c}$ from the worldsheet ghost fields. Physical vertices represent BRST cohomology classes; longitudinal changes of the polarization tensor are BRST-exact operators and decouple from physical scattering amplitudes. For the graviton, this string-state gauge redundancy becomes the linearized target-space diffeomorphism $h_{mn}\mapsto h_{mn}+{\partial }_m{\xi }_n+{\partial }_n{\xi }_m$.

The Polyakov path integral computes a closed-string scattering amplitude by inserting the external string vertex operators, integrating their unfixed positions and the inequivalent worldsheet moduli, and including the worldsheet ghost fields. On a Riemann sphere three positions are fixed by the residual conformal group. For example, contractions of tachyon vertex operators produce the Koba-Nielsen factor, whose position integral gives the Virasoro–Shapiro amplitude.

The string dual resonance property means that one crossing-symmetric scattering amplitude has equivalent expansions in the different channels: its poles exhibit intermediate string states in each channel, rather than separate channel contributions being added again. The scattering-amplitude factorization at a pole identifies the intermediate particle and its couplings. Set $8\pi T=1$, so ${\alpha }^{\prime }=4$ and the closed-string tachyon has $M^2=-1$. For four such external states the Mandelstam variables obey $s+t+u=-4$. The Virasoro–Shapiro amplitude has generic $s$-channel poles at $s=-1,0,1,\dots$, exactly the bosonic string mass spectrum in these units.

The massless pole gives an especially direct check. With overall normalization $C$ and $u=-4-s-t$, the Gamma function recurrence givesThusThe quadratic residue in $t$ contains a spin-two exchange. A scalar exchange alone could not give this angular dependence. Combined with the known massless spectrum and scattering-amplitude factorization, it identifies exchange of the graviton, with possible scalar contributions from the dilaton; the antisymmetric Kalb–Ramond field does not couple to two identical scalar tachyons here. Consequently the graviton participates in interactions, rather than being an isolated free state.

Decoupling longitudinal graviton polarizations forces a universal coupling to the conserved stress-energy tensor. Consistency of this massless spin-two gauge invariance extends the linearized coupling to the nonlinear dynamics of general relativity. At distances large compared with $\sqrt{{\alpha }^{\prime }}$, the gravitational sector of the effective action begins with the Einstein-Hilbert action, alongside dilaton and Kalb–Ramond field terms and higher-derivative string corrections. The infinite tower of string states supplies the short-distance completion of this quantum gravity expansion.

Interactions are organized by string worldsheet topology. A constant dilaton gives the string coupling $g_s=e^{{\Phi }_0}$, and a connected oriented surface of genus $g$ has Euler characteristic $\chi =2-2g$. Its weight is $g_s^{-\chi }$; with $n$ normalized external vertices,The sphere is tree order, the torus is one loop, and each additional handle adds a factor $g_s^2$. Each coefficient is an integral over worldsheet moduli, so this is string perturbation theory in $g_s$, with an independent low-energy expansion in ${\alpha }^{\prime }$.

For the one-loop vacuum contribution, torus modular invariance identifies $\tau$ with $(a\tau +b)/(c\tau +d)$, $ad-bc=1$. Integration is over the standard fundamental domain of the modular group,The potential short-proper-time region ${\tau }_2\to 0$, responsible for a point-particle ultraviolet divergence, is absent. It would count metrics already represented elsewhere in $\mathcal{F}$. Torus modular invariance is therefore the geometric reason for one-loop ultraviolet finiteness. More explicitly, in $D=26$ the vacuum integrand is proportional to $d^2\tau {\tau }_2^{-14}|\eta (\tau ){|}^{-48}$, with $\eta$ the Dedekind eta function; the complete expression is invariant under the modular group.

Ultraviolet finiteness does not make the bosonic one-loop vacuum energy finite. The remaining long-tube region ${\tau }_2\to \infty$ is an infrared divergence from the tachyon: $|\eta (\tau ){|}^{-48}\sim e^{4\pi {\tau }_2}$. It signals instability of the bosonic vacuum. Tachyon-free consistent backgrounds remove this particular obstruction, though other infrared effects must still be treated. The ultraviolet improvement comes from the extended string and the complete spectrum together with the worldsheet gauge identifications; the genus expansion remains a perturbative description of quantum gravity.

A three-dimensional Bravais lattice is the set of all integer combinations of three linearly independent primitive vectors. Its reciprocal lattice isand the first Brillouin zone is the Wigner-Seitz cell of the origin in reciprocal space.

For the stated FCC lattice, the primitive-cell volume isThe reciprocal-basis formula givesA reciprocal primitive cell, and hence the Brillouin zone, has volume

In the first Born approximation, with momentum transfer $q=k-k^{\prime }$, the scattering amplitude isFor the spherical shell, align the polar axis with $q$ and use its radial Dirac delta function:Therefore

Translation of each atomic potential multiplies its transform by a phase, so the crystal lattice structure factor givesFor an infinite lattice, the lattice Poisson summation formula turns the sum intoThus the amplitude vanishes away from reciprocal-lattice momentum transfers.

In elastic scattering, $|k|=|k^{\prime }|$ and henceThe two shortest relevant reciprocal-vector lengths of this FCC lattice areBoth are kinematically available when $k>2\pi /a$. Assigning their scattering angles ${\theta }_2$ and ${\theta }_1$, respectively, gives

In the classical experiment, particles whose impact parameter lies between $b$ and $b+db$ occupy an incident annulus of areaFor a spherically symmetric potential, azimuthal symmetry scatters this annulus into the solid-angle stripThus the central-potential classical differential cross-section isIf the relation between $b$ and $\theta$ has several branches, one sums this expression over every branch. The corresponding diagram consists of an incoming line offset by $b$ from the centre, an outgoing asymptote making angle $\theta$ with the incident direction, and the annulus and solid-angle strip just described.

For quantum scattering by a central potential, the far-field wavefunction iswhere $f(\theta )$ is the scattering amplitude. The incident plane wave has probability currentTo leading order in $1/r$, the scattered spherical wave has radial currentThe scattered rate through area $r^{2}d\Omega$ divided by the incident flux is therefore the quantum differential cross-section from a scattering amplitude:

For a real short-range spherically symmetric potential, the stationary wavefunction must solve the Schrodinger equation, be regular at the origin whenever the potential is finite there, and satisfy the scattering boundary conditionHere $e^{ikz}$ is the incident plane wave, the second term is an outgoing spherical wave, and its coefficient $f(\theta )$ is the scattering amplitude.

Comparing the stated asymptotic expression with the partial-wave expansion of a scattering amplitude giveswhere $P_l$ is a Legendre polynomial. The outgoing-to-incoming coefficient in channel $l$ isPartial-wave unitarity requiresor equivalentlyThe real scattering phase shift ${\delta }_l$ is defined modulo $\pi$ by

The scattering amplitude is defined byFor a central potential its partial-wave expansion isUsing orthogonality of the Legendre polynomials in ${\sigma }_T=\int |f{|}^{2}d\Omega$ gives

Let the Bravais lattice have primitive vectors $a_1,a_2,a_3$, and putThe reciprocal lattice has the basisThe scalar triple product and cross product identities giveConsequently every $q={\sum }_j{n}_j{b}_j$ satisfies $q\cdot l\in 2\pi \mathbb{Z}$ for every lattice point $l={\sum }_i{m}_i{a}_i$, and conversely these three conditions force the coefficients of $q$ in the reciprocal basis to be integers. This proves that the displayed vectors generate ${\Lambda }^{\ast }$.

In the Born approximation, suppose that the crystal potential is the sum of translates of one atomic potential $V_0$ over the finite set $S$ of lattice sites. Its Fourier transform at the momentum transfer $Q=k-k^{\prime }$ factors asThus the single-atom scattering amplitude is multiplied by the crystal lattice structure factor $\Delta (Q)$. Writingseparates the sum into three finite geometric series. For $l_i=-L_i/2,\dots ,L_i/2$ this givesWhen the $L_i$ are large, a factor has a sharp maximum when ${\alpha }_i=2\pi n_i$. All three factors are therefore simultaneously large precisely when $Q\in {\Lambda }^{\ast }$. These are the reciprocal-lattice peaks of a finite crystal.

For the stated Body-centered cubic lattice basis,the reciprocal-basis formula givesEquivalently,The shortest nonzero reciprocal vectors have squared Euclidean norm $2(2\pi /a{)}^2$, so

In elastic scattering, $|k|=|k^{\prime }|=k$. If $q=k-k^{\prime }\in {\Lambda }^{\ast }$ and $\theta$ is the angle between $k$ and $k^{\prime }$, Euclidean geometry givesThe first possible diffraction peak therefore occurs atas $ka\to \infty$, by the small-angle approximation. This is the elastic Bragg scattering condition for the shortest reciprocal-lattice vector.

The momentum-space Feynman rules for Phi cubed theory areEach vertex carries $(2\pi {)}^4{\delta }^{(4)}$ of momentum conservation, each independent loop momentum is integrated with $d^4\ell /(2\pi {)}^4$, and a graph is divided by its Feynman-diagram symmetry factor. External propagators are omitted from an amputated scattering amplitude.

SetThe integral equation becomesEvaluating it at the three delta functions givesThus

For $x>a$, all three Green-function terms are proportional to $e^{ikx}$. Hence

Now define $\gamma =ik{\hslash }^2/(\lambda m)$, so that $q=-1/\gamma$. If $M$ denotes the matrix multiplying $\mathbf{\Psi }$, direct evaluation givesThe displayed algebraic equation is therefore exactly the condition that the finite-dimensional scattering system become singular. Its solutions are poles of the analytically continued scattering amplitude.

To see the imaginary-axis poles directly, write $k=i\kappa$ with $\kappa >0$. When $a\to 0$, the equation reduces toand its nonzero root is $\gamma =-3$, orThis is the bound state of the coincident potential $-3\lambda \delta (x)$. When $a\to \infty$, the three centres decouple and the roots approachwith exponentially small splitting. A pole at $k=i\kappa$ has energy $E=-{\hslash }^2{\kappa }^2/(2m)$ and an exponentially decaying wavefunction, so these upper imaginary-axis singularities represent bound states.

For a barrier of radius $a$ and potential ${\hslash }^2{\gamma }^2/(2m)$, matching the regular inner hyperbolic-sine wave to the outer sine wave gives $a_s=a-\tanh (\gamma a)/\gamma$. The low-energy $s$-wave scattering amplitude is $-a_s$.

Near a one-particle pole, a scattering amplitude factors into amplitudes coupling each side to the intermediate state, summed over its physical polarizations. The pole position gives its squared mass, while its residue encodes its spin and couplings.

For a localized target and a plane wave incident along the $z$-axis, the large-radius scattering wavefunction has the formwhere $f$ is the scattering amplitude.

The S-matrix maps incoming asymptotic states to outgoing asymptotic states. Its connected transition matrix elements contain an overall energy-momentum Dirac delta function and an invariant scattering amplitude. In quantum field theory, the LSZ reduction formula obtains these elements from amputated field correlators.

For an unpolarized initial state, average the squared scattering amplitude over its equally populated spin states. With two incoming spin-one-half particles this gives a factor $1/(2\cdot 2)=1/4$. Unobserved final spins are summed rather than averaged.