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Crossing symmetry relates scattering amplitudes by analytic continuation exchanging incoming and outgoing particles. A symmetric four-point scalar amplitude has equivalent descriptions under permutations of the Mandelstam variables.

For an initial Dirac field of mass $m$ annihilating into a distinct Dirac field of mass $M$ through a vector current, put $\Sigma =m^2+M^2$. With Mandelstam variables satisfying $s+t+u=2\Sigma$, the spin average of the two contracted vector currents is $2F$, where $F=(\Sigma -t{)}^2+(\Sigma -u{)}^2+2s\Sigma =t^2+u^2+4s\Sigma -2{\Sigma }^2$. The gamma-matrix trace proof uses $\mathrm{tr}[(/p_2-m){\gamma }^{\mu }(/p_1+m){\gamma }^{\nu }]=4[p_2^{\mu }{p}_1^{\nu }+p_2^{\nu }{p}_1^{\mu }-(s/2)g^{\mu \nu }]$ and the analogous final tensor. Couplings, the mediator denominator and color charge factors multiply this common kinematic expression.

For an especially simple neutral current example, scatter a muon neutrino elastically from an initially stationary unpolarized electron:Here $T_e$ is the final electron kinetic energy, $E_{\nu }$ is the incident neutrino energy, and ${\theta }_W$ is the Weinberg angle. These are electron chiral couplings in the $g/\cos {\theta }_W$ vertex convention of neutral-current vector and axial couplings. The chosen muon neutrino elastic channel has no additional W boson exchange amplitude; the electron-neutrino elastic channel would have one. Other charged-current final states can become kinematically accessible at sufficiently high incident energies and are not part of this elastic model.

Use natural units $\hslash =c=1$, retain only tree-level Feynman diagrams, and take the contact limit $|q^2|\ll m_Z^2$ with $E_{\nu }\gg m_e$. Neglect electron-mass terms relative to $E_{\nu }$, radiative corrections and detector response. Then$G_F$ is the Fermi constant. The approximate endpoint $y=1$ replaces the exact elastic endpoint $y_{\max }=2E_{\nu }/(m_e+2E_{\nu })$. The omitted leading mass-dependent interference term in $d{\sigma }_{\nu }/dy$ is $-K{g}_L{g}_R{m}_{e}y/E_{\nu }$. Thus the approximation is not appropriate for arbitrarily soft neutrinos.

The shape follows from the two chiral contributions to the massless scattering amplitude. The spin sum gives terms proportional to $s^2$ and $u^2$ in Mandelstam variables; in this limit $u/s=-(1-y)$, giving the square $(1-y{)}^2$. The two electron chiralities do not interfere when the relevant mass corrections are dropped. Replacing the neutrino by its antineutrino exchanges which coupling multiplies each kinematic term. This is the same chiral angular weighting that appears in neutrino scattering on an antiquark, now with a target whose couplings are particularly simple.

Integrating the two elementary terms givesFor an original teaching example choose $s_W^2=1/4$, so $g_L=-1/4$ and $g_R=1/4$. This is a stipulated parameter, not a measurement of the Weinberg angle. ThenThe normalized probability density function is $p(y)=3[1+(1-y{)}^2]/4$, not the unnormalized differential cross section. Its endpoints have ratio two, and the fraction with $y>1/2$ is $13/32$. The antineutrino spectrum happens to coincide at this chosen parameter because $g_L^2=g_R^2$; this equality is not a general prediction for arbitrary Weinberg angles.

Unlike a visible electron recoil, the scattered neutrino is unrecorded. A measured recoil spectrum must therefore be compared with incident neutrino flux, energy distribution, acceptance, detection efficiency and backgrounds. A shape or rate tests a specified theoretical model only after these experimental ingredients have been supplied.

There is one tree-level Feynman diagram: annihilation of the up quark and its antiquark into an $s$-channel gluon, followed by production of the down quark and its antiquark. A gluon interaction vertex preserves quark flavor, so a $t$-channel line joining the incoming up quark to the outgoing down quark is impossible.

Let $i,j,k,l$ label the initial quark, initial antiquark, final quark, final antiquark color charges. Up to an irrelevant common sign, the scattering amplitude isThe Mandelstam variables are $s=(p_1+p_2{)}^2$, $t=(p_1-p_3{)}^2$, $u=(p_1-p_4{)}^2$, with $s+t+u=0$ in the zero-mass limit. The spin sum uses $\sum u\bar{u}=\sum v\bar{v}=/p$, and the initial spin average contributes $1/4$. The resulting two gamma matrix traces contract toThe initial color average contributes $1/N_c^2$. Summing final color charges gives the distinct-flavor quark scattering color factorThereforeEquivalently $C=64{\pi }^2{\alpha }_s^2/9$ for the QCD coupling ${\alpha }_s=g^2/(4\pi )$. There is no initial gluon spin average: the gluon is internal.

In pure QCD, different incoming and outgoing quark flavours allow only the tree-level $s$-channel gluon graph. A quark-gluon vertex preserves flavour, so no flavour-changing $t$-channel exchange is available. Label the incoming momenta $p_1,p_2$ and outgoing momenta $k_1,k_2$. For transparent spin traces, use the auxiliary metric $\eta =-g$ and its usual gamma matrices; the positive-energy Mandelstam variables areThis is a consistent change of metric notation, not a change of the physical invariants. Define the fixed-index colour tensor $C_{ji;kl}={\sum }_a(T^a{)}_{ji}(T^a{)}_{kl}$. Up to an irrelevant common phase,The indices $i,j$ label incoming colours and $k,l$ outgoing colours. Averaging over the two spins of each incoming particle and summing the outgoing spins givesThe gamma-matrix trace identity makes the contractionFor instance $p_1\cdot k_1=p_2\cdot k_2=-t/2$ and the crossed products are $-u/2$. Therefore, before the remaining colour sum and average,Here the bar averages spins only; it does not silently include the colour average requested next.

The different-flavour quark annihilation into a massive pair has only an $s$-channel gluon graph at tree level when the two quark flavours differ. Its amplitude contains one incoming and one outgoing colour current, with common magnitude factor $g_s^2/s$. On-shell current conservation removes the gauge-parameter-dependent longitudinal part of the gluon propagator. To display the Dirac signs transparently, evaluate spin traces with the auxiliary mostly-minus metric $\eta =-g$ and $\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2{\eta }^{\mu \nu }$; the invariants $s,t$ are then the usual positive-energy Mandelstam invariants, numerically identical to those defined in the paper.

The incoming colour average and outgoing colour sum giveAveraging over the two spins of each incident particle supplies an additional $1/4$. The physical spin sums are $\sum u\bar{u}=/p+M$ and $\sum v\bar{v}=/p-M$. ThereforeThe coefficient is equivalently $8{\pi }^2{\alpha }_s^2/(9s^2)$, with ${\alpha }_s=g_s^2/(4\pi )$.

The two top projectors cannot both contain $-M$. The printed expression has that error. It is not repaired by changing the metric convention consistently: a particle and antiparticle still require the appropriate distinct projectors. In the auxiliary convention, its outgoing tensor would obey $q_{\mu }{H}_{\mathrm{printed}}^{\mu \alpha }=8M^2{q}^{\alpha }$, rather than zero for $q=p_1+p_2$. At threshold it gives a zero contracted trace, while the correctly averaged vector-current production amplitude squared has a nonzero limit $4g_s^4/9$. The phase-space cross section still vanishes at threshold because the final velocity tends to zero. The colour and spin prefactor above is consequently the prefactor for the corrected trace; no overall constant can make the misprinted massive trace the physical answer for all kinematics.

For massless quark Compton scattering, the color average contributes one, since $(1/N_c){\sum }_{i,j}|{\delta }_{ji}{|}^2=1$. The spin average and incoming photon polarization average supply $1/(2\cdot 2)=1/4$. The Ward identities permit use of $-g_{\mu \nu }$ for each photon polarization completeness relation, and the massless fermion spin sum is ${\sum }_r{u}_r(p){\bar{u}}_r(p)=/p$.

LetReversing the gamma matrices order gives its Dirac-conjugate matrix ${\bar{A}}^{\mu \nu }$. The squared scattering amplitude isFor clarity evaluate the four traces before dividing by their denominators. The Mandelstam variables obey $s+t+u=0$, with $t=(p_1-p_3{)}^2$, $p_1\cdot p_2=s/2$, $p_1\cdot p_4=u(-1/2)$ and $p_1\cdot p_3=-t/2$. For the squared $s$-channel contribution, the gamma matrix trace identities implyHere $/q_s/p_1/q_s=2(q_s\cdot p_1)/q_s-q_s^2/p_1=s/p_2$ and $p_3\cdot p_2=-u/2$. For the squared $u$-channel contribution, $/q_u/p_1/q_u=-u/p_4$ givesOne interference trace reduces tosince $q_s\cdot q_u=(s+t+u)/2=0$. The other interference trace is its complex conjugate and also vanishes. ThereforeEquivalently $C=512{\pi }^2{\alpha }_{\mathrm{em}}^2/81$ in terms of the fine-structure constant. In the physical noncollinear region $s>0$, $u<0$, so the result is positive. The singular massless endpoint $u=0$ reflects a collinear divergence; it is not a negative squared scattering amplitude.

Use the Mandelstam variables $s=(p_1+p_2{)}^2$, $t=(p_1-p_3{)}^2$ and $u=(p_1-p_4{)}^2$. The external on shell relations are $p_1^2=p_2^2=m_e^2$ and $p_3^2=p_4^2=m_t^2$. Expanding the three Mandelstam variables and using four-momentum conservation, $p_3+p_4=p_1+p_2$, gives

Use the Feynman rules and Feynman gauge convention from part a. Put $s=q^2$, $t=r_t^2$, $u=r_u^2$, the Mandelstam variables, and define the three-boson tensor with incoming four-momenta $(q,-k_1,-k_2)$:The contributions, with external Dirac spinors included and both external gauge boson propagators amputated, areFor example, $i{\mathcal{M}}_t$ contains $(+ig{)}^{2}i=-i{g}^2$. In the annihilation channel, the current vertex, propagator, and three-boson vertex give $(+ig)(-i)g=g^2$, so division by the defining overall $i$ gives $-i{g}^2$. These relative signs are essential for the cancellation in part d. The antisymmetric colour factor makes the last expression symmetric under simultaneously exchanging $(a,\mu ,k_1)$ and $(b,\nu ,k_2)$, as required for identical outgoing gauge bosons. The on shell external Dirac spinors obey $(/p_1-m)u(p_1)=0$, $\bar{v}(p_2)(/p_2+m)=0$.

Label the incoming momenta and spin states $(p_1,r_1),(p_2,r_2)$ and the outgoing ones $(p_3,r_3),(p_4,r_4)$, with $p_1+p_2=p_3+p_4$. Write $u_i=u^{r_i}({\mathbf{p}}_i)$ and $v_i=v^{r_i}({\mathbf{p}}_i)$. The Mandelstam variables are $s=(p_1+p_2{)}^2$, $t=(p_1-p_3{)}^2$ and $u=(p_1-p_4{)}^2$. At lowest order, each connected Feynman diagram has two pseudoscalar Yukawa interaction vertices and one internal scalar propagator.

Define the scattering amplitude by $S_{fi}^{\mathrm{connected}}=i(2\pi {)}^4{\delta }^4(p_1+p_2-p_3-p_4)\mathcal{M}$. Fix the external Fock state ordering as $b_1^{†}{b}_2^{†}|0⟩$ to $b_3^{†}{b}_4^{†}|0⟩$ for the two-fermion process. The $t$ exchange connects $1$ to $3$ and $2$ to $4$; the $u$ exchange connects $1$ to $4$ and $2$ to $3$. The two Feynman vertices contribute $(-i\lambda {)}^2$, and the internal scalar propagator contributes $i$ divided by its denominator. Exchanging the two identical final fermions supplies a relative fermionic sign. ThusThis changes sign when the two outgoing labels are exchanged, as identical-fermion antisymmetry requires. There is no $s$ annihilation diagram for two incoming fermions: the interaction preserves Dirac fermion number conservation.

For the fermion-antifermion process take initial state $b_1^{†}{d}_2^{†}|0⟩$ and final state $b_3^{†}{d}_4^{†}|0⟩$. The $t$ exchange uses bilinears $({\bar{u}}_3{\gamma }^5{u}_1)({\bar{v}}_2{\gamma }^5{v}_4)$, and the $s$ annihilation uses $({\bar{v}}_2{\gamma }^5{u}_1)({\bar{u}}_3{\gamma }^5{v}_4)$. To fix the sign without a guess, expand the current $J=:\bar{\psi }{\gamma }^5\psi :$ in creation and annihilation operators. Its relevant terms have the signsThe antifermion sign of a normal-ordered bilinear makes the exchange contraction negative, whereas the product of annihilation and creation contractions is positive for the specified external ordering. Multiplying by $(-i\lambda {)}^{2}i=-i{\lambda }^2$ therefore givesA different common phase convention for an external state can reverse the overall amplitude, but never the relative sign of its two contributions. Both results are tree scattering with pseudoscalar exchange and are of order ${\lambda }^2$.

Use the printed interaction, so the Dyson series contributes $-ig$ per vertex before its spinor matrix. The momentum-space Feynman rules areEach interaction vertex has one pseudoscalar line and one incoming and one outgoing fermion-number line. Conserve four-momentum at every vertex, integrate each independent loop momentum with $d^4\ell /(2\pi {)}^4$, and divide by any Feynman-diagram symmetry factor. A closed fermion loop contributes an extra minus sign. The external wavefunctions are $u$ for incoming fermions, $\bar{u}$ for outgoing fermions, $\bar{v}$ for incoming antifermions and $v$ for outgoing antifermions; spinor matrices are multiplied in order along the fermion line. An external pseudoscalar has unit wavefunction factor in standard relativistic normalization. Relative fermionic signs also occur when the external contractions are permuted.

The leading scattering amplitudes are of order $g^2$. Label the incoming momenta $p_1,p_2$ and outgoing momenta $p_3,p_4$, all with positive energy and $p_i^2=m^2$. Use the Mandelstam variables $s=(p_1+p_2{)}^2$, $t=(p_1-p_3{)}^2$ and $u=(p_1-p_4{)}^2$. The tree scattering with pseudoscalar exchange is shown below; dashed edges are pseudoscalar propagators, and arrows indicate fermion-number flow, opposite to the motion of an antifermion.

Define the scattering amplitude by $⟨f|S-1|i⟩=i(2\pi {)}^4{\delta }^4(p_3+p_4-p_1-p_2)\mathcal{M}$. Use standard relativistic normalization for external states and spinors. To fix even the overall signs, suppress the positive state-normalization factors and write the creator order of the two-fermion states as $|i⟩=b_1^{†}{b}_2^{†}|0⟩$, $|f⟩=b_3^{†}{b}_4^{†}|0⟩$, and the fermion-antifermion states as $|i⟩=b_1^{†}{d}_2^{†}|0⟩$, $|f⟩=b_3^{†}{d}_4^{†}|0⟩$. Spin labels on the $u_i,v_i$ are implicit.

For two fermions, the direct $t$ graph connects $1\to 3$ and $2\to 4$, while the exchanged $u$ graph connects $1\to 4$ and $2\to 3$. The canonical anticommutation relations give signs $+$ and $-$ for these contractions in the chosen state order. Multiplication of the two vertices and the scalar propagator gives $(-ig{)}^{2}i=-i{g}^2$. HenceInterchanging the labels of either pair of identical external fermions reverses this amplitude, as required for the antisymmetric states of fermionic Fock space.

For a fermion and an antifermion, the graphs are $t$ exchange and $s$ annihilation. Their signs can be derived without guessing an analogy with the first process. The normal-ordered product of the pseudoscalar bilinear contains terms of the formsThe minus in the second term is the antifermion sign of a normal-ordered bilinear. Between the chosen external states, exchange has sign $-$ and pair annihilation followed by pair creation has sign $+$. ThereforeChanging an external-state phase can change the common overall sign; it cannot change the relative minus between the two channels. Both amplitudes use the coefficient exactly as printed. For the Hermitian convention $g=i{g}_P$ discussed in part (a), substitute $g^2=-g_P^2$ throughout, or equivalently use vertex $g_P{\gamma }^5$ from the outset.

Use a mostly-plus target Minkowski spacetime metric and restore the string tension by $T=(2\pi {\alpha }^{\prime }{)}^{-1}$. After Wick rotation, the Polyakov action on a Riemann surface isThe Polyakov path integral integrates over embeddings and worldsheet metrics, dividing by worldsheet diffeomorphisms and Weyl transformations. For a consistent flat bosonic string theory, take the critical dimension of the bosonic string $d=26$. Conformal gauge turns the embedding fields into free scalars; its Faddeev-Popov determinant is represented by the bc system. The remaining integrations are over worldsheet moduli and external insertion points.

A tachyon vertex operator isThe worldsheet Green function is $⟨X^a(z)X^b(w)⟩=-({\alpha }^{\prime }/2){\eta }^{ab}\log |z-w{|}^2$. The operator product expansion with the holomorphic stress-energy tensor gives conformal weights $({\alpha }^{\prime }{k}^2/4,{\alpha }^{\prime }{k}^2/4)$. An integrated string vertex operator must have weights $(1,1)$, so the mass-shell condition is $k^2=4/{\alpha }^{\prime }$, or tachyonic mass parameter $m^2=-4/{\alpha }^{\prime }$.

At lowest order in the string coupling, the worldsheet is a Riemann sphere. For four external states, three insertion points can be fixed by a Möbius transformation. In the worldsheet ghost field description these three vertices carry $c\bar{c}$, while the fourth is integrated. The three-point worldsheet ghost correlator gives the squared product of the three fixed-point separations, precisely the sphere gauge fixing for four string vertices factor. Thus fixing three points is a gauge choice, rather than deleting three integrations without their Jacobian.

The integral over the constant embedding worldsheet zero mode gives $(2\pi {)}^d{\delta }^{(d)}({\sum }_i{k}_i)$. The remaining Gaussian integral, using normal ordering to remove self-contractions, gives the Koba-Nielsen factorSet $(z_1,z_2,z_3,z_4)=(0,z,1,\infty )$; the vertex at infinity is defined with its conformal-weight factor. Let all momenta be incoming and defineThe mass-shell condition and momentum conservation imply $s+t+u=-16/{\alpha }^{\prime }$. Since ${\alpha }^{\prime }{k}_1\cdot k_2=-4-{\alpha }^{\prime }s/2$ and similarly for the pair $(2,3)$, the sphere tachyon position integral becomesHere $\mathcal{N}$ depends on the normalization of external string vertex operators; it does not affect the kinematic dependence.

To evaluate the remaining worldsheet modulus, put $a=-1-{\alpha }^{\prime }s/4$ and $b=-1-{\alpha }^{\prime }t/4$. The complex beta integral isFor example, this identity follows by writing the two powers as Schwinger-parameter gamma integrals, doing the two-dimensional Gaussian integral, and changing the two positive parameters to their sum and ratio. The complex beta integral first holds for $\mathrm{Re}a,\mathrm{Re}b>0$ and $\mathrm{Re}(a+b)<1$; the scattering result is its analytic continuation. The third gamma-function argument is $1-a-b=-1-{\alpha }^{\prime }u/4$. Absorbing $\pi$ into $\mathcal{N}$, the result is the tree-level four-tachyon amplitude, the Virasoro–Shapiro amplitude:It is symmetric in the three Mandelstam variables and has generic poles at $q=4(N-1)/{\alpha }^{\prime }$, matching the bosonic string mass spectrum. The formula is the leading term of string perturbation theory. Higher orders are constructed by the same Polyakov path integral prescription on higher-genus worldsheets, with four marked points and the corresponding ghost and worldsheet moduli measure; genus $g$ has coupling power $g_s^{2g+2}$.

Variation with respect to the Dirac adjoint, the Dirac field, and the real scalar field respectively givesThe derivative in the adjoint equation acts to the left; its sign follows by integrating $i\bar{\psi }{\gamma }^{\mu }{\partial }_{\mu }\delta \psi$ by parts. The Yukawa interaction supplies a spacetime-dependent effective fermion mass and a scalar source.

Use $/p={\gamma }^{\mu }{p}_{\mu }$ and the convention $S_{fi}^{\mathrm{conn}}=i(2\pi {)}^4{\delta }^4(p_f-p_i)\mathcal{M}$. The momentum-space Feynman rules are: a scalar internal line contributes $i/(p^2-m^2+i0)$; an oriented fermion line contributes $i(/p+m)/(p^2-m^2+i0)$; and each scalar-fermion vertex contributes $-ig$ times the identity in spinor space. Impose four-momentum conservation at each vertex. Incoming and outgoing fermions supply $u$ and $\bar{u}$, while incoming and outgoing antifermions supply $\bar{v}$ and $v$; scalar external legs supply one. Loop momenta are integrated with $d^{4}p/(2\pi {)}^4$, a closed fermion loop contributes a minus sign, and graph symmetry factors are included. Relative signs between distinct contractions of identical external fermions follow from their anticommutation relations. These specify the Feynman rules also beyond the tree approximation.

Label incoming momenta $p_1,p_2$ and outgoing momenta $p_3,p_4$, with all external particles on shell. Write the Mandelstam variables as $s=(p_1+p_2{)}^2$, $t=(p_1-p_3{)}^2$, $u=(p_1-p_4{)}^2$, and abbreviate $D_r=r-m^2+i0$. Spin indices on $u_i,v_i$ are implicit. The six required tree-level Feynman diagrams are:

For two incoming fermions, the two diagrams exchange a scalar in the $t$ and $u$ channels. With external state ordering $b_1^{†}{b}_2^{†}|0⟩$ and $b_3^{†}{b}_4^{†}|0⟩$, the result isEach scalar-exchange contraction has two factors $-ig$ and one scalar Feynman propagator. Exchanging the final fermions reverses the sign, as required by their identical-particle statistics.

For a fermion and antifermion, there is $t$-channel scalar exchange and $s$-channel annihilation. Take both initial and final states ordered as fermion creator followed by antifermion creator. ThenThe relative minus is not optional. One way to track it is the antifermion sign of a normal-ordered bilinear: the antifermion scattering part of $:\bar{\psi }\psi :$ is $-d^{†}d\bar{v}v$, whereas its annihilation part is $db\bar{v}u$. Thus the exchange contraction has the opposite fermionic sign to the annihilation contraction before multiplying by $(-ig{)}^{2}i/D_r$. A different overall phase convention for external states changes the common sign of this amplitude, but cannot change the relative sign.