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For a real shell potential $V(r)=V_0\delta (r-R)$, define $g=2m{V}_{0}R/{\hslash }^2$ and $\mathrm{sinc}u=\sin u/u$, with value one at zero. Angular integration of its three-dimensional Fourier transform gives $4\pi V_0{R}^2\mathrm{sinc}(qR)$, where the momentum transfer has wave-number magnitude $q=2k\sin (\theta /2)$. Substitution in the first Born approximation proves the displayed scattering amplitude. It is real. Its squared modulus is a nonzero order-$V_0^2$ differential scattering cross-section, whose optical theorem counterpart requires the order-$V_0^2$ imaginary part of the forward amplitude from the next Born term. Thus squaring the first-order amplitude does not supply a consistently unitary second-order amplitude.

The scattering amplitude at zero deflection angle. Its real part changes the phase of a coherent beam in matter; its imaginary part is related by the optical theorem to removal from that beam. Different forward amplitudes for a kaon and a neutral antikaon allow neutral kaon regeneration without fundamental CP violation.

Put $s=q^2>0$ and neglect electron and final-quark masses compared with $\sqrt{s}$. First fix a final color charge; its multiplicity will be restored for the inclusive rate. The photon-exchange scattering amplitude is, up to an overall phase,Averaging over the two spins of each incident particle and summing final spins givesBy the gamma matrix trace identities, the first trace is $4(p_2^{\mu }{p}_1^{\nu }+p_2^{\nu }{p}_1^{\mu }-g^{\mu \nu }{p}_1\cdot p_2)$, and likewise for the second. Contracting the two tensors cancels the terms proportional to $(p_1\cdot p_2)(k_1\cdot k_2)$, givingIn the centre-of-mass frame all four energies are $\sqrt{s}/2$. The first pair of scalar products equals $s(1-\cos \theta )/4$ each, and the second pair equals $s(1+\cos \theta )/4$ each. Hence $\overline{|\mathcal{M}{|}^2}=e^4{Q}_q^2(1+{\cos }^2\theta )$.

For clarity, the printed flux hint has a factor-of-four inconsistency: when $\mathcal{F}=4\sqrt{(p_1\cdot p_2{)}^2-m_1^2{m}_2^2}=2s$, the prefactor multiplying the spin-averaged amplitude is $1/\mathcal{F}$, not $1/(4\mathcal{F})$. Alternatively, $1/(4F_0)$ is correct if $F_0=\sqrt{(p_1\cdot p_2{)}^2-m_1^2{m}_2^2}=s/2$. Use this massless two-body scattering flux normalization, keeping the spin average separate.

To obtain the two-body Lorentz-invariant phase space directly, integrate the momentum-conserving spatial delta function, leaving ${\mathbf{k}}_2=-{\mathbf{k}}_1$. The remaining radial integral is $\int dk\delta (\sqrt{s}-2k)=1/2$, so $d{\Phi }_2/d\Omega =1/(32{\pi }^2)$. The relativistic scattering cross-section is consequentlySince $\int d\Omega (1+{\cos }^2\theta )=2\pi {\int }_{-1}^1(1+x^2)dx=16\pi /3$, the integrated electron-positron annihilation into a quark pair result, for this fixed color, isThe first two printed formulas are thus color-stripped. Distinct final colors are orthogonal states and must be summed, producing $N_c=3$. At leading order each accessible quark flavor contributes independently. Away from thresholds and narrow resonances, quark-hadron duality relates the inclusive parton rate to the observed inclusive electron-positron annihilation into hadrons, givingThis formula concerns the photon channel, assumes $\sqrt{s}$ is large compared with the relevant quark masses and the strong-interaction scale, and does not include Z boson exchange.

Beyond leading order, an exactly exclusive pair of colored quarks is not an observable asymptotic state because of confinement and hadronization. Moreover its virtual corrections have infrared divergences; real soft or collinear gluon emission cannot be excluded if a finite rate is wanted. The inclusive sum includes both the corrected two-parton channel and channels such as $q\bar{q}g$, with their degenerate soft and collinear limits. Real and virtual infrared divergences then cancel for an infrared-safe observable. A specified jet observable can also be meaningful if it defines an infrared-safe observable; a bare two-quark final state is not.

A convenient inclusive calculation uses the optical theorem. For the electromagnetic hadronic current $j_{\mu }={\sum }_f{Q}_f{\bar{q}}_f{\gamma }_{\mu }{q}_f$, defineThe hadronic electromagnetic-current spectral density is obtained from its discontinuity. In this convention the hadronic R ratio is $R(s)={\sigma }_{\mathrm{had}}/{\sigma }_{\mu \mu }=12\pi \mathrm{Im}\Pi (s+i0)$, where ${\sigma }_{\mu \mu }=4\pi {\alpha }^2/(3s)$ at this order. Cutting the forward amplitude sums all allowed hadronic intermediate states, so perturbative Quantum chromodynamics applied to this inclusive quantity implements the required sum of real and virtual contributions.

Finally, write $a={\alpha }_s({\mu }^2)$, $C=1.99-0.11n_f$, $L=\ln (s/{\mu }^2)$, and ${\beta }_0=11-2n_f/3$. The given truncated correction isDifferentiate with respect to $t=\ln {\mu }^2$, using the stated one-loop beta function $da/dt=-{\beta }_0{a}^2/(4\pi )$ and $dL/dt=-1$. The derivative of $a/\pi$ cancels the explicit logarithmic derivative at order $a^2$. Keeping the residual derivative of the truncated expression givesThus the principle of minimal sensitivity, applied to this expression with nonzero coupling and ${\beta }_0\ne 0$, selects the one-loop stationary scale for the hadronic annihilation correctionThis is a prescription based on the residual order-$a^3$ scale dependence of an order-$a^2$ approximation. At strictly order $a^2$, the derivative already vanishes for every renormalization scale; the prescription does not assert exact scale dependence of the observable. It is useful only while the selected running coupling remains perturbative. If $a=0$ or ${\beta }_0=0$, the stated derivative imposes no unique scale.

Expanding the modulus squared yields an outgoing-angle factor $e^{-i{\mathbf{k}}_f\cdot \mathbf{d}}$ and an incoming factor $e^{i{\mathbf{k}}_i\cdot \mathbf{d}}$, where $\mathbf{d}=\mathbf{r}-{\mathbf{r}}^{\prime }$. Choosing a polar axis along $\mathbf{d}$ givesConsequentlyThe apparent $d=0$ singularity is removable: $\sin (kd)/(kd)\to 1$. Exchanging $\mathbf{r}$ and ${\mathbf{r}}^{\prime }$ cancels the imaginary part, while nonnegativity follows directly from its origin as an integral of $|f^{(1)}{|}^2$. A complex absorbing potential would require $V({\mathbf{r}}^{\prime }{)}^{\ast }$ instead; the lossless optical theorem comparison below uses real $V$.

Define $A=m{g}^2/(2\pi {\hslash }^2)$, which has units of energy times length. Substituting the result of (b) into (c) givesAbsorb the real cutoff shift into the physical detuning,It follows thatUsing $k=\sqrt{2m{E}_k}/\hslash$, identifyHere $\gamma$ has units of square root of energy. This is the small-cutoff, renormalized resonant-level amplitude; terms vanishing with $k\delta$ have been omitted. Taking $\delta \to 0$ at fixed measured detuning requires the bare ${ϵ}_0$ to depend on the cutoff.

The sign is checked by $f_k^{-1}=-(E_k-{\widetilde{ϵ}}_0)/A-ik$, which has the elastic $s$-wave unitarity form. Thus $\mathrm{Im}{f}_k=k|f_k{|}^2$, consistent with the optical theorem and ${\sigma }_{\mathrm{tot}}=4\pi |f_k{|}^2$. If the renormalized level is positive, at $E_k={\widetilde{ϵ}}_0$ the amplitude is $i/k$ and the scattering cross-section reaches $4\pi /k^2$.

Integrate the squared partial-wave expansion of a scattering amplitude over the solid angle. Legendre polynomial orthogonality eliminates terms with unequal angular-momentum indices, givingAt zero scattering angle, $P_{\ell }(1)=1$ and $\mathrm{Im}(e^{i{\delta }_{\ell }}\sin {\delta }_{\ell })={\sin }^2{\delta }_{\ell }$. ConsequentlyThis is the optical theorem. In the present elastic problem the total and elastic cross sections coincide. For an absorptive interaction the inclusive total still obeys the optical theorem, but need not equal the integral of the elastic differential cross section.

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.

For an isotropic elastic amplitude normalized by an incident unit plane wave and outgoing $f{e}^{ikr}/r$, total cross section is $4\pi |f{|}^2$. The optical theorem gives the displayed identity. At a resonance where the real part of $1/f$ vanishes, $f=i/k$ and the cross section is $4\pi /k^2$.