Question 2.9: Elastic e^−π^+ Scattering (I) Determine the scattering ampli......
Elastic \mathrm{e}^{-} \pi^{+}Scattering (I)
Determine the scattering amplitude and explain the formal steps necessary to evaluate the cross section.
The graph for \mathrm{e}^{-} \pi^{+}scattering is of the following form (see Fig. 2.19). Most of the above notation is readily understood. Only the factor +\mathrm{i} in the transition current at the electron vertex, which has been denoted by (ie \gamma_{\mu} ), needs additional explanation. Reviewing our knowledge of QED, we start with the scattering amplitude (2.58),
\begin{aligned} S_{\mathrm{fi}}^{(1)} & =-\mathrm{i} \int \mathrm{d}^4 x \Psi_{\mathrm{f}}^{\dagger}\left(\mathrm{e}^{-} ; p^{\prime}, s^{\prime}\right) \hat{V} \Psi_{\mathrm{i}}\left(\mathrm{e}^{-} ; p, s\right) \\ & =-\mathrm{i} \int \mathrm{d}^4 x \Psi_{\mathrm{f}}^{\dagger}\left(\mathrm{e}^{-} ; p^{\prime}, s^{\prime}\right) \gamma^0 \gamma^0 \hat{V} \Psi_{\mathrm{i}}\left(\mathrm{e}^{-} ; p, s\right) \\ & =-\mathrm{i} \int \mathrm{d}^4 x \bar{\Psi}_{\mathrm{f}}\left(\mathrm{e}^{-} ; p^{\prime}, s^{\prime}\right)\left(-e \gamma^\mu A_\mu\right) \Psi_{\mathrm{i}}\left(\mathrm{e}^{-} ; p, s\right) \\ & =-\mathrm{i} \int \mathrm{d}^4 x J^\mu\left(\mathrm{e}^{-}\right) A_\mu, \end{aligned} (2.58)
i.e., with
S_{\mathrm{fi}}^{(1)}=-\mathrm{i} \int \mathrm{d}^{4} x j_{\mu}\left(\mathrm{e}^{-}\right) A^{\mu}, (1)
where according to (2.60)
J^\mu\left(\mathrm{e}^{-}\right)=\frac{(-e)}{V} \bar{u}_{\mathrm{f}}\left(p^{\prime}, s^{\prime}\right) \gamma^\mu u_{\mathrm{i}}(p, s) \mathrm{e}^{\mathrm{i}\left(p^{\prime}-p\right) \cdot x}, \quad N_{\mathrm{i}}=N_{\mathrm{f}}=\frac{1}{\sqrt{V}} (2.60)
j_{\mu}\left(\mathrm{e}^{-}\right)=(-e) N N^{\prime} \bar{u}\left(k^{\prime}, s^{\prime}\right) \gamma_{\mu} u(k, s) \mathrm{e}^{\mathrm{i}\left(k^{\prime}-k\right) \cdot x} (2)
denotes the electron transition current density. The electromagnetic fourpotential A^{\mu} in (1) is created by the pion \left(\pi^{+}\right)transition current density
j^{\mu}\left(\pi^{+}\right)=(+e) \bar{N} \bar{N}^{\prime}\left(p+p^{\prime}\right)^{\mu} \mathrm{e}^{\mathrm{i}\left(p^{\prime}-p\right) \cdot x} . (3)
According to (2.99)
A^\mu\left(\mathrm{K}^{+}\right)=-\frac{1}{q^2} j^\mu\left(\mathrm{K}^{+}\right)=-\frac{1}{q^2} e N_2 N_4\left(p_2+p_4\right)^\mu \mathrm{e}^{\mathrm{i} q \cdot x} . (2.99)
we have
A^{\mu}=-\frac{1}{q^{2}} j^{\mu}\left(\pi^{+}\right), (4)
with the four-momentum transfer
q=p^{\prime}-p=k-k^{\prime} . (5)
The scattering amplitude (1) in detail is then
In the last step we have introduced the invariant scattering amplitude
F_{s s^{\prime}}\left(k p ; k^{\prime} p^{\prime}\right)=(-e)\left[\bar{u}\left(k^{\prime}, s^{\prime}\right) \gamma_{\mu} u(k, s)\right]\left(-\frac{g^{\mu \nu}}{q^{2}}\right)(+e)\left[\left(p+p^{\prime}\right)_{\nu}\right] . (7)
Only the spin variables occur in addition. Now we see from (6) that the above Feynman rules yield the correct total sign for the scattering amplitude if a factor +\mathrm{i} is assigned to the vertex of the leptonic transition current. Figure 2.19 already contains this factor. Note that the spinor combinations \left(\bar{u} \gamma_{\mu} u\right) are the components of a four-vector. Contracting this vector with \left(p+p^{\prime}\right)_{\nu} yields a Lorentz scalar and therefore a Lorentz-invariant scattering amplitude. The derivation of the cross section consists in the same steps, which have been discussed in detail for \pi^{+} \mathrm{K}^{+}scattering. Employing the four-vectors
\begin{aligned} & k^{\mu}=(\omega, {k}), \quad k^{\prime \mu}=\left(\omega^{\prime}, {k}^{\prime}\right), \\ & p^{\mu}=(E, {p}), \quad p^{\prime \mu}=\left(E^{\prime}, {p}^{\prime}\right), & (8) \end{aligned}
we find the differential cross section to be (see equation (2.116))
\begin{aligned} \mathrm{d} \sigma & =P_{\mathrm{fi}} \frac{V^2}{2 E_1 2 E_2|\boldsymbol{v}|} \frac{V}{(2 \pi)^3} \frac{\mathrm{d}^3 p_3}{2 E_3} \frac{V}{(2 \pi)^3} \frac{\mathrm{d}^3 p_4}{2 E_4} \\ & =\frac{|F|^2}{2 E_1 2 E_2|\boldsymbol{v}|}(2 \pi)^4 \delta^4\left(p_3+p_4-p_1-p_2\right) \frac{\mathrm{d}^3 p_3}{(2 \pi)^3 2 E_3} \frac{\mathrm{d}^3 p_4}{(2 \pi)^3 2 E_4}. & (2.116) \end{aligned}
\mathrm{d} \sigma_{s s^{\prime}}=(2 \pi)^{4} \delta\left(k^{\prime}+p^{\prime}-k-p\right) \frac{\left|F_{s s^{\prime}}\right|^{2}}{2 E 2 \omega|{v}|} \frac{1}{(2 \pi)^{3}} \frac{\mathrm{d}^{3} k^{\prime}}{2 \omega^{\prime}} \frac{1}{(2 \pi)^{3}} \frac{\mathrm{d}^{3} p^{\prime}}{2 E^{\prime}}. (9)
The scattering amplitude F_{s s^{\prime}} can be easily evaluated if we insert the spinors (see (2.41))
\begin{aligned} u(p, s) & =\sqrt{E+m_0}\left(\begin{array}{c} \varphi^1 \\ \frac{\hat{\sigma} \cdot p}{E+m_0} \varphi^2 \end{array}\right), \quad s=1,2, \\ \varphi^1 & =\left(\begin{array}{l} 1 \\ 0 \end{array}\right), \quad \varphi^2=\left(\begin{array}{l} 0 \\ 1 \end{array}\right) . & (2.41) \end{aligned}
u(k, s)=\sqrt{w+m_{0}}\left(\begin{array}{c} \phi^{s} \\ \frac{\hat{\sigma} \cdot k}{\omega+m_{0}} \phi^{s} \end{array}\right) (10)
into (7). This procedure is quite tedious and, more importantly, does not yield the quantity observed in most of the actual experiments. The interesting quantity is the so-called nonpolarized cross section, which is obtained from (9) by averaging over the initial spins and summing over the final spins, i.e.,
\begin{aligned} \mathrm{d} \bar{\sigma} & =\frac{1}{2}\left(d \sigma_{\uparrow \uparrow}+\mathrm{d} \sigma_{\uparrow \downarrow}+\mathrm{d} \sigma_{\uparrow \downarrow}+\mathrm{d} \sigma_{\downarrow \downarrow}\right) \\ & =\frac{1}{2} \sum\limits_{s s^{\prime}} \mathrm{d} \sigma_{s s^{\prime}}. & (11) \end{aligned}
These lengthy summations need not be performed explicitly. Instead, employing Feynman’s trace techniques enables us to drastically simplify the spin summations (11).
