Question 2.12: Positron–Pion Scattering Show that the cross section for e^+......

The graph for \mathrm{e}^{+} \pi^{+}scattering is of the following form: The incoming positron with four-momentum k and spin s is described as an outgoing electron with four-momentum -k and spin -s. Correspondingly the outgoing positron with k^{\prime}, s^{\prime} can be interpreted as an incoming electron with -k^{\prime},-s^{\prime}. Therefore we have for the positron transition current

j_{\mu}\left(\mathrm{e}^{+}\right)=(-e) N N^{\prime} \bar{v}(k, s) \gamma_{\mu} v\left(k^{\prime}, s^{\prime}\right) \mathrm{e}^{-\mathrm{i}\left(-k^{\prime}+k\right) \cdot x}      (1)

where v\left(k^{\prime}, s^{\prime}\right) and \bar{v}\left(k^{\prime}, s^{\prime}\right) represent an incoming electron wave -k^{\prime},-s^{\prime} and an outgoing electron wave -k,-s, respectively. If we construct the scattering matrix element with j_{\mu}\left(\mathrm{e}^{+}\right), an additional minus sign is inserted according to the previously introduced rules. Everything else remains the same as in the \mathrm{e}^{-} \pi^{+} scattering discussed in Exercises 2.9 and 2.11. The evaluation of the cross section, too, is analogous to that of \mathrm{e}^{-} \pi^{+}scattering except for the spin average, where the sum

\sum\limits_{s} u(k, s) \bar{u}(k, s)=\left(\not k+m_{0}\right)       (2)

in the lepton tensor must be replaced by (see equations (2.164)

\sum\limits_s(u(k, s) \otimes \bar{u}(k, s))_{\beta \gamma}=\sum\limits_s u_\beta(k, s) \bar{u}_\gamma(k, s) \text {, }   (2.164)

and (20) of Exercise 2.10)

\sum\limits_{s} v(k, s) \bar{v}(k, s)=\left(\not k-m_{0}\right)     (3)

This calculation was made in Exercise 2.10 and leads to the trace

\operatorname{tr}\left\{\left(\not k^{\prime}-m_{0}\right) \gamma_{\mu}\left(\not k-m_{0}\right) \gamma_{\nu}\right\}      (4)

instead of

\operatorname{tr}\left\{\left(\not k^{\prime}+m_{0}\right) \gamma_{\mu}\left(\not k+m_{0}\right) \gamma_{\nu}\right\}.   (5)

Because of rules (2.156)

\operatorname{tr}{\not a \not b \not c \not d}  =4[(a \cdot b)(c \cdot d)+(a \cdot d)(b \cdot c)-(a \cdot c)(b \cdot d)],     (2.156)

we immediately recognize that these two traces are equal. Therefore the \mathrm{e}^{+} \pi^{+}cross section is to lowest order exactly equal to the \mathrm{e}^{-} \pi^{+}cross section. This result does not surprise us at all, since only the lepton charge changed its sign and the lowest-order cross section contains only the square of this charge.