= Solution
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>.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2008/iii/paper-48-yukawa-trees.png]
{title=The t and u exchanges for two fermions, and the t exchange and s annihilation for a fermion-antifermion pair, with external momentum and spin labels}
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 $b_1^\dagger b_2^\dagger|0\rangle$ to $b_3^\dagger b_4^\dagger|0\rangle$ 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>. Thus
$$
\boxed{\mathcal M_{\psi\psi}
=-\lambda^2\left[
\frac{(\bar u_3\gamma^5u_1)(\bar u_4\gamma^5u_2)}{t-\mu^2+i0}
-\frac{(\bar u_4\gamma^5u_1)(\bar u_3\gamma^5u_2)}{u-\mu^2+i0}
\right].}
$$
This 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 $b_1^\dagger d_2^\dagger|0\rangle$ and final state $b_3^\dagger d_4^\dagger|0\rangle$. The $t$ exchange uses bilinears $(\bar u_3\gamma^5u_1)(\bar v_2\gamma^5v_4)$, and the $s$ annihilation uses $(\bar v_2\gamma^5u_1)(\bar u_3\gamma^5v_4)$. 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 signs
$$
J\supset b^\dagger b\,\bar u\gamma^5u
-d^\dagger d\,\bar v\gamma^5v
+d b\,\bar v\gamma^5u
+b^\dagger d^\dagger\,\bar u\gamma^5v.
$$
The <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 $(-i\lambda)^2i=-i\lambda^2$ therefore gives
$$
\boxed{\mathcal M_{\psi\bar\psi}
=\lambda^2\left[
\frac{(\bar u_3\gamma^5u_1)(\bar v_2\gamma^5v_4)}{t-\mu^2+i0}
-\frac{(\bar v_2\gamma^5u_1)(\bar u_3\gamma^5v_4)}{s-\mu^2+i0}
\right].}
$$
A 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 $\lambda^2$.
Back to article page
