# Explicit derivation of the Feynman amplitude of $e^+e^-\rightarrow\mu^+\mu^-$

I'm trying to compute the Feynman amplitude of the process $$e^+(p_1,s_1)e^-(p_2,s_2)\rightarrow \mu^+(q_1,r_1)\mu^-(q_2,r_2),$$

considering as interaction Lagrangian $$\mathcal{L}_I=-\lambda_e\phi(x)\bar{\psi}(x)\psi(x)-\lambda_\mu\phi(x)\bar{\chi}(x)\chi(x),$$ where $\psi$ is the field of a scalar particle $H$, $\psi$ of $e$ and $\chi$ of $\mu$.

Using the Wick's theorem I get that the contribution to the transition amplitude is

$$S=-2\frac{\lambda_e\lambda_\mu}{2}\int d^4x_1 d^4 x_2[\phi(x_1)\phi(x_2)]\bar{\chi}(x_1)\chi(x_1) \bar{\psi}(x_2)\psi(x_2),$$

where the contracted term $[\phi(x_1)\phi(x_2)]=i\Delta_F(x_1-x_2)$.

Then $$S=-2\frac{\lambda_e\lambda_\mu}{2}\int d^4x_1 d^4 x_2\langle F|\bar{\chi}^-(x_1)\chi^-(x_1)|0\rangle\langle 0 |\bar{\psi}^+(x_2)\psi^+(x_2)| I \rangle i\Delta_F(x_1-x_2).$$

Writing $S=(2\pi)^4\delta^4(p_1+p_2-q_1-q_2)i\eta$, I get that $$i\eta=-\lambda_e\lambda_\mu \bar{u}'^{r_2}(q_2)v'^{r_1}(q_1)\bar{v}^{s_1}(p_1)u^{s_2}(p_2)\frac{i}{(p_1+p_2)^2-m^2+i\varepsilon}.$$

Now I'd like to do the square of the amplitude in the 16 possible helicity configurations in the center of mass ($\vec{p}_1=-\vec{p}_2$ and $\vec{q}_1=-\vec{q}_2$), but if I try to compute the product of the $u$ and $v$ spinors (even before doing the square), I get that they are all zero.

I have to do (and I did) the calculations using the spinors:

$$u^\pm(p)=\left( \begin{matrix} \sqrt{E\mp |\vec{p}|} \xi^\pm_p \\ \sqrt{E\pm |\vec{p}|} \xi^\pm_p \end{matrix}\right)$$

$$v^\pm(p)=\pm\left( \begin{matrix} \sqrt{E\pm |\vec{p}|} \xi^\mp_p \\ -\sqrt{E\mp |\vec{p}|} \xi^\mp_p \end{matrix}\right),$$

where

$$\xi^+_p=\left( \begin{matrix} cos\frac{\theta}{2} \\ e^{i\phi}sin\frac{\theta}{2} \end{matrix}\right)$$ and $$\xi^{-}_p=\left( \begin{matrix} -e^{-i\phi}sin\frac{\theta}{2}\\ cos\frac{\theta}{2} \end{matrix}\right).$$

I read on Peskin Schroeder that

$$u^{r \dagger}(\vec{p}) v^s (-\vec{p})=0$$

so it may seems that all amplitudes of that form are zero in the center of mass... What am I missing?

• This sounds like a "check-my-work" question, which are considered off-topic on this site. – Danu Oct 22 '15 at 19:41
• Sounds better now? – Charlie Oct 22 '15 at 19:51

## 1 Answer

Your mistake is hiding in plain sight! In your amplitude you have contributions of the form

$$\bar v ^{s_1} (\mathbf{p}) u ^ {s_2}(-\mathbf{p})$$

But you then try to use a formula for $v^\dagger$ rather than $\bar{v}$! Remember that $\bar v$ differs from $v^\dagger$ by a (crucial) factor of $\gamma^0$. This has the effect of making

$$\bar v ^{s_1} (\mathbf{p}) u ^ {s_2}(-\mathbf{p}) \neq 0$$

In fact the vanishing combination involving barred rather than daggered variables is

$$\bar v ^{s_1} (\mathbf{p}) u ^ {s_2}(\mathbf{p}) = 0$$

as proved on page 104 of these excellent notes.

• Yes, sorry, the thing with the $\dagger$ was just a guess. But then (I'll check it again) there's no reason for which the amplitude should be zero right? – Charlie Oct 22 '15 at 20:49
• Ok, I figured out the problem while re-thinking about daggered/barred spinors thanks to your answer: I may have done the calculations without the $\gamma_0$... so using daggered instead of barred – Charlie Oct 22 '15 at 20:57
• Yes - exactly: the amplitude should not be zero. – Edward Hughes Oct 22 '15 at 21:17
• Then I get that above all 16 helicity configurations only 4 are non vanishing: which seems to be right considering that the scalar has spin 0. But is it right that the amplitudes don't depend on the scattering angles but only on the modulus of the two momenta? – Charlie Oct 22 '15 at 21:19
• Are you sure your $|\mathbf{p}|$ in your definitions of $u$ and $v$ is correct? In general this is not right - see for example equation (4.115) here. Of course, the question might just want you to look at a special case. If you look at the general scenario and compute the unpolarized cross section, I would expect some angular dependence since you'd get terms looking like $p\cdot q$. See for instance this analogous QED example. – Edward Hughes Oct 22 '15 at 23:20