Unpolarized, Squared Matrix Element for e⁺e⁻ → μ⁺μ⁻

1. The Amplitude for $e^{+}{e}^{-}\to {\mu }^{+}{\mu }^{-}$

First, let's visualize the interaction. At the lowest order in perturbation theory, the process proceeds through the exchange of a single virtual photon. The electron and positron annihilate, creating a virtual photon which then decays into a muon-antimuon pair.

The lowest-order Feynman diagram for electron-positron annihilation to a muon-antimuon pair. The electron and positron annihilate into a virtual photon (wavy line) with momentum q, which then creates the muon-antimuon pair.

Let's label the four-momenta as follows:

• Incoming electron: $p_1$
• Incoming positron: $p_2$
• Outgoing muon: $p_1^{\prime }$
• Outgoing antimuon: $p_2^{\prime }$

The virtual photon carries the total initial momentum, $q=p_1+p_2$, which by momentum conservation equals the total final momentum, $p_1^{\prime }+p_2^{\prime }$. The squared momentum transfer is $q^2=(p_1+p_2{)}^2$, which in the center-of-mass frame is simply the square of the total collision energy, $s=E_{cm}^2$.

Using the Feynman rules, we can write down the invariant amplitude $\mathcal{M}$. A key difference from our last lesson is the presence of antiparticles, which are represented by $v$ and $\bar{v}$ spinors:

• Incoming positron ($p_2$): $\bar{v}(p_2)$
• Outgoing antimuon ($p_2^{\prime }$): $v(p_2^{\prime })$

The amplitude is formed by the electron current, the muon current, and the photon propagator that connects them.

As shown in the resource, the invariant amplitude $\mathcal{M}$ is given by:

Simplifying this and defining $s=q^2$, we get:

This expression is a function of the specific spin states of the four particles involved.

2. From Spinors to Traces: Handling Antiparticles

Just as in the previous lesson, we are interested in an unpolarized process. This requires us to average over the two initial electron and two initial positron spins (a factor of 1/4) and sum over all four final muon/antimuon spin states. The quantity to calculate is:

This looks like another complicated sum over spinor products. Fortunately, Casimir's trick works for antiparticles as well, but with a crucial modification. The spin-sum relation for antiparticle spinors is:

Notice the minus sign before the mass term, in contrast to the $/p+m$ relation for particle spinors.

To apply this, we first compute $|\mathcal{M}{|}^2=\mathcal{M}{\mathcal{M}}^{\ast }$. The complex conjugate of $\mathcal{M}$ is:

(We used the rule $(\bar{u}\Gamma v{)}^{\ast }=\bar{v}\bar{\Gamma }u$ and relabeled the dummy index from $\lambda$ to $\nu$).

Now, we can write the spin-summed squared amplitude as the product of two tensors, one for the electron-positron pair and one for the muon-antimuon pair. The following reading walks through this setup.

As derived in the text, the spin-averaged squared amplitude is:

where

• The electron tensor $L_{(e)}^{\lambda \nu }$ comes from averaging over initial spins: $$L_{(e)}^{\lambda \nu }=\sum _{s_1,s_2}[\bar{v}(p_2){\gamma }^{\nu }u(p_1)][\bar{u}(p_1){\gamma }^{\lambda }v(p_2)]=\text{Tr}\left[(/p_2-m_e){\gamma }^{\nu }(/p_1+m_e){\gamma }^{\lambda }\right]$$
• The muon tensor $L_{\lambda \nu }^{(\mu )}$ comes from summing over final spins: $$L_{\lambda \nu }^{(\mu )}=\sum _{s_1^{\prime },s_2^{\prime }}[\bar{u}(p_1^{\prime }){\gamma }^{\lambda }v(p_2^{\prime })][\bar{v}(p_2^{\prime }){\gamma }^{\nu }u(p_1^{\prime })]=\text{Tr}\left[(/p_1^{\prime }+M_{\mu }){\gamma }^{\lambda }(/p_2^{\prime }-M_{\mu }){\gamma }^{\nu }\right]$$

Our problem is once again reduced to the purely algebraic task of calculating gamma matrix traces.

3. The Trace Calculation

The calculation proceeds exactly as in the last lesson. We expand each trace, discard terms with an odd number of gamma matrices, and use the trace theorems for products of two and four gamma matrices.

Since your background is strong and you've seen this once, let's use the resources to guide the full calculation.

Let's summarize the key steps of the calculation you just reviewed:

1. Evaluate the muon trace $L_{\lambda \nu }^{(\mu )}=\text{Tr}\left[(/p_1^{\prime }+M_{\mu }){\gamma }^{\lambda }(/p_2^{\prime }-M_{\mu }){\gamma }^{\nu }\right]$:

   $$L_{\lambda \nu }^{(\mu )}=\text{Tr}[/p_1^{\prime }{\gamma }^{\lambda }/p_2^{\prime }{\gamma }^{\nu }]-M_{\mu }^2\text{Tr}[{\gamma }^{\lambda }{\gamma }^{\nu }]=4(p_1^{\prime \lambda }{p}_2^{\prime \nu }+p_1^{\prime \nu }{p}_2^{\prime \lambda }-(p_1^{\prime }\cdot p_2^{\prime }){\eta }^{\lambda \nu })-4M_{\mu }^2{\eta }^{\lambda \nu }$$

2. Evaluate the electron trace $L_{(e)}^{\lambda \nu }=\text{Tr}\left[(/p_2-m_e){\gamma }^{\nu }(/p_1+m_e){\gamma }^{\lambda }\right]$:

   $$L_{(e)}^{\lambda \nu }=\text{Tr}[/p_2{\gamma }^{\nu }/p_1{\gamma }^{\lambda }]-m_e^2\text{Tr}[{\gamma }^{\nu }{\gamma }^{\lambda }]=4(p_2^{\nu }{p}_1^{\lambda }+p_2^{\lambda }{p}_1^{\nu }-(p_1\cdot p_2){\eta }^{\nu \lambda })-4m_e^2{\eta }^{\nu \lambda }$$

3. Contract the tensors $L_{(e)}^{\lambda \nu }{L}_{\lambda \nu }^{(\mu )}$. This is the most tedious step, involving careful contraction of the Lorentz indices. The result, as shown in eq. (58) of the notes, is:

   $$L_{(e)}^{\lambda \nu }{L}_{\lambda \nu }^{(\mu )}=32(p_1^{\prime }\cdot p_1)(p_2^{\prime }\cdot p_2)+32(p_2^{\prime }\cdot p_1)(p_1^{\prime }\cdot p_2)+16s(M_{\mu }^2+m_e^2)$$

4. Assemble the final result. Plugging this back into our expression for $⟨|\mathcal{M}{|}^2⟩$:

   $$⟨|\mathcal{M}{|}^2⟩=\frac{e^4}{4s^2}\left[32(p_1^{\prime }\cdot p_1)(p_2^{\prime }\cdot p_2)+32(p_2^{\prime }\cdot p_1)(p_1^{\prime }\cdot p_2)+16s(M_{\mu }^2+m_e^2)\right]$$

   Which simplifies to:

   $$⟨|\mathcal{M}{|}^2⟩=\frac{8e^4}{s^2}\left[(p_1^{\prime }\cdot p_1)(p_2^{\prime }\cdot p_2)+(p_1^{\prime }\cdot p_2)(p_2^{\prime }\cdot p_1)+\frac{s}{2}(M_{\mu }^2+m_e^2)\right]$$

This is the unpolarized, squared matrix element for $e^{+}{e}^{-}\to {\mu }^{+}{\mu }^{-}$. It is expressed entirely in terms of Lorentz-invariant four-vector dot products, as expected.

For completeness, you might find it interesting to see an alternative approach. In the high-energy limit (where particle masses can be neglected), the calculation simplifies significantly. This is because only specific helicity combinations contribute, a feature related to the chiral nature of the QED interaction. The following resource demonstrates this method, calculating the amplitude for each allowed helicity combination and then summing and averaging.

This helicity amplitude method is often more intuitive than the brute-force trace technology, especially in gauge theories, but the trace method is more general as it works easily even when masses are not negligible.

Conclusion

In this lesson, we reinforced our command of trace technology by applying it to a new and important process, $e^{+}{e}^{-}\to {\mu }^{+}{\mu }^{-}$.

Key Takeaways:

• The procedure for calculating an unpolarized squared amplitude is general: write down $\mathcal{M}$, compute $⟨|\mathcal{M}{|}^2⟩=\frac{1}{2s_1+1}\frac{1}{2s_2+1}\sum |\mathcal{M}{|}^2$, and use Casimir's trick.
• Antiparticles are handled by using $v$ and $\bar{v}$ spinors, leading to the spin-sum relation $\sum v\bar{v}=/p-m$.
• The s-channel annihilation process $e^{+}{e}^{-}\to {\mu }^{+}{\mu }^{-}$ results in a squared matrix element structurally similar to, but distinct from, the t-channel scattering process $e^{-}{\mu }^{-}\to e^{-}{\mu }^{-}$ we studied previously.

Preview of the Next Lesson:
We now have the core dynamical information for this reaction locked in $⟨|\mathcal{M}{|}^2⟩$. The next step is to connect this to a measurable quantity. In the next lesson, we will use this result to derive the differential and total cross-section for $e^{+}{e}^{-}\to {\mu }^{+}{\mu }^{-}$ in the high-energy limit, which can be directly compared with experimental data.