Muon Pair Production Cross-Section

Hello! Welcome back to our study of Quantum Electrodynamics.

In the previous lesson, we successfully calculated the unpolarized, squared matrix element, $⟨ ∣ M ∣^{2} ⟩$ , for electron-positron annihilation into a muon-antimuon pair ( $e^{+} e^{-} \to μ^{+} μ^{-}$ ). This gave us the core dynamical information about the interaction, but it's not a directly measurable quantity.

Today, we will bridge the gap between theory and experiment. Our goal is to use the matrix element we derived to calculate the differential and total cross-sections for $e^{+} e^{-} \to μ^{+} μ^{-}$ in the high-energy limit. The cross-section is a measure of the probability of this interaction occurring and can be directly compared with data from particle colliders.

1. The Cross-Section: From Matrix Element to Observable

The general formula connecting the invariant matrix element $M$ to the cross-section $σ$ for a 2-to-n particle scattering process ( $p_{1}, p_{2} \to p_{3}, ..., p_{n + 2}$ ) is a cornerstone of QFT calculations.

P528 Notes #2: Calculating Stuff 1 Example

The document 'P528 Notes #2' provides a concise statement of this general formula in its appendix. This is our starting point.

Please review the formula for the 2→n scattering cross-section, given as equation (27) in 'Appendix A: Cross Section and Decay Rate Formulas'. Pay attention to the three main components: the flux factor (involving 'v'), the matrix element |M|², and the Lorentz-invariant phase space (LIPS) integral.

The general formula is:

σ = \frac{S}{4 E _{1} E _{2} v} \int f = 3 \prod n + 2 \frac{d ^{3} p _{f}}{( 2 π ) ^{3} 2 E _{f}} (2 π)^{4} δ^{(4)} (p_{1} + p_{2} - f \sum p_{f}) ∣ M ∣^{2}

where $v$ is the relative velocity of the incident particles and $S$ is a symmetry factor for identical particles in the final state.

While this formula is beautifully general and Lorentz invariant, it needs to be simplified for our specific case: a 2-to-2 process in the center-of-mass (CM) frame. This frame is particularly convenient as the total initial momentum is zero.

2. The Differential Cross-Section in the Center-of-Mass Frame

Let's specialize the general formula. For a 2-to-2 process ( $1 + 2 \to 3 + 4$ ), the phase space integral and flux factor simplify considerably.

Feynman Diagrams for Beginners

The document 'Feynman Diagrams for Beginners' provides an excellent, step-by-step derivation that simplifies the general formula for a 2-to-2 collision in the CM frame.

Please read Section 6.5, 'Integration over two-particle phase space'. Follow the argument from equation (65) to equation (71). The key steps are: Using the 3-momentum delta function to eliminate the integral over \mathbf{p}_4. Using the energy delta function to perform the integral over the magnitude of \mathbf{p}_3. Simplifying the flux factor in the CM frame. The final result, equation (71), gives the differential cross-section d\sigma/d\Omega in the CM frame.

As you've just seen, after integrating over the phase space, we arrive at a much more manageable formula for the differential cross-section in the CM frame:

\frac{d σ}{d Ω} = \frac{1}{64 π ^{2} E _{C M}^{2}} \frac{∣ p _{f} ∣}{∣ p _{i} ∣} ⟨ ∣ M ∣^{2} ⟩

Here, $E_{C M}$ is the total CM energy, $∣ p_{i} ∣$ is the magnitude of the momentum of either initial particle, and $∣ p_{f} ∣$ is the magnitude of the momentum of either final particle. For $e^{+} e^{-} \to μ^{+} μ^{-}$ , the final state particles ( $μ^{-}$ and $μ^{+}$ ) are not identical, so the symmetry factor $S = 1$ .

3. Combining Kinematics and Dynamics

Now we have all the pieces. We need to evaluate the kinematic factors and our matrix element $⟨ ∣ M ∣^{2} ⟩$ in the high-energy limit, where we can neglect the masses of both the electrons and muons ( $E ≫ m_{μ}, m_{e}$ ).

Kinematics

In the CM frame, we define the four-momenta as follows:

Electron $p_{1} = (E, 0, 0, E)$
Positron $p_{2} = (E, 0, 0, - E)$
Muon $p_{3} = (E, E sin θ, 0, E cos θ)$
Antimuon $p_{4} = (E, - E sin θ, 0, - E cos θ)$

Here, $E$ is the beam energy, so the total CM energy is $E_{C M} = 2 E$ . The scattering angle $θ$ is the angle between the outgoing muon ( $p_{3}$ ) and the incoming electron beam direction ( $p_{1}$ ). In this high-energy limit, energy conservation dictates that all particles have energy $E$ .

The kinematic factors in our $d σ / d Ω$ formula become:

$E_{C M}^{2} = (2 E)^{2} = 4 E^{2}$
$∣ p_{i} ∣ = ∣ p_{1} ∣ = E$
$∣ p_{f} ∣ = ∣ p_{3} ∣ = E$

So, the momentum ratio $∣ p_{f} ∣/∣ p_{i} ∣ = 1$ .

Dynamics: The Matrix Element

From the last lesson, the general unpolarized squared amplitude was:

⟨ ∣ M ∣^{2} ⟩ = \frac{8 e ^{4}}{s ^{2}} [(p_{3} \cdot p_{1}) (p_{4} \cdot p_{2}) + (p_{3} \cdot p_{2}) (p_{4} \cdot p_{1}) + \frac{s}{2} (M_{μ}^{2} + m_{e}^{2})]

(using $p_{3}, p_{4}$ for final momenta).

In the high-energy limit, we neglect the mass terms. We also evaluate the dot products using the CM frame kinematics:

$s = (p_{1} + p_{2})^{2} = (2 E, 0, 0, 0)^{2} = 4 E^{2}$
$p_{1} \cdot p_{3} = E^{2} - p_{1} \cdot p_{3} = E^{2} - E^{2} cos θ = E^{2} (1 - cos θ)$
$p_{2} \cdot p_{4} = E^{2} - p_{2} \cdot p_{4} = E^{2} - E^{2} cos θ = E^{2} (1 - cos θ)$
$p_{1} \cdot p_{4} = E^{2} - p_{1} \cdot p_{4} = E^{2} + E^{2} cos θ = E^{2} (1 + cos θ)$
$p_{2} \cdot p_{3} = E^{2} - p_{2} \cdot p_{3} = E^{2} + E^{2} cos θ = E^{2} (1 + cos θ)$

Substituting these into the expression for $⟨ ∣ M ∣^{2} ⟩$ :

⟨ ∣ M ∣^{2} ⟩ = \frac{8 e ^{4}}{( 4 E ^{2} ) ^{2}} [(E^{2} (1 - cos θ))^{2} + (E^{2} (1 + cos θ))^{2}]

= \frac{8 e ^{4}}{16 E ^{4}} [E^{4} (1 - 2 cos θ + cos^{2} θ) + E^{4} (1 + 2 cos θ + cos^{2} θ)]

= \frac{e ^{4}}{2} [2 + 2 cos^{2} θ] = e^{4} (1 + cos^{2} θ)

This is the elegantly simple form of the squared matrix element in the high-energy limit.

The Final Differential Cross-Section

Now, we plug everything into the formula for $d σ / d Ω$ :

\frac{d σ}{d Ω} = \frac{1}{64 π ^{2} ( 4 E ^{2} )} \frac{E}{E} \cdot e^{4} (1 + cos^{2} θ) = \frac{e ^{4}}{256 π ^{2} E ^{2}} (1 + cos^{2} θ)

It is conventional to express this using the fine-structure constant, $α = e^{2} /4 π$ . This gives $e^{4} = 16 π^{2} α^{2}$ .

\frac{d σ}{d Ω} = \frac{16 π ^{2} α ^{2}}{256 π ^{2} E ^{2}} (1 + cos^{2} θ) = \frac{α ^{2}}{16 E ^{2}} (1 + cos^{2} θ)

Since $s = 4 E^{2}$ , we can also write this in a Lorentz-invariant form (although $θ$ is frame-dependent):

\frac{d σ}{d Ω} = \frac{α ^{2}}{4 s} (1 + cos^{2} θ)

This famous result shows that the muons are preferentially produced in the forward ( $θ \approx 0$ ) and backward ( $θ \approx π$ ) directions, a characteristic signature of a spin-1 particle (the photon) exchange.

Test your understanding!

At what scattering angle $θ$ is the production of muons minimized? What is the ratio of the maximum production rate to the minimum production rate?

Show answer

The term $(1 + cos^{2} θ)$ is minimized when $cos^{2} θ = 0$ , which occurs at $θ = π /2$ (i.e., perpendicular to the beam line). The rate is maximized when $cos^{2} θ = 1$ , which occurs at $θ = 0$ and $θ = π$ . The ratio of maximum to minimum is $(1 + 1) / (1 + 0) = 2$ .

4. The Total Cross-Section

To find the total cross-section $σ$ , we simply integrate the differential cross-section over the entire solid angle $d Ω = d ϕ sin θ d θ$ .

σ = \int \frac{d σ}{d Ω} d Ω = \int_{0}^{2 π} d ϕ \int_{0}^{π} d θ sin θ [\frac{α ^{2}}{4 s} (1 + cos^{2} θ)]

The integral over $ϕ$ is trivial, giving a factor of $2 π$ .
$σ = \frac{2 π α ^{2}}{4 s} \int_{0}^{π} (1 + cos^{2} θ) sin θ d θ$
The integral over $θ$ can be solved with the substitution $u = cos θ$ , $d u = - sin θ d θ$ . The integration limits change from $0 \to π$ to $1 \to - 1$ .
$\int_{0}^{π} (1 + cos^{2} θ) sin θ d θ = \int_{1}^{- 1} (1 + u^{2}) (- d u) = \int_{- 1}^{1} (1 + u^{2}) d u$ $= [u + \frac{u ^{3}}{3}]_{- 1}^{1} = (1 + \frac{1}{3}) - (- 1 - \frac{1}{3}) = \frac{4}{3} + \frac{4}{3} = \frac{8}{3}$
Combining the pieces, we get the total cross-section:
$σ = \frac{2 π α ^{2}}{4 s} (\frac{8}{3}) = \frac{16 π α ^{2}}{12 s}$ $σ = \frac{4 π α ^{2}}{3 s}$

This is one of the classic, fundamental results of QED. It predicts that the total probability of producing a muon-antimuon pair falls off as $1/ s$ , or inversely with the square of the center-of-mass energy. This prediction has been verified with extraordinary precision in collider experiments.

Conclusion

In this lesson, we completed the journey from a Feynman diagram to a physically observable quantity. By combining the spin-averaged matrix element with the formula for the 2-to-2 cross-section, we derived both the angular distribution and the total production rate for $e^{+} e^{-} \to μ^{+} μ^{-}$ .

Key Takeaways:

The general formula for the cross-section connects dynamics ( $∣ M ∣^{2}$ ) with phase space and an initial flux factor.
In the CM frame, this simplifies to the differential cross-section $d σ / d Ω = \frac{1}{64 π ^{2} E _{C M}^{2}} \frac{∣ p _{f} ∣}{∣ p _{i} ∣} ⟨ ∣ M ∣^{2} ⟩$ .
For $e^{+} e^{-} \to μ^{+} μ^{-}$ in the high-energy limit, the differential cross-section is $\frac{d σ}{d Ω} = \frac{α ^{2}}{4 s} (1 + cos^{2} θ)$ .
The total cross-section is $σ = \frac{4 π α ^{2}}{3 s}$ , showing a characteristic $1/ s$ dependence.

This calculation is a template that you will see repeated for many other processes in QED and beyond.

Preview of the Next Lesson:
We will now move to another cornerstone process of QED: Compton scattering ( $e^{-} γ \to e^{-} γ$ ). This will present new challenges, as it involves an external photon in the initial state and requires us to sum over two different Feynman diagrams at tree-level. This will deepen our understanding of how to build and calculate amplitudes in QED.

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