Calculating the Matrix Element for e⁺e⁻ → γγ

Hello! Welcome back to our study of Quantum Electrodynamics.

In our last few lessons, we dissected the Compton scattering process, $e^{-} γ \to e^{-} γ$ . You calculated the unpolarized, squared matrix element and then used it to derive the famous Klein-Nishina formula for the differential cross-section. This demonstrated the full pipeline from Feynman diagrams to an observable quantity.

Today, we will apply this powerful machinery to another fundamental QED process: pair annihilation, $e^{+} e^{-} \to γ γ$ . Our goal is to calculate the unpolarized, squared matrix element for this interaction. This process is not only important in its own right but is also deeply related to Compton scattering through a concept called "crossing symmetry." You will see that the calculational techniques—applying Feynman rules, summing over spins and polarizations, and using gamma matrix trace technology—are precisely the ones you've been honing.

1. The Feynman Diagrams for Pair Annihilation

At the lowest order in perturbation theory (tree level), two diagrams contribute to pair annihilation. An incoming electron-positron pair annihilates, creating a virtual intermediate electron, which then emits the two final-state photons.

Feynman Diagrams for Electron-Positron Annihilation into Two Photons — The two tree-level Feynman diagrams for $e^+e^- \to \gamma\gamma$. The diagrams differ only in the order that the final-state photons are emitted from the internal fermion line. (Note: The diagram labels momenta as $p_1, p_2, k_1, k_2$, but we will use the notation $p_-, p_+, k_1, k_2$ to match our primary resource).

Let's define our four-momenta:

Incoming electron: $p_{-}$ (mass $m$ )
Incoming positron: $p_{+}$ (mass $m$ )
Outgoing photon 1: $k_{1}$ (polarization $ϵ_{1}$ )
Outgoing photon 2: $k_{2}$ (polarization $ϵ_{2}$ )

Following the Feynman rules, we can write down the matrix element, $M$ , which is the sum of the amplitudes for these two diagrams, $M_{1} + M_{2}$ . A detailed breakdown of this setup is presented in the following resource.

ANNIHILATION

The document 'ANNIHILATION' by Vadim Kaplunovsky provides a clear, step-by-step derivation for this process. We will use it as our guide. This first section sets up the amplitudes for the two diagrams.

Please read the first page of the document, down to equation (5). Pay close attention to: How the two diagrams translate into the expressions for (\mathcal{M}{\mu u 1}) and (\mathcal{M}{\mu u 2}). The use of the momentum-space propagator for the internal fermion line. The definition of the intermediate momenta, q and ilde{q}. The introduction of the variables t = q^2 and u = ilde{q}^2.

As you've just read, the total amplitude has the structure $M = ϵ_{1 μ}^{*} ϵ_{2 ν}^{*} M^{μν}$ , where $M^{μν} = M_{1}^{μν} + M_{2}^{μν}$ . The two components are:

M_{1}^{μν} = \frac{- e ^{2}}{t - m ^{2}} v ˉ (p_{+}) γ^{ν} (\neq q + m) γ^{μ} u (p_{-})

M_{2}^{μν} = \frac{- e ^{2}}{u - m ^{2}} v ˉ (p_{+}) γ^{μ} (\neq q ~ + m) γ^{ν} u (p_{-})

Here, $t = (p_{-} - k_{1})^{2}$ and $u = (p_{-} - k_{2})^{2}$ . These Lorentz-invariant quantities are two of the three Mandelstam variables, which are indispensable in high-energy physics. We are using them here as convenient shorthand; we will formalize them in our next lesson.

2. A Crucial Check: Gauge Invariance

A key consistency check for any QED amplitude involving external photons is that it must respect gauge invariance. This is mathematically expressed by the Ward identity. For our process, it requires that if we replace a photon's polarization vector $ϵ_{μ}$ with its momentum $k_{μ}$ , the amplitude must vanish. For the full amplitude, this means $k_{1 μ} M^{μν} = 0$ and $k_{2 ν} M^{μν} = 0$ .

One of the beautiful subtleties of gauge theories is that individual Feynman diagrams are often not gauge invariant by themselves. Only their sum is. Let's see how this works for pair annihilation.

ANNIHILATION

The 'ANNIHILATION' document continues with a demonstration of the Ward identity for our amplitude. This is a very important physical point.

Read the section 'Ward Identities' (pages 1-2). Follow the algebra showing that (k_{1\mu}\mathcal{M}^{\mu u}1 eq 0) and (k{1\mu}\mathcal{M}^{\mu u}_2 eq 0) individually, but their sum is zero. This cancellation is not accidental; it's a deep feature of the theory.

3. The Unpolarized Squared Amplitude

Our goal is to compute the unpolarized squared amplitude, which involves averaging over the initial electron and positron spins and summing over the final photon polarizations.

⟨ ∣ M ∣^{2} ⟩ = \frac{1}{4} spins \sum pols \sum ∣ M_{1} + M_{2} ∣^{2}

The sum over photon polarizations, thanks to the Ward identity, simplifies via $\sum_{λ} ϵ_{μ} (λ) ϵ_{ν}^{*} (λ) \to - g_{μν}$ . The average over fermion spins is handled, as before, by converting the expressions into traces using Casimir's trick: $\sum_{s} u_{s} (p) u ˉ_{s} (p) = \neq p + m$ and $\sum_{s} v_{s} (p) v ˉ_{s} (p) = \neq p - m$ .

Expanding the squared amplitude gives us three terms: one from the square of the first diagram ( $∣ M_{1} ∣^{2}$ ), one from the square of the second ( $∣ M_{2} ∣^{2}$ ), and an interference term ( $2 Re (M_{1} M_{2}^{*})$ ).

ANNIHILATION

The next section of the resource sets up the trace calculation for the unpolarized squared amplitude.

Read pages 3-4, from 'Summing over the Spins and Polarizations' down to equation (23). Focus on how equation (17), containing the spinors, becomes equation (20), which involves traces. Understand the structure of the three trace terms: A_{11}, A_{22}, and A_{12}.

This procedure leaves us with the task of calculating three large traces:

⟨ ∣ M ∣^{2} ⟩ = \frac{e ^{4}}{4} [\frac{A _{11}}{( t - m ^{2} ) ^{2}} + \frac{A _{22}}{( u - m ^{2} ) ^{2}} + \frac{2 A _{12}}{( t - m ^{2} ) ( u - m ^{2} )}]

where

$A_{11} = Tr [\dots]$ corresponds to $∣ M_{1} ∣^{2}$ .
$A_{22} = Tr [\dots]$ corresponds to $∣ M_{2} ∣^{2}$ .
$A_{12} = Tr [\dots]$ corresponds to the interference term.

4. Traceology: The Calculation

This is the most computationally intensive part of the lesson. We need to evaluate these traces, which involve products of up to eight gamma matrices. The process relies heavily on the gamma matrix identities you've used before.

The $A_{11}$ and $A_{22}$ Terms

Let's start by calculating $A_{11}$ , which comes from squaring the first diagram.

ANNIHILATION

The section 'Traceology 1' in the 'ANNIHILATION' PDF meticulously works through the A_{11} trace.

Read pages 4-6, covering the calculation of A_{11}. Follow the simplification steps, noting the use of identities like (\gamma_\mu ot{p} \gamma^\mu = -2 ot{p}) and how the dot products are expressed in terms of s, t, u and m. Note that s = (p_- + p_+)^2. We'll also use the relation s+t+u = \sum m_i^2 = 2m^2.

The result for $A_{11}$ is $8 (p_{+} \cdot q) (p_{-} \cdot q) - 4 (q^{2} - m^{2}) (p_{+} \cdot p_{-})$ , which simplifies to an expression in $t$ and $u$ .

Now for $A_{22}$ . We could repeat the entire calculation. However, notice that diagram 2 is just diagram 1 with the two photons exchanged ( $k_{1} \leftrightarrow k_{2}$ ). This symmetry is called crossing symmetry. In terms of our variables, this exchange corresponds to swapping $t \leftrightarrow u$ . Therefore, we can get $A_{22}$ for free simply by taking the expression for $A_{11}$ and swapping $t$ and $u$ . This is a powerful shortcut that highlights the underlying structure of the theory.

The Interference Term $A_{12}$

The interference term is often the most complicated. It requires evaluating a trace that mixes the components of both diagrams.

ANNIHILATION

The section 'Traceology 2' tackles the interference trace A_{12}.

Read pages 7-8 carefully. The algebra here is dense. The goal is not to reproduce every line instantly, but to understand the strategy: apply gamma matrix identities to reduce the length of the trace until it becomes manageable.

Test your understanding!

Why does the trace $A_{12}$ involve terms like $m^{2}$ and $m^{4}$ , which seem to suggest it would be small in the high-energy limit ( $s, ∣ t ∣, ∣ u ∣ ≫ m^{2}$ ), whereas the Compton scattering interference term did not?

Hint: Think about the structure of the gamma matrix products inside the trace for Compton scattering versus pair annihilation. What does the positron projection operator $(\neq p_{+} - m)$ introduce?

Show answer

In Compton scattering ( $e^{-} γ \to e^{-} γ$ ), both initial and final fermions are electrons, so their spin sums involve $(\neq p + m)$ . In pair annihilation, the initial state has an electron ( $\neq p_{-} + m$ ) and a positron ( $\neq p_{+} - m$ ).

The interference term $A_{12}$ involves a trace over a product containing both $(\neq p_{-} + m)$ and $(\neq p_{+} - m)$ . When you expand the trace, terms that are linear in mass (like $Tr [\dots m \dots]$ ) will have an odd number of gamma matrices and vanish. However, terms proportional to $m^{2}$ (from products like $(\dots m) (\dots - m)$ ) will have an even number of gamma matrices and survive. This is why the interference term for annihilation has a different mass dependence compared to Compton scattering.

5. The Final Result

Now, we collect our results for $A_{11}$ , $A_{22}$ , and $A_{12}$ and substitute them back into the expression for $⟨ ∣ M ∣^{2} ⟩$ . After some algebraic rearrangement, we arrive at the final answer.

ANNIHILATION

The final section, 'Annihilation Summary', brings everything together.

Read page 9 to see the final assembly of the terms. The algebra simplifies to a surprisingly compact and symmetric form.

The unpolarized, squared matrix element for $e^{+} e^{-} \to γ γ$ is:

⟨ ∣ M ∣^{2} ⟩ = 2 e^{4} [\frac{u - m ^{2}}{t - m ^{2}} + \frac{t - m ^{2}}{u - m ^{2}} + 2 m^{2} (\frac{1}{t - m ^{2}} + \frac{1}{u - m ^{2}}) - 2 m^{4} (\frac{1}{t - m ^{2}} + \frac{1}{u - m ^{2}})^{2}]

This expression is fully Lorentz invariant, depending only on the Mandelstam variables $t$ and $u$ , and the electron mass $m$ .

Conclusion

Excellent work! You have successfully calculated the unpolarized, squared matrix element for electron-positron annihilation into two photons. This is another classic QED calculation that showcases the power and consistency of the theory.

Key Takeaways:

Pair annihilation ( $e^{+} e^{-} \to γ γ$ ) is described by two tree-level diagrams, which are related by crossing symmetry.
Gauge invariance, in the form of the Ward identity, is a crucial constraint that is only satisfied by the sum of all contributing diagrams.
The calculation follows the established procedure: write the amplitude, square it, average/sum over spins and polarizations, and evaluate the resulting gamma matrix traces.
The final result can be expressed compactly and elegantly using Lorentz-invariant variables.

Preview of the Next Lesson:
In this calculation, we used the variables $t$ and $u$ as convenient shorthands. In the next lesson, we will give them, along with their partner $s$ , a formal introduction as the Mandelstam variables. We will explore their properties, their physical interpretations in different "channels," and how they provide a unified language for describing all 2-to-2 scattering processes, including Compton scattering and pair annihilation.

Recommended resources

1. The Feynman Diagrams for Pair Annihilation

2. A Crucial Check: Gauge Invariance

3. The Unpolarized Squared Amplitude

4. Traceology: The Calculation

The A11​ and A22​ Terms

The Interference Term A12​

5. The Final Result

Conclusion

The $A_{11}$ and $A_{22}$ Terms

The Interference Term $A_{12}$