Tree-Level Amplitudes in Quark-Quark Scattering

Hello! In our last lesson, we mastered the essential tools of QCD color algebra, learning how to compute the color factors associated with various interactions using SU(N) identities and the intuitive birdtrack notation. We now have all the components needed to perform a full calculation in QCD.

Today, we will assemble these pieces to achieve a major milestone: calculating the tree-level scattering amplitude for a representative QCD process, quark-quark scattering ( $q + q \to q + q$ ). This process is a cornerstone of our understanding of the strong force, analogous to electron-electron (Møller) scattering in QED, but with the rich complexity introduced by the non-Abelian nature of color. By the end of this lesson, you will have performed a complete perturbative QCD calculation from start to finish.

Setting up the Calculation: Quark-Quark Scattering

Let's consider the elastic scattering of two identical quarks:

q_{i} (p_{1}) + q_{j} (p_{2}) \to q_{i^{'}} (p_{1}^{'}) + q_{j^{'}} (p_{2}^{'})

Here, $p_{1}, p_{2}$ are the initial four-momenta, and $p_{1}^{'}, p_{2}^{'}$ are the final four-momenta. The indices $i, j, i^{'}, j^{'}$ represent the quark colors, which can change during the interaction.

Feynman Diagrams and Amplitude

At the lowest order in perturbation theory (tree level), this process is mediated by the exchange of a single gluon. Because the initial quarks are identical fermions, we must account for two possibilities that lead to the same final state:

t-channel: The quark at $p_{1}$ scatters to $p_{1}^{'}$ , exchanging a gluon with momentum $q = p_{1} - p_{1}^{'}$ .
u-channel: The quark at $p_{1}$ scatters to $p_{2}^{'}$ , exchanging a gluon with momentum $q^{'} = p_{1} - p_{2}^{'}$ .

According to Fermi-Dirac statistics, the total amplitude is the difference between these two contributions.

Feynman diagrams for quark-quark scattering, showing the t-channel and u-channel exchange of a gluon.

Figure 1: Tree-level Feynman diagrams for elastic quark-quark scattering. The t-channel diagram is on the left, and the u-channel diagram is on the right. Note the exchange of the final state particles in the u-channel diagram.

Applying the QCD Feynman rules we derived, we can write down the matrix element for each diagram. To simplify the kinematics, we use the Mandelstam variables:

$s = (p_{1} + p_{2})^{2}$ (center-of-mass energy squared)
$t = (p_{1} - p_{1}^{'})^{2}$ (momentum transfer squared for the t-channel)
$u = (p_{1} - p_{2}^{'})^{2}$ (momentum transfer squared for the u-channel)

The following reading provides a concise derivation of the amplitudes for these two diagrams.

HUGS 2022 Lecture 4: Introduction to Small-x Physics (Part 1)

Please read the section "Quark/Quark Scattering in Regge Kinematics" from the HUGS 2022 lectures by R. D. Ball, S. Carrazza, and A. Glazov. We will start by focusing on setting up the scattering amplitude.

Read the text from the beginning of Section 4.3 up to and including Equation (4.4c). Focus on how the Feynman rules are applied to construct the t-channel and u-channel amplitudes, \mathcal{M}_t and \mathcal{M}_u. Pay attention to the color indices and the minus sign for the u-channel amplitude.

As derived in the reading, the total amplitude $M = M_{t} + M_{u}$ is given by:

i M = i g^{2} (\frac{( t ^{a} ) _{i^{'} i} ( t ^{a} ) _{j^{'} j}}{t} [u ˉ_{1}^{'} γ^{μ} u_{1}] [u ˉ_{2}^{'} γ_{μ} u_{2}] - \frac{( t ^{a} ) _{j^{'} i} ( t ^{a} ) _{i^{'} j}}{u} [u ˉ_{1}^{'} γ^{μ} u_{2}] [u ˉ_{2}^{'} γ_{μ} u_{1}])

Note the structure:

Each term has a kinematic part involving the spinors ( $u ˉ, u$ ) and Dirac matrices ( $γ^{μ}$ ).
Each term has a color part involving the SU(3) generators ( $t^{a}$ ).
The denominators are the squared momentum transfers $t$ and $u$ from the gluon propagators.
A relative minus sign accounts for the fermionic exchange symmetry.

Calculating the Squared Amplitude

To compute a physical observable like a cross-section, we need the squared amplitude $∣ M ∣^{2}$ , averaged over the initial unobserved quantum numbers (spin and color) and summed over the final ones.

⟨ ∣ M ∣^{2} ⟩ = \frac{1}{4} \frac{1}{N _{c}^{2}} spins \sum colors \sum ∣ M_{t} + M_{u} ∣^{2}

Here, $1/4$ is for averaging over the two initial spins, and $1/ N_{c}^{2} = 1/9$ is for averaging over the $3 \times 3 = 9$ possible initial color combinations.

The full calculation involving the interference term $2 Re (M_{t} M_{u}^{*})$ is quite lengthy. For today's lesson, we will focus on calculating the $∣ M_{t} ∣^{2}$ term, as it demonstrates all the essential techniques. The other terms follow similar logic.

1. The Spin Sum (Kinematic Part)

The spin-dependent part of $∣ M_{t} ∣^{2}$ , after summing over initial and final spins, becomes a trace over Dirac matrices. Assuming massless quarks for high-energy scattering, this is a standard QED-like calculation:

spins \sum ∣ [u ˉ_{1}^{'} γ^{μ} u_{1}] [u ˉ_{2}^{'} γ_{μ} u_{2}] ∣^{2} = Tr (\neq p_{1}^{'} γ^{μ} \neq p_{1} γ^{ν}) Tr (\neq p_{2}^{'} γ_{μ} \neq p_{2} γ_{ν})

Using standard trace theorems ( $Tr (γ^{α} γ^{β} γ^{γ} γ^{δ}) = 4 (g^{α β} g^{γ δ} - g^{α γ} g^{β δ} + g^{α δ} g^{β γ})$ ), this evaluates to:

\dots = 32 [(p_{1} \cdot p_{2}) (p_{1}^{'} \cdot p_{2}^{'}) + (p_{1} \cdot p_{2}^{'}) (p_{2} \cdot p_{1}^{'})]

In terms of Mandelstam variables (and neglecting masses, so $s + t + u \approx 0$ ), this is:

\dots = 8 (s^{2} + u^{2})

2. The Color Sum

Now for the uniquely QCD part. We need to sum the color part of $∣ M_{t} ∣^{2}$ over all initial and final colors:

C_{t} = i, i^{'}, j, j^{'} \sum ∣ (t^{a})_{i^{'} i} (t^{a})_{j^{'} j} ∣^{2} = i, i^{'}, j, j^{'}, a, b \sum (t^{a})_{i^{'} i} (t^{a})_{j^{'} j} ((t^{b})_{i^{'} i} (t^{b})_{j^{'} j})^{*}

Recalling that $(t^{b})_{i^{'} i}^{*} = (t^{b})_{i i^{'}}$ , this becomes:

C_{t} = a, b \sum i, i^{'} \sum (t^{a})_{i^{'} i} (t^{b})_{i i^{'}} j, j^{'} \sum (t^{a})_{j^{'} j} (t^{b})_{j j^{'}}

Each term in parentheses is a trace:

C_{t} = a, b \sum Tr (t^{a} t^{b}) Tr (t^{a} t^{b})

From our last lesson, we know the fundamental normalization identity $Tr (t^{a} t^{b}) = \frac{1}{2} δ^{ab}$ . Substituting this in:

C_{t} = a, b \sum (\frac{1}{2} δ^{ab}) (\frac{1}{2} δ^{ab}) = \frac{1}{4} a \sum δ^{aa} = \frac{1}{4} (N_{c}^{2} - 1)

For QCD with $N_{c} = 3$ , this factor is $(3^{2} - 1) /4 = 2$ .

Test your understanding!

Let's consider scattering of two distinguishable quarks, for example $u + d \to u + d$ . In this case, the u-channel diagram is not possible because the quark flavor must be conserved at each vertex. The amplitude is just $M = M_{t}$ . The color structure is the same as before.

Calculate the color-averaged factor for this process. Remember, you must average over the $N_{c}^{2}$ initial states.

Show answer

The total color sum is $C_{t} = (N_{c}^{2} - 1) /4$ , which is 2 for SU(3). We need to average this over the $N_{c}^{2} = 9$ possible initial color states.

The color-averaged factor is:

⟨ C_{t} ⟩ = \frac{1}{N _{c}^{2}} \frac{N _{c}^{2} - 1}{4} = \frac{9 - 1}{4 \times 9} = \frac{8}{36} = \frac{2}{9}

So, for distinguishable quarks, the color factor is $2/9$ .

Assembling the Result

Now we can combine the pieces for $∣ M_{t} ∣^{2}$ .

⟨ ∣ M_{t} ∣^{2} ⟩ = \frac{1}{4 N _{c}^{2}} \times (Spin Sum) \times (Color Sum) \times \frac{g ^{4}}{t ^{2}}

⟨ ∣ M_{t} ∣^{2} ⟩ = \frac{1}{36} \times [8 (s^{2} + u^{2})] \times [2] \times \frac{g ^{4}}{t ^{2}} = \frac{4}{9} g^{4} \frac{s ^{2} + u ^{2}}{t ^{2}}

A similar calculation for the $∣ M_{u} ∣^{2}$ term yields $\frac{4}{9} g^{4} \frac{s ^{2} + t ^{2}}{u ^{2}}$ . The interference term turns out to be $\frac{8}{27} g^{4} \frac{s ^{2}}{t u}$ . Combining all these terms gives the full result for identical quark scattering:

⟨ ∣ M_{q q \to q q} ∣^{2} ⟩ = \frac{4}{9} g^{4} (\frac{s ^{2} + u ^{2}}{t ^{2}} + \frac{s ^{2} + t ^{2}}{u ^{2}}) - \frac{8}{27} g^{4} \frac{s ^{2}}{t u}

This result is a classic prediction of perturbative QCD and forms the basis for describing jet production in high-energy proton-proton collisions. The calculation beautifully demonstrates how the kinematic structure from QED is modified by the non-trivial color algebra of QCD.

High-Energy Behavior

The calculation is shown in full detail in the recommended reading, which also explores the high-energy (Regge) limit where $s ≫ - t$ . In this limit, the amplitude simplifies significantly.

HUGS 2022 Lecture 4: Introduction to Small-x Physics (Part 1)

Now, let's study the complete calculation and its physical implications in the high-energy limit.

Please read Section 4.3 thoroughly, from "High-Energy Kinematics" to the end of the section (Equation 4.19). Follow the steps for calculating the squared and averaged amplitude. Note how the high-energy (eikonal) approximation simplifies both the spinor traces and the final cross-section expression.

In the high-energy forward-scattering limit ( $s \to \infty$ , $t$ fixed), the $t$ -channel diagram dominates, and the cross-section becomes:

\frac{d σ}{d t} \propto \frac{α _{s}^{2}}{t ^{2}}

This is the QCD analogue of Rutherford scattering, but for the strong force between quarks.

Conclusion

Key Takeaways:

Tree-level quark-quark scattering in QCD involves both t-channel and u-channel gluon exchange diagrams due to the identical nature of quarks.
The total amplitude requires a relative minus sign between the diagrams, reflecting Fermi-Dirac statistics.
Calculating the squared amplitude involves two distinct steps: a spin sum (kinematic trace calculation similar to QED) and a color sum (a uniquely QCD calculation involving SU(3) generator identities).
The final amplitude contains a non-trivial color factor, which for $q q \to q q$ is composed of terms like $\frac{4}{9}$ and $\frac{8}{27}$ , arising from the structure of SU(3).

You have now successfully navigated a complete, non-trivial calculation in perturbative QCD, combining Feynman rules, Dirac algebra, and color algebra to derive a physical scattering amplitude.

Preview of the Next Lesson:

While tree-level calculations give us the leading-order picture, the most profound and unique features of QCD emerge at the loop level. In the next lesson, we will investigate one-loop corrections to the gluon propagator. This will lead us to calculate the one-loop beta function for QCD, a landmark result that reveals the property of asymptotic freedom—the surprising fact that the strong force becomes weaker at high energies.

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