Tree-Level φ⁴ Scattering Amplitude Calculation

Recapping the Feynman Rules for $\mathcal{M}$

Let's quickly recall the algorithm we derived for calculating $i\mathcal{M}$ in ${\varphi }^4$ theory:

1. Draw all connected, amputated Feynman diagrams for the process.
2. For each internal line with momentum $k$, write a propagator factor: $\frac{i}{k^2-m^2+iϵ}$.
3. For each vertex, write an interaction factor: $-i\lambda$.
4. Enforce momentum conservation at each vertex.
5. Integrate over any undetermined loop momenta: $\int \frac{d^{4}k}{(2\pi {)}^4}$.
6. Divide by the diagram's symmetry factor.
7. Sum the contributions from all diagrams.

1. The Tree-Level Diagram for $2\to 2$ Scattering

Let's apply these rules to the process $\varphi (p_1)+\varphi (p_2)\to \varphi (p_3)+\varphi (p_4)$.
We are interested in the tree-level amplitude, which means we only consider diagrams with no loops. In ${\varphi }^4$ theory, the fundamental interaction is a four-point vertex. Therefore, the lowest-order diagram that connects four external particles is just the vertex itself.

The diagram is simply four lines meeting at a point: two incoming lines carrying momenta $p_1$ and $p_2$, and two outgoing lines with momenta $p_3$ and $p_4$.

Let's calculate $i\mathcal{M}$ for this diagram:

• Diagrams: There is only one tree-level diagram.
• Internal lines: There are no internal lines, so no propagators.
• Vertices: There is one vertex, which gives a factor of $-i\lambda$.
• Momentum Conservation: The overall process conserves momentum, $p_1+p_2=p_3+p_4$. This is enforced at the single vertex.
• Loops: There are no loops to integrate over.
• Symmetry: The symmetry factor is 1.

Applying the rules, the calculation is remarkably simple:

This gives us the tree-level scattering amplitude:

The amplitude is a constant, independent of the particles' momenta or energies. This is a characteristic feature of a contact interaction at tree level.

2. A Deeper Look: The Role of Mandelstam Variables

The result $\mathcal{M}=-\lambda$ seems too simple. Does it capture the whole story of scattering? While this is the correct amplitude for the single diagram, in more complex theories, we would have multiple tree-level diagrams to sum. For instance, in a theory where particles interact by exchanging another particle, we would have diagrams like these:

Hypothetical s, t, and u-channel diagrams for 2→2 scattering mediated by an exchanged particle. Unlike our contact interaction, these diagrams have internal propagators, and their amplitudes would depend on momentum.

Each diagram corresponds to a different way the scattering can proceed kinetically:

• s-channel: The two incoming particles annihilate into a virtual particle, which then decays into the two outgoing particles.
• t-channel: Particle 1 "scatters off" particle 2, exchanging a virtual particle and becoming particle 3.
• u-channel: Similar to the t-channel, but particle 1 becomes particle 4.

Even though our ${\varphi }^4$ theory only has one tree-level diagram, it's incredibly useful to describe the kinematics of any $2\to 2$ process using variables that correspond to these channels. These are the Mandelstam variables, which are Lorentz-invariant scalars constructed from the four-momenta:

• s (center-of-mass energy squared):
  $s=(p_1+p_2{)}^2=(p_3+p_4{)}^2$
  This represents the total squared energy available in the center-of-mass frame. It is the momentum squared that would flow through the propagator in an s-channel diagram.

• t (momentum transfer squared):
  $t=(p_1-p_3{)}^2=(p_4-p_2{)}^2$
  This represents the squared four-momentum transferred from particle 1 to particle 3. It corresponds to the momentum flowing through a t-channel propagator.

• u (momentum transfer squared):
  $u=(p_1-p_4{)}^2=(p_3-p_2{)}^2$
  This is the squared four-momentum transferred if particle 1 scatters into particle 4. It corresponds to the momentum in a u-channel propagator.

Even for our contact interaction, where $\mathcal{M}$ is constant, we can still think of the full amplitude as a sum over these channels. It just so happens that in ${\varphi }^4$ theory, all channels contribute the same constant value, $\mathcal{M}(s,t,u)=-\lambda$. In more complex theories like QED, the amplitudes for different channels will have distinct dependencies on $s$, $t$, and $u$, such as ${\mathcal{M}}_t\propto 1/t$.

As you proved in the exercise, the Mandelstam variables for a $2\to 2$ process with identical masses $m$ satisfy:
$s+t+u=4m^2$
This means that the scattering amplitude $\mathcal{M}(s,t,u)$, though seemingly a function of three variables, is actually only a function of two independent kinematic variables (e.g., center-of-mass energy and scattering angle).

3. Calculating the Amplitude for a More Complex Theory

To see how these concepts come together in a more typical scenario, let's briefly consider a different theory—one with a three-point vertex $\frac{g}{3!}{\varphi }^3$ instead of a four-point one. In such a theory, $2\to 2$ scattering cannot happen at a single vertex. The lowest-order contributions must involve two vertices connected by an internal propagator.

The three distinct tree-level diagrams would be exactly the s, t, and u-channel diagrams shown previously. Let's calculate the amplitude for the t-channel diagram as an example.

• Diagram: $p_1$ and $p_2$ enter. $p_1$ goes to a vertex, emits a virtual particle, and becomes $p_3$. The virtual particle is absorbed by $p_2$, which becomes $p_4$.
• Vertices: Two vertices, giving a factor of $(-ig)\times (-ig)=-g^2$.
• Internal Line: One internal line (propagator). What is its momentum? By momentum conservation at the top vertex, the momentum flowing "down" is $k=p_1-p_3$. The propagator factor is $\frac{i}{k^2-m^2}=\frac{i}{(p_1-p_3{)}^2-m^2}$.
• Mandelstam $t$: Notice that the denominator is exactly $t-m^2$.

The amplitude for this t-channel diagram is:
$i{\mathcal{M}}_t=(-ig{)}^2\times \frac{i}{(p_1-p_3{)}^2-m^2}=-g^2\frac{i}{t-m^2}$
${\mathcal{M}}_t=\frac{-g^2}{t-m^2}$
Similarly, the s- and u-channel diagrams would give:
${\mathcal{M}}_s=\frac{-g^2}{s-m^2}\text{and}{\mathcal{M}}_u=\frac{-g^2}{u-m^2}$
The total tree-level amplitude would be the sum:
${\mathcal{M}}_{\text{total}}={\mathcal{M}}_s+{\mathcal{M}}_t+{\mathcal{M}}_u=-g^2\left(\frac{1}{s-m^2}+\frac{1}{t-m^2}+\frac{1}{u-m^2}\right)$

This example highlights why separating the amplitude into these channels is so powerful. It organizes the calculation and makes the physical processes (annihilation vs. exchange) manifest in the mathematical structure of the amplitude.

Conclusion

In this lesson, we performed our first full calculation of a scattering amplitude using the Feynman rules. We found that for ${\varphi }^4$ theory, the tree-level $2\to 2$ amplitude is simply a constant, $\mathcal{M}=-\lambda$. While this specific result is simple, the process forced us to introduce the Mandelstam variables ($s,t,u$), which provide a universal, Lorentz-invariant language for describing the kinematics of any two-body scattering process.

Key Takeaways:

• The tree-level $2\to 2$ scattering amplitude in ${\varphi }^4$ theory arises from a single, four-point contact interaction diagram, yielding $\mathcal{M}=-\lambda$.
• Mandelstam variables ($s,t,u$) are Lorentz-invariant kinematic variables that correspond to the squared four-momentum flowing in the s-channel, t-channel, and u-channel, respectively.
• For any $2\to 2$ scattering, $s+t+u={\sum }_i{m}_i^2$, meaning only two of the variables are independent.
• In theories with exchanged particles (like a hypothetical ${\varphi }^3$ theory), the total amplitude is the sum of contributions from each channel, $\mathcal{M}={\mathcal{M}}_s+{\mathcal{M}}_t+{\mathcal{M}}_u$, where each term typically includes a propagator that depends on the corresponding Mandelstam variable.

Preview of the Next Lesson:

We now have the invariant matrix element $\mathcal{M}$. But how does this relate to a physically measurable quantity like a cross-section? In the next lesson, we will connect $\mathcal{M}$ to the differential cross-section, $d\sigma /d\Omega$. We will derive the general formula and then apply it to our ${\varphi }^4$ theory result to calculate our first QFT prediction for an experiment.