Quantum Field Theory Course

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Part IV, Chapter 2 | Page 2 of 3

Deriving the Cross Section Formula

From the S-matrix to Fermi's Golden Rule in quantum field theory

2.7 The S-Matrix Decomposition

The S-matrix connects initial and final states. We separate the trivial (no-scattering) part:

$$S_{fi} = \langle f | S | i \rangle = \delta_{fi} + i(2\pi)^4 \delta^4(p_f - p_i) \mathcal{M}_{fi}$$

The first term represents no interaction. All the physics is in the invariant amplitude $\mathcal{M}_{fi}$, computed from Feynman diagrams.

Transition Probability

The transition probability involves squaring the S-matrix element. The square of the delta function requires careful treatment using the box normalization prescription:

$$\left[(2\pi)^4 \delta^4(p_f - p_i)\right]^2 = (2\pi)^4 \delta^4(p_f - p_i) \cdot VT$$

where V is the spatial volume and T the time interval. This gives a transition rate (probability per unit time):

$$\frac{|S_{fi}|^2}{T} = (2\pi)^4 \delta^4(p_f - p_i) \frac{|\mathcal{M}|^2}{V}$$

💡Why VT Appears

The delta function squared is an artifact of working with plane waves (infinite extent). In a finite box of volume V over time T, momentum eigenstates are discrete and the "squared delta" gives a factor of VT. This cancels in physical observables when we properly normalize.

2.8 Fermi's Golden Rule: Full Derivation

To obtain the cross section, we need the transition rate per unit volume, normalized by the incident flux and target density. Start from first-order time-dependent perturbation theory.

Step 1: Transition Amplitude

In the interaction picture, the first-order transition amplitude is:

$$\langle f | S^{(1)} | i \rangle = -i \int d^4x \, \langle f | \mathcal{H}_{\text{int}}(x) | i \rangle$$

Using plane-wave expansions for the initial and final states, the spatial integral gives a momentum-conserving delta function:

$$\int d^3x \, e^{i(\vec{p}_i - \vec{p}_f) \cdot \vec{x}} = (2\pi)^3 \delta^3(\vec{p}_i - \vec{p}_f)$$

Step 2: Transition Rate

The transition probability for a process happening in time T is:

$$P_{i \to f} = \left|\langle f | S | i \rangle\right|^2 = \frac{T}{V^{n-1}} (2\pi)^4 \delta^4(p_f - p_i) |\mathcal{M}|^2 \prod_{j} \frac{1}{2E_j}$$

where n is the total number of external particles and the $1/(2E_j)$ factors come from the relativistic normalization of states. The transition rate is:

$$\Gamma_{i \to f} = \frac{P_{i \to f}}{T}$$

Step 3: Sum over Final States

In the continuum limit, summing over final-state momenta:

$$\sum_{\vec{p}_f} \to V \int \frac{d^3p_f}{(2\pi)^3}$$

Combining everything for a 2 → n process, the differential rate is:

$$\boxed{d\Gamma = \frac{|\mathcal{M}|^2}{2E_1 \cdot 2E_2 \cdot V} \, d\Phi_n}$$

where the Lorentz-invariant phase space is:

$$d\Phi_n = (2\pi)^4 \delta^4\!\left(p_1 + p_2 - \sum_{j=1}^n p_j'\right) \prod_{j=1}^n \frac{d^3p_j'}{(2\pi)^3 \, 2E_j'}$$

2.9 From Rate to Cross Section

The cross section is the rate divided by the incident flux. For two colliding particles with momenta p₁ and p₂:

Incident Flux

The flux factor in the center-of-mass frame is:

$$F = \frac{|\vec{v}_1 - \vec{v}_2|}{V} = \frac{|\vec{p}_1/E_1 - \vec{p}_2/E_2|}{V}$$

In the CM frame where $\vec{p}_1 = -\vec{p}_2 = \vec{p}$:

$$F = \frac{|\vec{p}|}{V} \left(\frac{1}{E_1} + \frac{1}{E_2}\right) = \frac{|\vec{p}|(E_1 + E_2)}{V \cdot E_1 E_2} = \frac{|\vec{p}|\sqrt{s}}{V \cdot E_1 E_2}$$

where $\sqrt{s} = E_1 + E_2$ is the total CM energy. Combining with the rate formula, the cross section is:

$$\boxed{d\sigma = \frac{|\mathcal{M}|^2}{4 |\vec{p}_i| \sqrt{s}} \, d\Phi_n}$$

Lorentz-Invariant Form

Using the Lorentz-invariant Møller flux factor:

$$I = 4\sqrt{(p_1 \cdot p_2)^2 - m_1^2 m_2^2}$$

In the CM frame, $I = 4|\vec{p}_i|\sqrt{s}$, recovering the formula above. The invariant form is:

$$d\sigma = \frac{|\mathcal{M}|^2}{4\sqrt{(p_1 \cdot p_2)^2 - m_1^2 m_2^2}} \, d\Phi_n$$

💡Dimensions Check

Cross sections have dimensions of area: [σ] = length². In natural units, [σ] = 1/energy². Since |M|² is dimensionless for 2→2 scattering, and the flux factor contributes 1/energy² while phase space contributes (energy)⁰ for 2-body final states, we get [σ] ~ 1/s, as expected.

2.10 Two-Body Phase Space: Explicit Evaluation

For 2 → 2 scattering, we evaluate the two-body phase space explicitly. Start with:

$$d\Phi_2 = \frac{d^3p_3}{(2\pi)^3 2E_3} \frac{d^3p_4}{(2\pi)^3 2E_4} (2\pi)^4 \delta^4(p_1 + p_2 - p_3 - p_4)$$

Step-by-Step Evaluation

Step 1: Integrate over $\vec{p}_4$ using the 3-momentum delta function:

$$\int \frac{d^3p_4}{(2\pi)^3 2E_4} (2\pi)^3 \delta^3(\vec{p}_1 + \vec{p}_2 - \vec{p}_3 - \vec{p}_4) = \frac{1}{2E_4}\bigg|_{\vec{p}_4 = \vec{p}_1 + \vec{p}_2 - \vec{p}_3}$$

Step 2: Write $d^3p_3 = |\vec{p}_3|^2 d|\vec{p}_3| d\Omega$ and use the remaining energy delta function. In the CM frame ($\vec{p}_1 + \vec{p}_2 = 0$):

$$\delta(E_3 + E_4 - \sqrt{s}) = \frac{E_3 E_4}{|\vec{p}_f|(E_3 + E_4)} \delta(|\vec{p}_3| - |\vec{p}_f|)$$

where $|\vec{p}_f|$ is the magnitude of the final-state 3-momentum, determined by kinematics.

Step 3: Collecting all factors:

$$\boxed{d\Phi_2 = \frac{|\vec{p}_f|}{16\pi^2 \sqrt{s}} d\Omega}$$

This beautifully compact result, combined with the cross-section formula, gives:

$$\frac{d\sigma}{d\Omega} = \frac{|\mathcal{M}|^2}{64\pi^2 s} \frac{|\vec{p}_f|}{|\vec{p}_i|}$$

recovering the master formula from Page 1.

Key Concepts (Page 2)

  • • The S-matrix decomposes as S = 1 + iT, with the invariant amplitude M encoding all dynamics
  • • The squared delta function gives VT via box normalization — this cancels in physical rates
  • • Fermi's Golden Rule: dΓ = |M|²/(4E₁E₂V) dΦₙ
  • • Cross section = rate / flux: dσ = |M|²/(4|p⃗ᵢ|√s) dΦₙ
  • • Two-body phase space: dΦ₂ = |p⃗f|/(16π²√s) dΩ
  • • The Møller flux factor ensures Lorentz invariance
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