Cross sections
In quantum field theory we can predict matrix elements $\braket{\psi|\phi}$ which lead to probabilites of a state $\ket{\phi}$ evolving into another one $\bra{\psi}$. In the Heisenber picture the time evolution is absorbed into the operators.
In scattering experiments we consider intial and final states $\ket{i}$ and $\ket{f}$ for $t\to\pm\infty$ and assume that they are momentum eigenstates called asymptotic states. The probability of an initial state $\ket{i}$ evolving into a final state $\ket{f}$ can be described in terms of the Scattering matrix or S-matrix
Definition: S-matrix
The scattering matrix is defined as $$S_{fi} = \braket{f|S|i},$$ where $S$ is the time evolution operator in the Heisenberg picture, which projects from $t=-\infty$ to $t=\infty$.
The S-matrix can be calculated in quantum field theory. To compare calculations with experiments we need a measureable quantity
Definition: Cross section
The cross section $\sigma$ is a measure of the probability of incoming particles to be scattered to a certain final state. It can be written as $$\sigma = \frac{1}{T\cdot \Phi}N,$$ where $T$ is the timespan of the measurement, $\Phi$ is the flux of incoming particles and $N$ the measured number of scattered particles.
The cross section can be differential in terms of other kinematic variables. In that case one can measure how many final state particles are in a certain range of final states.
Definition: Luminosity
The Luminosity $\mathcal{L}$ is a defined by the number of scattering events over time and cross section $$\mathcal{L} = \frac{1}{\sigma}\frac{dN}{dt}$$ This is a accelerator specific quantity Experiments usually report the integrated Luminosity (over time) for a measurement
To compare a theoretical calculation with the number of events measured by an experiment calculates one uses $$N = \int d\sigma \times \int \mathcal{L} dt,$$ so that we
- calculate the diff. cross section $d\sigma$
- Multiply by integraded Luminosity
The cross section can be related to the S-matrix. We seperate the S-matrix into a free part ($1$) and an interacting part ($\mathcal{T}$)
Definition: Transfer matrix
The Transfer matrix $\mathcal{T}$ is the interacting part of the S-matrix and defined by $$S = 1 + i\mathcal{T}$$ The $i$ is convention
We can factor out momentum conservatoin to get the Matrix element $\mathcal{M}$
Definition: matrix element
The matrix element $\braket{f|\mathcal{M}|i}$ is nontrivial part of the interaction, defined by $$\mathcal{T} = (2\pi)^4 \delta^4\left(\sum p^\mu_i - \sum p^\mu_f\right) \mathcal{M}.$$
Cross sections can directly be computed when we know the matrix element $|\mathcal{M}|^2 = |\braket{f|\mathcal{M}|i}|^2$
Cross section formula
For an $2 \to n$ process with inital and final states $\ket{i}$ and $\ket{f}$ we have $$d\sigma = \frac{(2\pi)^4 |\mathcal{M}|^2}{4\sqrt{(p_1 \cdot p_2)^2 - m^2_1m^2_2}}d\Phi,$$ with the lorentz invariant phase space $\Phi$
Definition: Lorentz invariant phase space
The lorentz invariant phase space $\Phi$ is $$\Phi = \delta^4\left(\sum p^\mu_i - \sum p^\mu_f\right)\prod \frac{d^3p_j}{(2\pi)^3}\frac{1}{2E_j},$$ where the product runs over all final states $j$.
The derivation inserts the S-matrix-element $|\braket{f|S|i}|^2$ for the probability in the formula of the cross section
Derivation cross section formula
For a single particle the flux is its velocity over the Volume it inhabits $\Phi = \frac{|\vec{v}|}{V}$. The probability $dP$ for a process is $$dP = \frac{\braket{f|S|i}}{\braket{f|f}\braket{i|i}}d\Phi.$$ Note that intial and final states are not normalized to one, but rather $\braket{p|p}=2E_pV$, with the volume $V$, which can be found from the normalization of momentum eigenstates Pluggin all this into the original formula $$d\sigma = \frac{1}{T}\frac{1}{\Phi}dP,$$ will cancel out the Volume and lead to the given expression.
Example: $2\to 2$ scattering
For a $2\to 2$ scattering process $$p_1 + p_2 \to p_3 + p_4$$ We use the center of mass frame $\vec{p_1} = -\vec{p_2}$. The phase space becomes $$d\Phi = \delta^4(p^\mu_1 + p^\mu_2 - p^\mu_3 - p^\mu_4)\frac{d^3p_3}{(2\pi)^3}\frac{1}{2E_3}\frac{d^3p_4}{(2\pi)^3}\frac{1}{2E_4}.$$ We can execute the $\delta^3$ for the momentum part to integrate over $\vec{p}_4$. The resulting integral over $\vec{p}_3$ can formally be solved by substituion resulting in $$\left(\frac{d\sigma}{d\Omega}\right)_\text{CM} = \frac{1}{64\pi^2 E^2_\text{CM}}\frac{p_f}{p_i} |\mathcal{M}|^2 \Theta(E_\text{CM}-m_3-m_4)$$