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# QS 15: Relativistic scattering

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$\newcommand{\f}{{\mathcal F}} \newcommand{\intp}{\int \frac{{\rm d}^3p}{2p^0}} \newcommand{\intpp}{\int \frac{{\rm d}^3p'}{2{p'}^0}} \newcommand{\intx}{\int{\rm d}^3{\rm r}} \newcommand{\tp}{\otimes} \newcommand{\tpp}{\tp\cdots\tp} \newcommand{\kk}[1]{|#1\rangle} \newcommand{\bb}[1]{\langle #1} \newcommand{\dd}[1]{\delta_{#1}} \newcommand{\ddd}[1]{\delta^3(#1)} \newcommand{\vv}[1]{{\bf #1}} \newcommand{\molp}{\Omega^{(+)}}$

Relativistic scattering differs from non-relativistic scattering in following respect. The process of scattering involves {\em creation} or {\em annihilation} particles. Therefore the time evolution operator generators $H_0$ or $H$ are defined on a much larger Hilbert space called the Fock space.

Relativistic one-particle states are given as momentum space wave functions in the basis $\kk{\vv{p}\lambda}$ with normalization
\begin{eqnarray*} \bb{\vv{p}\lambda}\kk{\vv{p'}\lambda'}=2p^0\ddd{\vv{p}-\vv{p}'}
\dd{\lambda\lambda'} \end{eqnarray*}
where $p^0=\sqrt{\vv{p}^2c^2+m^2c^4}$.

Multi-particle states are defined by tensor products but have to be properly symmetrised (or anti-symmetrised) over identical particles.

We consider two colliding beams with momenta  sharply defined around $\vv{p}_1$ and $\vv{p}_2$ and spins $\sigma_1$ and $\sigma_2$ respectively.

The final state projection operator is taken as
\begin{eqnarray*} P_\xi=\int\frac{d^3\vv{k}_1}{2k_1^0}\cdots\frac{d^3\vv{k}_n}{2k_n^0}
\kk{\vv{k}_1\lambda_1\dots\vv{k}_n\lambda_n}
\bb{\vv{k}_1\lambda_1\dots\vv{k}_n\lambda_n}|
\end{eqnarray*}
The argument for non-relativistic colliding beams can be repeated almost step by step except that the total momentum delta function is not integrated and the transition amplitude matrix is defined by
\begin{eqnarray*}
\bb{\vv{k}_1\lambda_1\dots\vv{k}_n\lambda_n}|S
\kk{\vv{p}_1\sigma_1\vv{p}_2\sigma_2}
=\bb{\vv{k}_1\lambda_1\dots\vv{k}_n\lambda_n}
\kk{\vv{p}_1\sigma_1\vv{p}_2\sigma_2}\\
-2\pi i\delta^4(P_f-P_i)T(\vv{k}_1\lambda_1\dots\vv{k}_n\lambda_n;
\vv{p}_1\sigma_1\vv{p}_2\sigma_2)
\end{eqnarray*}

The cross section comes out in this case as
\begin{eqnarray*} \sigma &=& \frac{(2\pi)^4}{2p_1^02p_2^0}\frac{\hbar^2}{v_{\rm rel}}
\int\frac{d^3\vv{k}_1}{2k_1^0}\cdots\frac{d^3\vv{k}_n}{2k_n^0}
\delta^4(P_f-P_i)\times \\
&&|T(\vv{k}_1\lambda_1\dots\vv{k}_n\lambda_n;
\vv{p}_1\sigma_1\vv{p}_2\sigma_2)|^2
\end{eqnarray*}
where the relative velocity of particles in one beam relative to those in the other is
\begin{eqnarray*} v_{\rm rel}=c^2\left|\frac{\vv{p}_1}{p_1^0}-\frac{\vv{p}_2}{p_2^0}\right|
\end{eqnarray*}

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