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# QS 13: Momentum conservation : colliding beams

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

We now consider the case of a beam of particles colliding with a a bunch of particles more or less stationary (when it is called a {\bf target} and we say we are in the  Lab frame") or colliding with another beam as happens when we are in the center of mass frame".

The free momentum states basis in this case can be chosen to be normalised as
\begin{eqnarray*} \bb{\vv{p}_1\beta_1,\vv{p}_2\beta_2}
\kk{\vv{p'}_1\beta_1',\vv{p'}_2\beta_2'} =
\ddd{\vv{p}_1-\vv{p'}_1}\dd{\beta_1\beta'_1}
\ddd{\vv{p}_2-\vv{p'}_2}\dd{\beta_2\beta'_2}
\end{eqnarray*}
where the subscript 1 refers to beam particles and 2 to target particles.  Of course, going to a different frame of reference their roles may be switched, or they become two colliding beams so the names beam' and target' are more for convenience.

Let $\phi_{1i}\tp\phi_{2j}$ refer to the states of the (free) beam plus target bunches, there being $N_1$ particles in the beam and $N_2$ in the target. Let the corresponding densities (number of particles per unit volume) be $rho_1$ and $\rho_2$ respectively. The states $\phi_{1i}$ differ from each other by just the space translation by $\vv{r}_i$ of the identical state $\phi_1$ which has the sharp momentum $\vv{k}_1$ and a fixed value of $\gamma_1$
\begin{eqnarray*} \bb{\vv{p}_1\beta_1}\kk{\phi_{1i}} &=& \exp[-i\vv{p}_1.\vv{r}_i/\hbar]
\bb{\vv{p}_1\beta_1}\kk{\phi_{1}}\\
&=& \exp[-i\vv{p}_1.\vv{r}_i/\hbar]
f_1(\vv{p}_1)\dd{\beta_1\gamma_1} \end{eqnarray*}
Similarly, the states $\phi_{2j}$ are space translates of $\phi_2$ (characterized by momentum $\vv{k}_2$ and $\gamma_2$)  by positions $\vv{r}_j$,
\begin{eqnarray*} \bb{\vv{p}_2\beta_2}\kk{\phi_{2i}} &=& \exp[-i\vv{p}_2.\vv{r}_j/\hbar]
\bb{\vv{p}_2\beta_2}\kk{\phi_{2}}\\
&=& \exp[-i\vv{p}_2.\vv{r}_j/\hbar]
f_2(\vv{p}_2)\dd{\beta_2\gamma_2}. \end{eqnarray*}

The quantity whose expectation value is to be measured on these bunches is $B$ {\em which has the property that it conserves, that is, commutes with total linear momentum},
\begin{eqnarray*} \bb{\vv{p}_1\beta_1,\vv{p}_2\beta_2}|B
\kk{\vv{p'}_1\beta_1',\vv{p'}_2\beta_2'} &=&
\ddd{\vv{p}_1+\vv{p}_2-\vv{p'}_1-\vv{p'}_2}\times \\
&& b(\vv{p}_1\beta_1,\vv{p}_2\beta_2;\vv{p'}_1\beta_1',\vv{p'}_2\beta_2')
\end{eqnarray*}
The average value over the colliding bunch is, repeating the argument in a previous section (plane wave=beam of particles?) we get
\begin{eqnarray*} \langle B\rangle&=&\frac{1}{N_1N_2}\sum_{i,j}\bb{\phi_{1i}\phi_{2j}}|B
\kk{\phi_{1i}\phi_{2j}}\\
&=&\frac{\rho_1\rho_2}{N_1N_2}\int d^3\vv{r}_1\int d^3\vv{r}_2
\int d^3\vv{p}_1\int d^3\vv{p}_2
\int d^3\vv{p}_1'\int d^3\vv{p}_2'\\
&&\exp[i(\vv{p}_1-\vv{p}_1').\vv{r}_1/\hbar]
\exp[i(\vv{p}_2-\vv{p}_2').\vv{r}_2/\hbar] \times \\
&&f_1(\vv{p}_1)^*f_1(\vv{p}_1')
f_2(\vv{p}_2)^*f_2(\vv{p}_2')\times \\
&& \ddd{\vv{p}_1+\vv{p}_2-\vv{p'}_1-\vv{p'}_2}
b(\vv{p}_1\gamma_1,\vv{p}_2\gamma_2;
\vv{p'}_1\gamma_1,\vv{p'}_2\gamma_2).
\end{eqnarray*}
Now integration over $\vv{r}_1$ gives $(2\pi\hbar)^3\ddd{\vv{p}_1-\vv{p}_1'}$ which removes one integration on $\vv{p}_1'$. However, the momentum conserving delta function then puts $\vv{p}_2=\vv{p}_2'$. Therefore the $\vv{r}_2$ integration is vacuous and $\rho_2\int d^3\vv{r}_2=N_2$. Due to assumed sharp peaks in $f_1$ and $\f_2$ at $\vv{k}_1$ and $\vv{k}_2$ respectively, and assuming the absence of sharp peak at these values in $b$, we get
\begin{eqnarray}\langle B\rangle=\frac{\rho_1(2\pi\hbar)^3}{N_1}
b(\vv{k}_1\gamma_1,\vv{k}_2\gamma_2;\vv{k}_1\gamma_1,\vv{k}_2\gamma_2).
\end{eqnarray}

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