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{\bb}[1]{\langle #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'} =
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]
&=& \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]
&=& \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')
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
&=&\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}_2-\vv{p}_2').\vv{r}_2/\hbar] \times \\
f_2(\vv{p}_2)^*f_2(\vv{p}_2')\times \\
&& \ddd{\vv{p}_1+\vv{p}_2-\vv{p'}_1-\vv{p'}_2}
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}

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