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# QS 14: Non-relativistic colliding beams

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We are now ready to define cross section for colliding beams or, as explained above, beam on target. Two beams with sharp values of momenta $\vv{k}_1$ and $\vv{k}_2$ respectively and other quantum numbers $\gamma_1,\gamma_2$ (resp.) collide and scatter. We are interested in the final states with projection operators
\begin{eqnarray*} P_\xi=\kk{\xi_1\xi_2}\bb{\xi_1\xi_2}|=
\int_\Delta d^3\vv{k}_1'\int d^3\vv{k}_2'
\kk{\vv{k}_1'\gamma_1',\vv{k}_2'\gamma_2'}
\bb{\vv{k}_1'\gamma_1',\vv{k}_2'\gamma_2'}|
\end{eqnarray*}
The number of particles making a transition into these final states is
\begin{eqnarray*} n_{\xi_1\xi_2}=\frac{2}{\hbar}{\rm Im}
\sum_{i,j}\bb{\phi_{1i}\phi_{2j}}|B\kk{\phi_{1i}\phi_{2j}} \end{eqnarray*}
with $B$ as before given by
\begin{eqnarray*} B = {\molp}^\dagger P_\xi V\molp \end{eqnarray*}
The number becomes
\begin{eqnarray*} n_{\xi_1\xi_2}=\rho_1N_2(2\pi\hbar)^3\frac{2}{\hbar}{\rm Im}\,
b(\vv{k}_1\gamma_1,\vv{k}_2\gamma_2;\vv{k}_1\gamma_1,\vv{k}_2\gamma_2)
\end{eqnarray*}
where $b$ is defined by
\begin{eqnarray*} \bb{\vv{k}_1\gamma_1,\vv{k}_2\gamma_2}|B
\kk{\vv{k}_1''\gamma_1,\vv{k}_2''\gamma_2}=
\ddd{\vv{K}-\vv{K}''}
b(\vv{k}_1\gamma_1,\vv{k}_2\gamma_2;\vv{k}_1''\gamma_1,\vv{k}_2''\gamma_2)
\end{eqnarray*}

To calculate this we begin with the general expression
\begin{eqnarray*} \bb{\vv{k}_1\gamma_1,\vv{k}_2\gamma_2}|{\molp}^\dagger P_\xi V\molp
\kk{\vv{k}_1''\gamma_1,\vv{k}_2''\gamma_2}
\end{eqnarray*}
and put $\vv{k}_1''=\vv{k}_1,\vv{k}_2''=\vv{k}_2$ at the end.  Use the Lippmann-Schwinger equation to write matrix element of ${\molp}^\dagger$ in terms of complex conjugate of that of $V\molp$.  There are two momentum conserving delta functions, we can integrate over one by changing to variables
\begin{eqnarray*} \int_\Delta d^3\vv{k}_1'\int d^3\vv{k}_2'
=\int_\Delta d^3\vv{K'}\int d^3\vv{k}' \end{eqnarray*}
where $\vv{K}'=\vv{k}_1'+\vv{k}_2'$ is the total momentum and
$\vv{k}'=(m_1\vv{k}_2'-m_2\vv{k}_1')/(m_1+m_2)$ the relative momentum.
\begin{eqnarray*} E_{\vv{k}_1\vv{k}_2}=\frac{|\vv{k}_1|^2}{2m_1}+\frac{|\vv{k}_2|^2}{2m_2}
=\frac{|\vv{K}|^2}{2M}+\frac{|\vv{k}|^2}{2\mu}\equiv E_{\vv{K}}+e_{\vv{k}}
=E_{\vv{K}\vv{k}}
\end{eqnarray*}
with $M=m_1+m_2$ is the total mass and $\mu=m_1m_2/(m_1+m_2)$ the reduced mass. We get, separating into total and relative momenta,
\begin{eqnarray*} \bb{\vv{p}_1\beta_1,\vv{p}_2\beta_2}|V\molp
\kk{\vv{p'}_1\beta_1',\vv{p'}_2\beta_2'}\equiv
\ddd{\vv{P}-\vv{P'}}
T(\vv{p}\beta_1\beta_2,\vv{p'}\beta_1'\beta_2';\vv{P})
\end{eqnarray*}
Therefore,
\begin{eqnarray*} && \bb{\vv{k}_1\gamma_1,\vv{k}_2\gamma_2}|{\molp}^\dagger P_\xi V\molp
\kk{\vv{k}_1''\gamma_1,\vv{k}_2''\gamma_2} =\\
&& \ddd{\vv{K}-\vv{K}''} \times \\
&& T(\vv{k}'\gamma_1'\gamma_2',\vv{k}\gamma_1\gamma_2;\vv{K})^*
T(\vv{k}'\gamma_1'\gamma_2',\vv{k}''\gamma_1\gamma_2;\vv{K'}) \times \\
&&(E_{\vv{K}\vv{k}}-E_{\vv{K'}\vv{k'}}-i\epsilon)^{-1}.
\end{eqnarray*}
This defines $b$ in which we put $\vv{k}''=\vv{k}$. We can then take the imaginary part, which gives,
\begin{eqnarray*} n_{\xi_1\xi_2} &=& \rho_1N_2(2\pi\hbar)^3\frac{2\pi}{\hbar}\times \\
&&\int_\Delta
d^3\vv{k}'\delta(E_{\vv{K}\vv{k}}-E_{\vv{K'}\vv{k'}})
|T(\vv{k}'\gamma_1'\gamma_2',\vv{k}\gamma_1\gamma_2;\vv{K})|^2
\end{eqnarray*}
Because of $\vv{K}=\vv{K}'$ the energy corresponding to total momentum is already equal, so $E_{\vv{K}\vv{k}}-E_{\vv{K'}\vv{k'}}=e_{\vv{k}}-e_{\vv{k'}}$.  As, in the case of potential scattering, we change
\begin{eqnarray*} d^3\vv{k}=|\vv{k}|^2d|\vv{k}|d\Omega_\vv{k}=
\mu|\vv{k}|de_\vv{k}d\Omega_\vv{k} \end{eqnarray*}
this gives us the formula for cross section which is now defined as rate of transitions {\em per target particle}. The flux is now given by $\rho_1\times{\rm relative\ velocity}=\rho_1 |\vv{k}|/\mu$. We get
\begin{eqnarray}d\sigma =(2\pi)^4\mu^2\hbar^2
|T(\vv{k}'\gamma_1'\gamma_2',\vv{k}\gamma_1\gamma_2;\vv{K})|^2 d\Omega_\vv{k}'
\end{eqnarray}
{\bf This is the same formula as the one for the cross section by a fixed potential, with relative momentum $|\vv{k}|=|\vv{k}'|$ for the particle momentum and the reduced mass in place of particle mass. } In the center of mass frame, we must put $\vv{K}=0$.

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