# Absorption of pi mesons on deuterium nuclei

In the process we have

$$\begin{equation}\pi^{-}+d \longrightarrow n+n\end{equation}$$

and by the parity conservation

$$\begin{equation}\eta_{\pi} \eta_{d}(-1)^{\ell_{i}}=\eta_{n} \eta_{n}(-1)^{\ell_{f}}\end{equation}$$

so

$$\begin{equation}\eta_{\pi}=(-1)^{\ell_{j}-\ell_{i}}\end{equation}$$

usually in some books we can find that $$l_i=0$$. Could somebody explain why $$l_i=0$$? I have read books where they explain that this is because the pion is stationary and others where they say that the pion moves slowly. Can this process take place if the pion is not stopped? Does this happen naturally? Where does this absorbed pion come from?

The Deuteron is measured to have total spin $$I=1$$ and positive parity. Therefore the allowed values for its total angular momentum are $$l=0$$or $$l=2$$, s or d states respectively. The wave function can be written as $$\psi=a_s \psi_s + a_d \psi_d$$ with $$a_s^2\simeq 96\%$$. Therefore in most of the experiments it is assumed to have a s deuteron state.
• i know that $\eta_d=\eta_{p} \eta_{n}(-1)^{\ell_\text{deuterium}}$ where $\ell_\text{deuterium}=0,2$ in the deuterium, but i don't know in the case of the initial process $\eta_{\pi} \eta_{d}(-1)^{\ell_{i}}$ if it is correct suppose that $l_i=\ell_\text{deuterium}$ – Elpis Aug 25 at 18:12
• Let’s put in this way. In a first part of the experiment the pion is captured by the deuteron forming a “medic-deuteron” state. Then the meson goes to the lower (which is an l=0) level emitting a photon. Then the reaction occurs. In fact the total reaction experimentally studied Is $\pi^- + d\rightarrow n + n + \gamma$. In the end the reaction studied has $l_i=l_d$ – Lox Aug 25 at 19:33