Scattering amplitudes and bound states
Finally, our example can be used to illustrate a point which is very useful these days in the physics of high-energy particles. It has to do with a relationship between scattering amplitudes and bound states. Suppose we have discovered—through experiment and theoretical analysis—the way that pions scatter from protons. Then a new particle is discovered and someone wonders whether maybe it is just a combination of a pion and a proton held together in some bound state (in an analogy to the way an electron is bound to a proton to make a hydrogen atom). By a bound state we mean a combination which has a lower energy than the two free-particles. There is a general theory which says that a bound state will exist at that energy at which the scattering amplitude becomes infinite if extrapolated algebraically (the mathematical term is “analytically continued”) to energy regions outside of the permitted band. The physical reason for this is as follows. A bound state is a situation in which there are only waves tied on to a point and there’s no wave coming in to get it started, it just exists there by itself. The relative proportion between the so-called “scattered” or created wave and the wave being “sent in” is infinite. We can test this idea in our example. Let’s write our expression Eq. (13.37) for the scattered amplitude directly in terms of the energy $E$ of the particle being scattered (instead of in terms of $k$). Since Equation (13.30) can be rewritten as \begin{equation*} 2A\sin kb=\sqrt{4A^2-(E-E_0)^2}, \end{equation*}  the scattered amplitude is \begin{equation} \label{Eq:III:13:44} \beta=\frac{-F}{F-i\sqrt{4A^2-(E-E_0)^2}}. \end{equation}  From our derivation, this equation should be used only for real states—those with energies in the energy band, $E=E_0\pm2A$. But suppose we forget that fact and extend the formula into the “unphysical” energy regions where $\abs{E-E_0}>2A$. For these unphysical regions we can write2 \begin{equation*} \sqrt{4A^2-(E-E_0)^2}=i\sqrt{(E-E_0)^2-4A^2}. \end{equation*}  Then the “scattering amplitude,” whatever it may mean, is \begin{equation} \label{Eq:III:13:45} \beta=\frac{-F}{F+\sqrt{(E-E_0)^2-4A^2}}. \end{equation}  Now we ask: Is there any energy $E$ for which $\beta$ becomes infinite (i.e., for which the expression for $\beta$ has a “pole”)? Yes, so long as $F$ is negative, the denominator of Eq. (13.45) will be zero when \begin{equation*} (E-E_0)^2-4A^2=F^2, \end{equation*}  or when \begin{equation*} E=E_0\pm\sqrt{4A^2+F^2}. \end{equation*} The minus sign gives just the energy we found in Eq. (13.43) for the trapped energy. What about the plus sign? This gives an energy above the allowed energy band. And indeed there is another bound state there which we missed when we solved the equations of Eq. (13.28). We leave it as a puzzle for you to find the energy and amplitudes $a_n$ for this bound state. The relation between scattering and bound states provides one of the most useful clues in the current search for an understanding of the experimental observations about the new strange particles.