Two spin waves
Now we would like to discuss what happens if there are two down spins. Again we pick a set of base states. We’ll choose states in which there are down spins at two atomic locations, such as the state shown in Fig. 15–2. We can label such a state by the $x$-coordinates of the two sites with down spins. The one shown can be called $\ket{x_2,x_5}$. In general the base states are $\ket{x_n,x_m}$—a doubly infinite set! In this system of description, the state $\ket{x_4,x_9}$ and the state $\ket{x_9,x_4}$ are exactly the same state, because each simply says that there is a down spin at $4$ and one at $9$; there is no meaning to the order. Furthermore, the state $\ket{x_4,x_4}$ has no meaning, there isn’t such a thing. We can describe any state $\ket{\psi}$ by giving the amplitudes to be in each of the base states. Thus $C_{m,n}=\braket{x_m,x_n}{\psi}$ now means the amplitude for a system in the state $\ket{\psi}$ to be in a state in which both the $m$th and $n$th atoms have a down spin. The complications which now arise are not complications of ideas—they are merely complexities in bookkeeping. (One of the complexities of quantum mechanics is just the bookkeeping. With more and more down spins, the notation becomes more and more elaborate with lots of indices and the equations always look very horrifying; but the ideas are not necessarily more complicated than in the simplest case.)  The equations of motion of the spin system are the differential equations for the $C_{n,m}$. They are \begin{equation} \label{Eq:III:15:15} i\hbar\,\ddt{C_{n,m}}{t}=\sum_{i,j}(H_{nm,ij})C_{ij}. \end{equation} Suppose we want to find the stationary states. As usual, the derivatives with respect to time become $E$ times the amplitudes and the $C_{m,n}$ can be replaced by the coefficients $a_{m,n}$. Next we have to work out carefully the effect of $H$ on a state with spins $m$ and $n$ down. It is not hard to figure out. Suppose for a moment that $m$ and $n$ are far enough apart so that we don’t have to worry about the obvious trouble. The operation of exchange at the location $x_n$ will move the down spin either to the $(n+1)$ or $(n-1)$ atom, and so there’s an amplitude that the present state has come from the state $\ket{x_m,x_{n+1}}$ and also an amplitude that it has come from the state $\ket{x_m,x_{n-1}}$. Or it may have been the other spin that moved; so there’s a certain amplitude that $C_{m,n}$ is fed from $C_{m+1,n}$ or from $C_{m-1,n}$. These effects should all be equal. The final result for the Hamiltonian equation on $C_{m,n}$ is  \begin{equation} \label{Eq:III:15:16} Ea_{m,n}=-A(a_{m+1,n}+a_{m-1,n}+a_{m,n+1}+a_{m,n-1})+4Aa_{m,n}. \end{equation} 
  \begin{gather} \label{Eq:III:15:16} Ea_{m,n}=\\ -A(a_{m+1,n}\!+a_{m-1,n}\!+a_{m,n+1}\!+a_{m,n-1})\!+\!4Aa_{m,n}.\notag \end{gather}  
 This equation is correct except in two situations. If $m=n$ there is no equation at all, and if $m=n\pm1$, then two of the terms in Eq. (15.16) should be missing. We are going to disregard these exceptions. We simply ignore the fact that some few of these equations are slightly altered. After all, the crystal is supposed to be infinite, and we have an infinite number of terms; neglecting a few might not matter much. So for a first rough approximation let’s forget about the altered equations. In other words, we assume that Eq. (15.16) is true for all $m$ and $n$, even for $m$ and $n$ next to each other. This is the essential part of our approximation. Then the solution is not hard to find. We get immediately \begin{equation} \label{Eq:III:15:17} C_{m,n}=a_{m,n}e^{-iEt/\hbar}, \end{equation} with \begin{equation} \label{Eq:III:15:18} a_{m,n}=(\text{const.})\,e^{ik_1x_m}e^{ik_2x_n}, \end{equation} where \begin{equation} \label{Eq:III:15:19} E=4A-2A\cos k_1b-2A\cos k_2b. \end{equation}  Think for a moment what would happen if we had two independent, single spin waves (as in the previous section) corresponding to $k=k_1$ and $k=k_2$; they would have energies, from Eq. (15.12), of \begin{equation*} E_1 =(2A-2A\cos k_1b) \end{equation*} and \begin{equation*} E_2 =(2A-2A\cos k_2b). \end{equation*} Notice that the energy $E$ in Eq. (15.19) is just their sum, \begin{equation} \label{Eq:III:15:20} E=E_1+E_2. \end{equation} In other words we can think of our solution in this way. There are two particles—that is, two spin waves. One of them has a momentum described by $k_1$, the other by $k_2$, and the energy of the system is the sum of the energies of the two objects. The two particles act completely independently. That’s all there is to it. Of course we have made some approximations, but we do not wish to discuss the precision of our answer at this point. However, you might guess that in a reasonable size crystal with billions of atoms—and, therefore, billions of terms in the Hamiltonian—leaving out a few terms wouldn’t make much of an error. If we had so many down spins that there was an appreciable density, then we would certainly have to worry about the corrections. [Interestingly enough, an exact solution can be written down if there are just the two down spins. The result is not particularly important. But it is interesting that the equations can be solved exactly for this case. The solution is: \begin{equation} \label{Eq:III:15:21} a_{m,n}=\exp[ik_c(x_m+x_n)]\sin(k\abs{x_m-x_n}), \end{equation} with the energy \begin{equation} E=4A-2A\cos k_1b-2A\cos k_2b,\notag \end{equation} and with the wave numbers $k_c$ and $k$ related to $k_1$ and $k_2$ by \begin{equation} \label{Eq:III:15:22} k_1=k_c-k,\quad k_2=k_c+k. \end{equation} This solution includes the “interaction” of the two spins. It describes the fact that when the spins come together there is a certain chance of scattering. The spins act very much like particles with an interaction. But the detailed theory of their scattering goes beyond what we want to talk about here.]