Independent particles
In the last section we wrote down a Hamiltonian, Eq. (15.15), for a two-particle system. Then using an approximation which is equivalent to neglecting any “interaction” of the two particles, we found the stationary states described by Eqs. (15.17) and (15.18). This state is just the product of two single-particle states. The solution we have given for $a_{m,n}$ in Eq. (15.18) is, however, really not satisfactory. We have very carefully pointed out earlier that the state $\ket{x_9,x_4}$ is not a different state from $\ket{x_4,x_9}$—the order of $x_m$ and $x_n$ has no significance. In general, the algebraic expression for the amplitude $C_{m,n}$ must be unchanged if we interchange the values of $x_m$ and $x_n$, since that doesn’t change the state. Either way, it should represent the amplitude to find a down spin at $x_m$ and a down spin at $x_n$. But notice that (15.18) is not symmetric in $x_m$ and $x_n$—since $k_1$ and $k_2$ can in general be different. The trouble is that we have not forced our solution of Eq. (15.15) to satisfy this additional condition. Fortunately it is easy to fix things up. Notice first that a solution of the Hamiltonian equation just as good as (15.18) is \begin{equation} \label{Eq:III:15:23} a_{m,n}=Ke^{ik_2x_m}e^{ik_1x_n}. \end{equation}  It even has the same energy we got for (15.18). Any linear combination of (15.18) and (15.23) is also a good solution, and has an energy still given by Eq. (15.19). The solution we should have chosen—because of our symmetry requirement—is just the sum of (15.18) and (15.23): \begin{equation} \label{Eq:III:15:24} a_{m,n}=K[e^{ik_1x_m}e^{ik_2x_n}+e^{ik_2x_m}e^{ik_1x_n}]. \end{equation}  Now, given any $k_1$ and $k_2$ the amplitude $C_{m,n}$ is independent of which way we put $x_m$ and $x_n$—if we should happen to define $x_m$ and $x_n$ reversed we get the same amplitude. Our interpretation of Eq. (15.24) in terms of “magnons” must also be different. We can no longer say that the equation represents one particle with wave number $k_1$ and a second particle with wave number $k_2$. The amplitude (15.24) represents one state with two particles (magnons). The state is characterized by the two wave numbers $k_1$ and $k_2$. Our solution looks like a compound state of one particle with the momentum $p_1=\hbar k_1$ and another particle with the momentum $p_2=\hbar k_2$, but in our state we can’t say which particle is which. By now, this discussion should remind you of Chapter 4 and our story of identical particles. We have just been showing that the particles of the spin waves—the magnons—behave like identical Bose particles. All amplitudes must be symmetric in the coordinates of the two particles—which is the same as saying that if we “interchange the two particles,” we get back the same amplitude and with the same sign. But, you may be thinking, why did we choose to add the two terms in making Eq. (15.24). Why not subtract? With a minus sign, interchanging $x_m$ and $x_n$ would just change the sign of $a_{m,n}$ which doesn’t matter. But interchanging $x_m$ and $x_n$ doesn’t change anything—all the electrons of the crystal are exactly where they were before, so there is no reason for even the sign of the amplitude to change. The magnons will behave like Bose particles.2 The main points of this discussion have been twofold: First, to show you something about spin waves, and, second, to demonstrate a state whose amplitude is a product of two amplitudes, and whose energy is the sum of the energies corresponding to the two amplitudes. For independent particles the amplitude is the product and the energy is the sum. You can easily see why the energy is the sum. The energy is the coefficient of $t$ in an imaginary exponential—it is proportional to the frequency. If two objects are doing something, one of them with the amplitude $e^{-iE_1t/\hbar}$ and the other with the amplitude $e^{-iE_2t/\hbar}$, and if the amplitude for the two things to happen together is the product of the amplitudes for each, then there is a single frequency in the product which is the sum of the two frequencies. The energy corresponding to the amplitude product is the sum of the two energies. We have gone through a rather long-winded argument to tell you a simple thing. When you don’t take into account any interaction between particles, you can think of each particle independently. They can individually exist in the various different states they would have alone, and they will each contribute the energy they would have had if they were alone. However, you must remember that if they are identical particles, they may behave either as Bose or as Fermi particles depending upon the problem. Two extra electrons added to a crystal, for instance, would have to behave like Fermi particles. When the positions of two electrons are interchanged, the amplitude must reverse sign. In the equation corresponding to Eq. (15.24) there would have to be a minus sign between the two terms on the right. As a consequence, two Fermi particles cannot be in exactly the same condition—with equal spins and equal $k$’s. The amplitude for this state is zero. 