Spin waves
In Chapter 13 we worked out the theory for the propagation of an electron or of some other “particle,” such as an atomic excitation, through a crystal lattice. In the last chapter we applied the theory to semiconductors. But when we talked about situations in which there are many electrons we disregarded any interactions between them. To do this is of course only an approximation. In this chapter we will discuss further the idea that you can disregard the interaction between the electrons. We will also use the opportunity to show you some more applications of the theory of the propagation of particles. Since we will generally continue to disregard the interactions between particles, there is very little really new in this chapter except for the new applications. The first example to be considered is, however, one in which it is possible to write down quite exactly the correct equations when there is more than one “particle” present. From them we will be able to see how the approximation of disregarding the interactions is made. We will not, though, analyze the problem very carefully. As our first example we will consider a “spin wave” in a ferromagnetic crystal. We have discussed the theory of ferromagnetism in Chapter 36 of Volume II. At zero temperature all the electron spins that contribute to the magnetism in the body of a ferromagnetic crystal are parallel. There is an interaction energy between the spins, which is lowest when all the spins are down. At any nonzero temperature, however, there is some chance that some of the spins are turned over. We calculated the probability in an approximate manner in Chapter 36. This time we will describe the quantum mechanical theory—so you will see what you would have to do if you wanted to solve the problem more exactly. (We will still make some idealizations by assuming that the electrons are localized at the atoms and that the spins interact only with neighboring spins.) We consider a model in which the electrons at each atom are all paired except one, so that all of the magnetic effects come from one spin-$\tfrac{1}{2}$ electron per atom. Further, we imagine that these electrons are localized at the atomic sites in the lattice. The model corresponds roughly to metallic nickel.  We also assume that there is an interaction between any two adjacent spinning electrons which gives a term in the energy of the system \begin{equation} \label{Eq:III:15:1} E=-\sum_{i,j}K\FLPsigma_i\cdot\FLPsigma_j, \end{equation}  where $\FLPsigma$’s represent the spins and the summation is over all adjacent pairs of electrons. We have already discussed this kind of interaction energy when we considered the hyperfine splitting of hydrogen due to the interaction of the magnetic moments of the electron and proton in a hydrogen atom. We expressed it then as $A\FLPsigma_{\text{e}}\cdot\FLPsigma_{\text{p}}$. Now, for a given pair, say the electrons at atom $4$ and at atom $5$, the Hamiltonian would be $-K\FLPsigma_4\cdot\FLPsigma_5$. We have a term for each such pair, and the Hamiltonian is (as you would expect for classical energies) the sum of these terms for each interacting pair. The energy is written with the factor $-K$ so that a positive $K$ will correspond to ferromagnetism—that is, the lowest energy results when adjacent spins are parallel. In a real crystal, there may be other terms which are the interactions of next nearest neighbors, and so on, but we don’t need to consider such complications at this stage. With the Hamiltonian of Eq. (15.1) we have a complete description of the ferromagnet—within our approximation—and the properties of the magnetization should come out. We should also be able to calculate the thermodynamic properties due to the magnetization. If we can find all the energy levels, the properties of the crystal at a temperature $T$ can be found from the principle that the probability that a system will be found in a given state of energy $E$ is proportional to $e^{-E/\kappa T}$. This problem has never been completely solved. We will show some of the problems by taking a simple example in which all the atoms are in a line—a one-dimensional lattice. You can easily extend the ideas to three dimensions. At each atomic location there is an electron which has two possible states, either spin up or spin down, and the whole system is described by telling how all of the spins are arranged. We take the Hamiltonian of the system to be the operator of the interaction energy. Interpreting the spin vectors of Eq. (15.1) as the sigma-operators—or the sigma-matrices—we write for the linear lattice \begin{equation} \label{Eq:III:15:2} \Hop=\sum_n-\frac{A}{2}\,\FLPsigmaop_n\cdot\FLPsigmaop_{n+1}. \end{equation}  In this equation we have written the constant as $A/2$ for convenience (so that some of the later equations will be exactly the same as the ones in Chapter 13). Now what is the lowest state of this system? The state of lowest energy is the one in which all the spins are parallel—let’s say, all up.1 We can write this state as $\ket{\dotsb+\,+\,+\,+\dotsb}$, or $\ket{\text{gnd}}$ for the “ground,” or lowest, state. It’s easy to figure out the energy for this state. One way is to write out all the vector sigmas in terms of $\sigmaop_x$, $\sigmaop_y$, and $\sigmaop_z$, and work through carefully what each term of the Hamiltonian does to the ground state, and then add the results. We can, however, also use a good short cut. We saw in Section 12–2, that $\FLPsigmaop_i\cdot\FLPsigmaop_j$ could be written in terms of the Pauli spin exchange operator like this: \begin{equation} \label{Eq:III:15:3} \FLPsigmaop_i\cdot\FLPsigmaop_j=(2\Pop^{\text{spin ex}}_{ij}-1), \end{equation}  where the operator $\Pop^{\text{spin ex}}_{ij}$ interchanges the spins of the $i$th and $j$th electrons. With this substitution the Hamiltonian becomes \begin{equation} \label{Eq:III:15:4} \Hop=-A\sum_n(\Pop^{\text{spin ex}}_{n,n+1}-\tfrac{1}{2}). \end{equation}  It is now easy to work out what happens to different states. For instance if $i$ and $j$ are both up, then exchanging the spins leaves everything unchanged, so $\Pop_{ij}$ acting on the state just gives the same state back, and is equivalent to multiplying by $+1$. The expression $(\Pop_{ij}-\tfrac{1}{2})$ is just equal to one-half. (From now on we will leave off the descriptive superscript on the $\Pop$.)  For the ground state all spins are up; so if you exchange a particular pair of spins, you get back the original state. The ground state is a stationary state. If you operate on it with the Hamiltonian you get the same state again multiplied by a sum of terms, $-(A/2)$ for each pair of spins. That is, the energy of the system in the ground state is $-A/2$ per atom. Next we would like to look at the energies of some of the excited states. It will be convenient to measure the energies with respect to the ground state—that is, to choose the ground state as our zero of energy. We can do that by adding the energy $A/2$ to each term in the Hamiltonian. That just changes the “$\tfrac{1}{2}$” in Eq. (15.4) to “$1$.” Our new Hamiltonian is \begin{equation} \label{Eq:III:15:5} \Hop=-A\sum_n(\Pop_{n,n+1}-1). \end{equation}  With this Hamiltonian the energy of the lowest state is zero; the spin exchange operator is equivalent to multiplying by unity (for the ground state) which is cancelled by the “$1$” in each term. For describing states other than the ground state we will need a suitable set of base states. One convenient approach is to group the states according to whether one electron has spin down, or two electrons have spin down, and so on. There are, of course, many states with one spin down. The down spin could be at atom “4,” or at atom “5,” or at atom “6,” … We can, in fact, choose just such states for our base states. We could write them this way: $\ket{4}$, $\ket{5}$, $\ket{6}$, … It will, however, be more convenient later if we label the “odd atom”—the one with the down-spinning electron—by its coordinate $x$. That is, we’ll define the state $\ket{x_5}$ to be one with all the electrons spinning up except for the one on the atom at $x_5$, which has a down-spinning electron (see Fig. 15–1). In general, $\ket{x_n}$ is the state with one down spin that is located at the coordinate $x_n$ of the $n$th atom. What is the action of the Hamiltonian (15.5) on the state $\ket{x_5}$? One term of the Hamiltonian is say $-A(\Pop_{7,8}-1)$. The operator $\Pop_{7,8}$ exchanges the two spins of the adjacent atoms $7$, $8$. But in the state $\ket{x_5}$ these are both up, and nothing happens; $\Pop_{7,8}$ is equivalent to multiplying by $1$: \begin{equation*} \Pop_{7,8}\,\ket{x_5}=\ket{x_5}. \end{equation*} It follows that \begin{equation*} (\Pop_{7,8}-1)\,\ket{x_5}=0. \end{equation*} Thus all the terms of the Hamiltonian give zero—except those involving atom $5$, of course. On the state $\ket{x_5}$, the operation $\Pop_{4,5}$ exchanges the spin of atom $4$ (up) and atom $5$ (down). The result is the state with all spins up except the atom at $4$. That is \begin{equation*} \Pop_{4,5}\,\ket{x_5}=\ket{x_4}. \end{equation*} In the same way \begin{equation*} \Pop_{5,6}\,\ket{x_5}=\ket{x_6}. \end{equation*}  Hence, the only terms of the Hamiltonian which survive are $-A(\Pop_{4,5}-1)$ and $-A(\Pop_{5,6}-1)$. Acting on $\ket{x_5}$ they produce $-A\,\ket{x_4}+A\,\ket{x_5}$ and $-A\,\ket{x_6}+A\,\ket{x_5}$, respectively. The result is  \begin{equation} \label{Eq:III:15:6} \Hop\,\ket{x_5}=-A\sum_n(\Pop_{n,n+1}-1)\,\ket{x_5}= -A\{\ket{x_6}+\ket{x_4}-2\,\ket{x_5}\}. \end{equation} 
 \begin{align} \Hop\,\ket{x_5}&=-A\sum_n(\Pop_{n,n+1}-1)\,\ket{x_5}\notag\\[.5ex] \label{Eq:III:15:6} &=-A\{\ket{x_6}+\ket{x_4}-2\,\ket{x_5}\}. \end{align} 
 When the Hamiltonian acts on state $\ket{x_5}$ it gives rise to some amplitude to be in states $\ket{x_4}$ and $\ket{x_6}$. That just means that there is a certain amplitude to have the down spin jump over to the next atom. So because of the interaction between the spins, if we begin with one spin down, then there is some probability that at a later time another one will be down instead. Operating on the general state $\ket{x_n}$, the Hamiltonian gives \begin{equation} \label{Eq:III:15:7} \Hop\,\ket{x_n}=-A\{\ket{x_{n+1}}+\ket{x_{n-1}}-2\,\ket{x_n}\}. \end{equation} Notice particularly that if we take a complete set of states with only one spin down, they will only be mixed among themselves. The Hamiltonian will never mix these states with others that have more spins down. So long as you only exchange spins you never change the total number of down spins.  It will be convenient to use the matrix notation for the Hamiltonian, say $H_{n,m}\equiv\bracket{x_n}{\Hop}{x_m}$; Eq. (15.7) is equivalent to \begin{equation} \begin{aligned} H_{n,n}&=2A;\\[1ex] H_{n,n+1}&=H_{n,n-1}=-A;\\[1ex] H_{n,m}&=0,\text{ for }\abs{n-m}>1. \end{aligned} \label{Eq:III:15:8} \end{equation}  Now what are the energy levels for states with one spin down? As usual we let $C_n$ be the amplitude that some state $\ket{\psi}$ is in the state $\ket{x_n}$. If $\ket{\psi}$ is to be a definite energy state, all the $C_n$’s must vary with time in the same way, namely, \begin{equation} \label{Eq:III:15:9} C_n=a_ne^{-iEt/\hbar}. \end{equation} We can put this trial solution into our usual Hamiltonian equation \begin{equation} \label{Eq:III:15:10} i\hbar\,\ddt{C_n}{t}=\sum_mH_{nm}C_m, \end{equation} using Eq. (15.8) for the matrix elements. Of course we get an infinite number of equations, but they can all be written as \begin{equation} \label{Eq:III:15:11} Ea_n=2Aa_n-Aa_{n-1}-Aa_{n+1}. \end{equation} We have again exactly the same problem we worked out in Chapter 13, except that where we had $E_0$ we now have $2A$. The solutions correspond to amplitudes $C_n$ (the down-spin amplitude) which propagate along the lattice with a propagation constant $k$ and an energy \begin{equation} \label{Eq:III:15:12} E=2A(1-\cos kb), \end{equation} where $b$ is the lattice constant.   The definite energy solutions correspond to “waves” of down spin—called “spin waves.” And for each wavelength there is a corresponding energy. For large wavelengths (small $k$) this energy varies as \begin{equation} \label{Eq:III:15:13} E=Ab^2k^2. \end{equation} Just as before, we can consider a localized wave packet (containing, however, only long wavelengths) which corresponds to a spin-down electron in one part of the lattice. This down spin will behave like a “particle.” Because its energy is related to $k$ by (15.13) the “particle” will have an effective mass: \begin{equation} \label{Eq:III:15:14} m_{\text{eff}}=\frac{\hbar^2}{2Ab^2}. \end{equation} These “particles” are sometimes called “magnons.” 