The Hamiltonian of a spin one-half particle in a magnetic field
Now we would like to discuss a two-state system involving an object of spin one-half. Some of what we will say has been covered in earlier chapters, but doing it again may help to make some of the puzzling points a little clearer. We can think of an electron at rest as a two-state system. Although we will be talking in this section about “an electron,” what we find out will be true for any spin one-half particle. Suppose we choose for our base states $\ketsl{\slOne}$ and $\ketsl{\slTwo}$ the states in which the $z$-component of the electron spin is $+\hbar/2$ and $-\hbar/2$. These states are, of course, the same ones we have called $(+)$ and $(-)$ in earlier chapters. To keep the notation of this chapter consistent, though, we call the “plus” spin state $\ketsl{\slOne}$ and the “minus” spin state $\ketsl{\slTwo}$—where “plus” and “minus” refer to the angular momentum in the $z$-direction. Any possible state $\psi$ for the electron can be described as in Eq. (10.1) by giving the amplitude $C_1$ that the electron is in state $\ketsl{\slOne}$, and the amplitude $C_2$ that it is in state $\ketsl{\slTwo}$. To treat this problem, we will need to know the Hamiltonian for this two-state system—that is, for an electron in a magnetic field. We begin with the special case of a magnetic field in the $z$-direction. Suppose that the vector $\FLPB$ has only a $z$-component $B_z$. From the definition of the two base states (that is, spins parallel and antiparallel to $\FLPB$) we know that they are already stationary states with a definite energy in the magnetic field. State $\ketsl{\slOne}$ corresponds to an energy4 equal to $-\mu B_z$ and state $\ketsl{\slTwo}$ to $+\mu B_z$. The Hamiltonian must be very simple in this case since $C_1$, the amplitude to be in state $\ketsl{\slOne}$, is not affected by $C_2$, and vice versa: \begin{equation} \begin{aligned} i\hbar\,\ddt{C_1}{t}&=E_1C_1=-\mu B_zC_1,\\[2ex] i\hbar\,\ddt{C_2}{t}&=E_2C_2=+\mu B_zC_2. \end{aligned} \label{Eq:III:10:17} \end{equation}  For this special case, the Hamiltonian is \begin{alignat}{2} H_{11}&=-\mu B_z,&\quad H_{12}&=0,\notag\\[2ex] \label{Eq:III:10:18} H_{21}&=0,&\quad H_{22}&=+\mu B_z. \end{alignat}  So we know what the Hamiltonian is for the magnetic field in the $z$-direction, and we know the energies of the stationary states. Now suppose the field is not in the $z$-direction. What is the Hamiltonian? How are the matrix elements changed if the field is not in the $z$-direction? We are going to make an assumption that there is a kind of superposition principle for the terms of the Hamiltonian. More specifically, we want to assume that if two magnetic fields are superposed, the terms in the Hamiltonian simply add—if we know the $H_{ij}$ for a pure $B_z$ and we know the $H_{ij}$ for a pure $B_x$, then the $H_{ij}$ for both $B_z$ and $B_x$ together is simply the sum. This is certainly true if we consider only fields in the $z$-direction—if we double $B_z$, then all the $H_{ij}$ are doubled. So let’s assume that $H$ is linear in the field $\FLPB$. That’s all we need to be able to find the $H_{ij}$ for any magnetic field. Suppose we have a constant field $\FLPB$. We could have chosen our $z$-axis in its direction, and we would have found two stationary states with the energies $\mp\mu B$. Just choosing our axes in a different direction won’t change the physics. Our description of the stationary states will be different, but their energies will still be $\mp\mu B$—that is, \begin{equation} \begin{aligned} E_{\slI}&=-\mu\sqrt{B_x^2+B_y^2+B_z^2},\\[1ex] E_{\slII}&=+\mu\sqrt{B_x^2+B_y^2+B_z^2}. \end{aligned} \label{Eq:III:10:19} \end{equation}  The rest of the game is easy. We have here the formulas for the energies. We want a Hamiltonian which is linear in $B_x$, $B_y$, and $B_z$, and which will give these energies when used in our general formula of Eq. (10.3). The problem: find the Hamiltonian. First, notice that the energy splitting is symmetric, with an average value of zero. Looking at Eq. (10.3), we can see directly that that requires \begin{equation*} H_{22}=-H_{11}. \end{equation*}  (Note that this checks with what we already know when $B_x$ and $B_y$ are both zero; in that case $H_{11}=-\mu B_z$, and $H_{22}=\mu B_z$.) Now if we equate the energies of Eq. (10.3) with what we know from Eq. (10.19), we have \begin{equation} \label{Eq:III:10:20} \biggl(\!\frac{H_{11}\!-H_{22}}{2}\!\biggr)^2\!\!\!+\abs{H_{12}}^2= \mu^2(B_x^2\!+\!B_y^2\!+\!B_z^2). \end{equation}  (We have also made use of the fact that $H_{21}=H_{12}\cconj$, so that $H_{12}H_{21}$ can also be written as $\abs{H_{12}}^2$.) Again for the special case of a field in the $z$-direction, this gives \begin{equation*} \mu^2B_z^2+\abs{H_{12}}^2=\mu^2B_z^2. \end{equation*}  Clearly, $\abs{H_{12}}$ must be zero in this special case, which means that $H_{12}$ cannot have any terms in $B_z$. (Remember, we have said that all terms must be linear in $B_x$, $B_y$, and $B_z$.) So far, then, we have discovered that $H_{11}$ and $H_{22}$ have terms in $B_z$, while $H_{12}$ and $H_{21}$ do not. We can make a simple guess that will satisfy Eq. (10.20) if we say that \begin{align*} H_{11} &=-\mu B_z,\notag\\[2ex] H_{22} &=\mu B_z, \end{align*} and \begin{equation} \label{Eq:III:10:21} \quad\quad\abs{H_{12}}^2 =\mu^2(B_x^2+B_y^2). \end{equation}  And it turns out that that’s the only way it can be done! “Wait”—you say—“$H_{12}$ is not linear in $B$; Eq. (10.21) gives $H_{12}=$$\mu\sqrt{B_x^2+B_y^2}$.” Not necessarily. There is another possibility which is linear, namely, \begin{equation*} H_{12}=\mu(B_x+iB_y). \end{equation*}  There are, in fact, several such possibilities—most generally, we could write \begin{equation*} H_{12}=\mu(B_x\pm iB_y)e^{i\delta}, \end{equation*}  where $\delta$ is some arbitrary phase. Which sign and phase should we use?  It turns out that you can choose either sign, and any phase you want, and the physical results will always be the same. So the choice is a matter of convention. People ahead of us have chosen to use the minus sign and to take $e^{i\delta}=-1$. We might as well follow suit and write \begin{equation*} H_{12}=-\mu(B_x-iB_y),\quad H_{21}=-\mu(B_x+iB_y). \end{equation*}  (Incidentally, these conventions are related to, and consistent with, some of the arbitrary choices we made in Chapter 6.) The complete Hamiltonian for an electron in an arbitrary magnetic field is, then \begin{equation} \begin{alignedat}{2} H_{11}&=-\mu B_z,&\quad H_{12}&=-\mu(B_x-iB_y),\\[1ex] H_{21}&=-\mu(B_x+iB_y),&\quad H_{22}&=+\mu B_z. \end{alignedat} \label{Eq:III:10:22} \end{equation}  And the equations for the amplitudes $C_1$ and $C_2$ are \begin{equation} \begin{aligned} i\hbar\,\ddt{C_1}{t}&=-\mu[B_zC_1+(B_x-iB_y)C_2],\\[1ex] i\hbar\,\ddt{C_2}{t}&=-\mu[(B_x+iB_y)C_1-B_zC_2]. \end{aligned} \label{Eq:III:10:23} \end{equation}  So we have discovered the “equations of motion for the spin states” of an electron in a magnetic field. We guessed at them by making some physical argument, but the real test of any Hamiltonian is that it should give predictions in agreement with experiment. According to any tests that have been made, these equations are right. In fact, although we made our arguments only for constant fields, the Hamiltonian we have written is also right for magnetic fields which vary with time. So we can now use Eq. (10.23) to look at all kinds of interesting problems.