The spin matrices as operators
While we are on the subject of mathematical notation, we would like to describe still another way of writing things—a way which is used very often because it is so compact. It follows directly from the notation introduced in Chapter 8. If we have a system in a state $\ket{\psi(t)}$, which varies with time, we can—as we did in Eq. (8.34)—write the amplitude that the system would be in the state $\ket{i}$ at $t+\Delta t$ as \begin{equation*} \braket{i}{\psi(t+\Delta t)}=\sum_j \bracket{i}{U(t+\Delta t,t)}{j} \braket{j}{\psi(t)} \end{equation*} The matrix element $\bracket{i}{U(t+\Delta t,t)}{j}$ is the amplitude that the base state $\ket{j}$ will be converted into the base state $\ket{i}$ in the time interval $\Delta t$. We then defined $H_{ij}$ by writing \begin{equation*} \bracket{i}{U(t+\Delta t,t)}{j}=\delta_{ij}-\frac{i}{\hbar}\, H_{ij}(t)\,\Delta t, \end{equation*} and we showed that the amplitudes $C_i(t)=\braket{i}{\psi(t)}$ were related by the differential equations \begin{equation} \label{Eq:III:11:15} i\hbar\,\ddt{C_i}{t}=\sum_jH_{ij}C_j. \end{equation} If we write out the amplitudes $C_i$ explicitly, the same equation appears as \begin{equation} \label{Eq:III:11:16} i\hbar\,\ddt{}{t}\,\braket{i}{\psi}=\sum_jH_{ij}\braket{j}{\psi}. \end{equation} Now the matrix elements $H_{ij}$ are also amplitudes which we can write as $\bracket{i}{H}{j}$; our differential equation looks like this: \begin{equation} \label{Eq:III:11:17} i\hbar\,\ddt{}{t}\,\braket{i}{\psi}= \sum_j\bracket{i}{H}{j}\braket{j}{\psi}. \end{equation} We see that $-i/\hbar\,\bracket{i}{H}{j}\,dt$ is the amplitude that—under the physical conditions described by $H$—a state $\ket{j}$ will, during the time $dt$, “generate” the state $\ket{i}$. (All of this is implicit in the discussion of Section 8–4.)  Now following the ideas of Section 8–2, we can drop out the common term $\bra{i}$ in Eq. (11.17)—since it is true for any state $\ket{i}$—and write that equation simply as \begin{equation} \label{Eq:III:11:18} i\hbar\,\ddt{}{t}\,\ket{\psi}= \sum_jH\,\ket{j}\braket{j}{\psi}. \end{equation} Or, going one step further, we can also remove the $j$ and write \begin{equation} \label{Eq:III:11:19} i\hbar\,\ddt{}{t}\,\ket{\psi}=H\,\ket{\psi}. \end{equation} In Chapter 8 we pointed out that when things are written this way, the $H$ in $H\,\ket{j}$ or $H\,\ket{\psi}$ is called an operator. From now on we will put the little hat ($\op{\enspace}$) over an operator to remind you that it is an operator and not just a number. We will write $\Hop\,\ket{\psi}$. Although the two equations (11.18) and (11.19) mean exactly the same thing as Eq. (11.17) or Eq. (11.15), we can think about them in a different way. For instance, we would describe Eq. (11.18) in this way: “The time derivative of the state vector $\ket{\psi}$ times $i\hbar$ is equal to what you get by operating with the Hamiltonian operator $\Hop$ on each base state, multiplying by the amplitude $\braket{j}{\psi}$ that $\psi$ is in the state $j$, and summing over all $j$.”  Or Eq. (11.19) is described this way. “The time derivative (times $i\hbar$) of a state $\ket{\psi}$ is equal to what you get if you operate with the Hamiltonian $\Hop$ on the state vector $\ket{\psi}$.” It’s just a shorthand way of saying what is in Eq. (11.17), but, as you will see, it can be a great convenience.  If we wish, we can carry the “abstraction” idea one more step. Equation (11.19) is true for any state $\ket{\psi}$. Also the left-hand side, $i\hbar\,d/dt$, is also an operator—it’s the operation “differentiate by $t$ and multiply by $i\hbar$.”  Therefore, Eq. (11.19) can also be thought of as an equation between operators—the operator equation \begin{equation*} i\hbar\,\ddt{}{t}=\Hop. \end{equation*} The Hamiltonian operator (within a constant) produces the same result as does $d/dt$ when acting on any state. Remember that this equation—as well as Eq. (11.19)—is not a statement that the $\Hop$ operator is just the identical operation as $i\hbar\,d/dt$. The equations are the dynamical law of nature—the law of motion—for a quantum system.  Just to get some practice with these ideas, we will show you another way we could get to Eq. (11.18). You know that we can write any state $\ket{\psi}$ in terms of its projections into some base set [see Eq. (8.8)], \begin{equation} \label{Eq:III:11:20} \ket{\psi}=\sum_i\ket{i}\braket{i}{\psi}. \end{equation} How does $\ket{\psi}$ change with time? Well, just take its derivative: \begin{equation} \label{Eq:III:11:21} \ddt{}{t}\,\ket{\psi}=\ddt{}{t}\sum_i\ket{i}\braket{i}{\psi}. \end{equation} Now the base states $\ket{i}$ do not change with time (at least we are always taking them as definite fixed states), but the amplitudes $\braket{i}{\psi}$ are numbers which may vary. So Eq. (11.21) becomes \begin{equation} \label{Eq:III:11:22} \ddt{}{t}\,\ket{\psi}=\sum_i\ket{i}\,\ddt{}{t}\,\braket{i}{\psi}. \end{equation} Since we know $d\braket{i}{\psi}/dt$ from Eq. (11.16), we get  \begin{align*} \ddt{}{t}\,\ket{\psi}&=-\frac{i}{\hbar} \sum_i\ket{i}\sum_jH_{ij}\braket{j}{\psi}\\[1ex] &=-\frac{i}{\hbar}\sum_{ij}\ket{i}\bracket{i}{H}{j}\braket{j}{\psi}= -\frac{i}{\hbar}\sum_jH\,\ket{j}\braket{j}{\psi}. \end{align*} 
 \begin{align*} \ddt{}{t}\,\ket{\psi}&=-\frac{i}{\hbar} \sum_i\ket{i}\sum_jH_{ij}\braket{j}{\psi}\\[1ex] &=-\frac{i}{\hbar}\sum_{ij}\ket{i}\bracket{i}{H}{j}\braket{j}{\psi}\\[1ex] &=-\frac{i}{\hbar}\sum_jH\,\ket{j}\braket{j}{\psi}. \end{align*}  This is Eq. (11.18) all over again.  So we have many ways of looking at the Hamiltonian. We can think of the set of coefficients $H_{ij}$ as just a bunch of numbers, or we can think of the “amplitudes” $\bracket{i}{H}{j}$, or we can think of the “matrix” $H_{ij}$, or we can think of the “operator” $\Hop$. It all means the same thing.  Now let’s go back to our two-state systems. If we write the Hamiltonian in terms of the sigma matrices (with suitable numerical coefficients like $B_x$, etc.), we can clearly also think of $\sigma_{ij}^x$ as an amplitude $\bracket{i}{\sigma_x}{j}$ or, for short, as the operator $\sigmaop_x$. If we use the operator idea, we can write the equation of motion of a state $\ket{\psi}$ in a magnetic field as \begin{equation} \label{Eq:III:11:23} i\hbar\,\ddt{}{t}\,\ket{\psi}= -\mu(B_x\sigmaop_x+B_y\sigmaop_y+B_z\sigmaop_z)\,\ket{\psi}. \end{equation} When we want to “use” such an equation we will normally have to express $\ket{\psi}$ in terms of base vectors (just as we have to find the components of space vectors when we want specific numbers). So we will usually want to put Eq. (11.23) in the somewhat expanded form: \begin{equation} \label{Eq:III:11:24} i\hbar\ddt{}{t}\ket{\psi}\!=\!-\mu\! \sum_i(B_x\sigmaop_x\!+\!B_y\sigmaop_y\!+\!B_z\sigmaop_z)\ket{i} \braket{i}{\psi}. \end{equation}  Now you will see why the operator idea is so neat. To use Eq. (11.24) we need to know what happens when the $\sigmaop$ operators work on each of the base states. Let’s find out. Suppose we have $\sigmaop_z\,\ket{+}$; it is some vector $\ket{?}$, but what? Well, let’s multiply it on the left by $\bra{+}$; we have \begin{equation*} \bracket{+}{\sigmaop_z}{+}=\sigma_{11}^z=1 \end{equation*} (using Table 11–1). So we know that \begin{equation} \label{Eq:III:11:25} \braket{+}{?}=1. \end{equation} Now let’s multiply $\sigmaop_z\,\ket{+}$ on the left by $\bra{-}$. We get \begin{equation*} \bracket{-}{\sigmaop_z}{+}=\sigma_{21}^z=0; \end{equation*} so \begin{equation} \label{Eq:III:11:26} \braket{-}{?}=0. \end{equation} There is only one state vector that satisfies both (11.25) and (11.26); it is $\ket{+}$. We discover then that \begin{equation} \label{Eq:III:11:27} \sigmaop_z\,\ket{+}=\ket{+}. \end{equation} By this kind of argument you can easily show that all of the properties of the sigma matrices can be described in the operator notation by the set of rules given in Table 11–3.  If we have products of sigma matrices, they go over into products of operators. When two operators appear together as a product, you carry out first the operation with the operator which is farthest to the right. For instance, by $\sigmaop_x\sigmaop_y\,\ket{+}$ we are to understand $\sigmaop_x(\sigmaop_y\,\ket{+})$. From Table 11–3, we get $\sigmaop_y\,\ket{+}=i\,\ket{-}$, so \begin{equation} \label{Eq:III:11:28} \sigmaop_x\sigmaop_y\,\ket{+}=\sigmaop_x(i\,\ket{-}). \end{equation} Now any number—like $i$—just moves through an operator (operators work only on state vectors); so Eq. (11.28) is the same as \begin{equation*} \sigmaop_x\sigmaop_y\,\ket{+}=i\sigmaop_x\,\ket{-}=i\,\ket{+}. \end{equation*} If you do the same thing for $\sigmaop_x\sigmaop_y\,\ket{-}$, you will find that \begin{equation*} \sigmaop_x\sigmaop_y\,\ket{-}=-i\,\ket{-}. \end{equation*} Looking at Table 11–3, you see that $\sigmaop_x\sigmaop_y$ operating on $\ket{+}$ or $\ket{-}$ gives just what you get if you operate with $\sigmaop_z$ and multiply by $i$. We can, therefore, say that the operation $\sigmaop_x\sigmaop_y$ is identical with the operation $i\sigmaop_z$ and write this statement as an operator equation: \begin{equation} \label{Eq:III:11:29} \sigmaop_x\sigmaop_y=i\sigmaop_z. \end{equation} Notice that this equation is identical with one of our matrix equations of Table 11–2. So again we see the correspondence between the matrix and operator points of view. Each of the equations in Table 11–2 can, therefore, also be considered as equations about the sigma operators. You can check that they do indeed follow from Table 11–3. It is best, when working with these things, not to keep track of whether a quantity like $\sigma$ or $H$ is an operator or a matrix. All the equations are the same either way, so Table 11–2 is for sigma operators, or for sigma matrices, as you wish. 