The Pauli spin matrices
We continue our discussion of two-state systems. At the end of the last chapter we were talking about a spin one-half particle in a magnetic field. We described the spin state by giving the amplitude $C_1$ that the $z$-component of spin angular momentum is $+\hbar/2$ and the amplitude $C_2$ that it is $-\hbar/2$. In earlier chapters we have called these base states $\ket{+}$ and $\ket{-}$. We will now go back to that notation, although we may occasionally find it convenient to use $\ket{+}$ or $\ketsl{\slOne}$, and $\ket{-}$ or $\ketsl{\slTwo}$, interchangeably.  We saw in the last chapter that when a spin one-half particle with a magnetic moment $\mu$ is in a magnetic field $\FLPB=(B_x,B_y,B_z)$, the amplitudes $C_+$ ($=C_1$) and $C_-$ ($=C_2$) are connected by the following differential equations: \begin{equation} \begin{aligned} i\hbar\,\ddt{C_+}{t}&=-\mu[B_zC_+\!+(B_x\!-iB_y)C_-],\\[2ex] i\hbar\,\ddt{C_-}{t}&=-\mu[(B_x\!+iB_y)C_+\!-\!B_zC_-]. \end{aligned} \label{Eq:III:11:1} \end{equation} In other words, the Hamiltonian matrix $H_{ij}$ is \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:11:2} \end{equation} And Eqs. (11.1) are, of course, the same as \begin{equation} \label{Eq:III:11:3} i\hbar\,\ddt{C_i}{t}=\sum_jH_{ij}C_{j}, \end{equation} where $i$ and $j$ take on the values $+$ and $-$ (or $1$ and $2$).  The two-state system of the electron spin is so important that it is very useful to have a neater way of writing things. We will now make a little mathematical digression to show you how people usually write the equations of a two-state system. It is done this way: First, note that each term in the Hamiltonian is proportional to $\mu$ and to some component of $\FLPB$; we can then—purely formally—write that \begin{equation} \label{Eq:III:11:4} H_{ij}=-\mu[\sigma_{ij}^xB_x+\sigma_{ij}^yB_y+\sigma_{ij}^zB_z]. \end{equation} There is no new physics here; this equation just means that the coefficients $\sigma_{ij}^x$, $\sigma_{ij}^y$, and $\sigma_{ij}^z$—there are $4\times3=12$ of them—can be figured out so that (11.4) is identical with (11.2).  Let’s see what they have to be. We start with $B_z$. Since $B_z$ appears only in $H_{11}$ and $H_{22}$, everything will be O.K. if \begin{alignat*}{2} \sigma_{11}^z&=1,&\quad \sigma_{12}^z&=0,\\[2ex] \sigma_{21}^z&=0,&\quad \sigma_{22}^z&=-1. \end{alignat*} We often write the matrix $H_{ij}$ as a little table like this: \begin{equation*} H_{ij}= \!\!\!\raise 10 pt {\scriptstyle i\downarrow \,\raise 10 pt \scriptstyle j\rightarrow}\kern -13pt % ebook remove % ebook insert: \raise{12pt}{\scriptstyle i\downarrow \,\raise{16pt}{\scriptstyle j\rightarrow}\kern -20pt}} \begin{pmatrix} H_{11} & H_{12}\\[1ex] H_{21} & H_{22} \end{pmatrix}. \end{equation*}  For the Hamiltonian of a spin one-half particle in the magnetic field $B_z$, this is the same as \begin{equation*} H_{ij}= \!\!\!\raise 12 pt {\scriptstyle i\downarrow \,\raise 7 pt \scriptstyle j\rightarrow}\kern -13pt % ebook remove % ebook insert: \raise{12pt}{\scriptstyle i\downarrow \,\raise{16pt}{\scriptstyle j\rightarrow}\kern -20pt} \begin{pmatrix} -\mu B_z & -\mu(B_x-iB_y)\\[1ex] -\mu(B_x+iB_y) & +\mu B_z \end{pmatrix}. \end{equation*} In the same way, we can write the coefficients $\sigma_{ij}^z$ as the matrix \begin{equation} \label{Eq:III:11:5} \sigma_{ij}^z= \raise 10 pt {\scriptstyle i\downarrow \,\raise 7 pt \scriptstyle j\rightarrow}\kern -13pt % ebook remove % ebook insert: \raise{12pt}{\scriptstyle i\downarrow \,\raise{12pt}{\scriptstyle j\rightarrow}\kern -20pt} \begin{pmatrix} 1 & \phantom{-}0\\ 0 & -1 \end{pmatrix}. \end{equation}  Working with the coefficients of $B_x$, we get that the terms of $\sigma_x$ have to be \begin{equation*} \begin{alignedat}{2} \sigma_{11}^x&=0,&\quad \sigma_{12}^x&=1,\\[2ex] \sigma_{21}^x&=1,&\quad \sigma_{22}^x&=0. \end{alignedat} \end{equation*} Or, in shorthand, \begin{equation} \label{Eq:III:11:6} \sigma_{ij}^x= \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}. \end{equation}  Finally, looking at $B_y$, we get \begin{equation*} \begin{alignedat}{2} \sigma_{11}^y&=0,&\quad \sigma_{12}^y&=-i,\\[2ex] \sigma_{21}^y&=i,&\quad \sigma_{22}^y&=0; \end{alignedat} \end{equation*} or \begin{equation} \label{Eq:III:11:7} \sigma_{ij}^y= \begin{pmatrix} 0 & -i\\ i & \phantom{-}0 \end{pmatrix}. \end{equation} With these three sigma matrices, Eqs. (11.2) and (11.4) are identical. To leave room for the subscripts $i$ and $j$, we have shown which $\sigma$ goes with which component of $\FLPB$ by putting $x$, $y$, and $z$ as superscripts. Usually, however, the $i$ and $j$ are omitted—it’s easy to imagine they are there—and the $x$, $y$, $z$ are written as subscripts. Then Eq. (11.4) is written \begin{equation} \label{Eq:III:11:8} H=-\mu[\sigma_xB_x+\sigma_yB_y+\sigma_zB_z]. \end{equation} Because the sigma matrices are so important—they are used all the time by the professionals—we have gathered them together in Table 11–1. (Anyone who is going to work in quantum physics really has to memorize them.) They are also called the Pauli spin matrices after the physicist who invented them.  In the table we have included one more two-by-two matrix which is needed if we want to be able to take care of a system which has two spin states of the same energy, or if we want to choose a different zero energy. For such situations we must add $E_0C_+$ to the first equation in (11.1) and $E_0C_-$ to the second equation. We can include this in the new notation if we define the unit matrix “$1$” as $\delta_{ij}$, \begin{equation} \label{Eq:III:11:9} 1=\delta_{ij}= \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix}, \end{equation} and rewrite Eq. (11.8) as \begin{equation} \label{Eq:III:11:10} H=E_0\delta_{ij}-\mu(\sigma_xB_x+\sigma_yB_y+\sigma_zB_z). \end{equation} Usually, it is understood that any constant like $E_0$ is automatically to be multiplied by the unit matrix; then one writes simply \begin{equation} \label{Eq:III:11:11} H=E_0-\mu(\sigma_xB_x+\sigma_yB_y+\sigma_zB_z). \end{equation}  One reason the spin matrices are useful is that any two-by-two matrix at all can be written in terms of them. Any matrix you can write has four numbers in it, say, \begin{equation*} M= \begin{pmatrix} a & b\\ c & d \end{pmatrix}. \end{equation*} It can always be written as a linear combination of four matrices. For example, \begin{equation*} M=a\! \begin{pmatrix} 1&0\\ 0&0 \end{pmatrix} \!+b\! \begin{pmatrix} 0&1\\ 0&0 \end{pmatrix} \!+c\! \begin{pmatrix} 0&0\\ 1&0 \end{pmatrix} \!+d\! \begin{pmatrix} 0&0\\ 0&1 \end{pmatrix}\!. \end{equation*} There are many ways of doing it, but one special way is to say that $M$ is a certain amount of $\sigma_x$, plus a certain amount of $\sigma_y$, and so on, like this: \begin{equation*} M=\alpha1+\beta\sigma_x+\gamma\sigma_y+\delta\sigma_z, \end{equation*} where the “amounts” $\alpha$, $\beta$, $\gamma$, and $\delta$ may, in general, be complex numbers.  Since any two-by-two matrix can be represented in terms of the unit matrix and the sigma matrices, we have all that we ever need for any two-state system. No matter what the two-state system—the ammonia molecule, the magenta dye, anything—the Hamiltonian equation can be written in terms of the sigmas. Although the sigmas seem to have a geometrical significance in the physical situation of an electron in a magnetic field, they can also be thought of as just useful matrices, which can be used for any two-state problem.  For instance, in one way of looking at things a proton and a neutron can be thought of as the same particle in either of two states. We say the nucleon (proton or neutron) is a two-state system—in this case, two states with respect to its charge. When looked at that way, the $\ketsl{\slOne}$ state can represent the proton and the $\ketsl{\slTwo}$ state can represent the neutron. People say that the nucleon has two “isotopic-spin” states.  Since we will be using the sigma matrices as the “arithmetic” of the quantum mechanics of two-state systems, let’s review quickly the conventions of matrix algebra. By the “sum” of any two or more matrices we mean just what was obvious in Eq. (11.4). In general, if we “add” two matrices $A$ and $B$, the “sum” $C$ means that each term $C_{ij}$ is given by \begin{equation*} C_{ij}=A_{ij}+B_{ij}. \end{equation*} Each term of $C$ is the sum of the terms in the same slots of $A$ and $B$.  In Section 5–6 we have already encountered the idea of a matrix “product.” The same idea will be useful in dealing with the sigma matrices. In general, the “product” of two matrices $A$ and $B$ (in that order) is defined to be a matrix $C$ whose elements are \begin{equation} \label{Eq:III:11:12} C_{ij}=\sum_kA_{ik}B_{kj}. \end{equation} It is the sum of products of terms taken in pairs from the $i$th row of $A$ and the $j$th column of $B$. If the matrices are written out in tabular form as in Fig. 11-1, there is a good “system” for getting the terms of the product matrix. Suppose you are calculating $C_{23}$. You run your left index finger along the second row of $A$ and your right index finger down the third column of $B$, multiplying each pair and adding as you go. We have tried to indicate how to do it in the figure.  It is, of course, particularly simple for two-by-two matrices. For instance, if we multiply $\sigma_x$ times $\sigma_x$, we get \begin{equation*} \sigma_x^2=\sigma_x\cdot\sigma_x= \begin{pmatrix} 0&1\\ 1&0 \end{pmatrix} \cdot \begin{pmatrix} 0&1\\ 1&0 \end{pmatrix} = \begin{pmatrix} 1&0\\ 0&1 \end{pmatrix}, \end{equation*} which is just the unit matrix $1$. Or, for another example, let’s work out $\sigma_x\sigma_y$: \begin{equation*} \sigma_x\sigma_y= \begin{pmatrix} 0&1\\ 1&0 \end{pmatrix} \cdot \begin{pmatrix} 0&-i\\ i&\phantom{-}0 \end{pmatrix} = \begin{pmatrix} i&\phantom{-}0\\ 0&-i \end{pmatrix}. \end{equation*} Referring to Table 11–1, you see that the product is just $i$ times the matrix $\sigma_z$. (Remember that a number times a matrix just multiplies each term of the matrix.) Since the products of the sigmas taken two at a time are important—as well as rather amusing—we have listed them all in Table 11–2. You can work them out as we have done for $\sigma_x^2$ and $\sigma_x\sigma_y$.  There’s another very important and interesting point about these $\sigma$ matrices. We can imagine, if we wish, that the three matrices $\sigma_x$, $\sigma_y$, and $\sigma_z$ are analogous to the three components of a vector—it is sometimes called the “sigma vector” and is written $\FLPsigma$. It is really a “matrix vector” or a “vector matrix.” It is three different matrices—one matrix associated with each axis, $x$, $y$, and $z$. With it, we can write the Hamiltonian of the system in a nice form which works in any coordinate system: \begin{equation} \label{Eq:III:11:13} H=-\mu\FLPsigma\cdot\FLPB. \end{equation}  Although we have written our three matrices in the representation in which “up” and “down” are in the $z$-direction—so that $\sigma_z$ has a particular simplicity—we could figure out what the matrices would look like in some other representation.  Although it takes a lot of algebra, you can show that they change among themselves like the components of a vector. (We won’t, however, worry about proving it right now. You can check it if you want.) You can use $\FLPsigma$ in different coordinate systems as though it is a vector.  You remember that the $H$ is related to energy in quantum mechanics. It is, in fact, just equal to the energy in the simple situation where there is only one state. Even for two-state systems of the electron spin, when we write the Hamiltonian as in Eq. (11.13), it looks very much like the classical formula for the energy of a little magnet with magnetic moment $\FLPmu$ in a magnetic field $\FLPB$. Classically, we would say \begin{equation} \label{Eq:III:11:14} U=-\FLPmu\cdot\FLPB, \end{equation} where $\FLPmu$ is the property of the object and $\FLPB$ is an external field. We can imagine that Eq. (11.14) can be converted to (11.13) if we replace the classical energy by the Hamiltonian and the classical $\FLPmu$ by the matrix $\mu\FLPsigma$. Then, after this purely formal substitution, we interpret the result as a matrix equation. It is sometimes said that to each quantity in classical physics there corresponds a matrix in quantum mechanics. It is really more correct to say that the Hamiltonian matrix corresponds to the energy, and any quantity that can be defined via energy has a corresponding matrix.  For example, the magnetic moment can be defined via energy by saying that the energy in an external field $\FLPB$ is $-\FLPmu\cdot\FLPB$. This defines the magnetic moment vector $\FLPmu$. Then we look at the formula for the Hamiltonian of a real (quantum) object in a magnetic field and try to identify whatever the matrices are that correspond to the various quantities in the classical formula. That’s the trick by which sometimes classical quantities have their quantum counterparts.  You may try, if you want, to understand how a classical vector is equal to a matrix $\mu\FLPsigma$, and maybe you will discover something—but don’t break your head on it. That’s not the idea—they are not equal. Quantum mechanics is a different kind of a theory to represent the world. It just happens that there are certain correspondences which are hardly more than mnemonic devices—things to remember with. That is, you remember Eq. (11.14) when you learn classical physics; then if you remember the correspondence $\FLPmu\to\mu\FLPsigma$, you have a handle for remembering Eq. (11.13). Of course, nature knows the quantum mechanics, and the classical mechanics is only an approximation; so there is no mystery in the fact that in classical mechanics there is some shadow of quantum mechanical laws—which are truly the ones underneath. To reconstruct the original object from the shadow is not possible in any direct way, but the shadow does help you to remember what the object looks like. Equation (11.13) is the truth, and Eq. (11.14) is the shadow. Because we learn classical mechanics first, we would like to be able to get the quantum formula from it, but there is no sure-fire scheme for doing that. We must always go back to the real world and discover the correct quantum mechanical equations. When they come out looking like something in classical physics, we are in luck.  If the warnings above seem repetitious and appear to you to be belaboring self-evident truths about the relation of classical physics to quantum physics, please excuse the conditioned reflexes of a professor who has usually taught quantum mechanics to students who hadn’t heard about Pauli spin matrices until they were in graduate school. Then they always seemed to be hoping that, somehow, quantum mechanics could be seen to follow as a logical consequence of classical mechanics which they had learned thoroughly years before. (Perhaps they wanted to avoid having to learn something new.) You have learned the classical formula, Eq. (11.14), only a few months ago—and then with warnings that it was inadequate—so maybe you will not be so unwilling to take the quantum formula, Eq. (11.13), as the basic truth. 