The projection matrix for spin one
We would like now to use our knowledge of the hydrogen atom to do something special. We discussed in Chapter 5 that a particle of spin one which was in one of the base states ($+$, $0$, or $-$) with respect to a Stern-Gerlach apparatus of a particular orientation—say an $S$ apparatus—would have a certain amplitude to be in each of the three states with respect to a $T$ apparatus with a different orientation in space. There are nine such amplitudes $\braket{jT}{iS}$ which make up the projection matrix. In Section 5–7 we gave without proof the terms of this matrix for various orientations of $T$ with respect to $S$. Now we will show you one way they can be derived.  In the hydrogen atom we have found a spin-one system which is made up of two spin one-half particles. We have already worked out in Chapter 6 how to transform the spin one-half amplitudes. We can use this information to calculate the transformation for spin one. This is the way it works: We have a system—a hydrogen atom with the energy $+A$—which has spin one. Suppose we run it through a Stern-Gerlach filter $S$, so that we know it is in one of the base states with respect to $S$, say $\ket{+S}$. What is the amplitude that it will be in one of the base states, say $\ket{+T}$, with respect to the $T$ apparatus? If we call the coordinate system of the $S$ apparatus the $x,y,z$ system, the $\ket{+S}$ state is what we have been calling the state $\ket{+\,+}$. But suppose another guy took his $z$-axis along the axis of $T$. He will be referring his states to what we will call the $x',y',z'$ frame. His “up” and “down” states for the electron and proton would be different from ours. His “plus-plus” state—which we can write $\ket{+'\,+'}$, referring to the “prime” frame—is the $\ket{+T}$ state of the spin-one particle. What we want is $\braket{+T}{+S}$ which is just another way of writing the amplitude $\braket{+'\,+'}{+\,+}$.  We can find the amplitude $\braket{+'\,+'}{+\,+}$ in the following way. In our frame the electron in the $\ket{+\,+}$ state has its spin “up”. That means that it has some amplitude $\braket{+'}{+}_{\text{e}}$ of being “up” in his frame, and some amplitude $\braket{-'}{+}_{\text{e}}$ of being “down” in that frame. Similarly, the proton in the $\ket{+\,+}$ state has spin “up” in our frame and the amplitudes $\braket{+'}{+}_{\text{p}}$ and $\braket{-'}{+}_{\text{p}}$ of having spin “up” or spin “down” in the “prime” frame. Since we are talking about two distinct particles, the amplitude that both particles will be “up” together in his frame is the product of the two amplitudes, \begin{equation} \label{Eq:III:12:44} \braket{+'\,+'}{+\,+}= \braket{+'}{+}_{\text{e}}\braket{+'}{+}_{\text{p}}. \end{equation} We have put the subscripts e and p on the amplitudes $\braket{+'}{+}$ to make it clear what we were doing. But they are both just the transformation amplitudes for a spin one-half particle, so they are really identical numbers. They are, in fact, just the amplitude we have called $\braket{+T}{+S}$ in Chapter 6, and which we listed in the tables at the end of that chapter.  Now, however, we are about to get into trouble with notation. We have to be able to distinguish the amplitude $\braket{+T}{+S}$ for a spin one-half particle from what we have also called $\braket{+T}{+S}$ for a spin-one particle—yet they are completely different!  We hope it won’t be too confusing, but for the moment at least, we will have to use some different symbols for the spin one-half amplitudes. To help you keep things straight, we summarize the new notation in Table 12-4. We will continue to use the notation $\ket{+S}$, $\ket{\OS}$, and $\ket{-S}$ for the states of a spin-one particle.  With our new notation, Eq. (12.44) becomes simply \begin{equation*} \braket{+'\,+'}{+\,+}=a^2, \end{equation*} and this is just the spin-one amplitude $\braket{+T}{+S}$. Now, let’s suppose, for instance, that the other guy’s coordinate frame—that is, the $T$, or “primed,” apparatus—is just rotated with respect to our $z$-axis by the angle $\phi$; then from Table 6-2, \begin{equation*} a=\braket{+'}{+}=e^{i\phi/2}. \end{equation*} So from (12.44) we have that the spin-one amplitude is \begin{equation} \label{Eq:III:12:45} \braket{+T}{+S}=\braket{+'\,+'}{+\,+}= (e^{i\phi/2})^2=e^{i\phi}. \end{equation} You can see how it goes.  Now we will work through the general case for all the states. If the proton and electron are both “up” in our frame—the $S$-frame—the amplitudes that it will be in any one of the four possible states in the other guy’s frame—the $T$-frame—are \begin{equation} \begin{aligned} \braket{+'\,+'}{+\,+}&= \braket{+'}{+}_{\text{e}}\braket{+'}{+}_{\text{p}}= a^2,\\[1ex] \braket{+'\,-'}{+\,+}&= \braket{+'}{+}_{\text{e}}\braket{-'}{+}_{\text{p}}= ab,\\[1ex] \braket{-'\,+'}{+\,+}&= \braket{-'}{+}_{\text{e}}\braket{+'}{+}_{\text{p}}= ba,\\[1ex] \braket{-'\,-'}{+\,+}&= \braket{-'}{+}_{\text{e}}\braket{-'}{+}_{\text{p}}= b^2. \end{aligned} \label{Eq:III:12:46} \end{equation} We can, then, write the state $\ket{+\,+}$ as the following linear combination:  \begin{equation} \label{Eq:III:12:47} \ket{+\,+}=a^2\,\ket{+'\,+'}+ab\,\{\ket{+'\,-'}+\ket{-'\,+'}\}+ b^2\,\ket{-'\,-'}. \end{equation} 
 \begin{alignat}{3} \ket{+\,+}&=&\phantom{ab}&a^2&\,&\ket{+'\,+'}\notag\\[.5ex] &+&\,ab&\,\{&&\ket{+'\,-'}+\ket{-'\,+'}\;\}\notag\\[.5ex] \label{Eq:III:12:47} &+&&b^2&\,&\ket{-'\,-'}. \end{alignat}  Now we notice that $\ket{+'\,+'}$ is the state $\ket{+T}$, that $\{\ket{+'\,-'}+\ket{-'\,+'}\}$ is just $\sqrt{2}$ times the state $\ket{\OT}$—see (12.41)—and that $\ket{-'\,-'}=\ket{-T}$. In other words, Eq. (12.47) can be rewritten as \begin{equation} \label{Eq:III:12:48} \ket{+S}=a^2\,\ket{+T}+\sqrt{2}\,ab\,\ket{\OT}+b^2\,\ket{-T}. \end{equation} In a similar way you can easily show that \begin{equation} \label{Eq:III:12:49} \ket{-S}=c^2\,\ket{+T}+\sqrt{2}\,cd\,\ket{\OT}+d^2\,\ket{-T}. \end{equation} For $\ket{\OS}$ it’s a little more complicated, because \begin{equation*} \ket{\OS}=\frac{1}{\sqrt{2}}\,\{\ket{+\,-}+\ket{-\,+}\}. \end{equation*} But we can express each of the states $\ket{+\,-}$ and $\ket{-\,+}$ in terms of the “prime” states and take the sum. That is,  \begin{equation} \label{Eq:III:12:50} \ket{+\,-}= ac\,\ket{+'\,+'}+ad\,\ket{+'\,-'}+ bc\,\ket{-'\,+'}+bd\,\ket{-'\,-'} \end{equation} 
 \begin{alignat}{2} \ket{+\,-}&&=ac\,&\ket{+'\,+'}+ad\,\ket{+'\,-'}\notag\\[.5ex] \label{Eq:III:12:50} &&+\,bc\,&\ket{-'\,+'}+bd\,\ket{-'\,-'} \end{alignat}  and  \begin{equation} \label{Eq:III:12:51} \ket{-\,+}= ac\,\ket{+'\,+'}+bc\,\ket{+'\,-'}+ ad\,\ket{-'\,+'}+bd\,\ket{-'\,-'}. \end{equation} 
 \begin{alignat}{2} \ket{-\,+}&&=ac\,&\ket{+'\,+'}+bc\,\ket{+'\,-'}\notag\\[.5ex] \label{Eq:III:12:51} &&+\,ad\,&\ket{-'\,+'}+bd\,\ket{-'\,-'}. \end{alignat}  Taking $1/\sqrt{2}$ times the sum, we get  \begin{equation*} \ket{\OS}=\frac{2}{\sqrt{2}}\,ac\,\ket{+'\,+'}+ \frac{ad+bc}{\sqrt{2}}\,\{\ket{+'\,-'}+\ket{-'\,+'}\}+ \frac{2}{\sqrt{2}}\,bd\,\ket{-'\,-'}. \end{equation*} 
 \begin{alignat*}{3} \ket{\OS}&=&&\phantom{a}\frac{2}{\sqrt{2}}\,ac&\,&\ket{+'\,+'}\\[.75ex] &+&&\frac{ad+bc}{\sqrt{2}}&\{\,&\ket{+'\,-'}+\ket{-'\,+'}\,\}\\[.75ex] &+&&\phantom{a}\frac{2}{\sqrt{2}}\,bd&\,&\ket{-'\,-'}. \end{alignat*}  It follows that  \begin{equation} \label{Eq:III:12:52} \ket{\OS}=\sqrt{2}\,ac\,\ket{+T}+(ad+bc)\,\ket{\OT}+ \sqrt{2}\,bd\,\ket{-T}. \end{equation} 
 \begin{align} \ket{\OS}&=\sqrt{2}ac\,\ket{+T}\notag\\[.5ex] &+(ad+bc)\,\ket{\OT}\notag\\[.5ex] \label{Eq:III:12:52} &+\sqrt{2}\,bd\,\ket{-T}. \end{align} 
 We have now all of the amplitudes we wanted. The coefficients of Eqs. (12.48), (12.49), and (12.52) are the matrix elements $\braket{jT}{iS}$. Let’s pull them all together:  \begin{equation} \label{Eq:III:12:53} \braket{jT}{iS}= \kern -2.5ex \raise 23 pt {\scriptstyle jT\downarrow \,\raise 7 pt \scriptstyle iS\rightarrow}\kern -19pt % ebook remove % ebook insert: \kern -1.5ex \raise{40pt}{\scriptstyle jT\downarrow \,\raise{7pt}{\scriptstyle iS\rightarrow}}\kern -23pt \begin{pmatrix} a^2 & \sqrt{2}ac & c^2\\[1ex] \sqrt{2}ab & ad\!+\!bc & \sqrt{2}cd\\[1ex] b^2 & \sqrt{2}bd & d^2 \end{pmatrix} \!\! % ebook remove . \end{equation}  We have expressed the spin-one transformation in terms of the spin one-half amplitudes $a$, $b$, $c$, and $d$.  For instance, if the $T$-frame is rotated with respect to $S$ by the angle $\alpha$ about the $y$-axis—as in Fig. 5-6—the amplitudes in Table 12-4 are just the matrix elements of $R_y(\alpha)$ in Table 6-2. \begin{equation} \begin{alignedat}{2} a\;&=\cos\frac{\alpha}{2},&\quad b\;&=-\sin\frac{\alpha}{2},\\[1ex] c\;&=\sin\frac{\alpha}{2},&\quad d\;&=\cos\frac{\alpha}{2}. \end{alignedat} \label{Eq:III:12:54} \end{equation} Using these in (12.53), we get the formulas of (5.38), which we gave there without proof.  What ever happened to the state $\ketsl{\slIV}$?! Well, it is a spin-zero system, so it has only one state—it is the same in all coordinate systems. We can check that everything works out by taking the difference of Eq. (12.50) and (12.51); we get that \begin{equation*} \ket{+\,-}-\ket{-\,+}=(ad-bc)\{\ket{+'\,-'}-\ket{-'\,+'}\}. \end{equation*} But $(ad-bc)$ is the determinant of the spin one-half matrix, and so is equal to $1$. We get that \begin{equation*} \ketsl{\slIV'}=\ketsl{\slIV} \end{equation*} for any relative orientation of the two coordinate frames. 