Rotations about $\boldsymbol{x}$
You may be thinking: “This is getting ridiculous. What are they going to do next, $47^\circ$ around $y$, then $33^\circ$ about $x$, and so on, forever?” No, we are almost finished. With just two of the transformations we have—$90^\circ$ about $y$, and an arbitrary angle about $z$ (which we did first if you remember)—we can generate any rotation at all.  As an illustration, suppose that we want the angle $\alpha$ around $x$. We know how to deal with the angle $\alpha$ around $z$, but now we want it around $x$. How do we get it? First, we turn the axis $z$ down onto $x$—which is a rotation of $+90^\circ$ about $y$, as shown in Fig. 6-8. Then we turn through the angle $\alpha$ around $z'$. Then we rotate $-90^\circ$ about $y''$. The net result of the three rotations is the same as turning around $x$ by the angle $\alpha$. It is a property of space.  (These facts of the combinations of rotations, and what they produce, are hard to grasp intuitively. It is rather strange, because we live in three dimensions, but it is hard for us to appreciate what happens if we turn this way and then that way. Perhaps, if we were fish or birds and had a real appreciation of what happens when we turn somersaults in space, we could more easily appreciate such things.)  Anyway, let’s work out the transformation for a rotation by $\alpha$ around the $x$-axis by using what we know. From the first rotation by $+90^\circ$ around $y$ the amplitudes go according to Eq. (6.32). Calling the rotated axes $x'$, $y'$, and $z'$, the next rotation by the angle $\alpha$ around $z'$ takes us to a frame $x''$, $y''$, $z''$, for which \begin{equation*} C_+''=e^{i\alpha/2}C_+',\quad C_-''=e^{-i\alpha/2}C_-'. \end{equation*} The last rotation of $-90^\circ$ about $y''$ takes us to $x'''$, $y'''$, $z'''$; by (6.33), \begin{equation*} C_+'''=\frac{1}{\sqrt{2}}\,(C_+''-C_-''),\quad C_-'''=\frac{1}{\sqrt{2}}\,(C_+''+C_-''). \end{equation*} Combining these last two transformations, we get \begin{align*} C_+'''&=\frac{1}{\sqrt{2}}\, (e^{+i\alpha/2}C_+'-e^{-i\alpha/2}C_-'),\\ \\ C_-'''&=\frac{1}{\sqrt{2}}\, (e^{+i\alpha/2}C_+'+e^{-i\alpha/2}C_-'). \end{align*} Using Eqs. (6.32) for $C_+'$ and $C_-'$, we get the complete transformation: \begin{align*} C_+'''&=\tfrac{1}{2} \{e^{+i\alpha/2}(C_++C_-)-e^{-i\alpha/2}(-C_++C_-)\},\\ \\ C_-'''&=\tfrac{1}{2} \{e^{+i\alpha/2}(C_++C_-)+e^{-i\alpha/2}(-C_++C_-)\}. \end{align*} We can put these formulas in a simpler form by remembering that \begin{equation*} e^{i\theta}+e^{-i\theta}=2\cos\theta,\quad\text{and}\quad e^{i\theta}-e^{-i\theta}=2i\sin\theta. \end{equation*} We get \begin{equation} \left. \begin{alignedat}{3} C_+'''&=\phantom{i}\biggl(\!&\cos\frac{\alpha}{2}&\biggr)C_+\!+ i\biggl(\!&\sin\frac{\alpha}{2}&\biggr)C_-\\ \\[1ex] C_-'''&=i\biggl(\!&\sin\frac{\alpha}{2}&\biggr)C_+\!+ \phantom{i}\biggl(\!&\cos\frac{\alpha}{2}&\biggr)C_- \end{alignedat} \!\! % ebook remove \right\} \text{$\alpha$ about $x$}. \label{Eq:III:6:34} \end{equation} Here is our transformation for a rotation about the $x$-axis by any angle $\alpha$. It is only a little more complicated than the others. 