Rotation matrix for any spin
By now you can see, we hope, how important the idea of the angular momentum is in understanding atomic processes. So far, we have considered only systems with spins—or “total angular momentum”—of zero, one-half, or one. There are, of course, atomic systems with higher angular momenta. For analyzing such systems we would need to have tables of rotation amplitudes like those in Section 17-6. That is, we would need the matrix of amplitudes for spin $\tfrac{3}{2}$, $2$, $\tfrac{5}{2}$, $3$, etc. Although we will not work out these tables in detail, we would like to show you how it is done, so that you can do it if you ever need to.  As we have seen earlier, any system which has the spin or “total angular momentum” $j$ can exist in any one of $(2j+1)$ states for which the $z$-component of angular momentum can have any one of the discrete values in the sequence $j$, $j-1$, $j-2$, $\ldots\,$, $-(j-1)$, $-j$ (all in units of $\hbar$). Calling the $z$-component of angular momentum of any particular state $m\hbar$, we can define a particular angular momentum state by giving the numerical values of the two “angular momentum quantum numbers” $j$ and $m$. We can indicate such a state by the state vector $\ket{j,m}$. In the case of a spin one-half particle, the two states are then $\ket{\tfrac{1}{2},\tfrac{1}{2}}$ and $\ket{\tfrac{1}{2},-\tfrac{1}{2}}$; or for a spin-one system, the states would be written in this notation as $\ket{1,+1}$, $\ket{1,0}$, $\ket{1,-1}$. A spin-zero particle has, of course, only the one state $\ket{0,0}$.  Now we want to know what happens when we project the general state $\ket{j,m}$ into a representation with respect to a rotated set of axes. First, we know that $j$ is a number which characterizes the system, so it doesn’t change. If we rotate the axes, all we do is get a mixture of the various $m$-values for the same $j$. In general, there will be some amplitude that in the rotated frame the system will be in the state $\ket{j,m'}$, where $m'$ gives the new $z$-component of angular momentum. So what we want are all the matrix elements $\bracket{j,m'}{R}{j,m}$ for various rotations. We already know what happens if we rotate by an angle $\phi$ about the $z$-axis. The new state is just the old one multiplied by $e^{im\phi}$—it still has the same $m$-value. We can write this by  \begin{equation} \label{Eq:III:18:24} R_z(\phi)\,\ket{j,m}=e^{im\phi}\,\ket{j,m}. \end{equation}  Or, if you prefer,  \begin{equation} \label{Eq:III:18:25} \bracket{j,m'}{R_z(\phi)}{j,m}=\delta_{m,m'}e^{im\phi} \end{equation}  (where $\delta_{m,m'}$ is $1$ if $m'=m$, or zero otherwise).  For a rotation about any other axis there will be a mixing of the various $m$-states. We could, of course, try to work out the matrix elements for an arbitrary rotation described by the Euler angles $\beta$, $\alpha$, and $\gamma$. But it is easier to remember that the most general such rotation can be made up of the three rotations $R_z(\gamma)$, $R_y(\alpha)$, $R_z(\beta)$; so if we know the matrix elements for a rotation about the $y$-axis, we will have all we need.  How can we find the rotation matrix for a rotation by the angle $\theta$ about the $y$-axis for a particle of spin $j$? We can’t tell you how to do it in a basic way (with what we have had). We did it for spin one-half by a complicated symmetry argument. We then did it for spin one by taking the special case of a spin-one system which was made up of two spin one-half particles. If you will go along with us and accept the fact that in the general case the answers depend only on the spin $j$, and are independent of how the inner guts of the object of spin $j$ are put together, we can extend the spin-one argument to an arbitrary spin. We can, for example, cook up an artificial system of spin $\tfrac{3}{2}$ out of three spin one-half objects. We can even avoid complications by imagining that they are all distinct particles—like a proton, an electron, and a muon. By transforming each spin one-half object, we can see what happens to the whole system—remembering that the three amplitudes are multiplied for the combined state. Let’s see how it goes in this case.  Suppose we take the three spin one-half objects all with spins “up”; we can indicate this state by $\ket{+\,+\,+}$. If we look at this system in a frame rotated about the $z$-axis by the angle $\phi$, each plus stays a plus, but gets multiplied by $e^{i\phi/2}$. We have three such factors, so \begin{equation} \label{Eq:III:18:26} R_z(\phi)\,\ket{+\,+\,+}=e^{i(3\phi/2)}\,\ket{+\,+\,+}. \end{equation} Evidently the state $\ket{+\,+\,+}$ is just what we mean by the $m=+\tfrac{3}{2}$ state, or the state $\ket{\tfrac{3}{2},+\tfrac{3}{2}}$.  If we now rotate this system about the $y$-axis, each of the spin one-half objects will have some amplitude to be plus or to be minus, so the system will now be a mixture of the eight possible combinations $\ket{+\,+\,+}$, $\ket{+\,+\,-}$, $\ket{+\,-\,+}$, $\ket{-\,+\,+}$, $\ket{+\,-\,-}$, $\ket{-\,+\,-}$, $\ket{-\,-\,+}$, or $\ket{-\,-\,-}$. It is clear, however, that these can be broken up into four sets, each set corresponding to a particular value of $m$. First, we have $\ket{+\,+\,+}$, for which $m=\tfrac{3}{2}$. Then there are the three states $\ket{+\,+\,-}$, $\ket{+\,-\,+}$, and $\ket{-\,+\,+}$—each with two plusses and one minus. Since each spin one-half object has the same chance of coming out minus under the rotation, the amounts of each of these three combinations should be equal. So let’s take the combination \begin{equation} \label{Eq:III:18:27} \frac{1}{\sqrt{3}}\,\{ \ket{+\,+\,-}+\ket{+\,-\,+}+\ket{-\,+\,+}\} \end{equation} with the factor $1/\sqrt{3}$ put in to normalize the state. If we rotate this state about the $z$-axis, we get a factor $e^{i\phi/2}$ for each plus, and $e^{-i\phi/2}$ for each minus. Each term in (18.27) is multiplied by $e^{i\phi/2}$, so there is the common factor $e^{i\phi/2}$. This state satisfies our idea of an $m=+\tfrac{1}{2}$ state; we can conclude that  \begin{equation} \label{Eq:III:18:28} \frac{1}{\sqrt{3}}\,\{ \ket{+\,+\,-}+\ket{+\,-\,+}+\ket{-\,+\,+}\}= \ket{\tfrac{3}{2},+\tfrac{1}{2}}. \end{equation} 
 \begin{gather} \frac{1}{\sqrt{3}}\,\{ \ket{+\,+\,-}+\ket{+\,-\,+}+\ket{-\,+\,+}\}\notag\\ \label{Eq:III:18:28} =\ket{\tfrac{3}{2},+\tfrac{1}{2}}. \end{gather}  Similarly, we can write  \begin{equation} \label{Eq:III:18:29} \frac{1}{\sqrt{3}}\,\{ \ket{+\,-\,-}+\ket{-\,+\,-}+\ket{-\,-\,+}\}= \ket{\tfrac{3}{2},-\tfrac{1}{2}}, \end{equation} 
 \begin{gather} \frac{1}{\sqrt{3}}\,\{ \ket{+\,-\,-}+\ket{-\,+\,-}+\ket{-\,-\,+}\}\notag\\ \label{Eq:III:18:29} =\ket{\tfrac{3}{2},-\tfrac{1}{2}}, \end{gather}  which corresponds to a state with $m=-\tfrac{1}{2}$. Notice that we take only the symmetric combinations—we do not take any combinations with minus signs. They would correspond to states of the same $m$ but a different $j$. (It’s just like the spin-one case, where we found that $(1/\sqrt{2})\{\ket{+\,-}+\ket{-\,+}\}$ was the state $\ket{1,0}$, but the state $(1/\sqrt{2})\{\ket{+\,-}-\ket{-\,+}\}$ was the state $\ket{0,0}$.) Finally, we would have that \begin{equation} \label{Eq:III:18:30} \ket{\tfrac{3}{2},-\tfrac{3}{2}}=\ket{-\,-\,-}. \end{equation} We summarize our four states in Table 18–1.  Now all we have to do is take each state and rotate it about the $y$-axis and see how much of the other states it gives—using our known rotation matrix for the spin one-half particles. We can proceed in exactly the same way we did for the spin-one case in Section 12-6. (It just takes a little more algebra.) We will follow directly the ideas of Chapter 12, so we won’t repeat all the explanations in detail. The states in the system $S$ will be labelled $\ket{\tfrac{3}{2},+\tfrac{3}{2},S}=\ket{+\,+\,+}$, $\ket{\tfrac{3}{2},+\tfrac{1}{2},S}= (1/\sqrt{3})\{\ket{+\,+\,-}+\ket{+\,-\,+}+\ket{-\,+\,+}\}$, and so on. The $T$-system will be one rotated about the $y$-axis of $S$ by the angle $\theta$. States in $T$ will be labelled $\ket{\tfrac{3}{2},+\tfrac{3}{2},T}$, $\ket{\tfrac{3}{2},+\tfrac{1}{2},T}$, and so on. Of course, $\ket{\tfrac{3}{2},+\tfrac{3}{2},T}$ is the same as $\ket{+'\,+'\,+'}$, the primes referring always to the $T$-system. Similarly, $\ket{\tfrac{3}{2},+\tfrac{1}{2},T}$ will be equal to $(1/\sqrt{3})\{\ket{+'\,+'\,-'}+\ket{+'\,-'\,+'}+\ket{-'\,+'\,+'}\}$, and so on. Each $\ket{+'}$ state in the $T$-frame comes from both the $\ket{+}$ and $\ket{-}$ states in $S$ via the matrix elements of Table 12–4.  When we have three spin one-half particles, Eq. (12.47) gets replaced by  \begin{align} \ket{+\,+\,+}=&\;a^3\ket{+'\,+'\,+'}+a^2b\,\{\ket{+'\,+'\,-'}+ \ket{+'\,-'\,+'}+\ket{-'\,+'\,+'}\}\notag\\[1ex] &+\,ab^2\{\ket{+'\,-'\,-'}+\ket{-'\,+'\,-'}+ \ket{-'\,-'\,+'}\}\notag\\[1ex] \label{Eq:III:18:31} &+\,b^3\ket{-'\,-'\,-'}. \end{align} 
 \begin{align} &\ket{+\,+\,+}=\notag\\[1ex] &\phantom{+a^2b} a^3\ket{+'\,+'\,+'}\notag\\[.5ex] &+a^2b\{\ket{+'\,+'\,-'}\!+\!\ket{+'\,-'\,+'}+\ket{-'\,+'\,+'}\}\notag\\[.5ex] &+ab^2\!\{\ket{+'\,-'\,-'}\!+\!\ket{-'\,+'\,-'}+\ket{-'\,-'\,+'}\}\notag\\[.5ex] \label{Eq:III:18:31} &+\phantom{a\!\{}b^3\ket{-'\,-'\,-'}. \end{align}  Using the transformation of Table 12–4, we get instead of (12.48) the equation  \begin{align} \ket{\tfrac{3}{2},+\tfrac{3}{2},S}&= a^3\,\ket{\tfrac{3}{2},+\tfrac{3}{2},T}+ \sqrt{3}\,a^2b\,\ket{\tfrac{3}{2},+\tfrac{1}{2},T}\notag\\[1ex] \label{Eq:III:18:32} &\quad+\sqrt{3}\,ab^2\,\ket{\tfrac{3}{2},-\tfrac{1}{2},T} +b^3\,\ket{\tfrac{3}{2},-\tfrac{3}{2},T}. \end{align} 
 \begin{align} \ket{\tfrac{3}{2},+\tfrac{3}{2},S}&=a^3\,\ket{\tfrac{3}{2},+\tfrac{3}{2},T}\notag\\[.5ex] &+\sqrt{3}\,a^2b\,\ket{\tfrac{3}{2},+\tfrac{1}{2},T}\notag\\[.5ex] &+\sqrt{3}\,ab^2\,\ket{\tfrac{3}{2},-\tfrac{1}{2},T}\notag\\[.5ex] \label{Eq:III:18:32} &+b^3\,\ket{\tfrac{3}{2},-\tfrac{3}{2},T}. \end{align}  This already gives us several of our matrix elements $\braket{jT}{iS}$. To get the expression for $\ket{\tfrac{3}{2},+\tfrac{1}{2},S}$ we begin with the transformation of a state with two “$+$” and one “$-$” pieces. For instance,  \begin{align} \ket{+\,+\,-}&=a^2c\,\ket{+'\,+'\,+'}+a^2d\,\ket{+'\,+'\,-'}+ abc\,\ket{+'\,-'\,+'}\notag\\[1ex] &\quad+bac\,\ket{-'\,+'\,+'}+abd\,\ket{+'\,-'\,-'}+ bad\,\ket{-'\,+'\,-'}\notag\\[1ex] \label{Eq:III:18:33} &\quad+b^2c\,\ket{-'\,-'\,+'}+b^2d\,\ket{-'\,-'\,-'}. \end{align} 
 \begin{align} \ket{+\,+\,-}&=a^2c\,\ket{+'\,+'\,+'}\notag\\[.5ex] &+a^2d\,\ket{+'\,+'\,-'}\notag\\[.5ex] &+abc\,\ket{+'\,-'\,+'}\notag\\[.5ex] &+bac\,\ket{-'\,+'\,+'}+abd\,\ket{+'\,-'\,-'}\notag\\[.5ex] &+bad\,\ket{-'\,+'\,-'}\notag\\[.5ex] \label{Eq:III:18:33} &+b^2c\,\ket{-'\,-'\,+'}+b^2d\,\ket{-'\,-'\,-'}. \end{align}  Adding two similar expressions for $\ket{+\,-\,+}$ and $\ket{-\,+\,+}$ and dividing by $\sqrt{3}$, we find \begin{align} \ket{\tfrac{3}{2},+\tfrac{1}{2},S}&= \sqrt{3}\,a^2c\,\ket{\tfrac{3}{2},+\tfrac{3}{2},T}\notag\\[.5ex] &+(a^2d+2abc)\,\ket{\tfrac{3}{2},+\tfrac{1}{2},T}\notag\\[.5ex] &+(2bad+b^2c)\,\ket{\tfrac{3}{2},-\tfrac{1}{2},T}\notag\\[.5ex] \label{Eq:III:18:34} &+\sqrt{3}\,b^2d\,\ket{\tfrac{3}{2},-\tfrac{3}{2},T}. \end{align} Continuing the process we find all the elements $\braket{jT}{iS}$ of the transformation matrix as given in Table 18–2. The first column comes from Eq. (18.32); the second from (18.34). The last two columns were worked out in the same way.  (The coefficients $a$, $b$, $c$, and $d$ are given in Table 12–4.) Now suppose the $T$-frame were rotated with respect to $S$ by the angle $\theta$ about their $y$-axes. Then $a$, $b$, $c$, and $d$ have the values [see (12.54)] $a=$ $d=$ $\cos\theta/2$, and $c=$ $-b=$ $\sin\theta/2$. Using these values in Table 18–2 we get the forms which correspond to the second part of Table 17–2, but now for a spin $\tfrac{3}{2}$ system.  The arguments we have just gone through are readily generalized to a system of any spin $j$. The states $\ket{j,m}$ can be put together from $2j$ particles, each of spin one-half. (There are $j+m$ of them in the $\ket{+}$ state and $j-m$ in the $\ket{-}$ state.) Sums are taken over all the possible ways this can be done, and the state is normalized by multiplying by a suitable constant. Those of you who are mathematically inclined may be able to show that the following result comes out7:  \begin{align} &\bracket{j,m'}{R_y(\theta)}{j,m}= [(j+m)!(j-m)!(j+m')!(j-m')!]^{1/2}\notag\\[2ex] \label{Eq:III:18:35} &\kern{4.5em}\times\sum_k \frac{(-1)^{k+m-m'}(\cos\theta/2)^{2j+m'-m-2k}(\sin\theta/2)^{m-m'+2k}} {(m-m'+k)!(j+m'-k)!(j-m-k)!k!}, \end{align} 
 \begin{gather} \label{Eq:III:18:35} \bracket{j,m'}{R_y(\theta)}{j,m}=\\[1.5ex] [(j+m)!(j-m)!(j+m')!(j-m')!]^{1/2}\;\times\notag\\[1.25ex] \sum_k\!\! \frac{(-1)^{k+m-m'}\!(\cos\theta/2)^{2j+m'\!-m-2k}(\sin\theta/2)^{m-m'\!+2k}} {(m-m'+k)!(j+m'-k)!(j-m-k)!k!},\notag \end{gather}  where $k$ is to go over all values which give terms $\geq0$ in all the factorials.  This is quite a messy formula, but with it you can check Table 17–2 for $j=1$ and prepare tables of your own for larger $j$. Several special matrix elements are of extra importance and have been given special names. For example the matrix elements for $m=m'=0$ and integral $j$ are known as the Legendre polynomials and are called $P_j(\cos\theta)$: \begin{equation} \label{Eq:III:18:36} \bracket{j,0}{R_y(\theta)}{j,0}=P_j(\cos\theta). \end{equation} The first few of these polynomials are: \begin{align} \label{Eq:III:18:37} P_0(\cos\theta)&=1,\\[1ex] \label{Eq:III:18:38} P_1(\cos\theta)&=\cos\theta,\\[1ex] \label{Eq:III:18:39} P_2(\cos\theta)&=\tfrac{1}{2}(3\cos^2\theta-1),\\[1ex] \label{Eq:III:18:40} P_3(\cos\theta)&=\tfrac{1}{2}(5\cos^3\theta-3\cos\theta). \end{align} 