Angular momentum in quantum mechanics
We have already given you a relation between the magnetic moment and the angular momentum. That’s pleasant. But what do the magnetic moment and the angular momentum mean in quantum mechanics? In quantum mechanics it turns out to be best to define things like magnetic moments in terms of the other concepts such as energy, in order to make sure that one knows what it means. Now, it is easy to define a magnetic moment in terms of energy, because the energy of a moment in a magnetic field is, in the classical theory, $\FLPmu\cdot\FLPB$. Therefore, the following definition has been taken in quantum mechanics: If we calculate the energy of a system in a magnetic field and we find that it is proportional to the field strength (for small field), the coefficient is called the component of magnetic moment in the direction of the field. (We don’t have to get so elegant for our work now; we can still think of the magnetic moment in the ordinary, to some extent classical, sense.)  Now we would like to discuss the idea of angular momentum in quantum mechanics—or rather, the characteristics of what, in quantum mechanics, is called angular momentum. You see, when you go to new kinds of laws, you can’t just assume that each word is going to mean exactly the same thing. You may think, say, “Oh, I know what angular momentum is. It’s that thing that is changed by a torque.” But what’s a torque? In quantum mechanics we have to have new definitions of old quantities. It would, therefore, be legally best to call it by some other name such as “quantangular momentum,” or something like that, because it is the angular momentum as defined in quantum mechanics. But if we can find a quantity in quantum mechanics which is identical to our old idea of angular momentum when the system becomes large enough, there is no use in inventing an extra word. We might as well just call it angular momentum. With that understanding, this odd thing that we are about to describe is angular momentum. It is the thing which in a large system we recognize as angular momentum in classical mechanics.  First, we take a system in which angular momentum is conserved, such as an atom all by itself in empty space. Now such a thing (like the earth spinning on its axis) could, in the ordinary sense, be spinning around any axis one wished to choose. And for a given spin, there could be many different “states,” all of the same energy, each “state” corresponding to a particular direction of the axis of the angular momentum. So in the classical theory, with a given angular momentum, there is an infinite number of possible states, all of the same energy.  It turns out in quantum mechanics, however, that several strange things happen. First, the number of states in which such a system can exist is limited—there is only a finite number. If the system is small, the finite number is very small, and if the system is large, the finite number gets very, very large. Second, we cannot describe a “state” by giving the direction of its angular momentum, but only by giving the component of the angular momentum along some direction—say in the $z$-direction. Classically, an object with a given total angular momentum $J$ could have, for its $z$-component, any value from $+J$ to $-J$. But quantum-mechanically, the $z$-component of angular momentum can have only certain discrete values. Any given system—a particular atom, or a nucleus, or anything—with a given energy, has a characteristic number $j$, and its $z$-component of angular momentum can only be one of the following set of values: \begin{equation} \begin{aligned} j\hbar&\\ (j-1)\hbar&\\ (j-2)\hbar&\\ \vdots\phantom{)\hbar}&\\ -(j-2)\hbar&\\ -(j-1)\hbar&\\ -j\hbar&\\ \end{aligned} \label{Eq:II:34:23} \end{equation} The largest $z$-component is $j$ times $\hbar$; the next smaller is one unit of $\hbar$ less, and so on down to $-j\hbar$. The number $j$ is called “the spin of the system.” (Some people call it the “total angular momentum quantum number”; but we’ll call it the “spin.”)  You may be worried that what we are saying can only be true for some “special” $z$-axis. But that is not so. For a system whose spin is $j$, the component of angular momentum along any axis can have only one of the values in (34.23).  Although it is quite mysterious, we ask you just to accept it for the moment. We will come back and discuss the point later. You may at least be pleased to hear that the $z$-component goes from some number to minus the same number, so that we at least don’t have to decide which is the plus direction of the $z$-axis. (Certainly, if we said that it went from $+j$ to minus a different amount, that would be infinitely mysterious, because we wouldn’t have been able to define the $z$-axis, pointing the other way.)  Now if the $z$-component of angular momentum must go down by integers from $+j$ to $-j$, then $j$ must be an integer. No! Not quite; twice $j$ must be an integer. It is only the difference between $+j$ and $-j$ that must be an integer. So, in general, the spin $j$ is either an integer or a half-integer, depending on whether $2j$ is even or odd. Take, for instance, a nucleus like lithium, which has a spin of three-halves, $j=3/2$. Then the angular momentum around the $z$-axis, in units of $\hbar$, is one of the following: \begin{equation*} \begin{matrix} +3/2\phantom{.}\\ +1/2\phantom{.}\\ -1/2\phantom{.}\\ -3/2. \end{matrix} \end{equation*} There are four possible states, each of the same energy, if the nucleus is in empty space with no external fields. If we have a system whose spin is two, then the $z$-component of angular momentum has only the values, in units of $\hbar$, \begin{equation*} \begin{matrix} \phantom{-}2\phantom{.}\\ \phantom{-}1\phantom{.}\\ \phantom{-}0\phantom{.}\\ -1\phantom{.}\\ -2. \end{matrix} \end{equation*} If you count how many states there are for a given $j$, there are $(2j+1)$ possibilities. In other words, if you tell me the energy and also the spin $j$, it turns out that there are exactly $(2j+1)$ states with that energy, each state corresponding to one of the different possible values of the $z$-component of the angular momentum.  We would like to add one other fact. If you pick out any atom of known $j$ at random and measure the $z$-component of the angular momentum, then you may get any one of the possible values, and each of the values is equally likely. All of the states are in fact single states, and each is just as good as any other. Each one has the same “weight” in the world. (We are assuming that nothing has been done to sort out a special sample.) This fact has, incidentally, a simple classical analog. If you ask the same question classically: What is the likelihood of a particular $z$-component of angular momentum if you take a random sample of systems, all with the same total angular momentum?—the answer is that all values from the maximum to the minimum are equally likely. (You can easily work that out.) The classical result corresponds to the equal probability of the $(2j+1)$ possibilities in quantum mechanics.  From what we have so far, we can get another interesting and somewhat surprising conclusion. In certain classical calculations the quantity that appears in the final result is the square of the magnitude of the angular momentum $\FLPJ$—in other words, $\FLPJ\cdot\FLPJ$. It turns out that it is often possible to guess at the correct quantum-mechanical formula by using the classical calculation and the following simple rule: Replace $J^2=\FLPJ\cdot\FLPJ$ by $j(j+1)\hbar^2$. This rule is commonly used, and usually gives the correct result, but not always. We can give the following argument to show why you might expect this rule to work.  The scalar product $\FLPJ\cdot\FLPJ$ can be written as \begin{equation*} \FLPJ\cdot\FLPJ=J_x^2+J_y^2+J_z^2. \end{equation*} Since it is a scalar, it should be the same for any orientation of the spin. Suppose we pick samples of any given atomic system at random and make measurements of $J_x^2$, or $J_y^2$, or $J_z^2$, the average value should be the same for each. (There is no special distinction for any one of the directions.) Therefore, the average of $\FLPJ\cdot\FLPJ$ is just equal to three times the average of any component squared, say of $J_z^2$; \begin{equation*} \av{\FLPJ\cdot\FLPJ} = 3\av{J_z^2}. \end{equation*} But since $\FLPJ\cdot\FLPJ$ is the same for all orientations, its average is, of course, just its constant value; we have \begin{equation} \label{Eq:II:34:24} \FLPJ\cdot\FLPJ = 3\av{J_z^2}. \end{equation}  If we now say that we will use the same equation for quantum mechanics, we can easily find $\av{J_z^2}$. We just have to take the sum of the $(2j+1)$ possible values of $J_z^2$, and divide by the total number;  \begin{equation} \label{Eq:II:34:25} \av{J_z^2} = \frac {j^2+(j-1)^2+\dotsb+(-j+1)^2+(-j)^2} {2j+1}\,\hbar^2. \end{equation} 
 \begin{gather} \label{Eq:II:34:25} \av{J_z^2} =\\[1ex] \frac{j^2+(j-1)^2+\dotsb+(-j+1)^2+(-j)^2} {2j+1}\,\hbar^2.\notag \end{gather}  For a system with a spin of $3/2$, it goes like this:  \begin{equation*} \av{J_z^2} = \frac {(3/2)^2+(1/2)^2+(-1/2)^2+(-3/2)^2} {4}\,\hbar^2=\frac{5}{4}\,\hbar^2. \end{equation*}
  \begin{gather*} \av{J_z^2} =\\[1.25ex] \frac{(3/2)^2+(1/2)^2+(-1/2)^2+(-3/2)^2} {4}\,\hbar^2\\[1.25ex] \kern{-1.75ex}=\frac{5}{4}\,\hbar^2. \end{gather*}   We conclude that \begin{equation*} \FLPJ\cdot\FLPJ = 3\av{J_z^2} = 3\tfrac{5}{4}\hbar^2=\tfrac{3}{2}(\tfrac{3}{2}+1)\hbar^2. \end{equation*} We will leave it for you to show that Eq. (34.25), together with Eq. (34.24), gives the general result \begin{equation} \label{Eq:II:34:26} \FLPJ\cdot\FLPJ=j(j+1)\hbar^2. \end{equation} Although we would think classically that the largest possible value of the $z$-component of $\FLPJ$ is just the magnitude of $\FLPJ$—namely, $\sqrt{\FLPJ\cdot\FLPJ}$—quantum mechanically the maximum of $J_z$ is always a little less than that, because $j\hbar$ is always less than $\sqrt{j(j+1)}\hbar$. The angular momentum is never “completely along the $z$-direction.” 