What is J in the rigid rotor model?
TL;DR It depends on context and is unfortunately something that isn't going to change in the near future. However, the more you understand angular momentum, the easier it becomes to figure out the meaning from the context.



Quantum mechanical angular momentum is a tricky topic and it is not helped by some books which use confusing nomenclature. In QM, angular momentum is a vector quantity (much like in classical mechanics).* This is denoted with various letters;$^\dagger$ we will stick with J for the time being:

$$\text{angular momentum: }\mathbf{J} = (J_x, J_y, J_z)$$

Alternatively, you may also see vectors denoted with an arrow sign: $\vec{J}$.

All three components of the (QM) angular momentum are observable quantities. They are more commonly referred to as projections of the angular momentum onto an axis, which means the same thing. In any case, each one is associated with a Hermitian operator :

$$\text{ang. mom. projection operators: }\hat{J}_x, \hat{J}_y, \hat{J}_z$$

Apart from the $x$-, $y$-, and $z$-components of the angular momentum, the magnitude of the angular momentum is also an observable. This is similar to the length of the vector $\mathbf{J}$, and classically we would write $|\mathbf{J}|^2 = J_x^2 + J_y^2 + J_z^2$. The corresponding quantum mechanical operator is usually written without the boldface as

$$\text{ang. mom. magnitude operator: }\hat{J}{}^2 = \hat{J}{}_x^2 + \hat{J}{}_y^2 + \hat{J}{}_z^2$$

So, now, we have four unique operators: $\hat{J}{}^2, \hat{J}_x, \hat{J}_y, \hat{J}_z$. Because they do not all commute with one another, we usually focus on two of them, which are (conventionally) $\hat{J}{}^2$ and $\hat{J}_z$, and ignore the other two.

The quantum mechanical states of interest are those which are simultaneous eigenstates of these two operators, namely $\hat{J}{}^2$ and $\hat{J}_z$. It turns out that these eigenstates have the properties

\begin{align}
\hat{J}{}^2|\psi\rangle &amp;= s(s+1)\hbar^2 |\psi\rangle \\
\hat{J}_z|\psi\rangle &amp;= t\hbar |\psi\rangle \\
\end{align}

where $s$ is constrained to be a non-negative integer $0, 1, 2, \ldots$ and $t$ is constrained to be $-s, \ldots, -1, 0, 1, \ldots, s$. Different eigenstates will have different values of $s$ and $t$, which immediately suggests that we should label the eigenstates $|\psi\rangle$ by the values of $s$ and $t$ that apply to them. Constants such as $s$ and $t$ which can only take on certain values are called quantum numbers.

We can therefore associate the eigenstate $|\psi_{s,t}\rangle$ with the angular momentum $\sqrt{s(s+1)}\hbar$ (because the angular momentum squared was $s(s+1)\hbar^2$). We could write this as:

$$J_s = \sqrt{s(s+1)}\hbar$$

which says that the (magnitude of the) angular momentum $J$ of a state with quantum number $s$ is equal to $\sqrt{s(s+1)}\hbar$. In the rigid rotor system, we ignore $\hat{J}_z$ and instead look at another observable, which is the energy. The energy of a state with quantum number $s$ is:

$$E_s = \frac{\hbar^2}{2I}s(s+1)$$

where $I$ is the moment of inertia. So far, so good. But confusingly, instead of using innocuous labels like $s$, the convention is to use $J$ instead. And $t$ is replaced with $M$ (sometimes $M_J$), which at least is slightly less confusing.

\begin{align}
J_J &amp;= \sqrt{J(J+1)}\hbar \\
E_J &amp;= \frac{\hbar^2}{2I}J(J+1)
\end{align}

And therefore we have a situation where $J$ can mean several things, and it completely depends on context. To be extra clear, I'll colour the $J$ which means "the value of angular momentum" in red, and the $J$'s that refer to the quantum number in blue:

\begin{align}
\color{red}{J}_\color{blue}{J} &amp;= \sqrt{\color{blue}{J}(\color{blue}{J}+1)}\hbar \\
E_\color{blue}{J} &amp;= \frac{\hbar^2}{2I}\color{blue}{J}(\color{blue}{J}+1)
\end{align}

and the corresponding eigenvalue equations (recall that the operator for energy is always denoted $\hat{H}$, for the Hamiltonian):

\begin{align}
\hat{\color{green}{J}}{}^2|\color{blue}{J}\rangle &amp;= \color{blue}{J}(\color{blue}{J}+1)\hbar^2|\color{blue}{J}\rangle \\
\hat{H}|\color{blue}{J}\rangle &amp;= \frac{\hbar^2}{2I}\color{blue}{J}(\color{blue}{J}+1)|\color{blue}{J}\rangle
\end{align}

Here the green colour indicates an operator; the red in the previous equation was not an operator, but rather the quantity of the observable. Some texts will help you out by putting a hat on operators, but often it is omitted (either because authors are lazy or because it gets really cluttered after a while), so be careful!



* Technically a pseudovector.

$^\dagger$ $J$ is usually used for the total angular momentum of a system. In classical mechanics, there is only really one type of angular momentum, but in quantum mechanics there is orbital angular momentum and spin angular momentum.

Orbital angular momentum is the one which is much more closely related to the classical idea of angular momentum (for example, the orbital angular momentum operators are directly derived from $\mathbf{L} = \mathbf{r} \times \mathbf{p}$), and is typically denoted with $L$, not $J$. The quantum number is usually (helpfully) denoted $l$, although some books will also use $L$.

Spin angular momentum, on the other hand, is denoted with $S$. It has no classical analogue and has even weirder behaviour in that the associated quantum number $s$ can take on half-integer values, i.e. $0, \frac 1 2, 1, \frac 3 2, \ldots$

In quantum mechanics, the total angular momentum $J$ is the sum of both these two contributions. In the rigid rotor model we are not concerned with the spin of the particle(s), so $J$ is simply equal to the orbital angular momentum $L$, and there is no ambiguity about which is which. But if you start thinking about particles like nuclei or electrons, then it becomes important to distinguish between the two.