How do quantum numbers combine when two quantum systems combined?

Question

Quantum numbers are sometimes additive (e.g., baryon number, lepton number, $3$-component of the angular momentum, etc), sometimes multiplicative (e.g., parity), and at times, neither additive nor multiplicative (e.g., total angular momentum $j$, total isospin, etc).

What decides whether a quantum number will add, multiply or do something more complicated when two subsystems are combined to form a total system? Here, I am assuming that the subsystems have definite values of the quantum number under consideration.

Answer 1

I am quoting here from:

Patera, J. "The four sets of additive quantum numbers of SU(3)." Journal of Mathematical Physics 30.12 (1989): 2756-2762.

In quantum mechanics it is often convenient to distinguish three types of quantum numbers: additive, multiplicative, and the others. The additive ones are eigenvalues of operators of infinitesimal transformations, the multiplicative ones are eigenvalues of operators performing finite transformations, and the remaining ones are exemplified by the Casimir operators

Basically, the additive ones arise as a result of a separable action of an operator on each subspace. For instance, the total projection $$ L_z=L_z\otimes \mathbf{1}+ \mathbf{1}\otimes L_z := L_z^{(1)}+L_z^{(2)} \tag{1} $$ acts separately on states in the first subspace, leaving the 2nd subspace alone, and then acts on the second subspace, leaving the first one alone. Its eigenvalues on states of the type $\vert \ell_1m_1\rangle\vert \ell_2m_2\rangle$ (on composite states where both $L_z^{(1)}$ and $L_z^{(1)}$ are diagonal) will be the sum of eigenvalues in each subspace.

The operator $\mathbf{L}^2$ cannot be written the form $\mathbf{L}^2\otimes \mathbf{1}+ \mathbf{1}\otimes \mathbf{L}^2$ (because it is quadratic in the $L_i$'s) so its eigenvalues are no additive.

Finally, there are eigenvalues of finite transformations. On composite systems, finite transformation act by multiplication. For example, the parity transformation $\hat \Pi$ can be written as $\hat \Pi=\hat \Pi^{(1)}\otimes \hat \Pi^{(2)}$ so the parity of a composite system is the product of parities of the individual systems.

There's a nice connection between finite transformations and infinitesimal one. Consider for instance the finite rotation $$ R_z(\varphi)=e^{-i\varphi L_z}\otimes e^{-i\varphi L_z}\tag{2} $$ acting on $\vert \ell_1m_1\rangle\vert \ell_2m_2\rangle$. As a finite transformation, its eigenvalues are $e^{-i(m_1+m_2)\varphi}$. Because $\varphi$ is continuous, one can expand Eq.(2) near $\varphi=0$: \begin{align} e^{-i\varphi L_z}\otimes e^{-i\varphi L_z}& \approx (\mathbb{1}-i\varphi L_z)\otimes (\mathbb{1}-i\varphi L_z)\, ,\\ &= \mathbb{1}\otimes \mathbb{1}-i\varphi \left(L_z\otimes \mathbf{1}+ \mathbf{1}\otimes L_z \right) \end{align} so you can see how Eq.(1) arises as the linear part of the finite transformation of Eq.(2). Thus in this way additive quantum numbers are related to the expansion of continuous multiplicative ones when the expansion makes sense.

Obviously one cannot think of parity as continuous transformation so this is an example where one cannot find an additive version of the parity quantum number.

Answer 2

Mathematically, quantum numbers refer to irreducible representations (reps) of the (unitary) symmetry group. For example, parity corresponds to one-dimensional reps of $Z_2$, the baryon number/3-component of the angular momentum correspond to one-dimensional reps of U(1), and spin quantum number $j$ for SU(2). Let's say the symmetry group is $G$. When you combine two systems, each having a $G$ symmetry, we usually define the $G$ symmetry to act diagonally on both subsystems. In that case, reps of the whole system are tensor product of reps of the subsystems. When $G$ is Abelian (e.g. $Z_2$ or U(1)), all irreps are one-dimensional, so you can simply "add" or "multiply" (depending on whether you define the binary operation on the group additively or multiplicatively). When $G$ is non-Abelian (e.g. SU(2)), then in general tensor product of irreps decomposes into a direct sum of irreps, so they neither "add" or "multiply".