What is the difference between "additive" quantum numbers and "multiplicative" quantum numbers?
I think that this may have something to do with P and C Symmetry groups, but I may be mistaken.
I’m reading in "Halzen, F., and A. D. Martin. Quarks & Leptons: An Introductory Course in Modern Particle Physics. New York, NY: John Wiley & Sons, 1984. ISBN: 9780471887416", section 2.5.
Roughly, an additive quantum number is the log of a corresponding multiplicative quantum number.
Mathematically, this comes from the difference between the representations of a group and a Lie algebra; in the former, the natural operation is multiplication and in the latter it is addition. Many quantum numbers we care about come from continuous symmetry groups, which have Lie algebras, and so we use additive ones. Some, like parity, are discrete symmetries and the discrete groups don't have Lie algebras, forcing us to use the multiplicative ones.
If that was too mathematical and unclear, here's a nice example.
Consider a free particle living on a circle. The symmetry group is $U(1)$, corresponding to the translations. The natural quantum number to use is the eigenvalue of the momentum operator $\hat{p}$, which generates infinitesimal rotations. And if you have two particles, whose eigenvalues are $p_1$ and $p_2$, the total momentum in the system is $p_1+p_2$. This is, hopefully, obvious.
It is not, however, entirely trivial. Consider translating the state $|p_1,p_2\rangle$ by an angle $\theta$. The operator that does this is the product of the corresponding translation operators.
$$e^{i \hat{p}_1 \theta} e^{i \hat{p}_2 \theta} |p_1,p_2\rangle = e^{i (p_1+p_2) \theta} |p_1,p_2\rangle$$
Thus, despite the fact that translation is performed by the product of the translation operators, the eigenvalue is the sum of the eigenvalues. The reason? There's an infinitesimal translation operator, and two infinitesimal translations compose by adding. To see this clearly, consider the above case with $\theta \ll 1$ and expand the exponential.
Now, suppose an evil demon came along and discretised the circle into $N$ points. In this new system, there is no such thing as an infinitesimal rotation, since the closest point is an angle $2\pi/N$ away. Since there's no such operation, there's really no question of using its eigenvalue to label its states.
However, hopping is still a symmetry and therefore it furnishes a nice quantum number. Glancing at the above equation again, you can see that the eigenvalues of $e^{i \hat{p} \theta}$ got multiplied when we performed a joint rotation. Since there's no question of taking $\theta \ll 1$, there are no operators whose eigenvalues are additive quantum numbers. However, the finite translation operator gives us a multiplicative quantum number.
And finally, I should warn you that in general quantum numbers need neither add nor multiply in easy ways, if the symmetry group is not Abelian. The pain involved in addition of angular momentum is precisely because the corresponding group $\mathrm{SO}(3)$ is non-Abelian.
Kinetic energy of two free particles is additive: the total energy is just the sum of the individual energies: $$ K=K_1+K_2 $$
Another example is charge: the charge of a multiparticle system is the sum of the individual charges.
Parity is multiplicative: the parity of a two-particle system is the product of the parities of the inidividual particles: $$ \Pi=\Pi_1\Pi_2 $$
