Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
The (admittedly short) Wikipedia article explains it pretty well. Can you be more specific about what you want to know? – ACuriousMind Mar 29 at 15:28

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.

share|cite|improve this answer
Thank you so much for such an in-depth answer! So as far as the multiplicative quantum numbers are concerned, when we rotate the system by a finite amount.... The eigenvalue itself should stay the same, correct? Or should the eigenvalue change, as we rotate through all of the quantum eigenstates that are allowed? This concept is incredible to me. Thanks again! – Matt Steinberg Mar 30 at 9:10
Oh the eigenvalue doesn't change, and neither does the eigenstate. That's what it means to say that a state is an eigenstate: the action of the operator on that state is just multiplication by a number. – Ronak M Soni Mar 30 at 10:14
Ok, so if I understand you correctly, the reason as to why the "discretization" changes the roation to multiplicative quantum numbers is because we are rotating each state $p_1 and p_2$ by a different $\theta$ in order to reach the new angle ${2\pi}/N$? If so, then we would need to use different notation, wouldn't we? I'm sprry for me this is a very large conceptual leap that I'm still wrestling with. – Matt Steinberg Apr 2 at 7:01
$|p_1\rangle$ is the state of particle 1 and $|p_2\rangle$ is the state of particle 2. We rotate both particles by the same amount. The discretisation comes from the fact that only certain values of $\theta$, integral multiples of $2\pi/N$, are allowed rotations. Look at the equation above that has its own line; that equation describes the rotation of two particles by the same amount, and it's valid in both cases. The only thing that's different is the allowed values of $\theta$. – Ronak M Soni Apr 2 at 9:22

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: $$ \Pi=\Pi_1\Pi_2 $$

share|cite|improve this answer
So... Another words, a rotation operator in hilbert space is an example of a parity operator, and the hamiltonian can be considered an additive? – Matt Steinberg Mar 30 at 9:12
@MattSteinberg 1) not quite: rotations always leave an axis invariant (the rotation axis), while the parity operator only leaves the origin invariant; therefore, the partity operator is not a rotation (but is very related to rotations). 2) The Hamiltonian is additive for non-interacting particles. If the particles interact, the total Hamiltonian is the sum of the individual Hamiltonian plus the interaction term. – AccidentalFourierTransform Mar 30 at 15:56

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.