# What is a quantum number in a quantum field theory?

In non-relativistic quantum mechanics, quantum numbers are associated with eigenvalues of an operator. For example, $$\ell$$ is a quantum number associated with the eigenvalue $$\ell(\ell+1)\hbar^2$$ orbital angular momentum operator $$\textbf{L}^2$$. But in quantum field theory quantum numbers, as I understand, are not necessarily associated with eigenvalues of operators. For example, the color quantum number of Quantum Chromodynamics (QCD) is not associated with the eigenvalue of any operator.

So the question is: how should we think about quantum numbers in Quantum field theory?

• please be clear that quantum field theory is a meta-level of relativistic quantum mechanics,, it is based on the free particle solutions that describe the particles.. – anna v Sep 24 '18 at 18:05

Just like in quantum mechanics, in QFT the whole quantum field is described by just one state in a Hilbert space. Sometimes people say that a particle is in a certain "state," but this is an abuse of language. The whole field is in a particular "state." Particles are excitations of a field, just like how the quantum harmonic oscillator can have quantized excited energy levels. It's not like one particle is in a spin up "state" and another particle is in a spin down "state," for example. The whole field is in one state, and has both a spin up excitation and a spin down excitation. The Pauli exclusion principle, that no two fermions can be in the same "state" is really a property of what the quantum field Hilbert space is for a fermionic field.

So then what are "quantum numbers" in QFT? Certainly there must be some notion in which one electron can be "spin up" while another is "spin down." Really, quantum numbers in QFT are usually just field indices. These indices are present even at the classical level. For example, in a Dirac field there are four possible field indices at each point in space. They correspond to spin up electrons, spin down electrons, spin up positrons, and spin down positrons. (Usually they are not written in a basis where this is clear, but in principle this is why there are four degrees of freedom.) There is a sense in which there is a "classical Dirac field," and upon quantizing it as a fermionic field, these four field degrees of freedom can be associated with the four different types of particles you can find as excitations of your quantum field.

Likewise, you can consider a "classical" quark field and a "classical" gluon field. The classical quark field will have three extra indices corresponding to the three possible colors.

So, in conclusion, "quantum numbers" in QFT correspond to field indices that are even present in the classical field equivalents of your quantum field theory. Particles are excitations of the quantum fields, and you can have different types of excitations corresponding to each of these field indices.

No, quantum numbers are always associated with quantum operators. Specifically, if $$|\psi \rangle$$ has a value $$\lambda$$ for a quantum number, the associated operator $$Q$$ obeys $$Q|\psi \rangle = \lambda |\psi \rangle$$. Quantum field theory doesn't change this. For global symmetries a quantum operator is defined by Noether's theorem.

The case you're talking about is different because of gauge symmetry. The "total redness" or "total anti-blueness" of a state is not defined because it is gauge-dependent. Similarly there is no quantum operator.

• So color quantum number or electric charge are not quantum numbers?@knzhou – SRS Sep 24 '18 at 18:48
• @SRS Electric charge is a quantum number. – knzhou Sep 24 '18 at 19:02
• @SRS Color charge is also a quantum number, the point is that "red" is just a vector inside a particle color charge, namely the fundamental representation. The operator is the Casimir of the Lie algebra. – Ryan Thorngren Sep 24 '18 at 19:04
• But color is often referred as a quantum number in literature. Right? In fact, assignment of different color quantum numbers to quarks in a proton is such that the Pauli exclusion is respected. @knzhou – SRS Sep 26 '18 at 7:16
• @RyanThorngren There are two independent Casimir operators for $SU(3)$. Which one are you referring to? – SRS Sep 26 '18 at 7:20