When looking at the quark model we say we can write the wavefunction in 4 components, ie $$ |\Psi \rangle = |\Psi_{spatial} \rangle|\Psi_{spin} \rangle|\Psi_{flavour} \rangle|\Psi_{colour} \rangle \space \space $$

It isn't clear to me why we can assume that the quantum numbers of the state are completely separable like this. The most general state will be a linear combination of separable states, ie $|\Psi \rangle = C_{i j kl}|\Psi^{(i)}_{spatial} \rangle|\Psi^{(j)}_{spin} \rangle|\Psi^{(k)}_{flavour} \rangle|\Psi^{(l)}_{colour} \rangle \in \mathcal{H}= \mathcal{H_{spatial}} \otimes \mathcal{H_{spin}} \otimes \mathcal{H_{flavour}}\otimes\mathcal{H_{colour}}$

Why is the assumption of separability valid for the quark model?

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    $\begingroup$ Since quark model can mean a few different things (according to the Particle Data Group)... Is the assumption of separability a tenet of the quark model that you're referring to? I mean, are you asking for what purpose(s) this separability assumption is adequate, even though the exact QCD single-hadron states probably aren't separable in this way? $\endgroup$ Nov 14, 2021 at 2:58
  • $\begingroup$ Have you parsed out the complete set of commuting constants of the motion involved? $\endgroup$ Nov 14, 2021 at 3:08
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    $\begingroup$ @ChiralAnomaly So most notes that I have seen state this separability and then proceed to diagonalize each of the respective individual Hilbert spaces and combine the eigenstates after. I am asking if this is just an approximation or is there a fundamental mathematical or physical reason to expect the quantum numbers of the quark model to separate in this way. $\endgroup$ Nov 14, 2021 at 9:58
  • $\begingroup$ @CosmasZachos I am not sure what you mean by this $\endgroup$ Nov 14, 2021 at 9:59

1 Answer 1


Recall that flavor, color, and S are commuting operators. Ultimately, you may be interested in the hadron spectrum (the eigenstates of the strong Hamiltonian, which is symmetric, more or less, under the corresponding transformations generated by them), or couplings among such eigenstates.

The idea of the quark model is to build irreducible representations of the color, flavor, and rotation groups (exponentials of J, the coproduct of Ls with S) which describe hadrons. (Actually, the hadrons' color representations are trivial singlets). Further, one wishes to work out couplings dominating decays and splits of such hadrons.

One starts from single (factorized) quark wavefunctions, and proceed to compose (tensor) them together to build simplistic models of hadrons; these are, of course, entangled (not factorized!) to comport with the statistics of the constituent quarks (fermionic) and the composite hadrons (fermionic or bosonic). The simple unentangled building blocks, the irreducible representations of all of the above groups, are the quarks you use. The wavefunctions of the hadrons are as entangled as you contemplate, and fitting the sudoku involved together is a game of marvel.

Behold the composite wavefunction/state of the proton, suppressing the implied full antisymmetrization of color (so the full state is a singlet thereof),
$$ |p_\uparrow\rangle= \frac{1}{\sqrt {18}} [ 2| u_\uparrow d_\downarrow u_\uparrow \rangle + 2| u_\uparrow u_\uparrow d_\downarrow \rangle +2| d_\downarrow u_\uparrow u_\uparrow \rangle - | u_\uparrow u_\downarrow d_\uparrow\rangle -| u_\uparrow d_\uparrow u_\downarrow\rangle -| u_\downarrow d_\uparrow u_\uparrow\rangle -| d_\uparrow u_\downarrow u_\uparrow\rangle -| d_\uparrow u_\uparrow u_\downarrow\rangle -| u_\downarrow u_\uparrow d_\uparrow\rangle ]. $$ This is an eigenstate of the strong hamiltonian (which, however, has small isospin/flavor breaking terms in it). It is fully color and J invariant. It is not a string of separable states. It is a linear combination of such.

  • $\begingroup$ So if (under the naive quark model) a hadronic state is separable, perhaps like the pion, does this tell us something interesting about the particle? $\endgroup$ Nov 14, 2021 at 22:32
  • $\begingroup$ Interesting? Not sure. Does it? Spin singlet, L singlet, color singlet, isotriplet. $\endgroup$ Nov 14, 2021 at 22:53

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