If a baryon wavefunction is $\Psi = \psi_{spatial} \psi_{colour} \psi_{flavour} \psi_{spin}$, and we consider the ground state (L=0) only. We know that the whole thing has to be antisymmetric under the interchange of two quarks. We know that colour is antisymmetric (always colourless) and spatial is symmetric. Therefore, the combination of flavour and spin has to be symmetric.

That's fine, I understand that. However, I'm very uncertain about what the 'interchange of two quarks' actually means. Interchange how?

For the 'easy' example of |uuu>. How is the flavour symmetric?

Basically, what does it mean to be symmetric in the quark model?


1 Answer 1


Halzen, Martin: Quarks and Leptons has a good introduction into that matter in Chapter 2.

$|uuu\rangle$ is actually an abbreviation of the idea that we keep an eye on three quarks in particular order, call them "first", "second" and "third"; and then first quark is in $u$ flavour state, of $SU(2)$ or $SU(3)$ flavour group, that is, $$\left(\begin{array}{c}1\\0\end{array}\right)\qquad\text{or}\qquad\left(\begin{array}{c}1\\0\\0\end{array}\right),$$ and second and third quarks are in the same state (so we have to take a tensor product of three such columns, like a 3D matrix).

To check the symmetry of this state, we exchange the states of first and second quarks. (In terms of the 3D matrix, we transpose it on 1st and 2nd axes, in terms of the tensor product we permutate its indices.) Since we had (only a 2D slice is shown) $$\left(\begin{array}{cc}1&0\\0&0\end{array}\right)\qquad\text{or}\qquad\left(\begin{array}{ccc}1&0&0\\0&0&0\\0&0&0\end{array}\right),$$ our state is immune to such exchange, and we thus call it symmetric. Other possible outcomes would be antisymmetric, or neither of these two symmetries. For example, $|udu\rangle$ is neither symmetric nor antisymmetric, and $\tfrac{1}{\sqrt{2}}(|udu\rangle-|duu\rangle)$ is antisymmetric (with respect to two first quarks only, for simplicity).

If we take into account other variables than just flavour, then we should exchange all variables of two quarks in order to "interchange the quarks".

  • $\begingroup$ I'm not quite sure on SU(2) or the tensor product. How is that state immune to exchange? $\endgroup$ May 9, 2013 at 8:50
  • $\begingroup$ For quark flavours we use a fundamental representation of groups SU(2) or SU(3), which means flavour states (state vectors) are columns of 2 or 3 complex numbers. SU(2) acts on such columns with Pauli matrices, and SU(3) with Gell-Mann matrices respectively. Quarks is $u$-quarks if only the first component is non-zero, $d$-quark if 2nd, and $s$-quark if 3rd (heavier quarks usually don't count). Other states are superpositions of these. For two quarks you have to get 2x2 or 3x3 matrix of numbers to represent amplitudes for all possible combinations. ... $\endgroup$
    – firtree
    May 9, 2013 at 12:54
  • $\begingroup$ ... If these two quarks are in independent states, you simply take their columns of numbers, and fill the matrix with all their possible products. Here, since both quarks were (1 0 0), we had only one non-zero element in the matrix - (1,1)-th. Now, to exchange states of the quarks, we need to exchange matrix columns with rows and vice versa - every element switches with its reflection about the main diagonal. Our particular matrix stays the same under such exchange. Hence, we say that the state is unchanged. $\endgroup$
    – firtree
    May 9, 2013 at 13:01

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