# A question about quarks and quantum chromodynamics

Penrose writes the following on pg 648 of his book "Road to Reality"

How can we treat quarks as real particles, if they have the wrong spin-statistics relation? The way that this problem is dealt with, in the standard model, is to demand that each flavor of quark also comes in three (so called) colours, and that any actual particle, composed of quarks, must be completely anti-symmetrical in the colour degree of freedom. This anti-symmetry passes over to the quark states themselves, so that antisymmetry between individual fermionic quarks gets effectively converted into symmetry, in a three quark particle.

I don't understand this. Let us take a particle with spin $$\frac{3}{2}$$, composed of three "up" quarks. So we can write this as $$uuu$$. Now as each quark comes in a different colour, assuming these colours are red (R), blue (B) and green (G), we can write the particle as $$u_R u_B u_G$$. However, shouldn't this be symmetrical in the colours? Shouldn't this be equal to $$u_G u_B u_R$$, where I have permuted the colours?

However, as this is antisymmetrical in colours, we should have $$u_Ru_B u_G=-u_G u_B u_R$$. But this clearly does not happen. Where am I going wrong?

Let $$|0\rangle$$ be the vacuum state, and let $$q_C(f)$$ denote the operator that, when applied to any state, adds another quark with color $$C$$ and "spatial wavefunction" $$f$$. The statement that quarks are fermions means that these operators anticommute with each other.

Now consider the three-quark state $$|f,g,h\rangle := \sum_{\pi}(-1)^\pi q_{\pi(R)}(f)q_{\pi(G)}(g)q_{\pi(B)}(h)|0\rangle, \tag{1}$$ where the sum is over all permutations $$\pi$$ of the color indices, with a negative sign for odd permutations. This antisymmetrized sum over permutations makes the state a color-singlet, assuming that the color group is $$SU(3)$$. Spin indices are omitted because were assuming that they're all equal, so the total spin is $$3/2$$. All three quarks are assumed to be the same flavor.

The fact that the quark creation operators $$q_C(f)$$ anticommute with each other (because quarks are fermions) implies that the state $$|f,g,h\rangle$$ is symmetric with respect to permutations of $$f,g,h$$, as in $$|g,f,h\rangle = |f,g,h\rangle. \tag{2}$$ In detail: \begin{align} |g,f,h\rangle &:= \sum_{\pi}(-1)^\pi q_{\pi(R)}(g)q_{\pi(G)}(f)q_{\pi(B)}(h)|0\rangle \\ &= -\sum_{\pi}(-1)^\pi q_{\pi(G)}(f)q_{\pi(R)}(g)q_{\pi(B)}(h)|0\rangle \\ &= \sum_{\pi}(-1)^\pi q_{\pi(R)}(f)q_{\pi(G)}(g)q_{\pi(B)}(h)|0\rangle \\ &=: |f,g,h\rangle. \tag{3} \end{align} Narration:

• The overall negative sign on the second line comes from changing the order of the operators $$q_{\pi(R)}(g)$$ and $$q_{\pi(G)}(f)$$. This sign-change expresses the fact that quarks are fermions.

• The overall negative sign goes away on the third line because we permuted the color indices $$G$$ and $$R$$. This sign-change expresses the fact that the state is a color singlet (invariant under a color-$$SU(3)$$ transformation).

The conclusion is that the state $$|f,g,h\rangle$$ is color-antisymmetric but spatially symmetric, as claimed in the excerpt shown in the OP.

Qualifications:

• To be concise, I used the words "spatial wavefunction." That's ambiguous, but exactly I mean by it is not important here. The general idea is sufficient.

• The "valence quark" picture used here is good enough for addressing the question, but beware that $$|f,g,h\rangle$$ isn't really a single-baryon state. Real baryons also involve gluons, and they actually involve an indefinite number of quarks/antiquarks.