Meson build up from a quark-antiquark pair

[Pietjuh]

I have some questions about mesons. I don't really understand why they are build up from a quark-antiquark pair. I know from the theory that one can classify the mesons by considering the tensor product of the fundamental representation [3] and the representation [3'] (the prime for denoting anti-particles). But I'm wondering why you can't take the tensorproduct of two normal quarks, like [3]x[3] = [6]+[3'] to classify the mesons. Or can't you have a bound state between 2 normal quarks?

 

[SpaceTiger]

Or can't you have a bound state between 2 normal quarks?

A naive answer would be that a pair of quarks couldn't be color neutral, so if you believe hadrons respect color neutrality, then a diquark couldn't exist as an independent bound state.

 

[samalkhaiat]

I have some questions about mesons. I don't really understand why they are build up from a quark-antiquark pair. I know from the theory that one can classify the mesons by considering the tensor product of the fundamental representation [3] and the representation [3'] (the prime for denoting anti-particles). But I'm wondering why you can't take the tensorproduct of two normal quarks, like [3]x[3] = [6]+[3'] to classify the mesons. Or can't you have a bound state between 2 normal quarks?

The {6} in the flavour group SU(3) can not describe mesons. This is because, the states in the {6};

{6} = 1/2 [q(i)q(j) + q(j)q(i)], i = u,d and s

have non-zero baryon number;

$$B = \frac{1}{3} [N(q) - N(\bar{q})]$$.

However, you are pointing to one of many paradoxes in the simple quark model.
Even though hadrons are seen to be made of $q\bar{q}$ and $qqq$, and there is no evidevce for $qq$ and $qqqq$ bound states, still the simple quark model can not explain the absence of such hadron states with masses comparable to the observed particles.
After painting the quarks with three colours (the birth of the colour group SU(3) and its gauge theory QCD), it became possible to "resolve" the paradox by the so-called colour-singlet conjecture (we can only observe colourless hadrons ), This means that q, qq and qqqq can not be seen and, since $3 x \bar{3}$ and $3 x 3 x 3$ contain colour singlets, only $q\bar{q}$ and $qqq$ can bind into physically observable hadrons.

regards

sam

 

[Rade]

...only $q\bar{q}$ and $qqq$ can bind into physically observable hadrons.

Question. Does not $$\bar{q}$$$$\bar{q}$$$$\bar{q}$$ also form a hadron, of pure antimatter ?

 

[jtbell]

Yes indeed. A proton is $uud$, an antiproton is $\overline{u} \overline{u} \overline{d}$.

 

[samalkhaiat]

Question. Does not $$\bar{q}$$$$\bar{q}$$$$\bar{q}$$ also form a hadron, of pure antimatter ?

Yes, I should have said:

"only $q\bar{q}$ , $qqq$ and their conjugates can bind into physically observable hadrons"


regards

sam

 

[snow_in_quarks]

In the past two years, the existence of pentaquark(qqqqq_bar) was reported by several exp. groups,
eg. proton+proton -> (Sigma_+) + (Theta_+) .
Sigma_+ (uus) , Theta_+ (uudds_bar) ,
and Theta_+ decays to KN (nK_+ or pK_0) .
We say Theta_+ is produced, because KN forms a resonance as a peak
on the background.
We say Theta_+ is a pentaquark, because q_bar couldn't annihilate with
any other four quarks, and it couldn't be simplified as (qqq) state.

But to our surprise, later on several other exp. groups reported the null
results. So we shall wait for further decisive experiment.

 

[Rade]

I have read that a stable tetraquark is possible in theory when the wavefunction dynamics are viewed as interaction of two diquarks, thus |QQ$\overline{q}\overline{q}$|, where the two Q are high mass quarks, the two $\overline{q}$ are low mass. Stability results when mass ratio (M/m) is large enough--but I have no idea how large this enough must be. Apparently works best with lightest quark combinations (u,d,s). See Ader et al, Phys Rev D 25, 1982, 2370.