1
$\begingroup$

Suppose we only have two colors, for example, red (R) and blue (B) to construct the wave functions of baryons and mesons and that the color symmetry is SU (2) and not SU (3). In this situation, baryons would consist of systems bound quark and antiquark in a state of total neutral color. How can we determine the wave functions for baryons and for mesons if there is a color missing for baryons for example? Thank you.

| cite | improve this question | | | | |
$\endgroup$
  • 1
    $\begingroup$ Mesons would be quark-antiquark states; and baryons would be quark-quark ones with the two respective colors antisymmetrized, so with fractional charges, all else kept equal. Colorwise, anti-color doublets will resemble colored ones, since the conjugate doublet rep of SU(2) is equivalent to the doublet itself, but the baryons would still have nonzero baryon number, however you'd define it. So, what's this stuff about baryons being quark-antiquark?? $\endgroup$ – Cosmas Zachos Jun 16 '19 at 0:05
  • $\begingroup$ Excuse me, but I think I made a mistake on this part of baryons being quarks and antiquarks. Just to understand the subject, in this case then for the availability of existing SU(2) colors, would baryons be fermions or bosons? $\endgroup$ – Murillo de Godoy Jun 16 '19 at 0:31
  • 1
    $\begingroup$ @Phys_BR Anything made of an even number of fermions will be a boson, so baryons and mesons would both be bosons if the color symmetry were $SU(2)$. To put the comment by Cosmas Zachos into a larger context: if the gauge group were $SU(N)$, then a baryon would be "made of" $N$ quarks, and an antibaryon would be "made of" $N$ antiquarks. It's a fermion if $N$ is odd, and a boson if $N$ is even. $\endgroup$ – Chiral Anomaly Jun 16 '19 at 0:35
1
$\begingroup$

The snag in quark models is the baryon sector, where you need as many quarks in the state as colors, to get a color singlet antisymmetrizing them. So, N fermions for N colors, as @ChiralAnomaly reminds you.

So, for your 2 colors, your baryon is $\epsilon^{ab} q_a q_b=q_R q_B- q_B q_R $, hence now a boson (just as for all even - N color groups). The rest of the wave function will be spin-color symmetric, (with integral spin!), from Fermi statistics. You'd be advised to assign your quarks baryon number 1/2, then.

Your mesons will be automatically color singlets, $\bar q^a q^a$, but here the antiquark will be in the conjugate representation of the fundamental of SU(2), here, exceptionally, basically the same rep as the quark, with the identical color structure. (The Young tableaux for the fundamental and anti fundamental are just a box, here... You'll never see this again.) But, still, the baryon numbers of the mesons will be zero, so your won't have, e.g. a baryon decaying to two mesons...

| cite | improve this answer | | | | |
$\endgroup$
  • $\begingroup$ I am a bit curious, what is the fundamental difference between the 2-color Bosonic hadron you described and a usual hydrogen atom (which is also a bound state of two fermions)? I understand the constituents of the above two cases are not the same. The question is about whether there is a tangible characteristic distinguishing them (one is confined and the other is separable?). $\endgroup$ – MadMax Jun 17 '19 at 14:17
  • $\begingroup$ @MadMax Fundamental? Hard to define... It might well depend on your focus. Confinement of the former is striking, and strong interactions would lead to bosonic nuclei, unprotected against Bose condensation, among other things... $\endgroup$ – Cosmas Zachos Jun 17 '19 at 14:35
-4
$\begingroup$

I don't really think it's possible to have only two colors. The color symmetry of such particles consists of three colors, that's how the theory works.

Besides, how would you construct a symmetry with only two colors? Would the third color of baryons be a linear composition of the two colors?

| cite | improve this answer | | | | |
$\endgroup$
  • 3
    $\begingroup$ With gauge group $SU(N)$, any combination of $N$ quarks that is completely antisymmetric in the color indices is color-neutral (that is, invariant under an $SU(N)$ transformation), because an $SU(N)$ matrix has unit determinant. This works for any $N$, including $N=2$. $\endgroup$ – Chiral Anomaly Jun 16 '19 at 0:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.