2
$\begingroup$

QCD is built from the notion that Dirac's Lagrangian should be invariant under gauge colour transformations. Here, quarks are elements of $\psi_{\alpha,f,c}(x)$, where $\alpha$, $f$ and $c$ stand for Dirac spinor, flavour and colour indices, respectively.

Now, why do we say that quarks should be invariant under $SU(3)$ group local transformations in the fundamental representation:

\begin{equation} \psi(x)\longrightarrow\ V(x)\psi(x), \end{equation} with \begin{equation} V(x)\equiv\text{exp}[i\theta^a(x)t^a], \end{equation} in order to preserve colour gauge symmetry?

These transformations are multiplications of the $3\times3$ $t^a$ matrices with the $\psi(x)$, and there are 8 $t^a$ matrices. I understand that the colour degrees of freedom are 3 (red,green and blue) and that these spinors are complex-valued 4 component vectors, that is 8 real fields. Is the fact that there are 3 colour degrees of freedom related to the matrices being $3 \times 3$? And is the fact that there are 8 of these matrices related to the fact that there are 8 real fields in these Dirac spinors?

It has always seemed to me when building QCD, that saying that quarks live in the fundamental representation of $SU(3)$ is ad hoc, but do they have to be this way in order to have the quark's colour gauge symmetry?

Edit:

I am aware that the triplet colour assignment of quarks was so that these degrees of freedom would explain the observed hadrons as being composed of these particles. My question is, what is so special about the fundamental representation? In the relation between the transformation in the fundamental representation $V(x)$ and $\psi(x)$, i.e. the relation between the $t^a$ matrices and $\psi_{\alpha,f,c}(x)$, what is that imposes the desired colour gauge symmetry that we see expect to see in nature?

$\endgroup$
3
  • 5
    $\begingroup$ They don't have to be in the fundamental representation -- that's just how it turns out to be in the real world. $\endgroup$
    – knzhou
    Aug 5 at 3:47
  • 3
    $\begingroup$ Are you aware that as far back as 1968 scattering experiments revealed three pointlike charges inside protons? $\endgroup$
    – Ghoster
    Aug 5 at 5:38
  • 1
    $\begingroup$ Triplet quarks were proposed as mathematical props of SU(3) flavor representation theory: to make the group theory easier for the observed hadron representations, the octet, the decuplet, etc…. They were then observed/confirmed and assigned to a color SU(3) to firm up the consistency of a model, now confirmed to be a just-so law of nature. Can you focus your question? $\endgroup$ Aug 5 at 11:26

1 Answer 1

3
$\begingroup$

Now, why do we say that quarks should be invariant under $SU(3)$ group local transformations in the fundamental representation, \begin{equation} \psi(x)\longrightarrow\ V(x)\psi(x), \end{equation} with \begin{equation} V(x)\equiv\text{exp}[i\theta^a(x)t^a], \end{equation} in order to preserve colour gauge symmetry?

Quarks transform in the triplet representation of local (gauge) SU(3), as you described; they are not invariant under it. The Lagrangian that describes their couplings is invariant, through the magic of gauge couplings. Your text likely details how this works. It is an "accidental" fact of nature, like several such facts.

These transformations are multiplications of the $3\times3$ $t^a$ matrices with the $\psi(x)$, and there are 8 such $t^a$ matrices. I understand that the colour degrees of freedom are 3 (red,green and blue) and that these spinors are complex-valued 4 component vectors, that is 8 real fields.

You focussed on the wrong indices. The color indices in your notation are denoted by Latin indices, but here you garbled them: c, of which there are only 3, are fundamental representation color indices, suppressed in everything you have written here; ${\sf a}$, for which you should have used a different convention or font, are adjoint color indices, of which there are 8. When you reinsert the triplet indices into the $3\times 3$ matrices $t^{\sf a}_{bc}$ your above expression reads, in detail, $$ \psi_b(x)\longrightarrow\ (\exp[i\theta^{\sf a}(x)t^{\sf a}])_{bc}\psi_c(x), $$ $3\times 3$ matrices acting on complex 3-spinors to map them to complex 3-spinors.

Is the fact that there are 3 colour degrees of freedom related to the matrices being $3 \times 3$? And is the fact that there are 8 of these matrices related to the fact that there are 8 real fields in these Dirac spinors?

Yes, the 3 representation is the fundamental one for SU(3), the Lie group that the infinity of your Vs are elements of. The fact that there are 8 independent generators $t^{\sf a}$ is, in fact, related to the 8 arbitrary independent angles $\theta^{\sf a}$, each one corresponding to a different gluon. There are no 8 real fields in the quark... I'm not sure where you picked this one up.

If nature were different, the quark representation might be in a different color representation, of dimension d, and the $t^{\sf a}$ would be $d\times d$ matrices. But our world was experimentally confirmed to be in the 3, the fundamental, and you will see through asymptotic freedom how this is confirmed. It's a just-so thing...


Clarification Update in view of Comment

The entire discussion above considers only color indices c as active and involved in transformations, and the indices α and f, Dirac spinor and flavor, of the tensor product wf $\psi_{\alpha,f,c}(x)$, are fixed, constant and unchangeable in the color transformations discussed, hence irrelevant. The Lorentz, flavor, and color groups mutually commute, so $\psi_{\alpha,f,c}(x)$ here is considered as a complex 3-spinor $\psi_{c}(x)$, for any intents and purposes.

$\endgroup$
3
  • $\begingroup$ When I say 8 real fields I mean that there are 4 complex fields in the quark Dirac spinor, so 2 real fields for each one of these complex spinor elements. I was trying to relate the quark field fundamental transformation and colour gauge symmetry of the Lagrangian, to the fact that these transformations have 8 generators. But you mean that number of 8 generators is only related to the 8 different gluons? $\endgroup$
    – orochi
    Aug 6 at 2:45
  • 1
    $\begingroup$ Yes I do : color transformations leave Dirac indices alone, since color commutes with the Lorentz group. The two 8s are a coincidence. Consider different dimensions of space time or color groups…. $\endgroup$ Aug 6 at 5:57
  • 1
    $\begingroup$ I see. I thought that one of the reasons for the transformation in fundamental representation had something to do with the spinor structure. Writing this transformation with explicit indices definitely made this clear. Thanks! $\endgroup$
    – orochi
    Aug 6 at 16:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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