7
$\begingroup$

In the limit where the masses vanish, low energy QCD has a well known chiral symmetry (see http://arxiv.org/abs/hep-ph/0505265 for a very extensive review, and pg 19 for the section relevant for my question). The Lagrangian of interest is (I use 2 component notation),

\begin{equation} {\cal L} = \sum _{ i \in u, d , s}q _{ R , i} ^\dagger i \bar{\sigma} ^\mu D _\mu q _{ R , i} + q ^\dagger _{ L , i} i \bar{\sigma} ^\mu D _\mu q _{L , i} - \frac{1}{4} G _{ \mu \nu } G ^{ \mu \nu } \end{equation} Well below the weak scale we can rewrite the following as, \begin{equation} {\cal L} = \sum _{ \ell \in u _L , d _L , s _L , u _R , d _R , s _R }q ^\dagger _{ \ell } i \bar{\sigma} ^\mu D _\mu q _{ \ell } - \frac{1}{4} G _{ \mu \nu } G ^{ \mu \nu } \end{equation} Thus it seems that if we aren't worried about electroweak effects (and hence the covariant derivatives in the right handed fields are the same as in the left handed ones) then we in fact have an $ SU(6) \times U(1) _V $ symmetry, \begin{equation} \left( \begin{array}{c} u _L \\ d _L \\ s _L \\ u _R \\ d _R \\ s _R \end{array} \right) \rightarrow e ^{ i T _A \alpha _A }\left( \begin{array}{c} u _L \\ d _L \\ s _L \\ u _R \\ d _R \\ s _R \end{array} \right) \end{equation} instead of the more commonly discussed $ SU(3) \times SU(3) \times U(1) _V $. However, the fact that I've never seen this discuss makes me skeptical of its validity. Is the above a legitimate way to discuss chiral symmetry?

$\endgroup$
  • $\begingroup$ How would the $\mathrm{SU}(6)$ symmetry you see here act on the gauge fields? I'm not seeing this symmetry you seem to see, could you write down its transformations? (Also, there's some kind of trace missing on the $G$ for this to be an invariant lagrangian) $\endgroup$ – ACuriousMind Jun 22 '15 at 18:31
  • $\begingroup$ $\bar{q}_Li\gamma^\mu D_\mu q_R=0$. Your symmetry does not act on the fields. $\endgroup$ – Thomas Jun 22 '15 at 18:31
  • $\begingroup$ @ACuriousMind: I rewrote the question in two component notation to increase clarity and added in an explicit transformation. Please let me know if this isn't clear. $\endgroup$ – JeffDror Jun 22 '15 at 19:07
  • $\begingroup$ @Thomas: Perhaps my updated notation better explains what I had in mind. $\endgroup$ – JeffDror Jun 22 '15 at 19:08
  • $\begingroup$ I think the issue is that the quark masses break the chiral symmetry. The approximate chiral symmetry is still a good approximation for the up, down, and strange (ie we can approximate these as massless), the fact that we are approximating three quarks as massless is where the 3 in SU(3) comes from in chiral perturbation theory (note this is different from the existence of 3 colors, which is why the gauge group of QCD is SU(3)). $\endgroup$ – Andrew Aug 22 '15 at 2:32
2
$\begingroup$

Okay, I think I have an answer but I would interested to hear some feedback on this. I initially thought that by writing the left and right handed fields in 2 component notation ($q_L $ and $q_R^c$), they were 'equivalent' objects and can freely rotated between one another. However, I realized now this is not the case.

The left and right handed fields are in the fundamental and anti-fundamental representation of $SU(3)_C$ (due to the often omitted conjugate symbol on the right handed fields) and so can't be simply rotated between one another. This restricts left fields to rotate into left fields and right fields and right fields and hence the $SU(3) _L\times SU(3)_R $ instead of a rotation between all the fields ($SU(6)$).

Note that for the above discussion I was using common beyond Standard Model notation where the fundamental fields are all left-chiral fields and hence transform the same way (for details on this see hep-ph/9709356, bottom of pg 8). An alternative to this viewpoint is to define the fundamental fields as left and right chiral fields and then as @Thomas mentioned they transform the same way under $SU(3)_C$ but you can't rotate between them since they have different representations under the Lorentz group.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ $L$ and $R$ are in the same representation of $SU(3)_c$, but in different representations of the Lorentz group (there is something wrong with your Dirac operator, by the way, now that you wrote in 2-component notation), so your proposed flavor symmetry mixes different representations of $SO(3,1)$. $\endgroup$ – Thomas Jun 23 '15 at 0:40
  • $\begingroup$ @Thomas: Thanks for your feedback. I was using BSM notation (see update to answer), which I think resulted in the confusion. $\endgroup$ – JeffDror Jun 23 '15 at 1:06
  • 1
    $\begingroup$ Then you would still have to write something like $\sigma_2u_R^\dagger$ instead of $u_L$ in your 6-component vector. In any case, you violate either $SU(3)_c$ (and baryon number $U(1)_V$) or Lorentz invariance. $\endgroup$ – Thomas Jun 23 '15 at 15:29
  • $\begingroup$ I suspect any discussion of color, here, is irrelevant, as color is vectorlike, so L-R blind. But if you think you have a symmetry sending L to R and vice versa, consider the Noether currents of any such transformation defocussing off the flavor part. Do you see that any and all such currents would vanish from the chiral projectors annihilating each other? Yes, this transformation does not commute with the Lorentz group$\overline{\psi_L} \gamma^\mu ... \psi_R=0 $ , where ... suggests the irrelevant flavor matrices. There won't be a charge to achieve this. $\endgroup$ – Cosmas Zachos Mar 3 '16 at 15:57
  • $\begingroup$ So, to sum up, the symmetry of this one is SU(3)xSU(3)xP where P is the discrete, not continuous, symmetry, and lacks a Noether current. $\endgroup$ – Cosmas Zachos Mar 3 '16 at 16:32
2
$\begingroup$

On second thought, the reason the Noether currents and charges of all such continuous transformations vanish, as per my comments to your answer, @JeffDrorr, is that the transformation $\delta \psi_L = \theta \psi_R$ simply fails to exist, at all, and is made to appear to exist by the admitted lack of intuitive transparency of Dirac's notation!

Consider a free Lagrangian of a left-handed fermion, $ i\overline{\psi_L}~ \partial\!\!/ ~~\psi_L = i\overline{\psi_L}~P_R\partial\!\!/ ~ P_L\psi_L $, where the r.h.side is written in perverse redundancy, at no extra cost, by the idempotence of chiral projectors! Now, apply the L to R transformation the scheme is predicated on to the last fermion, to generate an "increment" $i\overline{\psi_L}~P_R\partial\!\!/ ~P_L \theta ~ \psi_R $, stillborn by the action of the projection operator. Well, given the left-hand projector preceding it, you never really transformed the field: the O(θ) term is not there, before we even go onto transforming the $\bar\psi$s, let alone a complement of a free right-handed fermion lagrangian whose right handed fermions would rotate to left handed ones. This is not really a symmetry under a transformation, it is invariance under stasis!

The internal symmetries combined with this fundamental non-transformation are tensor-multiplied red herrings. I confess such are the hardest problems to resolve, namely studying unicorns which do not exist, because they couldn't....

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ There is a further structural reason why your nontrivial blend of internal (flavor) symmetries and the Lorentz group can't possibly work, and that is the Coleman-Mandula theorem, which, however, is notoriously hard to check for excluding hidden loopholes... So it excludes the SU(6) I am commenting on below as a bona-fide symmetry, but that SU(6) is still a useful classification symmetry in that case--not yours.... $\endgroup$ – Cosmas Zachos Jul 29 '16 at 15:27

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.