# Breaking of $SU(3)$ symmetry by bi-fundamental representation

Are there any general theorems which fix the possible symmetry breaking patterns of Lie groups (such as $SU(3)$) by vacuum expectation values of fields in specific representations (such as the quark condensate in the bi-fundamental representation)?

Let me specify this question:

In the example of QCD, the approximate $SU(3)_L\times SU(3)_R$ flavor symmetry of the $(u,d,s)$ quarks is spontaneously broken down to $SU(3)_V$ by the quark-bilinear condensate, $|\langle\bar{q}q\rangle|$.

What kind of other symmetry breaking patterns, apart from $SU(3)\times SU(3)\rightarrow SU(3)$, are allowed for such bi-fundamental order parameters? Can such order parameters generally also break $SU(3)\times SU(3)\rightarrow U(1)$ or even $SU(3)\times SU(3)\rightarrow 1$, irrespectively of the given potential?

Edit in response to the comment by Cosmas Zachos:

I mentioned "irrespectively of the given potential" to emphasize that I am asking whether there is a general statement, which might forbid the existence of a bilinear order parameter, such as

$$|\langle\bar{x}x\rangle|= v \begin{pmatrix} a & b & 0 \\ c & d & 0 \\ 0 & 0 & 0 \end{pmatrix},$$

which breaks $SU(3)\times SU(3)\rightarrow U(1)$, or

$$|\langle\bar{x}x\rangle|= v \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix},$$

which breaks $SU(3)\times SU(3)$ completely for $a\neq ... \neq i$.

In other words, is there anything that prevents me from writing down a (probably non-renormalizable) potential that gives me these vevs and breaks $SU(3)\times SU(3)$ down to $U(1)$ or $1$?

• There is a quite nice review by Slansky: sciencedirect.com/science/article/pii/0370157381900922 (Unfortunately paywall, but you might have access if you e.g. are or know someone at a university.) – Toffomat Feb 8 '18 at 14:30
• Thanks for the reference. However, Slansky only treats the standard adjoint representations and not bi-fundamental representations when he discusses possible symmetry breaking patterns. – Thomas Feb 8 '18 at 15:19
• – Cosmas Zachos Feb 8 '18 at 16:37
• Do you know the answer of this physics.stackexchange.com/questions/387091/… by any chance? – annie heart Feb 18 '18 at 4:27

Asking for theorems might be overkill, in the absence of a firm grasp of the construction, and they would hinder such a grasp, rather than enhance it. Nevertheless, 30 years ago, technicolorists beat such theorems to death. In any case, chiral symmetry breaking dynamics particularities are not well-understood to the point of excluding potential condensation outcomes.

What I'll do here is illustrate the point and answer your questions for the simplest model that I assume you learned on week one of your course, based on just SU(2) and its generators, the half of the prosaic three Pauli matrices.

So consider your chiral model bifundamental, $$\Phi =\phi_0 {\mathbb 1}+ \vec{\phi} \cdot \vec{\sigma},$$ whose columns transform as $SU(2)_L$ doublets on the left, and its rows like $SU(2)_R$ doublets on the right.

A v.e.v. (condensate) matrix $\langle \Phi\rangle=M$ may or may not be left invariant under parts of the two SU(2) s. So your problem amounts to which v.e.v. s are left intact by which subgroups of the full six generator $SU(2)_L\times SU(2)_R$ group, $$e^{i\vec{\theta}_L \cdot \vec{\sigma}/2} M e^{-i\vec{\theta}_R \cdot \vec{\sigma}/2} =M ~, ~~\Longrightarrow \\ \vec{\theta}_L \cdot \vec{\sigma} ~M - M ~\vec{\theta}_R \cdot \vec{\sigma}=0 ~.$$

You already appreciated that the 3 axials are gone and broken, and at most the three diagonal generators ($SU(2)_V$ so, $\theta_L=\theta_R$) may survive by virtue of the condensate $$M= v {\mathbb 1},$$ since the identity commutes with all three diagonal (vector) generators, $$[\vec{\theta}\cdot \vec{\sigma} ,{\mathbb 1}]=0 ~.$$

What happens if we complicate M and tack onto the identity a further piece $$M= \vec{v}\cdot \vec{\sigma} ~? ~~~~\Longrightarrow\\ [\vec{\theta}\cdot \vec{\sigma} , \vec{v}\cdot \vec{\sigma}]=2i(\vec{\theta}\times \vec{v} )\cdot \vec{\sigma}~.$$

For the two θ directions (generators) perpendicular to $\vec v$, the cross product survives, so the r.h.s. cannot vanish, so these two generators are gone and broken. But for the direction parallel to $\vec v$, the corresponding generator survives, as the cross product collapses and the r.h.side vanishes to ensure invariance. So, indeed, the residual diagonal group is a vector U(1).

You cannot kill this U(1) with a condensate, as you know multiplication of the v.e.v. by a complex number and its conjugate will always preserve it. (So, forget about the singlet outcome.)

• You should be able to trivially generalize this to SU(3) which has several nifty SU(2) subgroups, as well as U(1) s, etc. It's really straightforward. (Hint: what does a term proportional to $\lambda_8$ do? This actually obtains in our real world: condensation, trailing vector strangeness explicit breaking; cf. §4.1.2 of Scherer 2002. Of course, in real life, isospin $\lambda_3$ is broken as well, and all three condensates differ.)

• NB in response to question's edit . No, the full matrix you wrote down does not break you down to a singlet... it can be written as an identity plus a linear combination of Gell-Mann matrices, so, then a generator in the Lie algebra of SU(3) : it then defines an unbroken direction commuting with it. Construction of potentials dialing-in specific triplets (~ diagonal 3×3 M s) is standard and routine.

• Thanks a lot for your detailed answer, that helped me a lot. If there is really no way of spontaneously breaking the residual U(1) symmetry with the bifundamental, which possibilities do we have to break this U(1)? – Thomas Feb 9 '18 at 16:16
• I don't know of anything... Remember, anything is invariant under multiplication by a (commuting) phase and its c.c. from the other side. – Cosmas Zachos Feb 9 '18 at 16:34
• If we assume for a second that the $(u,d,s)$ quarks are all massless and some bifundamental breaks the resulting exact $SU(3)\times SU(3)$ flavor symmetry, then wouldn't a residual $U(1)$ symmetry imply that at least one quark has to stay massless? How can such a strong physics statement be compatible with something as "unphysical" as a phase multiplication? – Thomas Feb 9 '18 at 16:48
• Oh, I see! You misread my point: I said "tack on" the extra piece to the standard, identity piece. Remember, we are not talking about the axials--they've all been broken right from the start, and all 3 quarks are massive--you accepted that. We were further fussing about the inequality of their masses, I thought... – Cosmas Zachos Feb 9 '18 at 16:54
• What is then the physical interpretation of this residual U(1) symmetry? If we break $SU(3)\times SU(3) \rightarrow U(1)\times U(1)\times U(1)$, aren't the $U(1)$s the individual quark (flavor) number symmetries? If one of the $U(1)$s needs to stay unbroken, does this have any consequence for the quark mass inequality and mixing? – Thomas Feb 9 '18 at 17:32

The phase structure of gauge theories as a function of $N_c$, $N_f$ (and the type of fermion representation), the theta angle, the chemical potential, and the temperature (and other control parameters) is very rich and not fully understood. There are some general statements.

1) Witten and Vafa showed that in QCD-like theories at zero theta angle and chemical potential vector-like symmetries (isospin, flavor, baryon number) do not break. This means that in $N_f=3$ QCD (at zero baryon density) the symmetry cannot break any further than $$SU(3)_L\times SU(3)_R\times U(1) \to SU(3) \times U(1)$$ because the remaining $SU(3)$ and $U(1)$ are vectorial.

2) 't Hooft argued that the anomalies in the low energy and microscopic theory must match. This is not quite powerful enough to fix the symmetry breaking pattern in QCD with three colors and three flavors, but it does fix the symmetry breaking in the large $N_c$ limit (modulo "reasonable" assumptions) to the conventional scenario $SU(3)\times SU(3)\times U(1) \to SU(3)\times U(1)$.

3) One can show that in QCD chiral symmetry breaking does imply $\langle\bar\psi\psi\rangle \neq 0$. In particular, one can rule out a scenario where chiral symmetry is broken, but $\langle\bar\psi\psi\rangle = 0$ and the order parameter is a suitable four-quark condensate, $\langle(\bar\psi\psi)^2\rangle \neq 0$.