# A SU groups in QCD

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1. Oct 22, 2016

### Phis

I am trying to learn about the various SU groups related to QCD. I have about 5 QFT and Particle physics books from my student library and written down about 20 pages of handwritten notes about specific parts of say generators, matrices, group properties etc. - but i dont really feel that I understand the deeper connection between all this. Can someone (preferably) in simple words and not too many equations, explain what relevant SU symmetries connect QCD and HOW?

To me it seems impossible to get a feeling of the big picture here. I feel that when i zoom in on a specific part it gets somewhat clear, but i need to understand the bigger picture. I am specifically interested in the connection between chiral symmetry - the left/right-handedness breakdown to vector/axial symmetry - what is happening here? I can understand that this can be done in SU(2) handeling only u and d quarks - or SU(3) handeling u, d, and s quarks. One is a better approximation than the other - again why?
How does this relate to the 8 Goldstone bosons (3 pions, 3 kaons, 2 eta) emerging when going from SU(3)_L x SU(3)_R to only SU(3)_V
The same can be said for the SU(2) "version" of this, where the breakdown gives rise to 3 Goldstone bosons (the 3 pions). I know that theory says they need to be massless, but how do we then use it as a legit model strong interactions?

i have read on the linear sigma model, eightfold way, chiral symmetry transformations, rotational transformations, gauge invariance among many things. What i really want is to connect some of the ideas. I know i am asking an extremely broad question, so feel free to choose any smaller domain you would like :-)

Thank you very much!

2. Oct 22, 2016

### Dr.AbeNikIanEdL

The $SU(3)_L \times SU(3)_R$ (or $SU(2)$) symmetry is exact for massless QCD. Now the masses of up and down quarks are smaller than that of the strange quark. Hence approximating only their masses by zero gives a better description.

You have a spontaneous symmetry breaking from $SU(3)_L \times SU(3)_R$ to just one $SU(3)_V$ group. This brings you from 16 generators (for two $SU(3)$ groups) to 8 Generators for just one $SU(3)$ group. By the Goldstone theorem, you get 8 massless Goldstone bosons (one for each broken generator). The same way you get 3 goldstone bosons by breaking $SU(2)_L \times SU(2)_R$ with 6 generators to 3 for just one $SU(2)$ group.

Finally, all the $SU(2)$ or $SU(3)$ symmetries are only approximate, because there are mass terms for quarks in QCD which explicitly break the symmetry. So the Goldstone theorem will only hold approximately, and the goldstone bosons become massive. Their masses stay however small, compared to e.g. nucleon masses, as you might expect as long as spontaneous symmetry breaking is the dominant effect.

3. Oct 23, 2016

### vanhees71

For a very nice introduction to chiral symmetry in the strong ineraction and chiral perturbation theory, see

https://arxiv.org/abs/nucl-th/9706075

4. Oct 23, 2016

### Phis

I can understand that when you work with a massless lagrangian, you don't get mixed terms with left and right-handed quarks - this ultimately means that the equations treat the two chiralities indenpendently. But what is different about the vector symmetry group $SU(3)_V$ that this breaks into. I can see in my books that you get a form of pairing of $SU(3)_{L+R}$ and $SU(3)_{L-R}$ where one rotates the two chiralities together or opposite, but im confused about what the vector symmetry in general means physically. How and why does it change anything?
I can see that one would also get an axial symmetry group and not only vector symmetry group, such that the whole symmetry break for lets say u and d quark-system goes from $SU(2)_L \times SU(2)_R \times U(1)_L \times U(1)_R$ to $SU(2)_{L+R} \times SU(2)_{L-R} \times U(1)_V \times U(1)_A$ how does this change things? I know that $SU(2)_A$ is chiral symmetry and $SU(2)_V$ is isospin symmetry, but what does it mean that it breaks down to this?

I have a bit of a stupid (i feel it should be obvious) question regarding the meaning of the x (direct product) between the groups in the notation $SU(3)_L \times SU(3)_R$. Why do we write it like a product of the group generators?

Now the 8 massless Goldstone bosons that are created, can achieve mass through explicit symmetry breaking and then using chiral pertubation theory we get small masses for the 8 mesons (pion, kaon, eta), or you can just leave it at spontanous symmetry breaking and achieve pseudo-scalar mesons that are massless. From what i understand, this way of understanding QCD is just as a more simple model to get the hadrons to form/appear in the first place (predict them so to speak) - because we of course can not and have not observed isolated quarks. But to do full on QCD with more correct results (not massless hadrons) )we can continue from the massless bosons to achieve mass by chiral pertubation theory. So is it correct that all this is, is just a "simple" way of using symmetry and quarks to explain or argue how and why simple mesons are created?

I know this is a lot of questions and maybe to overwhelming to explain - but it really helps my understanding and i am extremely grateful! This is a new area for me, and i am using a part of it on my bachelor thesis, but i have not had any QFT or advanced Particle Physics, only introductory nuclear and particle physics. I am starting later today to read up on the article you have posted vanhees71 - thank you very much. I have a lot of similar articles, but this one looks a bit simpler with a lot of text explanations, hoping to form an intuition for this.

Last edited: Oct 23, 2016
5. Oct 23, 2016

### vanhees71

First of all note that "$\mathrm{SU}(N)_A$", where $A=L-R$ denotes the transformations that are generated by the axial charges is a wrong expression since these transformations do not form an SU(N) subgroup of $\mathrm{SU}(N)_L \times \mathrm{SU}(N)_R$. The vector charges, however generate a subgroup $\mathrm{SU}(N)_V$. In the SU(2) model you can approximately assume $m_u \simeq m_d$, and then $\mathrm{SU}(2)_V$ (the isospin group) is a symmetry. In reality it's as much explicitly broken as chiral symmetry (a few MeV of current-quark masses vs. the typical hadronic scale of 1 GeV; formaly the cutoff-scale for chiral symmetry $4 \pi f_{\pi}$). I hope that answers the first question, and you never ever claim again that there's a group generated by the axial-vector charges! It's a coset!

Now concerning symmetry breaking. In the chiral limit, i.e., vanishing quark masses the physics of hadrons is still not explicitly symmetric under chiral rotations but only under isospin rotations. The reason is the formation of a quark condensate $\langle \bar{\psi} \psi \neq 0$, i.e., the QCD ground state is not symmetric under chiral rotations but only under isospin rotations although the dynamics is symmetric under the full chiral group. This implies that the symmetry is spontaneously broken to $\mathrm{SU}(N)_V$. This implies that you have massless Goldstone bosons. In the SU(2) models these are the pions, in the SU(3) model the pions, kaons and eta.

You are right, there are also of course the $\mathrm{U}(1)$ subgroups (scalar and axial). The axial U(1) is anomalously broken, i.e., it is not possible to find a regularization of the quantum corrections (e.g., the triangle diagram with one axial and two vector currents) that keeps both U(1)'s intact. You can choose which current should not be conserved, but gauge invariance forces you to break the axial symmetry. So there are only 8 but not 9 pseudo-Nambu-Goldstone pseudoscalars in the hadron spectrum.

Chiral symmetry in the light-quark sector (u, d, s) is used as a tool to build effective hadronic models as a description of QCD at low energies, where perturbation theory is not available since at low energy the QCD coupling constant becomes large ("asymptotic freedom").

6. Oct 25, 2016

### Phis

Alright i have read the article from arxiv that you linked to - and i must say that was exactly what i needed! That really helped on my understanding.

I tried to write down the basic structure of symmetries that describe the chiral symmetry break, and i now have some final questions:

We can have $SU(2)_V \times SU(2)_A$ that breaks into $SU(2)_{L+R} = SU(2)_V$

that arrises from the generators for isospin $\Lambda_{V} = \psi \rightarrow e^{-i \frac{\tau}{2} \theta} \psi$ and $\bar{\psi} \rightarrow e^{i \frac{\tau}{2} \theta} \bar{\psi}$

and axial $\Lambda_{A} = \psi \rightarrow e^{-i \gamma_{5} \frac{\tau}{2} \theta} \psi$ and $\bar{\psi} \rightarrow e^{-i \gamma_{5} \frac{\tau}{2} \theta} \bar{\psi}$

and then we can also have $SU(2)_L \times SU(2)_R \times U(1)_V \times U(1)_A$ that breaks into $SU(2)_{L+R} \times U(1)_V \times U(1)_A$ which basically says that in the beginning chiral symmetry treats left- and right-handed chiralities independentely but after the SSB the vector symmetry treats them equally - but still axial symmetry treats them differently. I understand the whole part of the Goldstone bosons, and the number of broken generators. My questions are only focused on the structure of the SU and U "representation" of the symmetry.

here the generators for $U(1)$ are $\psi \rightarrow e^{i \alpha \gamma^{5}} \psi$ and $\bar{\psi} \rightarrow \bar{\psi} e^{i \alpha \gamma^{5}}$

Where $U(1)_A$ is being broken on a quantum level due to the anomaly.

What is exactly happening to the treatment of how the symmetries treat the handedness, and why? And why do we have two different versions of the symmetry breaking - maybe they illustrate the same thing?

And as asked before, if anyone could answer the question about how to interpret the cross between two groups - why a direct product, what does it indicate? Why not direct sum or something else?

Thank you very much!

7. Oct 26, 2016

### vanhees71

Again, the expression $SU(2)_A$ is bad notation. It's NOT an SU(2) subgroup. The correct spontaneous-symmetry breaking pattern of the chiral isospin-symmetry is $\mathrm{SU}(2)_R \times \mathrm{SU}(2)_L \rightarrow \mathrm{SU}(2)_V$. Note thate $\mathrm{SU}(2)_V$ makes sense, because this is really a subgroup of the isospin group. You should also note that all these symmetries are explicitly broken by the u- and d-quark masses, which are however small compared to the scale $4\pi f_{\pi} \simeq 1 \; \mathrm{GeV}$.

For a detailed discussion about the $U(1)_A$ anomaly, see my QFT manuscript

http://th.physik.uni-frankfurt.de/~hees/publ/lect.pdf

8. Oct 26, 2016

### Phis

I can see that now from your paper that you have linked to. You explain it well on page 264. However in the paper that you linked to in post #3 it clearly uses that notation between equation 48 and 49 on page 9. But i am not saying that it is good notation, i was just confused because they used it.

And yes i have noted how everything gets explicitly broken with mass terms. In that sense the symmetries are only approximate.

9. Oct 26, 2016

### vanhees71

Argh. That's really bad :-(. Nevertheless overall it's a good review, from which I learnt a lot as a diploma student.