• Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.


New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback


(propose a free ad)

Site Statistics

204 submissions , 162 unreviewed
5,026 questions , 2,180 unanswered
5,344 answers , 22,685 comments
1,470 users with positive rep
815 active unimported users
More ...

  Confinement implies chiral symmetry breaking

+ 4 like - 0 dislike

Recently I've read that confinement implies chiral symmetry spontaneous breaking in QCD, and this fact follows formally from anomaly matching conditions. Could someone explain me how anonaly matching conditions imply CSSB from the confinement?

asked Feb 24, 2016 in Theoretical Physics by NAME_XXX (1,060 points) [ no revision ]

1 Answer

+ 2 like - 0 dislike

Well, now I have an answer. The sketch of proof is following.

Suppose we have confining theory with chiral fermions and gauge group $G_{\text{gauge}}$; it has global symmetry $G$, which has not gauge anomalies and chiral gauge anomalies $GG_{\text{gauge}}^{2}$, but has anomaly $G^{3}$, i.e., symmetric tensor
d_{abc}^{G} \equiv \text{tr}[[t_{a}^{G}, t_{b}^{G}]_{+}t_{c}^{G}] - L\leftrightarrow R,

which arises in the triangle diagram, is nonzero. Finally, suppose that $G$ isn't spontaneously broken.

By using anomaly matching condition we have the next statement: the confined sector of theory, which is represented by the set of bound states (which belong to the representation of $SU_{L}(3)\times SU_{R}(3)\times U_{V}(1)$), has to reproduce $d_{abc}^{G}$. Now we have to discover which massless states are realized in effective theory.

First, due to anomalous equation
(k_{1}+k_{2})_{\mu}\Gamma^{\mu \nu \rho}_{abc}(k_{1}, k_{2}, -(k_{1}+k_{2})) = -\frac{d_{abc}}{\pi^{2}}\epsilon^{\nu \lambda \rho \beta}k_{1\lambda}k_{2\beta},

on the 3-point function $\Gamma$,
\Gamma_{\mu \nu \rho}^{abc}(x, y, z) = \langle \left|T\left(J^{a}_{\mu}(x)J_{\nu}^{b}(y)J_{\rho}^{c}(z) \right) \right| \rangle

(here $J_{\mu}^{a}(x)$ is the current of symmetry $G$), we have the statement that $\Gamma$ has the pole structure at $k_{1} = k_{2} = 0$. So that in effective field theory only massless bound states make the contribution into anomaly.

Second, if we haven't spontaneous symmetry breaking, then only helicity 1/2 massless bound states may exist in confined theory. Really, massless helicity zero bound states may exist only when global symmetry becomes spontaneously broken, existence of massless helicity $> 1$ bound states is forbidden in Lorentz-invariant theory due to Weinberg-Witten theorem, while massless helicity 1 bound state can't exist (again, we need Lorentz invariant theory) because of transformation properties of chiral current $j_{\mu}^{a}$ under the small group of lightlike 4-vector (namely, Euclide group).

Third (we restrict themselves to the group $G\sim SU_{L}(n)\times SU_{R}(n)\times U_{V}(1)$ and $G_{\text{gauge}} \sim SU(N)$), due to confinement the only possible massless fermionic bound state is combined from $m_{L}, m_{R}$ particles and $\bar{m}_{L},\bar{m}_{R}$ antiparticles which numbers satisfy the condition
m_{L}+m_{R}-\bar{m}_{L}-\bar{m}_{R} = Nk, \quad k \in Z

Precisely, it belongs to the representation $(r,s)$, where $r$ defines the direct product of $m_{L}$ on $\bar{m}_{L}$ of representations of $SU(3)$, $s$ defines the direct product of $\bar{m}_{R}$ on $m_{R}$ of $SU(3)$, and the $U_{V}(1)$ charge is $Nk$. You may find out that these representations consist of hypothetical massless baryons for the case of QCD ($n = N = 3$).

Now we can write anomaly matching conditions for the QCD: since
d_{abc}(SU_{L/R}^{3}(3)) = 3\text{tr}[[t_{a}, t_{b}]_{+}, t_{c}], \quad d_{abc}(SU_{L/R}^{2}(3)U_{V}(1)) = 3\text{tr}[[t_{a}, t_{b}]_{+}] = \frac{3}{2}\delta_{ab},

we have that for the representation of massless fermion bound state with given generators $T_{a}$, integer quantities $l(r, s, k)$ and the dimension $d_{s}$ of the $s$ representation

\sum_{r, s, k>0}l(r,s,k)d_{s}\text{tr}^{r}[[T_{a}, T_{b}]_{+}T_{c}] = 3\text{tr}[[t^{G}_{a}, t^{G}_{b}]_{+}, t^{G}_{c}], $$

$$\sum_{r, s, k>0}l(r,s,k)d_{s}\text{tr}^{r}[[T_{a}, T_{b}]_{+}] = \frac{1}{2} \qquad (1)$$

It can be shown by explicit calculations (details are given in 't Hooft paper "Naturalness, chiral symmetry and spontaneous chiral symmetry breaking") that there don't exist integers $l$ which satisfy $(1)$. So we come to the statement that $G \sim SU_{L}(3)\times SU_{R}(3)\times U_{V}(1)$ must be spontaneously broken. We do not know which subgroup is broken and which isn't by using this argument (we only know that $U_{V}(1)$ is unbroken due to Vafa-Vitten theorem). However, we establish now that the confinement in QCD formally implies necessarity of chiral symmetry breaking. Finally, we know that for wide range of chiral effective field theories we can relate anomaly sector of underlying theories to the Wess-Zumino term given in terms of goldstone bosons through topological reasons.

answered Feb 27, 2016 by NAME_XXX (1,060 points) [ revision history ]
edited Feb 28, 2016 by NAME_XXX

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
To avoid this verification in future, please log in or register.

user contributions licensed under cc by-sa 3.0 with attribution required

Your rights