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Confinement implies chiral symmetry breaking

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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 (1,010 points) [ no revision ]

1 Answer

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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 (1,010 points) [ revision history ]
edited Feb 28, 2016 by NAME_XXX

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