Sufficient and Necessary Conditions for Chiral Symmetry Breaking

Question

In their 2005 paper, the authors write (just below eq. 3.19)

we see that a non-zero value of $F_0$ is a necessary and sufficient criterion for spontaneous chiral symmetry breaking. On the other hand, (...) a non-vanishing scalar quark condensate $\langle \bar qq\rangle$ is a sufficient (but not a necessary) condition for a spontaneous symmetry breakdown in QCD

Why is it that

• $F_0\neq 0$: necessary and sufficient for ChSB
• $\langle \bar qq\rangle\neq 0$: sufficient, but not necessary for ChSB

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Additional notes

• $F_0$ is a decay constant, as in $$ \langle 0|A^a_\mu |\pi^b\rangle = \text{i} F_0 \delta^{ab} p_\mu $$

• If I understand correctly,

$$ F_0\neq 0 \color{red}\Leftrightarrow \text{ChSB}\quad\text{and}\quad \langle \bar qq\rangle\neq 0 \color{red}\Rightarrow \text{ChSB} $$

Answer 1

Yes, you understand correctly, $$ F_0\neq 0 \qquad \color{red}\Leftrightarrow \qquad\chi SB \tag{↯} \\ \langle \bar qq\rangle\neq 0 \qquad\color{red}\Rightarrow \qquad\chi SB . $$

In the latter case, of course, a charge annihilating the vacuum in (3.18) dictates a vanishing r.h.s.

However, conversely, imagine a (complete, fantasy) contrapositive hypothetical, in which $$ \langle \bar qq\rangle = 0 $$ but, for whatever unrealistic reason, you also have $$ \langle \bar uu + \bar dd -2\bar ss \rangle =c\neq 0, \qquad \Longrightarrow \\ \langle \bar uu + \bar dd \rangle = \langle 2\bar ss \rangle +c, $$ which is to say $Q^A_8|0\rangle \neq 0$; so, for $c=-3 \langle \bar ss \rangle $ you may still have $\chi SB$, cf. (3.17), a=8.

For the former case, (↯), reversing the implications of the contrapositives and mindful that $F_0\sim v$ in the crucial (2.21) of section 2.2, an axial charge annihilating the vacuum will perforce dictate a vanishing $v \sim F_0$.

Conversely, the entire argument of (2.22) has no force behind it for vanishing $v \sim F_0$: you do not need massless states popped out of the vacuum by chiral charges, anymore. This is the "normal" chiral symmetry case where all fields have $\langle 0|\Phi ^b|0 \rangle =0$, so $\langle 0|Q_A^a|\phi^b \rangle =0$ for all a,b s, that is all Q s kill the vacuum (it is not as though they project it to spaces orthogonal to all states).

Answer 2

I think what people have in mind when the say that is the following:

1) (Spontaneous) Chiral symmetry breaking implies the existence of Goldstone modes, and the symmetries of the effective lagrangian then imply a non-zero coupling of the Goldstone mode to the axial current. $F_\pi$ then has to be non-zero.

2) We could imagine, however, that the order parameter is not $\langle \bar\psi\psi\rangle$, but some other operator sensitive to chiral symmetry, say $\langle \bar\psi\sigma\cdot G\psi\rangle$, or $\langle (\bar\psi\Gamma\psi)^2\rangle$, for a suitable matrix $\Gamma$.

3) The problem with this idea is that we need to explain what prevents $\langle \bar\psi\psi\rangle$ from acquiring a VEV once chiral symmetry breaking is broken by some other order parameter. A possible suggestion is that there is an unbroken discrete symmetry (a $Z_2$, for example) that prevents bilinears from acquiring a VEV, but allows four fermion operators. In QCD, this is known as the Stern scenario.

4) In the context of QCD, the Stern scenario was shown to be ruled out on general grounds, most recently using discrete anomaly matching. I am not aware of any theories that exhibit generalized chiral symmetry breaking, but there are cases where it is realized approximately, most notably in the CFL phase of high density QCD.

5) To summarize: $F_\pi\neq 0$ is indeed necessary and sufficient. In general, it is clear that $\langle\bar{q}q\rangle\neq 0$ is sufficient, but not that it is necessary. In QCD (and other gauge theories), however, we can show that it is also necessary.