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

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,

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

  • $\begingroup$ I can't follow your reasoning why/how $Q_A^a |0\rangle\neq 0$ leads to $\langle \bar qq\rangle=0$. In order to assume that $\langle \bar uu+\bar dd-2\bar ss\rangle=c\neq 0$, we'd have to assume a broken $Q_V^8$ from $SU(N)_V$..? $\endgroup$
    – ersbygre1
    Jul 29, 2019 at 6:04
  • 1
    $\begingroup$ Nononono... you are misreading the logic: S&S, sec 3.2.2, show you that $Q^a_A|0\rangle \neq 0$ allows a scalar condensate to be 0, it certainly does not lead to it. It does compel a nonvanishing asymmetric condensate, as stated! The logic flow point is, however unlikely or undesirable that is, the scalar condensate is up for grabs. $\endgroup$ Jul 29, 2019 at 15:15
  • $\begingroup$ See, that's the thing: I don't see where they show that $\langle S^0\rangle = \langle \bar qq\rangle = \langle \bar uu+\bar dd+\bar ss\rangle=0$ is allowed when $A_A^a|0\rangle \neq 0$. Could you elaborate on that? $\endgroup$
    – ersbygre1
    Jul 30, 2019 at 0:27
  • 1
    $\begingroup$ There is nothing they can do to restrict it! (Like they did (3.14) --> (3.15).) Eqn (3.11) gives no traction to the SU(3) generators, so you know nothing about the scalar condensate, at this level. This might be useful. $\endgroup$ Jul 30, 2019 at 1:00
  • $\begingroup$ Thank you, this is very helpful. I think I can narrow down my question: If we assume $\langle\bar qq\rangle=0$ in eq. (3.16), which must lead to $\langle \bar uu\rangle = \langle \bar dd\rangle = \langle \bar ss\rangle = 0$, it also leads to 0 on the r.h.s. of eq. (3.18). And now, either $P^a$ or $Q_A^a$ can have a non-zero matrix element with the vacuum and the massless state $\phi_a$. So even with $\langle \bar qq\rangle = 0$, we could have a non-zero $F_0$, which definitely implies SSB. Is this chain of thoughts valid? $\endgroup$
    – ersbygre1
    Jul 30, 2019 at 1:27

2 Answers 2


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).

  • $\begingroup$ I think you would have to explain what prevents $\langle\bar{q}q\rangle$ from acquiring a VEV. There is an old idea that it could be a discrete symmetry. $\endgroup$
    – Thomas
    Jun 7, 2019 at 16:06
  • $\begingroup$ @Thomas Sorry, this is resoundingly off the table in this question, a strictly technical issue of logical implications, not the essence of chiral symmetry. I am not addressing mechanisms of condensation or anything of the sort--I thought my dismissive comments of detail made that clear. I am only illustrating how the premise is unnecessary, despite its being sufficient. $\endgroup$ Jun 7, 2019 at 16:17
  • $\begingroup$ I'm not sure I'm getting your point. 1) There is no need to be dismissive, whether $\chi SB$ breaking implies a non-zero vev of $\bar{q}q$ is a legitimate question. 2) In general the answer would surely seem to be yes; once chiral symmetry breaking is broken, what would prevent $\bar{q}q$ from acquiring a vev? In QM and QFT, everything that is allowed is mandatory. 3) A possble answer is a discrete symmetry, like a $Z_2$. $\endgroup$
    – Thomas
    Jun 7, 2019 at 16:54
  • $\begingroup$ I am only answering the OP's question on logical implications, not what actually happens in QCD, or alternative mechanisms: this is not a forum for alternate $\chi SB$ scenarios and their causes and mechanisms. Perhaps you wish to ask a separate, disjoint question on that, but steering the OP's attention there might well not argue away his puzzlement. $\endgroup$ Jun 7, 2019 at 17:02
  • $\begingroup$ Maybe I am still misreading your argument. What is your claim? You argue that $\langle\bar{q}q\rangle \Rightarrow \chi SB$ but not $\langle\bar{q}q\rangle \Leftrightarrow \chi SB$? Or do you argue that indeed $\langle\bar{q}q\rangle \Rightarrow \chi SB$, but you have no opinion on the reverse implication? $\endgroup$
    – Thomas
    Jun 7, 2019 at 17:09

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.


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