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The chiral condensate serves as an order parameter for the chiral phase transition. Thus, it is a finite quantity in one phase and vanishes in the other phase. It is given as a vacuum expectation value of quark fields, e.g., $\langle \Omega |\bar\psi \psi | \Omega\rangle$. In QCD sum rules, for example for two quark flavors, it is related to the pion decay constant, pion mass, and quark masses (known as Gell-Mann--Oakes--Renner relation) via

$$\langle \Omega |\bar\psi \psi | \Omega\rangle = - \frac{\displaystyle f^2_\pi m^2_\pi}{\displaystyle 2(m_u + m_d)}$$

which are all positive quantities on the right-hand side such that the condensate must be negative.

Let me add the definition to be more precise:
The condensate is given as trace over the quark propagator ${\rm S}_q(x,y)= -i \langle \Omega |{\rm T} \{ \psi(x) \bar\psi(y) \}| \Omega\rangle$ such that

$$\langle \Omega |\bar\psi(x) \psi(0) | \Omega\rangle := -i \rm{Tr} \lim\limits_{x\to 0} S_q(x,0)\ .$$


How can a vacuum expectation value become a negative quantity?
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    $\begingroup$ You might, or might not, like Creutz 05. $\endgroup$
    – Cosmas Zachos
    Commented Dec 16, 2021 at 20:51
  • $\begingroup$ Crudely, the - sign comes from the transposition of the fermions in the trace of their propagator, cf this... evaluated with care. $\endgroup$
    – Cosmas Zachos
    Commented Dec 16, 2021 at 21:45
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    $\begingroup$ note that the condensate is basically $\langle v|\gamma^0|v\rangle$, where $|v\rangle=\psi|\Omega\rangle$. Hence, the sign depends on your gamma-matrix conventions: if $\gamma^0$ is positive definite, then the condensate will be positive, while if negative-definite, it will be negative. If you define Cliff as $\{\gamma^\mu,\gamma^\nu\}=s \eta^{\mu\nu}$, then the sign of the condensate will be $s\eta^{00}$. $\endgroup$
    – AccidentalFourierTransform
    Commented Dec 22, 2021 at 15:25
  • $\begingroup$ @AccidentalFourierTransform Thanks for your comment. But the sign of the chiral condensate should be (and I believe is) independent of the representation of the Clifford algebra. Often, an explicit negative value for the chiral condensate is quoted in other QCD approaches (e.g. see OPE in sum rules) which I try to understand why it is 'chosen' negative. To my understanding, it is not a matter of convention. Could it be that there are other low-energy theorems applied (G-parity etc) that fix/dictate the sign (from a physics point of view)? $\endgroup$
    – Bernd
    Commented Dec 22, 2021 at 15:57
  • $\begingroup$ @Bernd If you redefine $\gamma^0\to-\gamma^0$, the sign of the condensate changes. $\endgroup$
    – AccidentalFourierTransform
    Commented Dec 22, 2021 at 16:54

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