Theta-dependence of QCD quark condensate for $m=0$

Question

I'm trying to understand the $\theta$-dependence of the following expression for the quark condensate in QCD,

$$ \langle \bar{\psi}\psi\rangle = - \Sigma \cos(\theta)$$

taken from Eqs. (5) and (7) of the paper "Dirac spectrum and chiral condensate for QCD at fixed $\theta$-angle" by M. Kieburg et al. Here, $\theta$ is the QCD vacuum angle and $\Sigma$ is the absolute value of the chiral condensate in the limit $m= 0$ and $\theta = 0$.

The expression above implies that the quark condensate is independent of the bare quark mass $m$ but dependent on the vacuum angle $\theta$. However, for $m=0$, the $\theta$-angle can be rotated away by a chiral quark rotation. Thus, how is it possible that the expression above stays the same, as claimed in the said reference and also in Eq. (69) of the paper "Massive Schwinger model within mass perturbation theory" by C. Adam on the analogous Schwinger model? In other words, how is it possible that the condensate depends on the angle $\theta$, even though $\theta$ should become unphysical when $m=0$?

Edit, in response to the comments below: Physical observables in QCD should become independent of $\theta$ in the chiral limit. Thus, is $\langle \bar{\psi}\psi\rangle$ is not a physical quantity, but only $|\langle \bar{\psi}\psi\rangle|$ which enters the pion masses and is independent of $\theta$?

Answer 1

You have to make a distinction between bare mass and effective mass.

The $m=0$ case in the paper is referring to the bare mass of the quarks, while the chiral condensation $$\langle \bar{\psi}\psi\rangle = - \Sigma \cos(\theta)$$ will dynamically generate a non-zero effective mass for the quark via the Hartree self energy correction.

The $\theta$ parameter is surely relevant for the condensation, where the chiral symmetry is dynamically broken by the non-zero effective quark mass.

And FYI, there is actually a pseudo scalar cousin of the aforementioned condensation $$ \langle \bar{\psi}\gamma^5\psi\rangle = \Sigma \sin(\theta), $$ which is also $\theta$-dependent.

As you can see, if you perform an axial rotation to get rid off the pseudo scalar portion of the condensation, you end up with a correction to the QCD $\theta$ term via the chiral anomaly channel.

Answer 2

My comment about $m$ having to be non-zero in the equation $$ \langle \bar q q\rangle= -\Sigma \cos\theta $$ is incorrect. I was being an idiot especially as I have just finished writing up some notes on a toy model where this formula is relevent.

Consider $m=0$ limit of the mode expansion
$$ \langle \bar q(x)(1\pm \gamma_5) q(x)\rangle = \langle\sum_n \frac{\bar u_n(x)(1\pm \gamma_5)u_n(x)}{i\lambda_n +m} \rangle $$ coupled with the fermion determinant $$ {\rm Det}(iD+m)= m^{|\nu|} \prod_{\lambda_n \ne 0}(\lambda_n^2+|m|^2) $$ where $|\nu|= $ is the number of zero modes of the Dirac operator $D$. The $\langle\ldots\rangle$ on the RHS is the average over gauge configurations wighted by the fermion determinant and by the QCD $\theta$ term when present. When there is a net instanton number $\nu$ then there are at least $|\nu|$ zero modes and the weighting from the fermion determinant will be zero unless $\nu=\pm 1$. In this case the factor of $m$ from the determinant will cancel the $m$ in the denominator of the mode expansion and give a contribution $\bar u_0(x)(1\pm \gamma_5)u_0(x)$ to $\langle \bar q(x) (1\pm \gamma_5)q(x)\rangle$ even in the limit $m\to 0$. These $x$-dependent contributions will be centered near the instanton and will become constants when we average over intanton locations. The $\nu =+1$ contribution to $\langle \bar q(x) (1+ \gamma_5)q(x)\rangle $ will be weighted by $e^{i\theta}$ and the $\nu=-1$ contribution to $\langle \bar q(x) (1- \gamma_5)q(x)\rangle $ will be weighted by $e^{-i\theta}$ to give
$$ \langle \bar q q(x)\rangle=\langle \bar q(x)(1+\gamma_5) q(x)\rangle +\langle \bar q(x)(1-\gamma_5) q(x)\rangle= \# e^{i\theta}+ \# e^{-i\theta}= -\Sigma\cos \theta $$ Thus the $\langle \bar q q\rangle$ vev does depend on $\theta$ even in the massless case. The statement that no physics depends $\theta$ in the case is still OK (I think) because we cannot measure $\langle \bar q q(x)\rangle$ directly. All physical effects really come from two-point correlators $\langle \bar q q (x)\bar q q(y)\rangle$. The theta vacua are introduced to make the ground state clustering in that $$ \lim_{|x-y|\to \infty}\langle \bar q q (x)\bar q q(y)\rangle_\theta= \langle \bar q q\rangle_\theta \langle \bar q q\rangle_\theta. $$ Clustering does not hold when we work in a fixed instanton number sector, but determining $\langle \bar q q\rangle_\theta$ requires intergrating over all space-time so it is inaccessible to us and therefore unphysical. The limit $$ \lim_{|x-y|\to \infty}\langle \bar q q (x)\bar q q(y)\rangle_{\nu=0} $$ is physically accessible to us, has physics consequences, but is $\theta$ independent.

Incidently I think that an improved version of equation of equation (6) in the original paper you cite is $$ Z= \exp\{ \sqrt{m_1^2+m_2^2} \cos(\theta+\phi)V \Sigma\} $$ Here we have mass term $m_1+i\gamma_5 m_2$ with $$ m_1= m\cos\phi, \quad m_2=m\sin\phi $$ and the $\theta$ is the analogue of $\theta_{QCD}$. The adavantage of this expression is that it depends only on the combination of angles $\theta+\phi$, which is what we know from the form of fermion determinant.

Then using $\phi = \tan^{-1}(m_2/m_1)$ we find that $$ \frac 1 V \frac{\partial}{\partial m_1} \ln Z= \Sigma\cos\theta \\ \frac 1 V\frac{\partial}{\partial m_2} \ln Z= -\Sigma\sin\theta $$ and the condensate in this approximation (which holds only when $mV\Sigma\ll 1$ and so we are so near the line of eignvalues that the Banks-Casher formula is irrelevent) is always independent of $\phi$