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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 $\theta$?

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    $\begingroup$ What is $\Sigma$? (Also, please link to papers at least mentioning their names and authors, so that they can be reconstructed in case of link rot, and please link to arXiv abstract pages rather than directly to the PDF) $\endgroup$ – ACuriousMind Jul 26 at 14:44
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    $\begingroup$ You should read arxiv.org/pdf/1407.8393.pdf first. This explain what depends on $m$ and what does not. Thanks for asking the question though. $\endgroup$ – mike stone Jul 26 at 15:16
  • $\begingroup$ Thanks for the reference. The authors give the same expression as above in Eq. (2) but do not discuss the $\theta$-dependence of this expression for $m=0$. If $\theta$ is unphysical, how can $\langle\bar{\psi}\psi\rangle$ depend on it? That's what puzzles me. I've tried to phrase my question above more clearly. $\endgroup$ – Thomas Jul 26 at 15:32
  • $\begingroup$ The reference I cited makes clear that the formula $-\Sigma \cos\theta$ only applies when $m\ne0$. $\endgroup$ – mike stone Jul 26 at 16:54
  • $\begingroup$ Do you know the formula for $m=0$? The paper I cited above claims that $\Sigma\cos\theta$ is also given for $m=0$. $\endgroup$ – Thomas Jul 26 at 18:00
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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 $\phi$

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  • $\begingroup$ Thanks a lot for your answer! Are you sure that the full condensate is proportional to $\cos\theta$ and not only the real part? Equation (11) of arXiv:1408.1189 argues that the full condensate is proportional to $i\theta$, thus the real part contains $\cos\theta$ and the imaginary part $\sin\theta$. In total, physics (e.g. the meson mass) depends only on the absolute value of $\langle\bar{\psi}\psi\rangle$, see Eq. (6) of the scholarpedia article on chiral perturbation theory. $\endgroup$ – Thomas Jul 27 at 23:12
  • $\begingroup$ Also, for $m=0$ we have an anomalous $U(1)$ symmetry in QCD, similar to the Peccei-Quinn symmetry in the axion case. This symmetry gets spontaneously broken, giving rise to the $\eta'$ meson, similar to the axion in the latter case. From the equations of motion for the $\eta'$ meson, we see that the $\theta$-angle gets absorbed in the $\eta'$ field. The minimum of the potential is at $\theta=0$, thus $m=0$ dynamically enforces $\theta=0$ in the vacuum. Do you agree? $\endgroup$ – Thomas Jul 27 at 23:29
  • $\begingroup$ @Thomas. I don't think that $\theta$ is forced to be zero, it just becomes irrelevent. This is the "vacuum freezing" mecanism of Kogut and Susskind. The same thing happens in the massless Schwinger model where the bosonized verion has an induced mass term $(e^2/\pi)(\theta-\theta_\infty)^2$ for the analogue of the $\eta'$. A global chiral transform $\theta\to \theta+a$ has no effect as $\theta$ and $\theta_\infty$ rotate togther, but a local change in $\theta$ costs energy. $\endgroup$ – mike stone Jul 28 at 12:34
  • $\begingroup$ Okay, thanks! Do you agree that the full condensate is proportional to $i\theta$ and not to $\cos\theta$, because it's just a phase? $\endgroup$ – Thomas Jul 29 at 11:24
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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.

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  • $\begingroup$ Thanks for your answer, but I don't fully understand it. If the bare quark mass is zero, the $\theta$-angle can be rotated away and becomes unphysical, independently of the effective quark mass. You're saying that the $\theta$-parameter is relevant for condensation, but quarks condense even if the hard mass is zero, which should make $\theta$ unphysical. I still don't understand how an unphysical parameter can enter the expression for the quark condensate for $m_{\rm bare}=0$, this was my question. $\endgroup$ – Thomas Jul 26 at 16:12
  • $\begingroup$ @Thomas, the condensation is PHYSICAL. To make it scalar like (rotating away the pseudo scalar portion), you wind up with a specific $\theta$. $\endgroup$ – MadMax Jul 26 at 16:16
  • $\begingroup$ Yes, the condensation is physical, but the parameter $\theta$ should be unphysical for $m_{\rm bare}=0$. So what do you mean with "winding up" here? $\endgroup$ – Thomas Jul 26 at 16:18
  • $\begingroup$ @Thomas, if the condensation is physical, the parameter $\theta$ is physical, capiche? $\endgroup$ – MadMax Jul 26 at 16:21
  • $\begingroup$ @Thomas, probably you need some background on the premise: "If the bare quark mass is zero, the $\theta$-angle can be rotated away". Why? It is because a scalar non-zero mass will be $\theta$-rotated into a "complex" (scalar + pseudo-scalar) mass, while zero mass doesn't have this issue. The situation is exactly the same here with the condensation by just replacing "mass" with "condensation" in the above argument! $\endgroup$ – MadMax Jul 26 at 16:44

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