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In this Review paper in p.1462, bottom left: Rev.Mod.Phys.80:1455-1515,2008 -- Color superconductivity in dense quark matter

It says that "There is an associated gauge-invariant 6-quark order parameter with the flavor and color structure of two Lambda baryons, $$ \langle\Lambda\Lambda\rangle $$ where this order parameter distinguishes the color flavor locking (CFL) phase from the quark gluon plsma QGP.

I suppose that it means the 6 quark condensate is $$ \bigl\langle(\epsilon^{abc}\epsilon_{ijk}\psi^a_i\psi^b_j\psi^c_k) (\epsilon^{a'b'c'}\epsilon_{i'j'k'}\psi'^a_i\psi'^b_j\psi'^c_k)\bigr\rangle, $$

  1. but how does this distinguish CFL from QGP?

  2. Is this operator precise? And is this gauge invariant under SU(3)???

  3. It is a Lorentz scalar or pseudo scalar?

It seems that the claim is not clear.

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  • $\begingroup$ It looks gauge invariant, because $\epsilon$ tensor is SU(3) invariant. About the Lorentz structure, there is a problem with your expression. You have $6$ $\psi$ fields and no $\bar \psi$. Therefore I think this correlation function vanishes. Perhaps you meant something like $\bar \psi^3 \psi^3$. In this case you still need to specify what you do with bispinor indices of $\psi$ and $\bar \psi$. $\endgroup$
    – Blazej
    Dec 25, 2017 at 2:23
  • $\begingroup$ can we show ϵ tensor is SU(3) singlet? $\endgroup$ Dec 25, 2017 at 2:36
  • $\begingroup$ Yes, the $\epsilon$ tensor is how one constructs a singlet out of fundamentals. Georgi's group theory book might be a useful place to check this out if it isn't familiar. $\endgroup$ Dec 25, 2017 at 3:18

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  1. It breaks $U(1)_B$, and therefore distinguished QGP from CFL.

  2. Yes, this is a gauge invariant operator.

  3. This is a Lorentz scalar if the spinors are contracted appropriately, for example $$ \phi \sim \epsilon_{\alpha\alpha'}\epsilon_{\beta\gamma}\epsilon_{\beta'\gamma'} (\psi_\alpha\psi_\beta\psi_\gamma)(\psi_{\alpha'}\psi_{\beta'}\psi_{\gamma'}) $$ In 4-component notation this can be written in terms of a (positive parity) baryon current $$ \phi \sim \Psi C\gamma_5 \Psi, \qquad\Psi_\alpha = \psi_\alpha (\psi C\gamma_5\psi) $$

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  • $\begingroup$ thanks +1, is this 6-quark term related to breaking the axial $Z_6$ of QGP to the $Z_2$ vector symmetry (same as axial) of CFL? or am I wrong??? $\endgroup$ Dec 25, 2017 at 2:30
  • $\begingroup$ how do one contracts? you said: "This is a Lorentz scalar if the spinors are contracted appropriately." Do we need some gamma matrices? $\endgroup$ Dec 25, 2017 at 2:31
  • $\begingroup$ also, how about this one? physics.stackexchange.com/questions/376203 $\endgroup$ Dec 25, 2017 at 2:44
  • $\begingroup$ Instantons break $U(1)_A$ to $Z_6$ in both the QGP and CFL. CFL further breaks $Z_6$ to $Z_2$, but this is not seen from this order parameter. One way to see this is to observe that in CFL chiral symmetry is broken, and $\langle \bar\psi_L\psi_R\rangle$ is not zero. $\endgroup$
    – Thomas
    Dec 25, 2017 at 3:19
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    $\begingroup$ This is meant to be 2-component spinor notation. Then $\Psi_\alpha=\epsilon_{\beta\gamma}\psi_\alpha\psi_\beta\psi_\gamma$ is a spin 1/2 nucleon field (because $\Phi=\epsilon_{\beta\gamma}\psi_\beta\psi_\gamma$ is a spin 0 diquark). Taking a spin singlet combination of two nucleon fields gives a scalar order parameter. $\endgroup$
    – Thomas
    Mar 15, 2018 at 0:31

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