# gauge-invariant 6-quark order parameter

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.

• 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 $\bar \psi$. – Blazej Dec 25 '17 at 2:23
• can we show ϵ tensor is SU(3) singlet? – ann marie cœur Dec 25 '17 at 2:36
• 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. – David Schaich Dec 25 '17 at 3:18

1. It breaks $U(1)_B$, and therefore distinguished QGP from CFL.
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 $$\phi \sim \Psi C\gamma_5 \Psi, \qquad\Psi_\alpha = \psi_\alpha (\psi C\gamma_5\psi)$$
• 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??? – ann marie cœur Dec 25 '17 at 2:30
• 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. – Thomas Dec 25 '17 at 3:19
• 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. – Thomas Mar 15 '18 at 0:31