2
$\begingroup$

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.

Improve this question
$\endgroup$
3
  • $\begingroup$ 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? $\endgroup$
    – ann marie cœur
    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. $\endgroup$
    – David Schaich
    Dec 25, 2017 at 3:18

1 Answer 1

Reset to default
2
$\begingroup$

  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 $$ \phi \sim \Psi C\gamma_5 \Psi, \qquad\Psi_\alpha = \psi_\alpha (\psi C\gamma_5\psi) $$

Improve this answer
$\endgroup$
16
  • $\begingroup$ 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??? $\endgroup$
    – ann marie cœur
    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? $\endgroup$
    – ann marie cœur
    Dec 25, 2017 at 2:31
  • $\begingroup$ also, how about this one? physics.stackexchange.com/questions/376203 $\endgroup$
    – ann marie cœur
    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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.