gauge-invariant 6-quark order parameter

Question

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.

This post imported from StackExchange Physics at 2020-11-05 13:26 (UTC), posted by SE-user annie marie heart

Comments

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 $\bar \psi$.

This post imported from StackExchange Physics at 2020-11-05 13:26 (UTC), posted by SE-user Blazej

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can we show ϵ tensor is SU(3) singlet?

This post imported from StackExchange Physics at 2020-11-05 13:26 (UTC), posted by SE-user annie marie heart

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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.

This post imported from StackExchange Physics at 2020-11-05 13:26 (UTC), posted by SE-user David Schaich

Answer 1

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

This post imported from StackExchange Physics at 2020-11-05 13:26 (UTC), posted by SE-user Thomas

Comments

@ Thomas, I am also curious about the L and R handness indices and properties of parity here. Thanks a lot.

This post imported from StackExchange Physics at 2020-11-05 13:26 (UTC), posted by SE-user annie marie heart

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@ Thomas : In particular, I thought we are considering the Dirac spinor, how does $$\epsilon_{\alpha\alpha'} (\psi_\alpha )(\psi_{\alpha'})$$ , $$Ψα=ϵ_{βγ}ψ_αψ_βψ_γ$$ represent which part of Dirac spinor? Or should the baryon composed by three of 2-component Weyl spinor $(\psi_\alpha )$ instead of 4-component Dirac spinor? Thanks!

This post imported from StackExchange Physics at 2020-11-05 13:26 (UTC), posted by SE-user annie marie heart

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@ Thomas, also how is the parity $P$ property of the condensate? (under $L \to R$?) Are there choices?

This post imported from StackExchange Physics at 2020-11-05 13:26 (UTC), posted by SE-user annie marie heart

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@annie heart Added a postscript

This post imported from StackExchange Physics at 2020-11-05 13:26 (UTC), posted by SE-user Thomas

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can it be negative parity, too? accepted, but please clarufy.

This post imported from StackExchange Physics at 2020-11-05 13:26 (UTC), posted by SE-user annie marie heart

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@ Thomas, one more clarification, in this $\epsilon_{\alpha\alpha'}\epsilon_{\beta\gamma}\epsilon_{\beta'\gamma'} (\psi_\alpha\psi_\beta\psi_\gamma)(\psi_{\alpha'}\psi_{\beta'}\psi_{\gamma'})$, must I have $(\psi_\alpha\psi_\beta\psi_\gamma)$ as the 1st baryon and $(\psi_{\alpha'}\psi_{\beta'}\psi_{\gamma'})$ as the 2nd baryon?

This post imported from StackExchange Physics at 2020-11-05 13:26 (UTC), posted by SE-user annie marie heart

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Can I have instead $$\epsilon_{\alpha\alpha'}\epsilon_{\beta \beta'}\epsilon_{\gamma\gamma'} (\psi_\alpha\psi_\beta\psi_\gamma)(\psi_{\alpha'}\psi_{\beta'}\psi_{\gamma'})$$ while the color/flavor indices are given as above? (p.s. am going to accept you as the answer after this)

This post imported from StackExchange Physics at 2020-11-05 13:26 (UTC), posted by SE-user annie marie heart