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Chiral symmetry and quark condensate 
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#1
Dec1112, 10:41 AM

P: 305

I'm studying chiral symmetry in QCD. I understand that in order for a spontaneous symmetry breaking to occur, there must be some state with a vacuum expectation value different from zero. My question is: can someone prove that is the chiral symmetry is an exact symmetry of the QCD then necessarily:
$$\langle 0  \bar{\psi}\psi 0\rangle = 0$$ In understand that this has to be derived from the invariance of the vacuum but I can't prove it explicitely. Thanks 


#2
Dec1112, 05:49 PM

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PF Gold
P: 2,606

[tex]\psi = \begin{pmatrix} u \\ d \end{pmatrix},[/tex] where we can consider the up and down quarks [itex]u,d[/itex] as Dirac spinors. Then we can write the chiral symmetry as [tex]\psi \rightarrow \exp\left[i\gamma^5 \left( \vec{\theta}\cdot\vec{\tau}\right) \right] \psi,[/tex] where the [itex]\vec{\tau}[/itex] are the generators of [itex]SU(2)[/itex] flavor transformations. Some algebra will show that the kinetic term [itex] \bar{\psi} \gamma^\mu\partial_\mu \psi[/itex] is chiral invariant, but [itex] \bar{\psi} \psi[/itex] is not, because the [itex]\gamma^5[/itex] in the chiral transformation anticommutes with the factor of [itex]\gamma^0[/itex] in the Dirac conjugate. 


#3
Dec1212, 01:46 AM

P: 305

Ok, I knew that. Actually my question was a little bit different (or maybe it's the same thing but I can't see it ). I was asking: IF the theory is invariant under chiral symmetry (i.e. the vacuum state is invariant) how can I show that necessarily [itex]\langle \bar{\psi}\psi\rangle=0[/itex]??



#4
Dec1212, 10:07 AM

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PF Gold
P: 2,606

Chiral symmetry and quark condensate



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