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Chiral symmetry and quark condensate

by Einj
Tags: chiral, condensate, quark, symmetry
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Einj
#1
Dec11-12, 10:41 AM
P: 324
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
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fzero
#2
Dec11-12, 05:49 PM
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Quote Quote by Einj View Post
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
It's actually easy to see that [itex]\langle 0 | \bar{\psi}\psi |0\rangle[/itex] is not invariant under chiral symmetry. For QCD with one flavor,

[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.
Einj
#3
Dec12-12, 01:46 AM
P: 324
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]??

fzero
#4
Dec12-12, 10:07 AM
Sci Advisor
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P: 2,602
Chiral symmetry and quark condensate

Quote Quote by Einj View Post
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]??
Zero is the only value that is invariant under the chiral symmetry. You can be as explicit as you like by picking a one-parameter transformation and using the [itex]u,d[/itex] parameterization. Show that, if [itex]\rho = \langle \bar{\psi}\psi\rangle_0[/itex], then [itex]\delta\rho \neq 0[/itex] unless [itex]\rho =0[/itex].


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