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In Peskin and Schroeder (page 669), and other references, the heuristic reasoning for why one would expect chiral symmetry breaking at low energies is that quark masses are small and hence it's not very energetically costly to create quark-antiquark pairs from the vacuum. These are then generated with zero net linear and angular momentum which forces the pair to have a net chirality and this breaks the chiral symmetry.

Where does the requirement of low energy appear in this line of reasoning?

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  • $\begingroup$ Chiral symmetry breaking is an example of spontaneous symmetry breaking, which usually appears at low energy, because the shape of the potential depends on the energy order, see this wiki image $\endgroup$ – Trimok Oct 31 '13 at 10:45
  • $\begingroup$ Thanks, can this be made explicit though? Is there some calculation of the effective action which shows that this is true in the case of pions? $\endgroup$ – user26866 Oct 31 '13 at 17:00
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    $\begingroup$ I only found this interesting evolution of the potential parameters page $16-20$ of this ref. The most interesting diagram is page $18$ top right ($a_1$). If $a_1<0$ (see equation ($18$) page $17$ with the definitions of the parameters of the potential), we have the Mexican hat form of the potential. So, below $530$ Mev, we have chiral symmetry breaking. $\endgroup$ – Trimok Oct 31 '13 at 18:35
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It is not necessary to go into detail about the form of the potential or any kind of calculation in order to understand this heuristic line of reasoning. My understanding of the statement is the following:

By conservation of momentum/energy, we know that in order to create a particle-antiparticle pair, its total momentum/energy has to be present in some other form. For example, a photon of sufficient momentum can decay into a quark and an antiquark.

As a consequence, whether such a pair is created or not depends both on the energy present and the mass of the particles created. When one just looks at spontaneous (not explicit) symmetry breaking, and this is what Peskin and Schroeder do in the passage you cited, the quark masses are equal to zero (This is also referred to as the chiral limit of QCD). Therefore little energy is required in order to create such a a quark-antiquark pair. Thus, such processes can be expected to contribute to a nonvanishing quark condensate, given by

$$\langle0|\bar{q}q|0\rangle,$$

already at low energies.

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    $\begingroup$ Yes, but I'm effectively asking about why the condensate vanishes at higher energies. I understand that for zero (or very small) quark masses you will get this condensate at low energies. I don't understand,however, why it's only a low energy phenomenon. Your line of reasoning would seem to work at higher energies, as well. $\endgroup$ – user26866 Nov 4 '13 at 15:24
  • $\begingroup$ As I understand it, a restriction to low energies is not part of this heuristic argument, at leas not in Peskin and Schroeder's version. It is just argued that it should be there - and not why it should vanish at higher energies. $\endgroup$ – Frederic Brünner Nov 4 '13 at 15:50

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