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Strong chiral symmetry (SSB?)

by ChrisVer
Tags: chiral, strong, symmetry
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ChrisVer
#1
Apr9-14, 04:42 PM
P: 1,014
Why is the chiral symmetry breakdown determined for the vector/axial current as:

[itex] V = \frac{m_{π^{+}}-m_{π^{0}}}{m_{π^{0}}+m_{π^{+}}}\approx 0.01 [/itex]
[itex] A= \frac{m_{π^{+}}-m_{f^{0}}}{m_{f^{0}}+m_{π^{+}}}\approx 1[/itex]
?
why do we choose the difference between the pion+ (~140MeV) and pion0 (~135MeV) for the vector current or the difference between the pion(~140MeV) and the f0 meson (~900MeV) for the axial? Also why take that formulas for those asymmetries?
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The_Duck
#2
Apr9-14, 05:07 PM
P: 877
If a symmetry is not spontaneously broken, then we expect to find groups of particles with the same mass that are related to each other by that symmetry. For example, the lightest three mesons are the three pions. They all have approximately same mass--##V## in your post is small--and they are related to each other by isospin rotations. This is because isospin is conserved to pretty good accuracy.

If the chiral version of isospin were conserved, we would expect to find groups of particles with the same mass related to each other by chiral isospin rotations. Chiral transformations, among other things, transform pseudoscalars into scalars. So we would expect there to be some scalar meson degenerate with the pions, which are pseudoscalars. There's no such meson; the ##f_0## is one of the lightest scalar mesons and it is much heavier than the pions, as the large value of ##A## in your post shows. So chiral isospin rotations must be spontaneously broken.


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