Following the approach of Weinberg's book to discuss the chiral symmetry, at a certain point he says

If the $\rm SU(2)\times SU(2)$ symmetry is exact and unbroken, then this would require any one-hadron state $|h\rangle$ to be degenerate with another state $\vec{X}|h\rangle$ of opposite parity and equal spin, baryon number and strangeness. No such parity doubling is seen in the hadron spectrum, so we are forced to conclude that if the chiral symmetry is a god approximation t all, then it must be spontaneously broken to its isotopic spin $\rm SU(2)$ subgroup.

where $\{\vec{X}\}_i$ is the charge associated to the conserved axial-current $\vec{A}^\mu = i\bar{q}\gamma^\mu\gamma_5\vec{t}q$.

Can anyone explain better why the symmetry is broken and why it is broken in the isotopic spin group and not in another group? Or maybe if anyone can suggest some notes.

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    $\begingroup$ What about the given explanation is unclear to you? The unbroken symmetry predicts the wrong hadron spectrum, so it must be broken. $\endgroup$ – ACuriousMind Mar 3 '16 at 15:03
  • $\begingroup$ I would like to understand how to "see" that it predicts the wrong hadron spectrum because in this way I don't understand why $\vec{X}$ should send $|h>$ to another state of opposite parity etc... $\endgroup$ – apt45 Mar 3 '16 at 15:03
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    $\begingroup$ That's the "then" part of the first sentence. States related by a symmetry cannot differ in energy, but we do not observe this degeneracy experimentally. $\endgroup$ – ACuriousMind Mar 3 '16 at 15:05
  • $\begingroup$ it's still unclear for me in this way. Using the definition of X⃗ how do you show that $\vec{X}|h>$ it's a state of opposite parity and same others quantum numbers? $\endgroup$ – apt45 Mar 3 '16 at 15:34
  • $\begingroup$ You do appreciate that $\vec{A}^\mu$ flips the parity of the state, because of the γ5, but the isocurrent $\vec{V}^\mu = i\bar{q}\gamma^\mu\vec{t}q$ doesn't? $\endgroup$ – Cosmas Zachos Jul 23 '16 at 13:17

In simpler terms, if you take $SU(2)_L \times SU(2)_R$ symmetry, then X is the charge associated with the $SU(2)_R$ part.. so acting on a hadron state $|h>$ (with only chiral left symmetry) with X will lead to the creation of a new state $X|h>$ which will have similar features for both the right handed interaction as well as for the left handed interaction. Now since this state is related by a symmetry, it has to be degenerate, so accordingly, its mass spectrum will be the same as that of $|h>$, which is not true experimentally. Therefore you need to give it a higher mass, and as a result you would need to break the symmetry spontaneously from $SU(2)_L \times SU(2)_R$ to $SU(2)_L$.


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