Chiral Symmetry and Isospin Symmetry

Let us examine the part of ${\cal L}_{\rm QCD}$ that describes the $u$ (up), $d$ (down) quarks and gluons only in the limit of $m_{u,d}~=~0$,

 

$${\cal L}_{\rm QCD}~\rightarrow~ \overline{Q} i{\rm D}\hskip-0.65em /\hskip0.2emQ \ -\ {1\over 4} G_{\mu\nu}^A G^{A\ \mu\nu}$$

$$=~\overline{Q}_L i{\rm D}\hskip-0.65em /\hskip0.2emQ_L\ +\ \overl... ...{\rm D}\hskip-0.65em /\hskip0.2emQ_R \ -\ {1\over 4} G_{\mu\nu}^A G^{A\ \mu\nu}$$

$$Q_L~=~\left({u_L\atop d_L}\right) \ \ ,\ \ Q_R~=~\left({u_R\atop d_R}\right) \ \ \ .$$ (3)

This lagrange density is clearly invariant under the global transformations

$$Q_L\rightarrow L Q_L\ \ \ ,\ \ \ Q_R\rightarrow R Q_R \ \ \ ,$$ (4)

where $L$ and $R$ are independent SU(2) unitary transformation matrices, $L^\dagger L~=~R^\dagger R~=~1$. If we choose $L~=~R~=~V$, this corresponds to transforming the left and right handed fields by the same ``amount''. If the world ``looked'' the same no matter what we choose for $V$ then we would say that the world was invariant under isospin transformations (I will use $V$ to denote the transformation in this section. Be aware that I will change notation in the near future for ease). In fact, isospin is found to be a very good symmetry of nature, that is to say that the world only changes a little bit under isospin transformations.

It turns out that when we consider the non-zero values of the quark masses $m_u\sim 5\ {\rm MeV}$ and $m_d\sim 10\ {\rm MeV}$, isospin symmetry is explicitly broken,

$$\overline{q}_L\gamma_\mu q_L\rightarrow \overline{q}_L L^\dagger \gamma_\mu L q_L\ =\ \overline{q}_L\gamma_\mu q_L$$

$$\overline{q}_R\gamma_\mu q_R\rightarrow \overline{q}_R R^\dagger \gamma_\mu R q_R\ =\ \overline{q}_R\gamma_\mu q_R$$

$$\overline{q}_R m_Q q_L\rightarrow \overline{q}_R R^\dagger m_Q L q_L \ne \overline{q}_R m_Q q_L \ \ \ ,$$ (5)

with $m_Q=diag(m_u,m_d)$. Given that the $u$ and $d$ masses differ by a factor of two it might first appear that isospin will be badly broken. However this is not the case. The typical scale for the size of hadronic interaction strengths is set by the scale at which spontaneous symmetry breaking occurs, i.e. when

$$SU(2)_L~\otimes~S(2)_R~\rightarrow~SU(2)_V \ \ \ ,$$ (6)

due to $g_s$ becoming large. This scale is called the chiral symmetry breaking scale and is

$$\Lambda_\chi~\sim~1~{\rm GeV} \ \ \ .$$ (7)

Therefore, we conclude that the appropriate small dimensionless parameter that describes isospin breaking is

$$\left(m_d-m_u\right)/\Lambda_\chi\sim 5\times 10^{-3} \ \ \ .$$ (8)

Thus, isospin symmetry is an accidental symmetry of nature. It is an accident because the scale of chiral symmetry breaking and the mechanism that produces the quarks masses (the Higgs mechanism) are unrelated.