- #1

CAF123

Gold Member

- 2,889

- 88

## Main Question or Discussion Point

The quark sector of the QCD lagrangian can be written as (restricting to two flavours) $$\mathcal L = \sum_{i=u,d} \bar q_i ( i \gamma_{\mu} D^{\mu} + m) q_i .$$ Write ##q = (u d)^T## and $$M = \begin{pmatrix} m_u & 0 \\ 0 & m_d \end{pmatrix}$$

Given that the masses of the u and d quarks are much less than the hadronic confinement scale of QCD, we can consider ##m_u \approx m_d =m ## so that M is proportional to identity matrix. In this way, we have a U(2) symmetry ##q' = U q## where U is a U(2) transformation, ##U = \exp(i\alpha_i \sigma_i)##. (I think it's correct to say that this U(2) symmetry is a vector U(2) symmetry because using Noether's theorem we get vector currents as conserved quantities arising from this U(2) symmetry) Using properties of U(2), this is then equivalent to the direct product of vector U(1) and vector SU(2). In this form, the larger symmetry group U(2) from making the masses of u and d degenerate encompasses the global phase redefinition of the quark fields (which we had prior to assuming degeneracy of quark masses) and SU(2) is denoted an isospin symmetry.

Did I understand that part correctly? If we now make the masses of the quarks to be zero, then the symmetry of QCD is even larger and it turns out we can write the lagrangian as a sum of two terms, each involving only left and right handed fields. Therefore we have two independent U(2) symmetries, ##U(2)_L## and ##U(2)_R## forming the symmetry group ##U(2)_L \otimes U(2)_R##. My first question is many books say this symmetry is ##SU(2)_L \otimes SU(2)_R##. Why is this? We can write ##U(2)_L \otimes U(2)_R = SU(2)_L \otimes SU(2)_R \otimes U(1)_V \otimes U(1)_A## which also incorporate the previously discussed vector U(1) symmetry and the new axial U(1) symmetry now present given M=0.

My second question is if someone could explain this statement from Wikipedia (and also mentioned in many books):

Thanks!

Given that the masses of the u and d quarks are much less than the hadronic confinement scale of QCD, we can consider ##m_u \approx m_d =m ## so that M is proportional to identity matrix. In this way, we have a U(2) symmetry ##q' = U q## where U is a U(2) transformation, ##U = \exp(i\alpha_i \sigma_i)##. (I think it's correct to say that this U(2) symmetry is a vector U(2) symmetry because using Noether's theorem we get vector currents as conserved quantities arising from this U(2) symmetry) Using properties of U(2), this is then equivalent to the direct product of vector U(1) and vector SU(2). In this form, the larger symmetry group U(2) from making the masses of u and d degenerate encompasses the global phase redefinition of the quark fields (which we had prior to assuming degeneracy of quark masses) and SU(2) is denoted an isospin symmetry.

Did I understand that part correctly? If we now make the masses of the quarks to be zero, then the symmetry of QCD is even larger and it turns out we can write the lagrangian as a sum of two terms, each involving only left and right handed fields. Therefore we have two independent U(2) symmetries, ##U(2)_L## and ##U(2)_R## forming the symmetry group ##U(2)_L \otimes U(2)_R##. My first question is many books say this symmetry is ##SU(2)_L \otimes SU(2)_R##. Why is this? We can write ##U(2)_L \otimes U(2)_R = SU(2)_L \otimes SU(2)_R \otimes U(1)_V \otimes U(1)_A## which also incorporate the previously discussed vector U(1) symmetry and the new axial U(1) symmetry now present given M=0.

My second question is if someone could explain this statement from Wikipedia (and also mentioned in many books):

*'The quark condensate spontaneously breaks the ## SU(2)_{L}\times SU(2)_{R}## down to the diagonal vector subgroup ##SU(2)_V,## known as isospin.'*

Thanks!