# Tag Info

1

There is no need of starting with $SU(2)$ symmetry and then extending it to $SU(2)_L\otimes SU(2)_R$. In fact, the moment you write down the Lagrangian, the symmetry is by default $U(2)_L\otimes U(2)_R$ with $U(2)_L$ acting on the left handed quark doublet $(u , d)_L$ and $U(2)_R$ acting on the right handed quark doublet $(u , d)_R$, which can be extended ...

2

Question 1 First of all, when you discuss chiral symmetry spontaneous breaking, you need to assume pure QCD theory. QCD lagrangian with $u-,d-,s-$quarks (they have relatively small masses in compare with $b, t, c$-quarks) has the form  \tag 1 L_{QCD} = \bar{q}_{i}i\gamma_{\mu}D^{\mu}_{ij}q_{j} - \frac{1}{4}G_{\mu \nu}^{a}G^{\mu \nu}_{a} - ...

5

Spin is determined from the representation of the Lorentz group the quantum field transforms in. The projective finite-dimensional representations of the Lorentz group are labeled by two half-integers $(s_1,s_2)$. The spin of a field is the sum $s = s_1+s_2$. For example, a scalar transforms in $(0,0)$, a vector field in $(\frac{1}{2},\frac{1}{2})$, a Dirac ...

Top 50 recent answers are included