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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 ...
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 $$\tag 1 L_{QCD} = \bar{q}_{i}i\gamma_{\mu}D^{\mu}_{ij}q_{j} - \frac{1}{4}G_{\mu \nu}^{a}G^{\mu \nu}_{a} - ... 1 As your QFT theory text should tell you, for an action invariant under G, addition of a Higgs potential only invariant under its subgroup H will spontaneously break the generators in G/H. You ought to do due diligence and study and understand and reproduce all examples of elementary classics such as Ling-Fong Li, PhysRev D9 (1974) 1723. There are, of ... 0 There is a simple proof that the cancellation is impossible (at least unless you are willing to add to the classical a term proportional to \hbar), I am reformulating an answer by @Qmechanic in a simpler language: The anomaly, or the measure variation which is the typical source of the anomaly, contributes a term which is independent of \hbar, while any ... 4 Comments to the question (v2): Traditionally, the classical action S sits in the Boltzmann factor \exp\left[\frac{i}{\hbar} S\right] behind an inverse power of \hbar in the path integral, while the path integral measure is independent of \hbar. In the conventional way of counting, we say that the Jacobian J from the path integral measure is a ... 0 The early Universe is largely in a state of thermal equilibrium, after inflation the Universe has a large temperature allowing for all degrees of freedom to obtain equilibrium. This can be seen from the Boltzmann equation, which describes how the number density changes in the expanding Universe. Equilibrium is obtained by a particle species if \Gamma \gg ... 3 This is called the Fabri-Picasso theorem. Their argument requires both the vacuum and the charge Q to be translationally invariant: P |0\rangle = 0, and [P, Q] = 0. The argument goes as follows: Since the charge originates from a symmetry, then according to Noether's theorem:$$Q = \int d^3x j_0(x) Consider the correlation function of the charge ...