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8

The point is that eq. (1.35) should hold off-shell to have a symmetry, while eq. (1.37) may only hold on-shell. [The term on-shell (in this context) means that the Euler-Lagrange equations are satisfied. See also this Phys.SE post.] In other words: On-shell, the action will only change with at most a boundary term for any infinitesimal variation, whether ...

2

Consider an element $g$ of the symmetry group. Say $g$ is represented by a unitary operator on the Hilbertspace $$T_g = \exp(tX)$$ with generator $X$ and some parameter $t$. It acts on an operator $\phi(y)$ by conjugation $$(g\cdot\phi)(y) = T_g^{-1}\phi(y) T_g = e^{-tX}\phi(y) e^{tX} = \big[ 1 + t[X,\cdot]+\mathcal{O}(t^2)\big]\phi(y)$$ On the other ...

3

"Derivation" of Baryon Number Conservation - Consider the QCD Lagrangian (density) $$\mathcal{L} = \bar{\psi}(i\gamma^\mu D_\mu - m)\psi - \frac{1}{4}G^a_{\mu\nu}G_a^{\mu\nu}$$ where the symbols have their usual meaning. This is invariant under $U(1)$, which is nothing but a multiplication of $\psi$ by a global phase factor $e^{i \theta}$. This is ...

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