Isotopic invariance I am reading something related to Isotopic invariance. As i have read, Isotopic invariance of the interaction between nucleons (proton and nucleon) is just approximate since masses of proton and neutron are slightly different. This difference in mass is due to the electromagnetic interaction. (Ok, I understand this argument). 
To illustrate that the electromagnetic interaction is not isotopically invariant, they write down the Lagrangian for the EM interaction as follows:
$$L_{em}=-e\bar{p}\gamma^{\mu}pA_\mu=-\frac{1}{2}e\bar{N}\gamma^{\mu}(1+\sigma_3)NA_\mu$$
where: $N=(p\quad n)^T$ is the isospinor (spinor in the isotopic space) of nucleon.
and explain that the appearance of $\sigma_3$, not $\vec{\sigma}$, in the Lagrangian above causes the EM interaction to be not isotopically invariant. My question is: Why does the appearance of $\sigma_3$, not $\vec{\sigma}$, in the Lagrangian above make the EM interaction not isotopically invariant?
 A: In order to check the invariance, you have to perform an isospin transformation and see if the lagrangian is still the same. A general isospin rotation is given by the $SU(2)$ matrix:
$$U = e^{-i (\alpha_1 \sigma_1 + \alpha_2 \sigma_2 + \alpha_3 \sigma_3)}$$
where $\alpha_{1,2,3}$ are real numbers. $U$ has the usual $SU(2)$ properties $U^\dagger U = 1$. Now let's rotate $N \to N' = UN$ and see the transformation of the lagrangian:
$$L_{em} \to L_{em'} = -\frac{1}{2} \bar{N} U^\dagger \gamma^\mu (1+\sigma_3) U N A_\mu = -\frac{1}{2} \bar{N} \gamma^\mu (1+ U^\dagger \sigma_3 U) N A_\mu$$ 
In order to be invariant, one should have $U^\dagger \sigma_3 U = \sigma_3$ but it is not true in general because the Pauli matrices don't commute together. So the Isospin cannot be conserved, explaining for example why $\pi^0 \to \gamma \gamma$ can occur with E.M. ($I(\pi^0)=1$, while $I(\gamma)=0$). Notice however that if you consider only a rotation about the third axis reducing $U$ to $U= e^{-i (\alpha_3 \sigma_3)}$, the lagrangian is invariant. Thus, EM conserved the 3rd component of the isospin as in the decay of the $\pi^0$ since $I_3(\pi^0) = I_3(\gamma) = 0$.
