$SU(2)$ isospin invariance of the nucleon-pion system

Question

I'm reading "Dynamics of the Standard Model,Cambridge monographs on particle physics, nuclear physics and cosmology" and in page 10 it says that the response of an individual pion component to an isospin transformation can be found by multiplying $ \boldsymbol{\tau} \cdot\boldsymbol{\pi}'=U\boldsymbol{\tau}\cdot\boldsymbol{\pi}U^{\dagger}$ by $\tau^i$ and taking the trace, getting:

$\pi{'}^i=R^{ij}(\boldsymbol{ \alpha})\pi^j,R^{ij}(\boldsymbol{ \alpha})=\frac{1}{2}Tr(\tau^iU\tau^jU^{\dagger}) $.

Note.($\boldsymbol{\pi}$ is the pion triplet, we are looking at a SU(2) transformation with $U=exp(-i\boldsymbol{\tau\alpha}/2)$, for any $\alpha^i, i=1,2,3$)

With the result given by the book:

$\pi{'}^i=R^{ij}(\boldsymbol{ \alpha})\pi^j,R^{ij}(\boldsymbol{ \alpha})=\frac{1}{2}Tr(\tau^iU\tau^jU^{\dagger}) $. I can obtain this expression

How can I obtain the expression for an infinitesimal transformation?:

$\hat{\pi}{'}\ ^i=\pi^i-\epsilon^{ijk}\pi^j\alpha^k(x)$

here is what the book says:

TLDR:I want to arrive with the 3.10 equation at the second part of equation 3.11, in order to compute the final lagrangian.

Answer 1

The book coddles you and only bothers to give you the infinitesimal expression for this adjoint action, which is the only thing it needs for that purpose.

It is but a bland plug-in: $$ R^{ij}(\boldsymbol{ \alpha})=\frac{1}{2}\operatorname{Tr}(\tau^i \exp(-i\boldsymbol{\tau\cdot\alpha}/2)\tau^j \exp(i\boldsymbol{\tau\cdot\alpha}/2)) ~~~\leadsto \\ =\frac{1}{2}\operatorname{Tr}\bigl (\tau^i (1-i\tau^k \alpha^k/2 +...) \tau^j (1+i\tau^m \alpha^m/2+...) \bigr ),$$ where the ellipses denote terms of higher order in α, to be ignored in the infinitesimal truncation denoted by the caret. It then trivially flows that the above rotation matrix collapses to $$ =\frac{1}{2}\operatorname{Tr}\bigl (\tau^i \tau^j -{i\over 2} \alpha^k \tau^i [ \tau^k , \tau^j] + ... \bigr )=\delta^{ij}+\alpha^k \epsilon ^{kji} +... ~. $$ Consequently, multiplying $\pi^j$ with this truncated rotation matrix, $$\hat{\pi}{'}\ ^i=\pi^i-\epsilon^{ijk}\pi^j\alpha^k ,$$ as you might expect for an isovector.

The finite rotation, beyond the above infinitesimal truncation, is, as expected, the Rodrigues rotation formula, $ \pi ' \ ^i=\pi^i\cos \alpha -\epsilon^{ijk}\pi^j n^k \sin\alpha - n^i~~ n^j \pi^j (1-\cos\alpha)$, where $\boldsymbol{\alpha}\equiv \alpha \boldsymbol{n}$ for a unit direction vector n.