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

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.

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.
• I started bt replacing with the exponencials, but and then obtaining the second line, I now understand how to get the final result. Isn't there an i factor missing in the second line before the $\tau^m$ Mar 28, 2021 at 20:55