$SU(2)$ adjoint and fundamental transformations

Question

I have to demonstrate that the transformation from $\mathbf{\pi}=(\pi_1,\pi_2,\pi_3) \rightarrow U \mathbf{\pi}$ with $U= \exp(i \alpha^a \mathbb{T}^a)$ and $\mathbb{T}^a$ the adjoint representation of $SU(2)$ is equivalent to the transformation $U \frac{\pi^a\tau^a}{2}U^{-1}$ with $U = \exp(i\alpha^a\tau^a/2)$. I can easiliy do it at infinitesimal level (show that they transform as vectors) but i don't know how at "global" level. The adjoint is equivalent to the fundamental of $SO(3)$ so is just a rotation in space, but with the other one?

Answer 1

This is the standard buildup of the SU(2) adjoint representation out of two fundamental ones, ${\mathbf 2}\otimes {\mathbf 2}={\mathbf 1} \oplus {\mathbf 3} $. Most physics students will never do more elaborate Kronecker product compositions of representations explicitly like this, and will read the answers off textbook appendices, or use clever indirect methods.

Because of the normalization of the fundamental rep generators which halve the Pauli matrices, it pays to redefine the rotation angle α in terms of a half-angle a, $$ \vec{\alpha}\cdot \vec{\tau} /2\equiv \hat{n}\cdot \vec{\tau} a, $$ so that $$ \vec{\alpha}\cdot\vec{\alpha}= 4a, \qquad U=e^{ia~\hat{n}\cdot \vec{\tau}}. $$

Proceed to slug through the Kronecker product using the simple expression for the Pauli vector exponential and the simple multiplication rule of the Pauli matrices: $$ U \frac{\vec{\pi}\cdot\vec{\tau}}{2} U^\dagger= (\mathbb{1} \cos a+i \hat{n}\cdot \vec{\tau} \sin a ) \frac{\vec{\pi}\cdot\vec{\tau}}{2} (\mathbb{1} \cos a-i \hat{n}\cdot \vec{\tau} \sin a ) \\ = \frac{\vec{\pi}\cdot\vec{\tau}}{2} \cos^2 a+ \frac{ i\sin a \cos a }{2} [\hat{n}\cdot \vec{\tau}, \vec{\pi}\cdot\vec{\tau}]+ \frac{\sin^2 a }{2} \hat{n}\cdot \vec{\tau} ~ \vec{\pi}\cdot\vec{\tau} ~ \hat{n}\cdot \vec{\tau} \\ =\frac{\vec{\tau}}{2}\cdot \Bigl( \vec{\pi} \cos (2a) - \hat{n}\times \vec{\pi} \sin(2a) +\hat{n} ~ \hat{n}\cdot \vec{\pi} (1-\cos(2a))\Bigr).$$

The expression in the parenthesis is recognized as the Rodrigues rotation of the 3-vector $\vec{\pi}$ around the axis $\hat{n}$ now by the full angle, $|\alpha|=2a$. Presumably you have experienced this feature before in transitioning from spin 1/2 to spin 1, so
$$ ...=\frac{\vec{\tau}}{2}\cdot \Bigl( \exp (i\vec{\alpha}\cdot \vec{\mathbb{T}})~\vec{\pi}\Bigr). $$