Given the transformation of $SU(2)$ triplet $\vec{\phi}$ how to find the transformation of ${\Phi}\equiv\vec{\phi}\cdot\vec{\tau}$?

Question

Given the transformation of a $SU(2)$ triplet $\vec\phi$ $$\phi\to \exp{(-i\vec{T}\cdot\vec{\theta})}~\vec\phi\tag{1}$$ (in the question here by @physicslover) how does obtain the transformation of $\Phi\equiv\vec{\phi}\cdot\vec{\tau}$ given by $$\Phi\to e^{i\vec{\tau}\cdot\vec{\theta}/2}\Phi e^{-i\vec{\tau}\cdot\vec{\theta}/2}?\tag{2}$$ For a reference, look at the second equation of the answer by Cosmas Zachos.

The transformation by @physicslover was in terms of the $3\times 3$ matrix representations of the $\rm {SU(2)}$ generators $\vec{T}$. Then oracularly it became a transformation in terms of $\vec{\tau}$. Is there a systematic way to derive (2) from (1)?

Answer 1

Relation between rotation matrix $\,\mathcal{R}(\boldsymbol{\theta})\,$, $2\times2$ hermitian traceless Pauli matrices $\,\boldsymbol{\sigma}\,$ and $3\times 3$ hermitian traceless matrices $\,\boldsymbol{\mathcal{S}}$ \begin{equation} \boxed{\:\:\:\mathbf{x}'\boldsymbol{=}\boldsymbol{e}^{\boldsymbol{i}\,\boldsymbol{\theta}\boldsymbol{\cdot}\boldsymbol{\mathcal{S}}}\mathbf{x} \quad \boldsymbol{\Longleftrightarrow}\quad \left(\mathbf{x}'\boldsymbol{\cdot}\boldsymbol{\sigma}\right)\boldsymbol{=}\boldsymbol{e}^{\boldsymbol{- }\boldsymbol{i}\,\boldsymbol{\theta}\boldsymbol{\cdot}\boldsymbol{\sigma}/2}\left(\mathbf{x}\boldsymbol{\cdot}\boldsymbol{\sigma}\right)\boldsymbol{e}^{\boldsymbol{+ }\boldsymbol{i}\,\boldsymbol{\theta}\boldsymbol{\cdot}\boldsymbol{\sigma}/2}\vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}}\:\:\:\:} \tag{$\mathcal{R}$}\label{R} \end{equation} \begin{equation} \boldsymbol{\sigma}\boldsymbol{=}\left(\sigma_1,\sigma_2,\sigma_3\right)\,, \quad \sigma_1= \begin{bmatrix} 0 & \!\!\hphantom{\boldsymbol{-}}1 \vphantom{\tfrac{a}{b}}\\ 1 & \!\!\hphantom{\boldsymbol{-}}0\vphantom{\tfrac{a}{b}} \end{bmatrix} \quad \sigma_2= \begin{bmatrix} 0 & \!\!\boldsymbol{-} i \vphantom{\tfrac{a}{b}}\\ i & \!\!\hphantom{\boldsymbol{-}} 0\vphantom{\tfrac{a}{b}} \end{bmatrix} \quad \sigma_3= \begin{bmatrix} 1 & \!\!\hphantom{\boldsymbol{-}} 0 \vphantom{\frac{a}{b}}\\ 0 & \!\!\boldsymbol{-} 1\vphantom{\frac{a}{b}} \end{bmatrix} \tag{$\boldsymbol{\sigma}$}\label{s} \end{equation} \begin{equation} \boldsymbol{\mathcal{S}}\boldsymbol{=}\left(\mathcal{S}_1,\mathcal{S}_2,\mathcal{S}_3\right)\,, \quad \mathcal{S}_1\boldsymbol{=} \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & i \\ 0 & \!\!\!-\!i & 0 \end{bmatrix} \quad \mathcal{S}_2\boldsymbol{=} \begin{bmatrix} 0 & 0 & \!\!-\!i \\ 0 & 0 & 0 \\ i & 0 & 0 \end{bmatrix} \quad \mathcal{S}_3\boldsymbol{=} \begin{bmatrix} \:0 & i & 0 \\ \!\!-\!i & 0 & 0 \\ \:0 & 0 & 0 \end{bmatrix} \tag{$\boldsymbol{\mathcal{S}}$}\label{S} \end{equation} $=================================================$

Any vector in $\mathbb{R}^3$ can be represented by a $2\times2$ hermitian traceless matrix and vice versa. So, there exists a bijection (one-to-one and onto correspondence) between $\mathbb{R}^3$ and the space of $2\times2$ hermitian traceless matrices, let it be $\mathbb{H}$ : \begin{equation} \mathbf{x}=(x_1,x_2,x_3)\in \mathbb{R}^3\;\boldsymbol{\longleftrightarrow} \; \mathrm X= \begin{bmatrix} x_3 & x_1 \boldsymbol{-}ix_2 \\ x_1 \boldsymbol{+}ix_2 & \boldsymbol{-}x_3 \end{bmatrix} \in \mathbb{H} \tag{01}\label{01} \end{equation} From the usual basis of $\mathbb{R}^3$ \begin{equation} \mathbf{e}_{1}=\left(1,0,0\right),\quad \mathbf{e}_{2}=\left(0,1,0\right),\quad \mathbf{e}_{3}=\left(0,0,1\right) \tag{02}\label{02} \end{equation} we construct a basis for $\mathbb{H}$ \begin{align} \mathbf{e}_1 &= (1,0,0)\qquad \boldsymbol{\longleftrightarrow} \qquad \sigma_1= \begin{bmatrix} 0 & \!\!\hphantom{\boldsymbol{-}}1 \vphantom{\tfrac{a}{b}}\\ 1 & \!\!\hphantom{\boldsymbol{-}}0\vphantom{\tfrac{a}{b}} \end{bmatrix} \tag{03.1}\label{03.1}\\ \mathbf{e}_2 &= (0,1,0)\qquad \boldsymbol{\longleftrightarrow} \qquad \sigma_2= \begin{bmatrix} 0 & \!\!\boldsymbol{-} i \vphantom{\tfrac{a}{b}}\\ i & \!\!\hphantom{\boldsymbol{-}} 0\vphantom{\tfrac{a}{b}} \end{bmatrix} \tag{03.2}\label{03.2}\\ \mathbf{e}_3 &= (0,0,1)\qquad \boldsymbol{\longleftrightarrow} \qquad \sigma_3= \begin{bmatrix} 1 & \!\!\hphantom{\boldsymbol{-}} 0 \vphantom{\frac{a}{b}}\\ 0 & \!\!\boldsymbol{-} 1\vphantom{\frac{a}{b}} \end{bmatrix} \tag{03.3}\label{03.3} \end{align} where $\:\boldsymbol{\sigma}\equiv(\sigma_{1},\sigma_{2},\sigma_{3})\:$ the Pauli matrices.

Suppose now that the vector $\:\mathbf{x}=(x_1,x_2,x_3)\:$ is rotated around an axis with unit vector $\:\mathbf{n}=(\rm n_1,n_2,n_3)$ through an angle $\theta$ \begin{equation} \mathbf{x}'\boldsymbol{=} \mathcal{R}\,\mathbf{x}\boldsymbol{=}\cos\theta \;\mathbf{x}\boldsymbol{+}(1\boldsymbol{-}\cos\theta)\;(\mathbf{n}\boldsymbol{\cdot}\mathbf{x})\;\mathbf{n}\boldsymbol{+}\sin\theta\;(\mathbf{n}\boldsymbol{\times}\mathbf{x}) \tag{04}\label{04} \end{equation} The exponential expression of the rotation matrix $\:\mathcal{R}\boldsymbol{=}\boldsymbol{e}^{\boldsymbol{i}\,\boldsymbol{\theta}\boldsymbol{\cdot}\boldsymbol{\mathcal{S}}}\:$ is derived in APPENDIX B.

Now, let to the vectors $\:\mathbf{x},\mathbf{x}^{\prime}\:$ correspond the matrices \begin{align} \mathrm X & \boldsymbol{\equiv} \mathbf{x}\boldsymbol{\cdot} \boldsymbol{\sigma} \hphantom{\boldsymbol{'}}\boldsymbol{=} x_1\sigma_1\boldsymbol{+}x_2\sigma_2\boldsymbol{+}x_3\sigma_3\boldsymbol{=} \begin{bmatrix} x_3 & x_1\boldsymbol{-}ix_2\\ x_1\boldsymbol{+}ix_2 & \boldsymbol{-}x_3 \end{bmatrix} \tag{05.1}\label{05.1}\\ \mathrm X' & \boldsymbol{\equiv} \mathbf{x}' \boldsymbol{\cdot} \boldsymbol{\sigma} \boldsymbol{=} x'_1\sigma_1\boldsymbol{+}x'_2\sigma_2\boldsymbol{+}x'_3\sigma_3 \boldsymbol{=} \begin{bmatrix} x'_3 & x'_1\boldsymbol{-}ix'_2\\ x'_1\boldsymbol{+}ix'_2 & \boldsymbol{-}x'_3 \end{bmatrix} \tag{05.2}\label{05.2} \end{align}

Taking the inner product of \eqref{04} with $\boldsymbol{\sigma}$ \begin{equation} (\mathbf{x}'\boldsymbol{\cdot}\boldsymbol{\sigma}) \boldsymbol{=} \cos\theta(\mathbf{x}\boldsymbol{\cdot}\boldsymbol{\sigma})\boldsymbol{+}(1\boldsymbol{-}\cos\theta)(\mathbf{n}\boldsymbol{\cdot}\mathbf{x})(\mathbf{n}\boldsymbol{\cdot}\boldsymbol{\sigma})\boldsymbol{+}\sin\theta\bigl[(\mathbf{n}\boldsymbol{\times}\mathbf{x})\boldsymbol{\cdot}\boldsymbol{\sigma})\bigr] \tag{07}\label{07} \end{equation}
we have \begin{equation} \mathrm X' \boldsymbol{=} \cos\theta \;\mathrm X\boldsymbol{+}(1\boldsymbol{-}\cos\theta)(\mathbf{n}\boldsymbol{\cdot}\mathbf{x})\mathrm N\boldsymbol{+}\sin\theta\bigl[(\mathbf{n}\boldsymbol{\times}\mathbf{x})\boldsymbol{\cdot}\boldsymbol{\sigma})\bigr] \tag{08}\label{08} \end{equation} where \begin{equation} \mathrm N \boldsymbol{\equiv} \mathbf{n}\boldsymbol{\cdot}\boldsymbol{\sigma}\boldsymbol{=} \begin{bmatrix} \rm n_3 & \rm n_1\boldsymbol{-}i\rm n_2\\ \rm n_1\boldsymbol{+}i\rm n_2&\boldsymbol{-}\rm n_3 \end{bmatrix} \tag{09}\label{09} \end{equation} Now \begin{equation} \bigl[(\mathbf{n}\times\mathbf{x})\boldsymbol{\cdot}\boldsymbol{\sigma}\bigr] \boldsymbol{=}\frac{(\mathbf{n}\boldsymbol{\cdot}\boldsymbol{\sigma})(\mathbf{x}\boldsymbol{\cdot}\boldsymbol{\sigma})\boldsymbol{{-}}(\mathbf{x }\boldsymbol{\cdot}\boldsymbol{\sigma})(\mathbf{n}\boldsymbol{\cdot}\boldsymbol{\sigma})}{2i}\boldsymbol{=}i\frac{\rm XN\boldsymbol{-}NX}{2} \tag{10}\label{10} \end{equation} and \begin{equation} (\mathbf{n}\boldsymbol{\cdot}\mathbf{x})\,I =\frac{(\mathbf{n}\boldsymbol{\cdot}\boldsymbol{\sigma})(\mathbf{x}\boldsymbol{\cdot}\boldsymbol{\sigma})\boldsymbol{+}(\mathbf{x}\boldsymbol{\cdot}\boldsymbol{\sigma})(\mathbf{n}\boldsymbol{\cdot}\boldsymbol{\sigma})}{2}=\frac{\rm NX\boldsymbol{+}XN}{2} \tag{11}\label{11} \end{equation} For derivation of equations \eqref{10} and \eqref{11} see equations \eqref{A-05} and \eqref{A-06} in APPENDIX A.

Inserting expressions \eqref{10} and \eqref{11} in \eqref{08} \begin{equation} \mathrm X' \boldsymbol{=} \cos\theta \;\mathrm X\boldsymbol{+}\left(1\boldsymbol{-}\cos\theta\right)\left(\frac{\rm NX\boldsymbol{+}XN}{2}\right)\mathrm N\boldsymbol{+}i\sin\theta\left(\frac{\rm XN\boldsymbol{-}NX}{2}\right) \tag{12}\label{12} \end{equation} Replacing \begin{equation} \cos\theta =\cos^2\frac{\theta}{2}-\sin^2\frac{\theta}{2}=1-2\sin^2\frac{\theta}{2}, \qquad \sin\theta=2 \sin \frac{\theta}{2}\cos\frac{\theta}{2} \tag{13}\label{13} \end{equation} and using the property \begin{equation} \mathrm N^2=(\mathbf{n}\boldsymbol{\cdot}\boldsymbol{\sigma})^2=(\mathbf{n}\boldsymbol{\cdot}\boldsymbol{\sigma})(\mathbf{n}\boldsymbol{\cdot}\boldsymbol{\sigma})=(\mathbf{n}\boldsymbol{\cdot}\mathbf{n})I=\left\|\mathbf{n}\right\|^2I=I \tag{14}\label{14} \end{equation} we have \begin{align} \mathrm X'& \boldsymbol{=} \left(\cos^2\frac{\theta}{2}\right)\mathrm X+\left(\sin^2\frac{\theta}{2}\right)\mathrm N\mathrm X\mathrm N\boldsymbol{+}i \left(\sin\frac{\theta}{2} \cos\frac{\theta}{2}\right)(\mathrm X\mathrm N-\mathrm N\mathrm X) \nonumber\\ &= \left(I\cos\frac{\theta}{2}-i\mathrm N\sin\frac{\theta}{2} \right)\mathrm X\cos\frac{\theta}{2}+i\left( I\cos\frac{\theta}{2}-i\mathrm N\sin\frac{\theta}{2} \right)\mathrm X\mathrm N\sin\frac{\theta}{2} \nonumber\\ & = \left( I\cos\frac{\theta}{2}-i\mathrm N\sin\frac{\theta}{2} \right)\left(\mathrm X\cos\frac{\theta}{2}+i\mathrm X\mathrm N\sin\frac{\theta}{2} \right) \nonumber\\ & =\left( I\cos\frac{\theta}{2}-i\mathrm N\sin\frac{\theta}{2}\right)\;\mathrm X\;\left( I\cos\frac{\theta}{2}+i\mathrm N\sin\frac{\theta}{2}\right) \tag{15}\label{15} \end{align} so

\begin{equation} \mathrm X'=\left[I\cos\frac{\theta}{2}-i(\mathbf{n}\boldsymbol{\cdot}\boldsymbol{\sigma})\sin\frac{\theta}{2} \right]\;\mathrm X\;\left[I\cos\frac{\theta}{2}+i(\mathbf{n}\boldsymbol{\cdot}\boldsymbol{\sigma})\sin\frac{\theta}{2} \right] \tag{16}\label{16} \end{equation} or inserting back the definitions of $\;\mathrm X,\mathrm X'\;$ \begin{equation} (\mathbf{x}'\boldsymbol{\cdot}\boldsymbol{\sigma})=\left[I\cos\frac{\theta}{2}-i(\mathbf{n}\boldsymbol{\cdot}\boldsymbol{\sigma})\sin\frac{\theta}{2} \right]\;(\mathbf{x}\boldsymbol{\cdot}\boldsymbol{\sigma})\;\left[I\cos\frac{\theta}{2}+i(\mathbf{n}\boldsymbol{\cdot}\boldsymbol{\sigma})\sin\frac{\theta}{2} \right] \tag{17}\label{17} \end{equation} Equation \eqref{16} is written in compact form \begin{equation} \mathrm X'=U\;\mathrm X\;U^{\boldsymbol{*}} \tag{18}\label{18} \end{equation}
with \begin{equation} U\equiv I\cos\frac{\theta}{2}\boldsymbol{-}i(\mathbf{n}\boldsymbol{\cdot}\boldsymbol{\sigma})\sin\frac{\theta}{2} \tag{19}\label{19} \end{equation} and hermitian conjugate \begin{equation} U^{\boldsymbol{*}}=I\cos\frac{\theta}{2}\boldsymbol{+}i(\mathbf{n}\boldsymbol{\cdot}\boldsymbol{\sigma})\sin\frac{\theta}{2} \tag{20}\label{20} \end{equation} The exponential expression of the special unitary matrix $\:U\boldsymbol{=}\boldsymbol{e}^{\boldsymbol{- }\boldsymbol{i}\,\boldsymbol{\theta}\boldsymbol{\cdot}\boldsymbol{\sigma}/2}\:$ is derived in APPENDIX C, see equation \eqref{C-05.1}.

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^{APPENDIX A}

^{If to $\;\mathbf{x}\boldsymbol{=}(x_1,x_2,x_3),\mathbf{y}\boldsymbol{=}(y_1,y_2,y_3)\in \mathbb{R}^3\;$ there correspond respectively the $2\times2$ hermitian traceless matrices \begin{equation} \mathrm X\boldsymbol{=} \left(\mathbf{x}\boldsymbol{\cdot}\boldsymbol{\sigma}\right)\boldsymbol{=} x_1\sigma_1\boldsymbol{+}x_2\sigma_2\boldsymbol{+}x_3\sigma_3\boldsymbol{=} \begin{bmatrix} x_3 & x_1 \boldsymbol{-}ix_2 \\ x_1 \boldsymbol{+}ix_2 & \boldsymbol{-}x_3 \end{bmatrix} \tag{A-01}\label{A-01} \end{equation} and \begin{equation} \mathrm Y\boldsymbol{=} \left(\mathbf{y}\boldsymbol{\cdot}\boldsymbol{\sigma}\right)\boldsymbol{=}y_1\sigma_1\boldsymbol{+}y_2\sigma_2\boldsymbol{+}y_3\sigma_3\boldsymbol{=} \begin{bmatrix} y_3 & y_1 \boldsymbol{-}iy_2 \\ y_1 \boldsymbol{+}iy_2 & \boldsymbol{-}y_3 \end{bmatrix} \tag{A-02}\label{A-02} \end{equation} then using the properties of the Pauli matrices $\:\sigma_{k}\:$ we have \begin{equation} \mathrm X\mathrm Y\boldsymbol{=} \left(\mathbf{x}\boldsymbol{\cdot}\boldsymbol{\sigma}\right)\left(\mathbf{y}\boldsymbol{\cdot}\boldsymbol{\sigma}\right)\boldsymbol{=} \left(\mathbf{x}\boldsymbol{\cdot}\mathbf{y}\right)\mathrm I\boldsymbol{+} i\bigl[(\mathbf{x}\times\mathbf{y})\boldsymbol{\cdot}\boldsymbol{\sigma}\bigr] \tag{A-03}\label{A-03} \end{equation} and so \begin{equation} \mathrm Y\mathrm X\boldsymbol{=} \left(\mathbf{y}\boldsymbol{\cdot}\boldsymbol{\sigma}\right)\left(\mathbf{x}\boldsymbol{\cdot}\boldsymbol{\sigma}\right)\boldsymbol{=} \left(\mathbf{x}\boldsymbol{\cdot}\mathbf{y}\right)\mathrm I\boldsymbol{-} i\bigl[(\mathbf{x}\times\mathbf{y})\boldsymbol{\cdot}\boldsymbol{\sigma}\bigr] \tag{A-04}\label{A-04} \end{equation} Subtracting and adding \eqref{A-03},\eqref{A-04} we have respectively \begin{equation} \left[\mathrm X,\mathrm Y\right]\boldsymbol{=}\mathrm X\mathrm Y\boldsymbol{-}\mathrm Y\mathrm X \boldsymbol{=} \left(\mathbf{x}\boldsymbol{\cdot}\boldsymbol{\sigma}\right)\left(\mathbf{y}\boldsymbol{\cdot}\boldsymbol{\sigma}\right)\boldsymbol{-}\left(\mathbf{y}\boldsymbol{\cdot}\boldsymbol{\sigma}\right)\left(\mathbf{x}\boldsymbol{\cdot}\boldsymbol{\sigma}\right)\boldsymbol{=} 2 i\bigl[(\mathbf{x}\times\mathbf{y})\boldsymbol{\cdot}\boldsymbol{\sigma}\bigr] \tag{A-05}\label{A-05} \end{equation} and \begin{equation} \{\mathrm X,\mathrm Y\}\boldsymbol{=}\mathrm X\mathrm Y\boldsymbol{+}\mathrm Y\mathrm X \boldsymbol{=} \left(\mathbf{x}\boldsymbol{\cdot}\boldsymbol{\sigma}\right)\left(\mathbf{y}\boldsymbol{\cdot}\boldsymbol{\sigma}\right)\boldsymbol{+}\left(\mathbf{y}\boldsymbol{\cdot}\boldsymbol{\sigma}\right)\left(\mathbf{x}\boldsymbol{\cdot}\boldsymbol{\sigma}\right)\boldsymbol{=} 2 \left(\mathbf{x}\boldsymbol{\cdot}\mathbf{y}\right)\mathrm I \tag{A-06}\label{A-06} \end{equation}}

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^{APPENDIX B}

^{In order to see the relation between the starting equation \eqref{04} of the answer and the starting equation (1) of the question we must express the rotation matrix $\;\mathcal{R}\;$ of equation \eqref{04} in exponential form. Although this form is given in many textbooks we'll derive it in the following for completeness.}

^{In equation \eqref{04}, repeated here \begin{equation} \mathbf{x}'\boldsymbol{=} \mathcal{R}\,\mathbf{x}\boldsymbol{=}\cos\theta \;\mathbf{x}\boldsymbol{+}(1\boldsymbol{-}\cos\theta)\;(\mathbf{n}\boldsymbol{\cdot}\mathbf{x})\;\mathbf{n}\boldsymbol{+}\sin\theta\;(\mathbf{n}\boldsymbol{\times}\mathbf{x}) \tag{04}\label{repeat 04} \end{equation} the term $\:(\mathbf{n}\boldsymbol{\cdot}\mathbf{x})\;\mathbf{n}\:$ is the component of $\;\mathbf{x}\;$ parallel to $\;\mathbf{n}$. Given that $\:(\mathbf{n}\boldsymbol{\times}\mathbf{x})\boldsymbol{\times}\mathbf{n}\:$ is the component of $\;\mathbf{x}\;$ normal to $\;\mathbf{n}\;$ \begin{equation} (\mathbf{n}\boldsymbol{\cdot}\mathbf{x})\;\mathbf{n}\boldsymbol{=}\mathbf{x}\boldsymbol{-}(\mathbf{n}\boldsymbol{\times}\mathbf{x})\boldsymbol{\times}\mathbf{n}=\mathbf{x}\boldsymbol{+}\mathbf{n}\boldsymbol{\times}(\mathbf{n}\boldsymbol{\times}\mathbf{x}) \tag{B-01}\label{B-01} \end{equation} Inserting this expression in \eqref{repeat 04} \begin{equation} \mathcal{R}\,\mathbf{x}\boldsymbol{=}\mathbf{x}\boldsymbol{+}(1\boldsymbol{-}\cos\theta)\;\left[\mathbf{n}\boldsymbol{\times}(\mathbf{n}\boldsymbol{\times}\mathbf{x})\right]\boldsymbol{+}\sin\theta\;(\mathbf{n}\boldsymbol{\times}\mathbf{x}) \tag{B-02}\label{B-02} \end{equation} If we define the following real anti-symmetric transformation
\begin{equation} S\,\mathbf{x}\boldsymbol{=}\mathbf{n}\boldsymbol{\times}\mathbf{x}\boldsymbol{=} \begin{bmatrix} \hphantom{-}0 & -\rm n_3 & \hphantom{-}\rm n_2\hphantom{-}\vphantom{\frac12} \\ \hphantom{-} \rm n_3 & \hphantom{-} 0 & -\rm n_1\hphantom{-}\vphantom{\dfrac12} \\ -\rm n_2 & \hphantom{-} \rm n_1 & \hphantom{-}0\hphantom{-}\vphantom{\frac12} \end{bmatrix} \begin{bmatrix} x_1\vphantom{\frac12} \\ x_2\vphantom{\dfrac12} \\ x_3\vphantom{\frac12} \end{bmatrix} \tag{B-03}\label{B-03} \end{equation} or formally \begin{equation} S\boldsymbol{\equiv}\mathbf{n}\boldsymbol{\times}\boldsymbol{=} \begin{bmatrix} \hphantom{-}0 & -\rm n_3 & \hphantom{-}\rm n_2\hphantom{-}\vphantom{\frac12} \\ \hphantom{-} \rm n_3 & \hphantom{-} 0 & -\rm n_1\hphantom{-}\vphantom{\dfrac12} \\ -\rm n_2 & \hphantom{-} \rm n_1 & \hphantom{-}0\hphantom{-}\vphantom{\frac12} \end{bmatrix} \tag{B-04}\label{B-04} \end{equation} then \begin{equation} \mathbf{n}\boldsymbol{\times}(\mathbf{n}\boldsymbol{\times}\mathbf{x})\boldsymbol{=}S\left(S\,\mathbf{x}\right)\boldsymbol{=}S^2\mathbf{x} \tag{B-05}\label{B-05} \end{equation} that is formally \begin{equation} S^2\boldsymbol{=}\mathbf{n}\boldsymbol{\times}\text{(}\mathbf{n}\boldsymbol{\times}\boldsymbol{=} \begin{bmatrix} \rm n^2_1-1 & \rm n_1\rm n_2 & \rm n_1\rm n_3\vphantom{\frac12} \\ \rm n_2\rm n_1 & \rm n^2_2-1 & \rm n_2\rm n_3\vphantom{\dfrac12} \\ \rm n_3\rm n_1 & \rm n_3\rm n_2 & \rm n^2_3-1\vphantom{\frac12} \end{bmatrix} \tag{B-06}\label{B-06} \end{equation} So from \eqref{B-02},\eqref{B-04} and \eqref{B-06} \begin{equation} \mathcal{R}\boldsymbol{=} I \boldsymbol{+}S\,\sin\theta\boldsymbol{-}S^2(\cos\theta\boldsymbol{-}1) \tag{B-07}\label{B-07} \end{equation} Now, after proving the following properties of $\,S$ \begin{align} S & \boldsymbol{=}\boldsymbol{-}S^3\boldsymbol{=}\boldsymbol{+}S^5\boldsymbol{=}\boldsymbol{-}S^7\boldsymbol{=}\cdots\boldsymbol{=}(\boldsymbol{-}1)^k S^{2k\boldsymbol{+}1}\,, \quad k \in \mathbb{N} \tag{B-08.1}\label{B-08.1}\\ S^2 & \boldsymbol{=}\boldsymbol{-}S^4\boldsymbol{=}\boldsymbol{+}S^6\boldsymbol{=}\boldsymbol{-}S^8\boldsymbol{=}\cdots\boldsymbol{=}(\boldsymbol{-}1)^k S^{2k\boldsymbol{+}2}\,, \quad k \in \mathbb{N} \tag{B-08.2}\label{B-08.2} \end{align} and given the infinite series expansions of the trigonometric functions \begin{align} \cos \theta & \boldsymbol{=} 1\boldsymbol{-}\frac{\theta^2}{2!}\boldsymbol{+}\frac{\theta^4}{4!}\boldsymbol{+}\cdots\boldsymbol{=} \sum^{\infty}_{k=0}\frac{(\boldsymbol{-}1)^k\theta^{2k}}{(2k)!} \tag{B-09.1}\label{B-09.1}\\ \sin \theta & \boldsymbol{=} \theta\boldsymbol{-}\frac{\theta^3}{3!}\boldsymbol{+}\frac{\theta^5}{5!}\boldsymbol{+}\cdots\boldsymbol{=} \sum^{\infty}_{k=0}\frac{(\boldsymbol{-}1)^k\theta^{2k\boldsymbol{+}1}}{(2k\boldsymbol{+}1)!} \tag{B-09.2}\label{B-09.2} \end{align} equation \eqref{B-07} yields \begin{align} \mathcal{R} & \boldsymbol{=} I \boldsymbol{+}S\,\sin\theta\boldsymbol{-}S^2(\cos\theta\boldsymbol{-}1) \nonumber\\ & \boldsymbol{=} I \boldsymbol{+}S\left(\theta\boldsymbol{-}\frac{\theta^3}{3!} \boldsymbol{+}\frac{\theta^5}{5!} \boldsymbol{+}\cdots\right)\boldsymbol{-}S^{2}\left(\boldsymbol{-}\frac{\theta^2}{2!} \boldsymbol{+}\frac{\theta^4}{4!}\boldsymbol{-}\frac{\theta^6}{6!} \boldsymbol{+}\cdots\right) \nonumber\\ & \boldsymbol{=}I\boldsymbol{+}(\theta S)\boldsymbol{+}\frac{(\theta S)^2}{2!}\boldsymbol{+}\frac{(\theta S)^3}{3!}\boldsymbol{+}\frac{(\theta S)^4}{4!}\boldsymbol{+}\frac{(\theta S)^5}{5!}\boldsymbol{+}\cdots \nonumber\\ & \boldsymbol{=}\sum^{\boldsymbol{\infty}}_{k\boldsymbol{=}0}\frac{(\theta S)^{k}}{k!}\boldsymbol{=}\boldsymbol{e}^{\theta S} \tag{B-10}\label{B-10} \end{align} that is \begin{equation} \mathcal{R}\boldsymbol{=}\boldsymbol{e}^{\theta S} \tag{B-11}\label{B-11} \end{equation} Now defining the 3-vector of reals \begin{equation} \boldsymbol{\theta}\boldsymbol{=}\theta\,\mathbf{n}\boldsymbol{=}\left(\theta \rm n_1,\theta \rm n_2,\theta \rm n_3\right) \tag{B-12}\label{B-12} \end{equation} and the 3-vector of real anti-symmetric matrices \begin{equation} \boldsymbol{\mathcal{K}}\boldsymbol{=}\left(\mathcal{K}_1,\mathcal{K}_2,\mathcal{K}_3\right)\,, \quad \mathcal{K}_1\boldsymbol{=} \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & \!\!\!-\!1 \\ 0 & 1 & 0 \end{bmatrix} \quad \mathcal{K}_2\boldsymbol{=} \begin{bmatrix} \:0 & 0 & 1 \\ \:0 & 0 & 0 \\ \!-\!1 & 0 & 0 \end{bmatrix} \quad \mathcal{K}_3\boldsymbol{=} \begin{bmatrix} 0 & \!\!\!-\!1 & 0 \\ 1 & 0 & 0 \\ 0 &0 & 0 \end{bmatrix} \tag{B-13}\label{B-13} \end{equation} we have the expression \begin{equation} \mathcal{R}\boldsymbol{=}\boldsymbol{e}^{\boldsymbol{\theta}\boldsymbol{\cdot}\boldsymbol{\mathcal{K}}} \tag{B-14}\label{B-14} \end{equation} To be in agreement with the exponent in equation (1) of the question we express the exponent in \eqref{B-14} in terms of the following $\;3\times 3\;$ hermitian traceless matrices \begin{equation} \boldsymbol{\mathcal{S}}\boldsymbol{=}\left(\mathcal{S}_1,\mathcal{S}_2,\mathcal{S}_3\right)\,, \quad \mathcal{S}_1\boldsymbol{=} \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & i \\ 0 & \!\!\!-\!i & 0 \end{bmatrix} \quad \mathcal{S}_2\boldsymbol{=} \begin{bmatrix} 0 & 0 & \!\!-\!i \\ 0 & 0 & 0 \\ i & 0 & 0 \end{bmatrix} \quad \mathcal{S}_3\boldsymbol{=} \begin{bmatrix} \:0 & i & 0 \\ \!\!-\!i & 0 & 0 \\ \:0 & 0 & 0 \end{bmatrix} \tag{B-15}\label{B-15} \end{equation} so that \begin{equation} \mathcal{R}\boldsymbol{=}\boldsymbol{e}^{\boldsymbol{i}\,\boldsymbol{\theta}\boldsymbol{\cdot}\boldsymbol{\mathcal{S}}} \tag{B-16}\label{B-16} \end{equation}}

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^{APPENDIX C}

^{The special unitary matrix of equation \eqref{19}, repeated here for convenience \begin{equation} U\boldsymbol{\equiv }I\cos\frac{\theta}{2}\boldsymbol{-}i(\mathbf{n}\boldsymbol{\cdot}\boldsymbol{\sigma})\sin\frac{\theta}{2}\boldsymbol{=}I\cos\frac{\theta}{2}\boldsymbol{-}i\mathrm N\sin\frac{\theta}{2} \tag{19}\label{repeat19} \end{equation} could be expressed in exponential form.}

^{From equation \eqref{14} \begin{equation} \mathrm N^2\boldsymbol{=}(\mathbf{n}\boldsymbol{\cdot}\boldsymbol{\sigma})^2\boldsymbol{=}I \tag{14}\label{repeat14} \end{equation} we have \begin{equation} \mathrm N^{2k}\boldsymbol{=}(\mathbf{n}\boldsymbol{\cdot}\boldsymbol{\sigma})^{2k}\boldsymbol{=}I \quad \text{and} \quad \mathrm N^{2k\boldsymbol{+}1}\boldsymbol{=}(\mathbf{n}\boldsymbol{\cdot}\boldsymbol{\sigma})^{2k\boldsymbol{+}1}\boldsymbol{=}\mathrm N \tag{C-01}\label{C-01} \end{equation} Using the infinite series expansions of the trigonometric functions \eqref{B-09.1} and \eqref{B-09.2} for $\;\left(\theta/2\right)$ \begin{align} \cos \left(\frac{\theta}{2}\right) & \boldsymbol{=} \sum^{\infty}_{k=0}\frac{(\boldsymbol{-}1)^k}{(2k)!}\left(\frac{\theta}{2}\right)^{2k} \tag{C-02.1}\label{C-02.1}\\ \sin\left(\frac{\theta}{2}\right) & \boldsymbol{=} \sum^{\infty}_{k=0}\frac{(\boldsymbol{-}1)^k}{(2k\boldsymbol{+}1)!}\left(\frac{\theta}{2}\right)^{2k\boldsymbol{+}1} \tag{C-02.2}\label{C-02.2} \end{align} we have \begin{align} U & \boldsymbol{=}I\cos\frac{\theta}{2}\boldsymbol{-}i(\mathbf{n}\boldsymbol{\cdot}\boldsymbol{\sigma})\sin\frac{\theta}{2}\boldsymbol{=}\mathrm N^{2}\cos\frac{\theta}{2}\boldsymbol{-}i\mathrm N\sin\frac{\theta}{2} \nonumber\\ & \boldsymbol{=} \mathrm N^{2}\Biggl[\sum^{\infty}_{k=0}\frac{(\boldsymbol{-}1)^k}{(2k)!}\left(\frac{\theta}{2}\right)^{2k}\Biggr]\boldsymbol{-}i\mathrm N\Biggl[\sum^{\infty}_{k=0}\frac{(\boldsymbol{-}1)^k}{(2k\boldsymbol{+}1)!}\left(\frac{\theta}{2}\right)^{2k\boldsymbol{+}1} \Biggr] \nonumber\\ &\boldsymbol{=} \Biggl[\sum^{\infty}_{k=0}\frac{1}{(2k)!}\left(\boldsymbol{-}i\frac{\theta\mathrm N}{2}\right)^{2k}\Biggr]\boldsymbol{+}\Biggl[\sum^{\infty}_{k=0}\frac{1}{(2k\boldsymbol{+}1)!}\left(\boldsymbol{-}i\frac{\theta\mathrm N}{2}\right)^{2k\boldsymbol{+}1} \Biggr] \nonumber\\ & \boldsymbol{=}\sum^{\boldsymbol{\infty}}_{m\boldsymbol{=}0}\frac{1}{m!}\left(\boldsymbol{-}i\frac{\theta\mathrm N}{2}\right)^{m}\boldsymbol{=}\exp\left(\boldsymbol{-}i\frac{\theta\mathrm N}{2}\right) \tag{C-03}\label{C-03} \end{align} From definitions \eqref{09} and \eqref{B-12} \begin{equation} \theta\mathrm N\boldsymbol{=}\theta(\mathbf{n}\boldsymbol{\cdot}\boldsymbol{\sigma})\boldsymbol{=}(\theta\mathbf{n}\boldsymbol{\cdot}\boldsymbol{\sigma})=(\boldsymbol{\theta}\boldsymbol{\cdot}\boldsymbol{\sigma}) \tag{C-04}\label{C-04} \end{equation} so \begin{align} U &\boldsymbol{=}I\cos\frac{\theta}{2}\boldsymbol{-}i(\mathbf{n}\boldsymbol{\cdot}\boldsymbol{\sigma})\sin\frac{\theta}{2}\boldsymbol{=}\boldsymbol{e}^{\boldsymbol{- }\boldsymbol{i}\,\boldsymbol{\theta}\boldsymbol{\cdot}\boldsymbol{\sigma}/2} \tag{C-05.1}\label{C-05.1}\\ U^{\boldsymbol{*}}&\boldsymbol{=}I\cos\frac{\theta}{2}\boldsymbol{+}i(\mathbf{n}\boldsymbol{\cdot}\boldsymbol{\sigma})\sin\frac{\theta}{2}\boldsymbol{=}\boldsymbol{e}^{\boldsymbol{+ }\boldsymbol{i}\,\boldsymbol{\theta}\boldsymbol{\cdot}\boldsymbol{\sigma}/2} \tag{C-05.2}\label{C-05.2} \end{align}}

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Answer 2

I guess you know where the exponential comes from. A Lie group element close to 1 (and any Lie group element for most Lie groups of interest) can be written as the exponential of a Lie algebra element.

Let's start abstractly, with a Lie group $G$ with Lie algebra $\mathfrak g$. Then $G$ acts on $\mathfrak g$ trough the derivative of conjugation in $G$, which for matrix groups is simply $g\cdot X := gXg^{-1}$, where $g\in G\subset\text{GL}(n,\mathbb C)$ and $X\in\mathfrak g\subset\mathcal{M}(n,\mathbb C)$. This makes $\mathfrak g$ into a representation of $G$ called the adjoint representation.

Let $m$ be the dimension of $G$, hence also of $\mathfrak g\cong\mathbb R^m$. Choosing a basis of $X$ gives us a representation of $G$ on $\mathbb R^m$ by $m\times m$ matrices. This representation can also be obtained as the exponential of a Lie algebra representation, which has the advantage that you can work easily with generators. A general element of $G$ in this representation can then be written as the exponential of an $m\times m$ matrix (the $i$ in the exponent is conventional in physics).

Now let's go to $G = \text{SU}(2)$. Its Lie algebra $\mathfrak{su}(2)$ is spanned by the Pauli matrices, giving you the isomorphism between $\mathfrak{su}(2)$ and $\mathbb R^3$ (and likewise between the complexified Lie algebra $\mathfrak{su}(2)_{\mathbb C}$ and $\mathbb C^3$).

Writing an element of $\mathbb R^3$ as $\phi$, the corresponding element of $\mathfrak{su}(2)$ is $\phi\cdot\tau$, a linear combination of the Pauli matrices, while $T\cdot\theta$ is a linear combination of their images in $\mathcal M(3,\mathbb R)$.

Summarizing, you have the action of $\text{SU}(2)$ on $\mathfrak{su}(2)$, the Lie algebra of Hermitian matrices by conjugation. If you choose the basis of Pauli matrices for the latter, you obtain an explicit isomorphism of $\mathfrak{su}(2)$ with $\mathbb R^3$. You then write the action in this new basis to get (1) from (2), and likewise for the other way around.

Answer 3

The other answer framed the generic problem for you, very soundly. But SU(2) is privileged in that you can write the exact group transformations explicitly. The moment you adopted the adjoint language by defining ${\bf \Phi}$, you have adopted (2) for some $\theta$, and derived, with, e.g., this answer, or your favorite introductory group theory text, $$ {\boldsymbol \Phi}\equiv{\boldsymbol \phi}\cdot {\boldsymbol\tau} \mapsto e^{i{\boldsymbol \theta}\cdot {\boldsymbol \tau}} {\boldsymbol \phi}\cdot {\boldsymbol\tau} e^{-i{\boldsymbol \theta}\cdot {\boldsymbol \tau}} = {\boldsymbol \phi}\cdot ( e^{i{\boldsymbol \theta}\cdot {\boldsymbol \tau}} {\boldsymbol \tau} e^{-i{\boldsymbol \theta}\cdot {\boldsymbol \tau}} )= ({\boldsymbol \phi} -2{\boldsymbol\theta}\times{\boldsymbol \phi+... })\cdot {\boldsymbol \tau}= \\ = {\boldsymbol \phi}\cdot \bigl ( \boldsymbol{\tau} ~ \cos (2\theta)+ \boldsymbol{ \hat{\theta} }\times \boldsymbol{\tau} ~\sin (2\theta)+ \boldsymbol{\hat{\theta}} ~ \boldsymbol{\hat{\theta}} \cdot \boldsymbol{\tau} ~ (1-\cos (2\theta))\bigr ) \\ ={\boldsymbol \phi}'\cdot {\boldsymbol\tau} ~. $$

That is, as per the conventions you adopted, with $\vec \theta=\hat \theta ~\theta$, $$ \vec{\phi '}=\bigl(\cos (2\theta) ~ {\mathbb 1} - \sin (2\theta)~ { \hat{\theta} }\times ~ +(1-\cos (2\theta)) {\hat{\theta}} ~ {\hat{\theta}} \cdot ~ \bigr )\vec{\phi}= \exp (-i2\vec\theta \cdot \vec T)~ \vec \phi, $$ the standard Rodrigues rotation formula for vectors. I have used the double angle convention for simplicity in transforming the doublet: so, physical , triplet vector, rotations by $\theta$ amount to expressions with $2\theta$ here. Conversion of cross product and dyadic dotting operations on a vector to 3×3 matrix notation there is also standard, given $\vec \theta \times \vec \phi=i\vec \theta \cdot \vec T ~\vec \phi $.

Study the formula, standard in college courses, by inspecting the Lie generator pieces, the lowest order in $\theta$.

That is to say, each and every step of this manifestly reverses to produce (2) from (1).