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A vector $\boldsymbol{r}$ in $\mathbb{R}^3$ transforms under rotation $\boldsymbol{A}$ to $\boldsymbol{r}'=\boldsymbol{Ar}$. It is equivalent to an SU(2) "rotation" as $$\left( \boldsymbol{r}'\cdot\boldsymbol{\sigma} \right) = \boldsymbol{h} \left( \boldsymbol{r}\cdot\boldsymbol{\sigma} \right) \boldsymbol{h}^{-1},$$ where $\boldsymbol{h}$ is the counterpart of $\boldsymbol{A}$ in SU(2) given by the homomorphism between these two groups.

Now the question is, what would be the equivalent transformation in SU(2) of the rotation of a matrix in $\mathbb{R}^3$? In other words, what is the equivalent in SU(2) of $\boldsymbol{M}'=\boldsymbol{A}\boldsymbol{M}\boldsymbol{A}^{-1}$.

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I am a little confused about what you're asking. You presumably know that there is a double cover $\phi : \text{SU}(2) \to \text{SO}(3)$. Given any action of $\text{SO}(3)$ on any kind of thing, pulling back along $\phi$ gives you an action of $\text{SU}(2)$ on that thing (an element $g \in \text{SU}(2)$ acts by however $\phi(g)$ acts). –  Qiaochu Yuan Jun 20 '12 at 7:15
    
@QiaochuYuan: I didn't learn the concept of "pullback". Can you elaborate? –  C.R. Jun 20 '12 at 7:33
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All I mean is that if $S$ is a set (e.g. a vector space) and $\rho : \text{SO}(3) \to \text{Aut}(S)$ is an action of $\text{SO}(3)$ on that set (e.g. a linear representation on a vector space) then $\rho \circ \phi : \text{SU}(2) \to \text{Aut}(S)$ is the corresponding action of $\text{SU}(2)$. –  Qiaochu Yuan Jun 20 '12 at 8:11
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@Qiaochu: I think the OP is asking for an explicit formula for $\rho \circ \phi$ where $S = \mathcal{M}_{3\times 3}$ and $\rho(A)M = AMA^{-1}$. –  Willie Wong Jun 20 '12 at 8:15

1 Answer 1

Firstly, we need to map $\mathbb{R}^3$ to the representation space $V$ for $\mathrm{SU}(2)$. One possible map is given by the following formula: $$\begin{pmatrix} x \\ y \\ z \end{pmatrix} \mapsto x \mathbf{I} + y \mathbf{J} + z \mathbf{K}$$ where \begin{align} \mathbf{I} & = \begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix} & \mathbf{J} & = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} & \mathbf{K} & = \begin{pmatrix} 0 & i \\ i & 0 \end{pmatrix} \end{align}

$\mathrm{SU}(2)$ acts on $V$ by conjugation: so for each $X$ in $V$ and each $A$ in $\mathrm{SU}(2)$, the ordinary matrix product $A X A^{-1}$ is in $V$. This is linear in $X$ and is indeed a linear representation of $\mathrm{SU}(2)$. Indeed, if $$A = \begin{pmatrix} r e^{i \theta} & s e^{-i \phi} \\ -s e^{i \phi} & r e^{-i \theta} \end{pmatrix}$$ where $r, s, \theta, \phi$ are real numbers and $r^2 + s^2 = 1$, then $A \in \mathrm{SU}(2)$, and \begin{align} A \mathbf{I} A^{-1} & = (r^2 - s^2) \mathbf{I} + 2 r s \sin (\theta - \phi) \mathbf{J} - 2 r s \cos (\theta - \phi) \mathbf{K} \\ A \mathbf{J} A^{-1} & = 2 r s \sin (\theta + \phi) \mathbf{I} + (r^2 \cos 2 \theta + s^2 \cos 2 \phi) \mathbf{J} + (r^2 \sin 2 \theta - s^2 \sin 2 \phi) \mathbf{K} \\ A \mathbf{K} A^{-1} & = 2 r s \cos (\theta + \phi) \mathbf{I} - (r^2 \sin 2 \theta + s^2 \sin 2 \phi) \mathbf{J} + (r^2 \cos 2 \theta - s^2 \cos 2 \phi) \mathbf{K} \end{align} Thus, the induced action of $\mathrm{SU}(2)$ on $\mathbb{R}^3$ is given by the group homomorphism below, $$\begin{pmatrix} r e^{i \theta} & s e^{-i \phi} \\ -s e^{i \phi} & r e^{-i \theta} \end{pmatrix} \mapsto \begin{pmatrix} r^2 - s^2 & 2 r s \sin (\theta + \phi) & 2 r s \cos (\theta + \phi) \\ 2 r s \sin (\theta - \phi) & r^2 \cos 2 \theta + s^2 \cos 2 \phi & -r^2 \sin 2 \theta - s^2 \sin 2 \phi \\ -2 r s \cos (\theta - \phi) & r^2 \sin 2 \theta - s^2 \sin 2 \phi & r^2 \cos 2 \theta - s^2 \cos 2 \phi \end{pmatrix}$$ and one may verify that the RHS is a matrix in $\mathrm{SO}(3)$.

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You are just re-expressing what I already know, that $\left( \boldsymbol{r}'\cdot\boldsymbol{\sigma} \right) = \boldsymbol{h} \left( \boldsymbol{r}\cdot\boldsymbol{\sigma} \right) \boldsymbol{h}^{-1}$. –  C.R. Jun 20 '12 at 10:47
    
@KarsusRen: Then what would you like to know? –  Zhen Lin Jun 20 '12 at 11:14
    
I stated clearly in my question: what is the equivalent transformation in SU(2) of the rotation of a linear operator (matrix) instead of vector in $\mathbb{R}^3$? –  C.R. Jun 20 '12 at 11:54
    
Just conjugate by the corresponding $\mathrm{SO}(3)$ matrix. There isn't a nice way of representing it as a matrix formula in terms of the $\mathrm{SU}(2)$ matrix. (Think about it: if a vector in $\mathbb{R}^3$ corresponds to a matrix in $V$, then a matrix would have to correspond to some hideous rank-3 tensor!) –  Zhen Lin Jun 20 '12 at 13:22

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