# How to prove $e^{+i(\theta/2)(\hat{n}\cdot \sigma)}\sigma e^{-i(\theta/2)(\hat{n}\cdot \sigma)} = e^{\theta \hat{n}\cdot J}\sigma$?

Disclaimer: I'm sure this has been asked 100 times before, but I can't find the question asked or answered quite like this. If there are specific duplicates that could give me a simple satisfactory answer, in the manner I'm looking for, I would appreciate being directed towards them.

In 3D space ($$\mathbb{R}^3$$) we have the rotation matrices which are elements of $$SO(3)$$, the group of orthogonal ($$R^T R = I$$) matrices with unit determinant ($$\text{det}(R)=+1$$). From the fact that $$R$$ is orthogonal, it can be proven that $$R$$ can be written as $$R = e^A$$ where $$A$$ is skew-symmetric ($$A+A^T=0$$). There is a basis of skew-symmetric matrices \begin{align*} J_x =& \begin{pmatrix}0 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \end{pmatrix}\\ J_y =& \begin{pmatrix}0 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 0 \end{pmatrix}\\ J_z =& \begin{pmatrix}0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}\\ \end{align*} with commutation relations (I think these are somehow determined from $$R^TR=I$$ but I'm not totally sure how) $$[J_i, J_j] = \epsilon_{ijk} J_k$$ and it can be proven that we can write $$R = e^{\theta \hat{n}\cdot J}$$ where $$J=(J_x, J_y, J_z)$$, $$\hat{n}$$ is a unit vector, and $$\theta$$ is a rotation angle. In this case $$R$$ implements a rotation by angle $$\theta$$ about axis $$\hat{n}$$. In Lie-theoretic language we would say $$SO(3)$$ is a Lie group and the skew-symmetric basis $$J_i$$ spans the Lie algebra $$\mathfrak{so}(3)$$. The formula above indicates that the Lie Group $$SO(3)$$ can be generated by the Lie algebra $$\mathfrak{so}(3)$$ via the exponential map.

In complex 2D space ($$\mathbb{C}^2$$) (~spinor space modulo projections) we have spinor rotation matrices $$U$$ which are elements of $$SU(2)$$, the group of unitary ($$U^{\dagger}U=I$$) matrices with unit determinant ($$\det(U) = +1$$). from the fact that $$U$$ is unitary, it can be proven that $$U$$ can be written as $$U = e^{-iH}$$ where $$H$$ is Hermitian ($$H^{\dagger}=H$$). There is a basis of Hermitian $$2\times 2$$ matrices called the Pauli matrices: \begin{align*} \sigma_x =& \begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}\\ \sigma_y =& \begin{pmatrix}0 & -i \\ i & 0\end{pmatrix}\\ \sigma_z =& \begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}\\ \end{align*} which have the commutation relations $$[\sigma_i, \sigma_j] = i 2\epsilon_{ijk} \sigma_k$$ and it can be proven that we can write $$U = e^{-i(\theta/2)(\hat{n}\cdot\sigma)}$$ where $$\sigma = (\sigma_x, \sigma_y, \sigma_z)$$, $$\hat{n}$$ is a unit vector and $$\theta$$ is a rotation angle. In Lie-theoretic language we would say $$SU(2)$$ is a Lie group and the Hermitian basis $$\sigma_i$$ spans the Lie algebra $$\mathfrak{su}(2)$$. The formula above indicates that the Lie group $$SU(2)$$ can be generated by the Lie algebra $$\mathfrak{su}(2)$$ via the exponential map.

The stories above were almost the exact same stories, but I left one fact out of the second story. I left out any interpretation of $$\theta$$ and $$\hat{n}$$. There is, apparently, a sense in which the unitary $$U$$ implements a real-space ($$\mathbb{R}^3$$) rotation by angle $$\theta$$ about vector $$\hat{n}$$. But this is a bit strange because $$U$$ acts on elements of the spinor space $$\mathbb{C}^2$$.

The above stories are two parallel stories about two different Lie groups being generated by their Lie algebras. This is fine. But there is a connection between the two stories given by the equation in the question: $$\tag{*} U^{\dagger}\sigma U = e^{i(\theta/2)(\hat{n}\cdot\sigma)} \sigma e^{-i(\theta/2)(\hat{n}\cdot\sigma)} = e^{\theta(\hat{n}\cdot J)} \sigma = R\sigma$$ On the left-most equation the $$U$$ conjugates each of the three elements of the operator vector $$\sigma$$ to give three new elements whereas on the right side the $$R$$ matrix mixes up the three elements of the operator vector $$\sigma$$. This equation is significant to me because it says that if we act the Unitary operator $$U$$ on $$|\psi\rangle$$ then the Bloch vector $$\langle \sigma \rangle = \langle \psi|\sigma | \psi \rangle \in \mathbb{R}^3$$ rotates by angle $$\theta$$ about vector $$\hat{n}$$.

The question is, how is $$(*)$$ proven given the setup I gave above?

I know it is related to the fact that $$SU(2)$$ is a 2-dimensional (spin-1/2) representation of $$SO(3)$$. I think this is related to the fact that the commutation relations for $$\mathfrak{so}(3)$$ and $$\mathfrak{su}(2)$$ look very similar. In fact, I believe this similarity allows there to be a Lie algebra isomorphism between $$\mathfrak{so}(3)$$ and $$\mathfrak{su}(2)$$ which allows it to be such that $$SU(2)$$ is a representation of $$SO(3)$$. But these are just a lot of words.. I'm seeking, if possible, a simple algebraic proof of $$(*)$$.

I also think the answer is very related to the so-called Adjoint representation. I think a satisfactory answer could explain exactly what is implied by the Adjoint representation in this context (along with the isomorphisms between $$\mathfrak{so}(3)$$ and $$\mathfrak{su}(2)$$) and how that results in a proof of $$(*)$$.

If possible I would prefer algebraic answers that rely minimally, or not at all, on the specific matrix representations given.

Pointing out any inaccuracies in my statements would also be very much appreciated.

Physically, you can solve the Heisenberg equations of motion. It is equivalent to all the general arguments of the adjoint action, but at least it cuts to the chase and gives your formula explicitly. For $$\theta=0$$, both agree, you just need to check that the derivatives agree. I will use the notion for all operator: $$O_\theta = U^\dagger OU$$ The LHS gives: \begin{align} \frac{d}{d\theta} \sigma_\theta &= \frac i2[n\cdot\sigma,\sigma]_\theta \\ &= (n\times\sigma)_\theta \\ &= n\times\sigma_\theta \end{align} or in index notation: \begin{align} \frac{d}{d\theta} \sigma_k(\theta) &= \frac i2[n_i\sigma_i,\sigma_k](\theta)\\ &= \frac i2(2in_i\epsilon_{ikj}\sigma_j)(\theta)\\ &= -n_i\epsilon_{ikj}\sigma_j(\theta)\\ &= (n\times\sigma(\theta))_k \end{align}

If you are familiar with solid mechanics, you will recognise the equation of uniform circular motion. You can therefore solve for $$\sigma_\theta$$ using the rotation matrix. Formally, it is the same equation of motion for $$R\sigma$$: $$\frac d{d\theta}R\sigma = n\times R\sigma \\ (n\cdot J)_{jk} = -n_i\epsilon_{ijk} \\ ((n\cdot J)\sigma)_k = -n_i\epsilon_{ikj}\sigma_j = (n\times\sigma)_k$$ if you view $$in$$ as your angular velocity vector (if you are familiar with solid mechanics). $$\sigma_\theta$$ and $$R\sigma$$ obey the same equations of motion and have the same initial condition, they are therefore equal for all times.

• Showing the two sides solve the same differential equation with the same initial conditions seems like a satisfactory answer to my question. But I don't understand a lot of the steps you're showing here. I think I follow $d_{\theta}\sigma_{\theta} \propto [n\cdot \sigma, \sigma]_{\theta}$. That this is proportional to $(n\times \sigma)_{\theta}$ is not obvious to me. And then neither of the two steps for the right-hand-side are clear to me. Could you spell these out in more detail? I'm curious if they are proven algebraically or if they are specific to these specific Lie groups/algebras. Commented Jun 26 at 14:01
• Ok, actually I've worked out the expressions you have for the left hand side. But I think you have an extra factor of $i$ in the latter two lines. The $i$ in the first line gets canceled by an $i$ that arises in converting the commutator to a cross product. I'm now trying to understand why $(n\times J)A = n\times A$ then I think I will follow this answer. Commented Jun 26 at 14:13
• Ok, so it seems like one essential element of this proof is that $(J_i)_{jk} = -\epsilon_{ijk}$. That is, this proof requires using the explicit forms for $J$. But these forms only look this way for $SO(3)$, not for $SO(n)$ I don't think. I think this gets at the uniqueness of $SO(3)$ for this particular expression that is spelled out in physics.stackexchange.com/a/679357/128186 Commented Jun 26 at 14:29
• right, but that characterization of $J$ only holds for $\mathfrak{so}(3)$, I think the structure coefficients are a lot more complicated for $\mathfrak{so}(n)$. Right? Commented Jun 26 at 14:38
• Commented Jun 26 at 14:44

You appear to have two unrelated gaps, the final one being the rotation formula for the vector (triplet) representation, the celebrated Rodrigues rotation formula. $$\vec \sigma$$ rotates like a vector. Adjoint indices rotate like a vector in all representations!! This is standard textbook stuff, so I'll assume it.

Your question then is straightforward to answer via direct calculation, since the 2$$\times$$2 Pauli vector exponentials have an explicit and tractable form, $$e^{i\theta n\cdot \sigma/2}= {\mathbb I} \cos(\theta/2) +in\cdot \sigma \sin(\theta/2) .$$ Thus, $$e^{i(\theta/2)(\hat{n}\cdot\vec \sigma)} \vec \sigma e^{-i(\theta/2)(\hat{n}\cdot\vec \sigma)}\\ = \Bigl ( \cos(\theta/2) +i\hat n\cdot\vec \sigma \sin(\theta/2)\Bigr ) \sigma^k\Bigl ( \cos(\theta/2) -i\hat n\cdot\vec \sigma \sin(\theta/2)\Bigr )\\ =\cos\theta~\sigma^k + \sin\theta ~\epsilon _{kjm} n^j \sigma^m -(\cos\theta -1) n^k n^m \sigma^m \\ =\cos\theta~\vec \sigma +\sin\theta ~\hat n \times \vec \sigma + (1-\cos\theta )\hat n ~~\hat n \cdot \vec \sigma ,$$ by elementary (tedious) Pauli matrix algebra.

But this is just the celebrated Euler-Rodrigues vector rotation formula linked and explained above, and written in terms of matrix exponentials later in the article. Note how the prefactor trigonometric function coefficients have doubled the angle and rearranged themselves in the respective expressions. Dozens of students miss the magic and get confused by this and attempt to "correct" formulas which are already correct!

The (antisymmetric) logarithm of a rotation matrix is is found here, so (Tr$$R -1)/2=\cos\theta$$, etc., so $$R=e^{\theta \hat n\cdot \vec J}$$.

• In the context of isospin, subject to the same group, you can observe this maneuver here. Commented Jun 26 at 21:19
• You also appear to be interested in (*) for the fundamental and adjoint representations of $SU(N)$, a far broader and more abstract question, a question bypassing SO(N)s altogether... There is a third apparent gap confusing groups and representations. Commented Jun 26 at 21:56
• I upvoted this answer since relating both the LHS and RHS of $(*)$ to the Rodrigues rotation formula seems tractable and is along the lines of what I was asking. In the question (before you last comment about my third gap) you mentioned I appeared to have two gaps but you only detailed the final one. What's the first gap? I'm sure you're right about my gaps in knowledge! These are what I'm trying to fill in! I'm not so familiar with "adjoint stuff" that you say is textbook material. It's likely this is some of what I need to elarn... Commented Jun 26 at 22:57
• The first gap was the answer: explicit calculation in terms of Pauli vectors… Commented Jun 27 at 1:20

Your equation $$(*)$$ is a bit hand-wavy since it doesn't hold exactly (on the r.h.s. you have a vector in $$\mathbb{R}^3$$ and on the l.h.s. a vector of Pauli matrices), but it's clear what you mean. I think what you actually are trying to show though is that

$$U\rho U^{\dagger} = U\frac{1}{2}(I + \vec{r}\cdot \vec{\sigma}) U^{\dagger} = \frac{1}{2}(I + R_{\vec{n}}(\theta)\vec{r}\cdot \vec{\sigma}), \qquad U = \mathrm{exp}[-i\frac{\theta}{2}\vec{n}\cdot\vec{\sigma}]$$

A crude explicit way of doing this is by starting for example with a z-rotation. Let $$U_z(\theta) = e^{-i\frac{\theta}{2}\sigma_z}$$, then

\begin{aligned} U_z(\theta) \sigma_x U_z(\theta)^{\dagger} &= \cos^2\frac{\theta}{2} \sigma_x + i\sin\frac{\theta}{2}\cos\frac{\theta}{2}\sigma_x\sigma_z - i\sin\frac{\theta}{2}\cos\frac{\theta}{2}\sigma_z\sigma_x + \sin^2 \frac{\theta}{2} \sigma_z\sigma_x\sigma_z\\ &=\cos\theta \sigma_x + \sin\theta \sigma_y \end{aligned}

In the same way you find that

\begin{aligned} U_z(\theta) \sigma_y U_z(\theta)^{\dagger} &= \cos\theta \sigma_y - \sin\theta\sigma_x\\ U_z(\theta) \sigma_z U_z(\theta)^{\dagger} &= \sigma_z \end{aligned}

Therefore $$U_z(\theta)\frac{1}{2}(I + \vec{r}\cdot \vec{\sigma}) U_z(\theta)^{\dagger} = \frac{1}{2}(I + \vec{r}'\cdot \vec{\sigma})$$ with

$$\vec{r}' = \begin{bmatrix} \cos\theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{bmatrix} \vec{r} = \mathrm{exp}[\theta J_z]\vec{r}$$

If you do the same for rotations about one or two more axes, and use the fact that you can decompose any rotation into three rotations about the main axes, you get the general statement.

$$\newcommand{\bl}[1]{\boldsymbol{#1}} \newcommand{\e}{\bl=} \newcommand{\p}{\bl+} \newcommand{\m}{\bl-} \newcommand{\mb}[1]{\mathbf {#1}} \newcommand{\mc}[1]{\mathcal {#1}} \newcommand{\mr}[1]{\mathrm {#1}} \newcommand{\mf}[1]{\mathfrak{#1}} \newcommand{\gr}{\bl>} \newcommand{\les}{\bl<} \newcommand{\greq}{\bl\ge} \newcommand{\leseq}{\bl\le} \newcommand{\il}[1]{\:#1\:} \newcommand{\plr}[1]{\left(#1\right)} \newcommand{\blr}[1]{\left[#1\right]} \newcommand{\brclr}[1]{\left\{#1\right\}} \newcommand{\clr}[1]{\left\{#1\right\}} \newcommand{\vlr}[1]{\left\vert#1\right\vert} \newcommand{\Vlr}[1]{\left\Vert#1\right\Vert} \newcommand{\lara}[1]{\left\langle#1\right\rangle} \newcommand{\lav}[1]{\left\langle#1\right|} \newcommand{\vra}[1]{\left|#1\right\rangle} \newcommand{\lavra}[2]{\left\langle#1|#2\right\rangle} \newcommand{\lavvra}[3]{\left\langle#1\right|#2\left|#3\right\rangle} \newcommand{\vp}{\vphantom{\dfrac{a}{b}}} \newcommand{\Vp}[1]{\vphantom{#1}} \newcommand{\hp}[1]{\hphantom{#1}} \newcommand{\tl}[1]{\tag{#1}\label{#1}}$$

$$\bl\S$$ A. The main task

For the rotation of a 3-vector $$\:\mb x\e\plr{x_1,x_2,x_3}\bl\in \mathbb R^3\:$$ around a unit vector $$\:\mb n\e\plr{n_1,n_2,n_3}\:$$ through an angle $$\:\theta\:$$ we have(1) $$$$\mb x' \e \mr R\mb x \e \plr{\cos\theta}\mb x\p\plr{1\m\cos\theta}\plr{\mb n\bl\cdot\mb x}\mb n\p \sin\theta\plr{\mb n\bl\times\mb x} \tl{A-01}$$$$ so $$\:\mr R\:$$ being the $$\:3\times 3\:$$ rotation matrix \begin{align} \!\!\!\!\!\!\!\!\mr R\plr{\mb n,\theta} &\e \begin{bmatrix} R_{11} & R_{12} & R_{13}\vp\\ R_{21} & R_{22} & R_{23}\vp\\ R_{31} & R_{32} & R_{33}\vp \end{bmatrix} \e \begin{bmatrix} \bl R_1\vp\\ \bl R_2\vp\\ \bl R_3\vp \end{bmatrix} \tl{A-02}\\ &\e \begin{bmatrix} \cos\theta\p\plr{1\m\cos\theta}n^2_1 & \plr{1\m\cos\theta}n_1n_2\m n_3\sin\theta & \plr{1\m\cos\theta}n_1n_3\p n_2\sin\theta\vp\\ \plr{1\m\cos\theta}n_2n_1\p n_3\sin\theta & \cos\theta\p\plr{1\m\cos\theta}n^2_2 & \plr{1\m\cos\theta}n_2n_3\m n_1\sin\theta\vp\\ \plr{1\m\cos\theta}n_3n_1\m n_2\sin\theta & \plr{1\m\cos\theta}n_3n_2\p n_1\sin\theta & \cos\theta\p\plr{1\m\cos\theta}n^2_3\vp \end{bmatrix} \nonumber \end{align} For the representation of the rotation \eqref{A-01} by special unitary matrices $$\:\mr U\bl\in \mr{SU}\!\plr{2}\:$$ we proceed as follows : Any real 3-vector $$\:\mb x\e\plr{x_1,x_2,x_3}\bl\in \mathbb R^3\:$$ is represented uniquely by an hermitian traceless $$\:2\times 2\:$$ complex matrix $$\:\mr X\:$$ $$$$\mb x \e \begin{bmatrix} x_1\vp\\ x_2\vp\\ x_3\vp \end{bmatrix} \bl\in \mathbb R^3 \quad \bl {\m\!\!\!\m\!\!\!\m\!\!\!\longrightarrow} \quad \mr X\e \begin{bmatrix} x_3 & x_1\m i x_2\Vp{\dfrac{\tfrac{a}{b}}{\tfrac{c}{d}}}\\ x_1\p i x_2 & \m x_3\Vp{\dfrac{\tfrac{a}{b}}{\tfrac{c}{d}}} \end{bmatrix} \bl\in \mathbb H \tl{A-03}$$$$ Here we use the symbol $$\:\mathbb H\:$$ for the linear space of hermitian traceless $$\:2\times 2\:$$ complex matrices.

After a not so easy elaboration, see $$\bl\S$$ B, the rotation \eqref{A-01} is expressed via this representation of vectors by the equation below $$$$\mr X' \e \begin{bmatrix} x'_3 & x'_1\m i x'_2\Vp{\dfrac{\tfrac{a}{b}}{\tfrac{c}{d}}}\\ x'_1\p i x'_2 & \m x_3\Vp{\dfrac{\tfrac{a}{b}}{\tfrac{c}{d}}} \end{bmatrix}\e \mr U \begin{bmatrix} x_3 & x_1\m i x_2\Vp{\dfrac{\tfrac{a}{b}}{\tfrac{c}{d}}}\\ x_1\p i x_2 & \m x_3\Vp{\dfrac{\tfrac{a}{b}}{\tfrac{c}{d}}} \end{bmatrix} \mr U^{\bl *}\e \mr U \mr X \mr U^{\bl *} \nonumber$$$$ that is $$$$\mr X' \e \mr U \mr X \mr U^{\bl *} \tl{A-04}$$$$ where \begin{align} \mr U\hp{^{\bl *}} & \e \begin{bmatrix} \cos\dfrac{\theta}{2}\m i n_3\sin\dfrac{\theta}{2} & \m\sin\dfrac{\theta}{2}\plr{n_2\p i n_1}\Vp{\dfrac{\tfrac{a}{b}}{\dfrac{c}{d}}}\\ \sin\dfrac{\theta}{2}\plr{n_2\m i n_1} & \cos\dfrac{\theta}{2}\p i n_3\sin\dfrac{\theta}{2}\Vp{\dfrac{\tfrac{a}{b}}{\tfrac{c}{d}}} \end{bmatrix} \bl\in\mr{SU}\!\plr{2} \tl{A-05a}\\ \mr U^{\bl *} & \e \begin{bmatrix} \cos\dfrac{\theta}{2}\p i n_3\sin\dfrac{\theta}{2} & \sin\dfrac{\theta}{2}\plr{n_2\p i n_1}\Vp{\dfrac{\tfrac{a}{b}}{\dfrac{c}{d}}}\\ \m\sin\dfrac{\theta}{2}\plr{n_2\m i n_1} & \cos\dfrac{\theta}{2}\m i n_3\sin\dfrac{\theta}{2}\Vp{\dfrac{\tfrac{a}{b}}{\tfrac{c}{d}}} \end{bmatrix} \bl\in\mr{SU}\!\plr{2} \tl{A-05b} \end{align} with property $$$$\mr U\mr U^{\bl *}\e \mr I \e \mr U^{\bl *}\mr U \tl{A-06}$$$$

Using the usual basis $$\:\{\mb e_1,\mb e_2,\mb e_3\}\:$$ of $$\:\mathbb R^3\:$$ the representation \eqref{A-03} induces a basis in the space $$\:\mathbb H\:$$ of the hermitian traceless $$\:2\times 2\:$$ complex matrices \begin{align} \mb e_1 &\e \begin{bmatrix} \:1 \:\vp\\ 0\vp\\ 0\vp \end{bmatrix} \quad \bl {\m\!\!\!\m\!\!\!\m\!\!\!\longrightarrow} \quad \sigma_1\e \begin{bmatrix} \: 0 \: & \hp\m 1 \: \vp\\ \: 1\: & \hp\m0 \:\vp \end{bmatrix} \tl{A-07.1}\\ \mb e_2 &\e \begin{bmatrix} \:0 \:\vp\\ 1\vp\\ 0\vp \end{bmatrix} \quad \bl {\m\!\!\!\m\!\!\!\m\!\!\!\longrightarrow} \quad \sigma_2\e \begin{bmatrix} \: 0 \: & \m i \: \vp\\ \: i \: & \hp\m 0 \:\vp \end{bmatrix} \tl{A-07.2}\\ \mb e_3 &\e \begin{bmatrix} \:0 \:\vp\\ 0\vp\\ 1\vp \end{bmatrix} \quad \bl {\m\!\!\!\m\!\!\!\m\!\!\!\longrightarrow} \quad \sigma_3\e \begin{bmatrix} \: 1 \: & \hp\m 0 \: \vp\\ \: 0 \: & \m 1 \:\vp \end{bmatrix} \tl{A-07.3} \end{align} that is the Pauli matrices $$\:\{\sigma_1,\sigma_2,\sigma_3\}$$. We define the 3-vector Pauli operator $$$$\bl \sigma \e \begin{bmatrix} \:\sigma_1 \:\vp\\ \sigma_2\vp\\ \sigma_3\vp \end{bmatrix} \tl{A-08}$$$$ and use it to express alternatively the following representations \begin{align} \mb x\hp' &\e x_1\mb e_1\p x_2\mb e_2\p x_3\mb e_3 \quad \bl {\m\!\!\!\m\!\!\!\m\!\!\!\longrightarrow} \quad \mr X\hp'\e x_1\sigma_1\p x_2\sigma_2\p x_3\sigma_3\e \plr{\mb x\bl\cdot\bl \sigma} \tl{A-09a}\\ \mb x' &\e x'_1\mb e_1\p x'_2\mb e_2\p x'_3\mb e_3 \quad \bl {\m\!\!\!\m\!\!\!\m\!\!\!\longrightarrow} \quad \mr X'\e x'_1\sigma_1\p x'_2\sigma_2\p x'_3\sigma_3\e \plr{\mb x'\!\bl\cdot\bl \sigma} \tl{A-09b} \end{align} and \begin{align} \mr U\hp{^{\bl *}} & \e \begin{bmatrix} \cos\dfrac{\theta}{2}\m i n_3\sin\dfrac{\theta}{2} & \m\sin\dfrac{\theta}{2}\plr{n_2\p i n_1}\Vp{\dfrac{\tfrac{a}{b}}{\dfrac{c}{d}}}\\ \sin\dfrac{\theta}{2}\plr{n_2\m i n_1} & \cos\dfrac{\theta}{2}\p i n_3\sin\dfrac{\theta}{2}\Vp{\dfrac{\tfrac{a}{b}}{\tfrac{c}{d}}} \end{bmatrix} \e \blr{\mr I\cos\dfrac{\theta}{2}\m i\plr{\mb n\bl\cdot\bl \sigma}\sin\dfrac{\theta}{2}} \e e^{\m\frac12 i\,\theta\plr{\mb n\bl\cdot\bl \sigma}} \tl{A-10a}\\ \mr U^{\bl *} & \e \begin{bmatrix} \cos\dfrac{\theta}{2}\p i n_3\sin\dfrac{\theta}{2} & \sin\dfrac{\theta}{2}\plr{n_2\p i n_1}\Vp{\dfrac{\tfrac{a}{b}}{\dfrac{c}{d}}}\\ \m\sin\dfrac{\theta}{2}\plr{n_2\m i n_1} & \cos\dfrac{\theta}{2}\m i n_3\sin\dfrac{\theta}{2}\Vp{\dfrac{\tfrac{a}{b}}{\tfrac{c}{d}}} \end{bmatrix} \e \blr{\mr I\cos\dfrac{\theta}{2}\p i\plr{\mb n\bl\cdot\bl \sigma}\sin\dfrac{\theta}{2}} \e e^{\p\frac12 i\,\theta\plr{\mb n\bl\cdot\bl \sigma}} \tl{A-10b} \end{align} Using above expressions equation \eqref{A-04} yields $$$$\plr{\mb x'\!\bl\cdot\bl \sigma} \e e^{\m\frac12 i\,\theta\plr{\mb n\bl\cdot\bl \sigma}}\plr{\mb x\bl\cdot\bl \sigma}e^{\p\frac12 i\,\theta\plr{\mb n\bl\cdot\bl \sigma}} \tl{A-11}$$$$ Inverting we have $$$$\plr{\mb x\!\bl\cdot\bl \sigma} \e e^{\p\frac12 i\,\theta\plr{\mb n\bl\cdot\bl \sigma}}\plr{\mb x'\bl\cdot\bl \sigma}e^{\m\frac12 i\,\theta\plr{\mb n\bl\cdot\bl \sigma}} \tl{A-12}$$$$ Since $$\:\mb x\e \mr R^{\m 1} \mb x'\e\mr R^{\bl\top} \mb x'\:$$ where $$\:\mr R\:$$ the $$\:3\times 3\:$$ rotation matrix of equation \eqref{A-02} we have $$$$e^{\p\frac12 i\,\theta\plr{\mb n\bl\cdot\bl \sigma}}\plr{\mb x'\bl\cdot\bl \sigma}e^{\m\frac12 i\,\theta\plr{\mb n\bl\cdot\bl \sigma}}\e \plr{\blr{\mr R^{\bl\top} \mb x'}\bl\cdot\bl \sigma\vp} \tl{A-13}$$$$ If in equation \eqref{A-13} we take successively $$\:\mb x'\e\mb e_1,\mb x'\e\mb e_2,\mb x'\e\mb e_3\:$$ we have respectively \begin{align} e^{\p\frac12 i\,\theta\plr{\mb n\bl\cdot\bl \sigma}}\,\sigma_1\, e^{\m\frac12 i\,\theta\plr{\mb n\bl\cdot\bl \sigma}} &\e \plr{\blr{\mr R^{\bl\top} \mb e_1}\bl\cdot\bl \sigma\vp}\e\plr{\bl R_1\bl\cdot\bl \sigma} \tl{A-14.1}\\ e^{\p\frac12 i\,\theta\plr{\mb n\bl\cdot\bl \sigma}}\,\sigma_2\, e^{\m\frac12 i\,\theta\plr{\mb n\bl\cdot\bl \sigma}} &\e \plr{\blr{\mr R^{\bl\top} \mb e_2}\bl\cdot\bl \sigma\vp}\e\plr{\bl R_2\bl\cdot\bl \sigma} \tl{A-14.2}\\ e^{\p\frac12 i\,\theta\plr{\mb n\bl\cdot\bl \sigma}}\,\sigma_3\, e^{\m\frac12 i\,\theta\plr{\mb n\bl\cdot\bl \sigma}} &\e \plr{\blr{\mr R^{\bl\top} \mb e_3}\bl\cdot\bl \sigma\vp}\e\plr{\bl R_3\bl\cdot\bl \sigma} \tl{A-14.3} \end{align} so in one equation $$$$e^{\p\frac12 i\,\theta\plr{\mb n\bl\cdot\bl \sigma}}\, \begin{bmatrix} \:\sigma_1 \:\vp\\ \sigma_2\vp\\ \sigma_3\vp \end{bmatrix} \, e^{\m\frac12 i\,\theta\plr{\mb n\bl\cdot\bl \sigma}} \e \begin{bmatrix} \:\bl R_1\bl\cdot\bl \sigma \:\vp\\ \bl R_2\bl\cdot\bl \sigma\vp\\ \bl R_3\bl\cdot\bl \sigma\vp \end{bmatrix}\e \begin{bmatrix} R_{11} & R_{12} & R_{13}\vp\\ R_{21} & R_{22} & R_{23}\vp\\ R_{31} & R_{32} & R_{33}\vp \end{bmatrix} \begin{bmatrix} \:\sigma_1 \:\vp\\ \sigma_2\vp\\ \sigma_3\vp \end{bmatrix} \tl{A-15}$$$$ that is $$$$e^{\p\frac12 i\,\theta\plr{\mb n\bl\cdot\bl \sigma}}\, \bl\sigma \, e^{\m\frac12 i\,\theta\plr{\mb n\bl\cdot\bl \sigma}} \e \mr R\plr{\mb n,\theta}\bl\sigma \tl{A-16}$$$$

$$\bl{=\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!=}$$

$$\bl\S$$ B. The Unitary Representation of rotations in $$\mathbb R^3$$

From the equation of rotation $$\:\mr R\:$$ in $$\:\mathbb R^3$$ which relates a real 3-vector $$\:\mb x\:$$ and its image $$\:\mb x'\e \mr R\mb x\:$$ we'll produce the equation which relates the corresponding hermitian traceless $$\:2\times 2\:$$ complex matrices $$\:\mr X\e\plr{\mb x\bl\cdot\bl \sigma}\:$$ and $$\:\mr X'\e\plr{\mb x'\bl\cdot\bl \sigma}$$.

Taking the inner product of equation \eqref{A-01}, repeated here for convenience, $$$$\mb x' \e \mr R\mb x \e \plr{\cos\theta}\mb x\p\plr{1\m\cos\theta}\plr{\mb n\bl\cdot\mb x}\mb n\p \sin\theta\plr{\mb n\bl\times\mb x} \tl{B-01}$$$$ with the 3-vector Pauli operator $$\:\bl\sigma$$, see equation \eqref{A-08}, we have
$$$$\plr{\mb x'\bl\cdot\bl \sigma} \e \plr{\cos\theta}\plr{\mb x\bl\cdot\bl \sigma}\p\plr{1\m\cos\theta}\plr{\mb n\bl\cdot\mb x}\plr{\mb n\bl\cdot\bl \sigma}\p \sin\theta\blr{\plr{\mb n \bl\times \mb x\Vp{A^2_3}} \bl\cdot \bl \sigma \vp} \tl{B-02}$$$$ so $$$$\mr X' \e \plr{\cos\theta}\mr X\p\plr{1\m\cos\theta}\plr{\mb n\bl\cdot\mb x}\mr N \p \sin\theta\blr{\plr{\mb n \bl\times \mb x\Vp{A^2_3}} \bl\cdot \bl \sigma \vp} \tl{B-03}$$$$ where $$\:\mr N\:$$ the hermitian $$\:2\times 2\:$$ complex matrix corresponding to the unit vector $$\:\mb n$$ $$$$\mr N\e \plr{\mb n\bl\cdot\bl \sigma}\e \begin{bmatrix} n_3 & n_1\m i n_2\Vp{\dfrac{\tfrac{a}{b}}{\tfrac{c}{d}}}\\ n_1\p i n_2 & \m n_3\Vp{\dfrac{\tfrac{a}{b}}{\tfrac{c}{d}}} \end{bmatrix} \tl{B-04}$$$$ Using identities \eqref{D-05}, \eqref{D-06} with $$\:\mb p\e\mb n\:$$ and $$\:\mb q\e\mb x \:$$ we have respectively \begin{align} \blr{\plr{\mb n \bl\times \mb x\Vp{A^2_3}} \bl\cdot \bl \sigma \vp} &\e \dfrac{\plr{\mb n \bl\cdot \bl \sigma}\plr{\mb x \bl\cdot \bl \sigma}\m \plr{\mb x \bl\cdot \bl \sigma}\plr{\mb n \bl\cdot \bl \sigma}}{2\,i}\e i\,\dfrac{\mr X\mr N \m \mr N\mr X}{2} \tl{B-05}\\ &\nonumber\\ \plr{\mb n\bl\cdot\mb x}\mr I & \e \dfrac{\plr{\mb n \bl\cdot \bl \sigma}\plr{\mb x \bl\cdot \bl \sigma}\p \plr{\mb x \bl\cdot \bl \sigma}\plr{\mb n \bl\cdot \bl \sigma}}{2}\e \dfrac{\mr N\mr X \p \mr X\mr N}{2} \tl{B-06} \end{align} Inserting expressions \eqref{B-05}, \eqref{B-06} in \eqref{B-03} gives $$$$\mr X' \e \plr{\cos\theta}\mr X\p\plr{1\m\cos\theta}\plr{\dfrac{\mr N\mr X \p \mr X\mr N}{2}}\mr N \p i\sin\theta\plr{\dfrac{\mr X\mr N \m \mr N\mr X}{2}} \tl{B-07}$$$$ Replacing $$$$\cos\theta\e \cos^2\dfrac{\theta}{2} \m \sin^2\dfrac{\theta}{2}\e 1\m 2\sin^2\dfrac{\theta}{2}\,,\qquad \sin\theta\e 2\sin\dfrac{\theta}{2}\cos\dfrac{\theta}{2} \tl{B-08}$$$$ and using the property $$$$\mr N^2 \e \plr{\mb n \bl\cdot \bl \sigma}^2 \e \plr{\mb n \bl\cdot \bl \sigma}\plr{\mb n \bl\cdot \bl \sigma}\e\plr{\mb n \bl\cdot \mb n}\mr I\e\Vlr{\mb n}^2 \mr I \e \mr I \tl{B-09}$$$$ we have $$$$\begin{split} \mr X' & \e \plr{\cos^2\dfrac{\theta}{2}}\mr X \p \plr{\sin^2\dfrac{\theta}{2}}\mr N\mr X\mr N \p i \plr{\sin\dfrac{\theta}{2}\cos\dfrac{\theta}{2}}\plr{\mr X\mr N \m \mr N\mr X}\\ & \e \plr{\mr I\cos\dfrac{\theta}{2}\m i\,\mr N \sin\dfrac{\theta}{2}}\plr{\cos\dfrac{\theta}{2}}\mr X \p i\plr{\mr I\cos\dfrac{\theta}{2}\m i\,\mr N \sin\dfrac{\theta}{2}}\plr{\sin\dfrac{\theta}{2}}\mr X\mr N\\ & \e \plr{\mr I\cos\dfrac{\theta}{2}\m i\,\mr N \sin\dfrac{\theta}{2}}\plr{\mr X\cos\dfrac{\theta}{2}\p i\,\mr X\mr N \sin\dfrac{\theta}{2}}\\ & \e \plr{\mr I\cos\dfrac{\theta}{2}\m i\,\mr N \sin\dfrac{\theta}{2}}\mr X\plr{\mr I\cos\dfrac{\theta}{2}\p i\,\mr N \sin\dfrac{\theta}{2}}\\ \end{split} \tl{B-10}$$$$ so $$$$\mr X' \e \blr{\mr I\cos\dfrac{\theta}{2}\m i\plr{\mb n \bl\cdot \bl \sigma}\sin\dfrac{\theta}{2}}\mr X\blr{\mr I\cos\dfrac{\theta}{2}\p i\plr{\mb n \bl\cdot \bl \sigma} \sin\dfrac{\theta}{2}} \tl{B-11}$$$$ that is $$$$\mr X' \e \mr U \mr X\mr U^{\bl *} \tl{B-12}$$$$ where $$$$\mr U \e \mr I\cos\dfrac{\theta}{2}\m i\plr{\mb n \bl\cdot \bl \sigma}\sin\dfrac{\theta}{2} \tl{B-13}$$$$

$$\bl{=\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!=}$$

$$\bl\S$$ C. Exponential Form of the Unitary Representation

The unitary representation, see equation \eqref{B-13}, that is $$$$\mr U \e \mr I\cos\dfrac{\theta}{2}\m i\plr{\mb n \bl\cdot \bl \sigma}\sin\dfrac{\theta}{2} \tl{C-01}$$$$ could take exponential form, since the matrix $$\:\mr N \e \plr{\mb n \bl\cdot \bl \sigma}\:$$ satisfies equation \eqref{B-09}, repeated here for convenience $$$$\mr N^2 \e \plr{\mb n \bl\cdot \bl \sigma}^2 \e \mr I \tl{C-02}$$$$ and more generally for our purpose $$$$\begin{split} \mr N^{2k}\hp{^{\p 1}} & \e \plr{\mb n \bl\cdot \bl \sigma}^{2k}\hp{^{\p 1}}\e \mr I\\ \mr N^{2k\p 1} & \e \plr{\mb n \bl\cdot \bl \sigma}^{2k\p 1}\,\e \mr N\,, \qquad k\bl\in \mathbb N\\ \end{split} \tl{C-03}$$$$ Now, for any $$\:x\bl\in \mathbb R\:$$, the trigonometric functions $$\:\cos x\:$$ and $$\:\sin x\:$$ have the following infinite series expressions \begin{align} \cos x & \e 1\m\dfrac{x^2}{2!}\p\dfrac{x^4}{4!}\m\cdots \e \sum\limits_{k\e 0}^{\infty}\dfrac{\plr{\m 1}^k x^{2k}}{\plr{2k}!} \tl{C-04}\\ \sin x & \e x\m\dfrac{x^3}{3!}\p\dfrac{x^5}{5!}\m\cdots \e \sum\limits_{k\e 0}^{\infty}\dfrac{\plr{\m 1}^k x^{2k\p 1}}{\plr{2k\p 1}!} \tl{C-05} \end{align} which by the properties of $$\:\mr N \e \plr{\mb n \bl\cdot \bl \sigma}\:$$, equation \eqref{C-03}, give \begin{align} \mr I\cos x & \e \sum\limits_{k\e 0}^{\infty}\dfrac{\plr{\m 1}^k\plr{x\mr N}^{2k}}{\plr{2k}!}\e \cos\plr{x\mr N} \tl{C-06}\\ \mr N\sin x & \e \sum\limits_{k\e 0}^{\infty}\dfrac{\plr{\m 1}^k\plr{x\mr N}^{2k\p 1}}{\plr{2k\p 1}!}\e \sin\plr{x\mr N} \tl{C-07} \end{align} Setting $$\:x\e \m\theta/2\:$$ and $$\:\mr N \e \plr{\mb n \bl\cdot \bl \sigma}\:$$ in above equations, the expression \eqref{C-01} takes the form $$$$\mr U \e \cos\blr{\m\dfrac{1}{2}\,\theta\plr{\mb n \bl\cdot \bl \sigma}}\p i\sin\blr{\m\dfrac{1}{2}\,\theta\plr{\mb n \bl\cdot \bl \sigma}} \tl{C-08}$$$$ that is $$$$\mr U \e e^{\m\frac12\, i\,\theta\,\plr{\mb n\bl\cdot\bl \sigma}} \tl{C-09}$$$$

$$\bl{=\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!=}$$

$$\bl\S$$ D. Useful Identities with Pauli matrices

Consider two real 3-vectors $$\:\mb p, \mb q \bl\in \mathbb R^3\:$$ and their hermitian traceless $$\:2\times 2\:$$ complex matrices representatives $$\:\mr P,\mr Q \bl\in \mathbb H\:$$ respectively, see equation \eqref{A-03}, \begin{align} \mb p & \e \begin{bmatrix} p_1\vp\\ p_2\vp\\ p_3\vp \end{bmatrix}\quad \bl {\m\!\!\!\m\!\!\!\m\!\!\!\longrightarrow} \quad \mr P\e \begin{bmatrix} p_3 & p_1\m i p_2\Vp{\dfrac{\tfrac{a}{b}}{\tfrac{c}{d}}}\\ p_1\p i p_2 & \m p_3\Vp{\dfrac{\tfrac{a}{b}}{\tfrac{c}{d}}} \end{bmatrix} \e \plr{\mb p \bl\cdot \bl \sigma} \tl{D-01a}\\ \mb q & \e \begin{bmatrix} q_1\vp\\ q_2\vp\\ q_3\vp \end{bmatrix}\quad \bl {\m\!\!\!\m\!\!\!\m\!\!\!\longrightarrow} \quad \mr Q\e \begin{bmatrix} q_3 & q_1\m i q_2\Vp{\dfrac{\tfrac{a}{b}}{\tfrac{c}{d}}}\:\\ q_1\p i q_2 & \m q_3\Vp{\dfrac{\tfrac{a}{b}}{\tfrac{c}{d}}} \end{bmatrix} \e \plr{\mb q \bl\cdot \bl \sigma} \tl{D-01b} \end{align} Then at first $$$$\begin{split} \mr P\mr Q &\e \begin{bmatrix} p_3 & p_1\m i p_2\Vp{\dfrac{\tfrac{a}{b}}{\tfrac{c}{d}}}\\ p_1\p i p_2 & \m p_3\Vp{\dfrac{\tfrac{a}{b}}{\tfrac{c}{d}}} \end{bmatrix} \begin{bmatrix} q_3 & q_1\m i q_2\Vp{\dfrac{\tfrac{a}{b}}{\tfrac{c}{d}}}\:\:\\ q_1\p i q_2 & \m q_3\Vp{\dfrac{\tfrac{a}{b}}{\tfrac{c}{d}}} \end{bmatrix}\\ & \e \begin{bmatrix} \plr{\mb p \bl\cdot \mb q} \p i\plr{\mb p \bl\times \mb q}_3 & i\plr{\mb p \bl\times \mb q}_1 \p \plr{\mb p \bl\times \mb q}_2\Vp{\dfrac{\tfrac{a}{b}}{\tfrac{c}{d}}}\\ i\plr{\mb p \bl\times \mb q}_1 \m \plr{\mb p \bl\times \mb q}_2 & \plr{\mb p \bl\cdot \mb q} \m i\plr{\mb p \bl\times \mb q}_3\Vp{\dfrac{\tfrac{a}{b}}{\tfrac{c}{d}}} \end{bmatrix}\\ &\e \plr{\mb p \bl\cdot \mb q}\mr I\p i \begin{bmatrix} \plr{\mb p \bl\times \mb q}_3 & \plr{\mb p \bl\times \mb q}_1 \m i\plr{\mb p \bl\times \mb q}_2\Vp{\dfrac{\tfrac{a}{b}}{\tfrac{c}{d}}}\\ \plr{\mb p \bl\times \mb q}_1 \p i\plr{\mb p \bl\times \mb q}_2 &\m \plr{\mb p \bl\times \mb q}_3\Vp{\dfrac{\tfrac{a}{b}}{\tfrac{c}{d}}} \end{bmatrix}\\ & \e \plr{\mb p \bl\cdot \mb q\Vp{A^2_3}}\mr I\p i \blr{\plr{\mb p \bl\times \mb q\Vp{A^2_3}} \bl\cdot \bl \sigma \vp}\\ \end{split} \tl{D-02}$$$$ that is $$$$\boxed{\:\:\mr P\mr Q \e \plr{\mb p \bl\cdot \bl \sigma}\plr{\mb q \bl\cdot \bl \sigma}\e \plr{\mb p \bl\cdot \mb q\Vp{A^2_3}}\mr I\p i \blr{\plr{\mb p \bl\times \mb q\Vp{A^2_3}} \bl\cdot \bl \sigma \vp}\Vp{\dfrac{\dfrac{a}{b}}{\dfrac{c}{d}}}\:\:} \tl{D-03}$$$$ Interchanging $$\:\mr P,\mr Q\:$$ $$$$\mr Q\mr P \e \plr{\mb q \bl\cdot \bl \sigma}\plr{\mb p \bl\cdot \bl \sigma}\e \plr{\mb p \bl\cdot \mb q\Vp{A^2_3}}\mr I\m i \blr{\plr{\mb p \bl\times \mb q\Vp{A^2_3}} \bl\cdot \bl \sigma \vp} \tl{D-04}$$$$ Subtraction of equations \eqref{D-03}, \eqref{D-04} yields for the commutator $$$$\boxed{\:\:\blr{\mr P,\mr Q\vp}\e \mr P\mr Q\m\mr Q\mr P \e \plr{\mb p \bl\cdot \bl \sigma}\plr{\mb q \bl\cdot \bl \sigma}\m \plr{\mb q \bl\cdot \bl \sigma}\plr{\mb p \bl\cdot \bl \sigma} \e 2\, i\blr{\plr{\mb p \bl\times \mb q\Vp{A^2_3}} \bl\cdot \bl \sigma \vp}\Vp{\dfrac{\dfrac{a}{b}}{\dfrac{c}{d}}}\:\:} \tl{D-05}$$$$ while adding for the anticommutator yields $$$$\boxed{\:\:\brclr{\mr P,\mr Q\vp} \e\mr P\mr Q\p\mr Q\mr P\e \plr{\mb p \bl\cdot \bl \sigma}\plr{\mb q \bl\cdot \bl \sigma}\p \plr{\mb q \bl\cdot \bl \sigma}\plr{\mb p \bl\cdot \bl \sigma} \e 2 \plr{\mb p \bl\cdot \mb q\Vp{A^2_3}}\mr I\Vp{\dfrac{\dfrac{a}{b}}{\dfrac{c}{d}}}\:\:} \tl{D-06}$$$$

(1) See Rotation of a vector.