quantum-mechanics – Understanding Rotations of Quantum States: $SO(3)$ versus $SU(2)$

group-representationsgroup-theorylie-algebraquantum mechanicsrepresentation-theory

I'm trying to understand the relationship between rotations in "real space" and in quantum state space. Let me explain with this example:

Suppose I have a spin-1/2 particle, lets say an electron, with spin measured in the $z+$ direction. If I rotate this electron by an angle of $\pi$ to get the spin in the $z-$ direction, the quantum state "rotates" half the angle $(\pi/2)$ because of the orthogonality of the states $|z+\rangle$ and $|z-\rangle$. I think this isn't very rigorous, but is this way of seeing it correct?

I searched for how to derive this result, and started to learn about representations. I read about $SO(3)$ and $SU(2)$ and their relationship, but it's still unclear to me. I found this action of $SU(2)$ on spinors:

$$ \Omega(\theta, \hat n) = e^{-i \frac{\theta}{2}(\hat n\cdot \vec\sigma)}, $$
where $\hat n$ is a unitary $3D$ vector, $\vec\sigma = (\sigma_1,\sigma_2,\sigma_3)$ is the Pauli vector and $\sigma_i$ are the Pauli matrices. I see the factor of $\frac{1}{2}$ on the rotation angle $\theta$, but where does it come from?

I saw $[\sigma_i,\sigma_j] = 2i\epsilon_{ijk}\sigma_k $, and making $ X_j = -\frac{i}{2}\sigma_j $ the commutator becomes $[X_i,X_j] = \epsilon_{ijk}X_k$, which is the commutator of the $\mathfrak{so}(3)$ Lie algebra, isn't it? So when I compute the exponential $$e^{\theta(\hat n\cdot X)} = e^{-i \frac{\theta}{2}(\hat n\cdot \vec\sigma)}.$$ I get my result, and it seems like it's a rotation, but I read that it isn't an $SO(3)$ representation. So where do rotations appear?

However, my central question is: How can I demonstrate that a rotation on "our world" generates a rotation of quantum states, and how do I use that to show the formula for rotations on quantum states? And how I do it for higher spin values? I'm really new on this topic, and it was hard to formulate this question, so feel free to ask me for a better explanation or to clear any misconception.

Best Answer

  1. On one hand, if $\vec{\alpha}\in \mathbb{R}^3$ denotes a 3D rotation vector$^1$, then the corresponding rotation matrix $$R(\vec{\alpha})~=~\exp(i\vec{\alpha}\cdot \vec{L})~\in~ SO(3)~ \subseteq ~{\rm Mat}_{3\times 3}(\mathbb{R}). \tag{1}$$ Here $iL_j\in so(3) \subseteq {\rm Mat}_{3\times 3}(\mathbb{R})$ are three $so(3)$ Lie algebra generators, defined as $$\begin{align} i(L_j)_{k\ell} ~=~& \epsilon_{jk\ell} ,\cr j,k,\ell~\in~& \{1,2,3\}, \cr \epsilon_{123}~=~&1,\tag{1'}\end{align}$$ and fulfilling the $so(3)$ Lie bracket relations $$\begin{align} [L_j, L_k] ~=~& i\sum_{\ell=1}^3\epsilon_{jk\ell} L_{\ell}, \cr j,k,\ell~\in~& \{1,2,3\}.\end{align} \tag{1"}$$

  2. On the other hand, the corresponding $SU(2)$ matrix is$^2$ $$ X(\vec{\alpha})~=~\exp(\frac{i}{2}\vec{\alpha}\cdot \vec{\sigma})~\in~ SU(2)~ \subseteq ~{\rm Mat}_{2\times 2}(\mathbb{C}), \tag{2}$$ where $\sigma_{\ell}$ are the Pauli matrices.

  3. The two matrix representations (1) & (2) are related via $$\begin{align} X(\vec{\alpha}) \sigma_k X(\vec{\alpha})^{-1}~=~& \sum_{j=1}^3\sigma_j R(\vec{\alpha})^j{}_k, \cr R(\vec{\alpha})^j{}_k~=~&\frac{1}{2}{\rm Tr}(\sigma^jX(\vec{\alpha}) \sigma_k X(\vec{\alpha})^{-1}), \cr j,k~\in~& \{1,2,3\}. \end{align}\tag{3}$$

  4. The relation (3) exposes the fact that the adjoint representation $$ {\rm Ad}:~ SU(2) \quad\to\quad SO(su(2))~\cong~ SO(3),\tag{4} $$ given by $$ \begin{align} X\quad \mapsto&\quad{\rm Ad}(X)\sigma~:= ~X\sigma X^{-1}, \cr X~\in~& SU(2),\cr \sigma~\in~& su(2)\cr ~:=~&\{ \sigma\in {\rm Mat}_{2\times 2}(\mathbb{C})\mid \sigma^{\dagger}=\sigma~\wedge~{\rm tr}(\sigma)=0\}\cr ~=~&{\rm span}_{\mathbb{R}}\{\sigma_1,\sigma_2,\sigma_3\},\cr &||\sigma||^2 ~=~\det(\sigma),\cr &{\rm Ad}(\pm {\bf 1}_{2 \times 2})~=~{\bf 1}_{su(2)}, \end{align}\tag{4'} $$ is a surjective 2:1 Lie group homomorphism between $SU(2)$ and $SO(3)$, i.e. $$ SU(2)/\mathbb{Z}_2~\cong~SO(3). \tag{4"}$$

  5. See also this related Phys.SE post, which explains the half-angle appearance in eq. (2).

References:

  1. G. 't Hooft, Introduction to Lie Groups in Physics, lecture notes, chapters 3 & 6. The pdf file is available here.

--

$^1$ The direction of a rotation vector $\vec{\alpha}\in \mathbb{R}^3$ is parallel to the rotation axis, while the length $|\vec{\alpha}|$ denotes the counter-clockwise rotation angle (as seen from the tip of the rotation vector).

$^2$ The notation and conventions in this Phys.SE answer follows Ref. 1.

Related Question