Can classical physics be described through spinor representations?

Question

The homomorphism between $SO(3)$ and $SU(2)/\mathbb{Z}_2$ suggests that using $SO(3)$ to describe the real world (by real I mean classical) is simply a convention and that the exact same physics can be described using spinors (using a $2$ dimensional complex vector space instead of a $3$ dimensional real vector space).

My confusion is the following. For $G_i \in SU(2)$, we can write

$$ SU(2)/\mathbb{Z}_2 = \Big\{\pm G_1,\;\pm G_2,\;\pm G_3,\dots\Big\}\tag{I.} $$

where $G_i$ are $2\times 2$ matrices and therefore act on a $2$ dimensional vector space say $V$. For $H_i \in SO(3)$, $H_i$ are $3\times 3$ matricies acting on a $3$ dimensional real vector space say $U$. I understand that

$$SU(2)/\mathbb{Z}_2 \cong SO(3) \tag{II.}$$

which shows that the $3$ dimensional representation of $SO(3)$ is $SU(2)/\mathbb{Z}_2$, however, in equation (I.), I have explicitly constructed a $2$ dimensional representation and it is not clear how the vector spaces are related. I should in theory be able to map a vector from $U\rightarrow V$ but I'm not sure how to do this.

When people talk about spinors, are they talking about the vectors acted on by the fundamental representation of $SU(2)$ and not $SU(2)/\mathbb{Z}_2$? If this is the case, does that mean that spinors are only relevant in quantum systems i.e. where the projective representation can be considered due to the inner product of two states being invariant under phase?

EDIT:

Perhaps this is a possible way of solving.

It has been pointed out in the answer section that one can write $$G_i = \exp \imath \alpha_k^{(i)} \sigma_k/2$$ $$H_i = \exp \imath \alpha_k^{(i)} L_k$$ where $L_k$ are the generators forming the Lie algebra of the group $SO(3)$, $\sigma_k$ are the Pauli matricies which are the generators forming the Lie algebra of the group $SU(2)$ and $\alpha^{(i)}_k$ is an element of $\mathbb{R}^3$.

If I have the constants $$\alpha^{(1)}_k = (1,0,0)$$ corresponding to an element $H_1$ of $SO(3)$ (where $H_1$ rotates around the $x$ axis), the vector $$X= \begin{pmatrix} 1 \\ 0 \\0 \end{pmatrix} \in U$$

should be left invariant under the transformation $H_1$ since the vector along the axis of rotation is always left invariant. If I can show that the following

1. There is exactly one eigenvector of $SU(2)$ for each $\alpha$ therefore the vector $U$ of $SO(3)$ that is invariant under a particular element of $SO(3)$ gets mapped to the invariant vector of $SU(2)$ for the same $\alpha$.

then would that be a correct way of mapping a vector from $U\rightarrow V$?

Answer 1

1. There is not a linear map$^1$ $$\mathbb{R}^3~\cong~ \color{red}{U}\quad\to\quad \color{red}{V}~\cong~\mathbb{C}^2\tag{1}$$ of vector spaces, but instead a surjective 2:1 Lie group homomorphism $$ SU(2)~\cong~SU(\color{red}{V})\quad\to\quad SO(\color{red}{U})~\cong~SO(3).\tag{2} $$ For details, see e.g. my Phys.SE answer here.

2. However, there is a non-linear map $$\begin{align}\color{red}{V}~\cong~\mathbb{C}^2~\ni~\psi\quad\mapsto\quad\sigma~=~&\psi\psi^{\dagger} - \psi^{\dagger}\psi{\bf 1}_{2\times 2}\cr ~\in~&su(2)~\stackrel{(4)}{\cong}~\mathbb{R}^3~\cong ~\color{red}{U}\end{align}\tag{3}$$ if we use a bijective isometry (4) from 3D space $(\mathbb{R}^{3},||⋅||^2)$ to the space of $2\times 2$ traceless Hermitian matrices $(su(2),\det(⋅))$, $$\begin{align} \mathbb{R}^3 ~\cong ~& su(2)\cr ~:=~&\{\sigma\in {\rm Mat}_{2\times 2}(\mathbb{C}) \mid \sigma^{\dagger}=\sigma \wedge {\rm tr}(\sigma)=0\} \cr ~=~& {\rm span}_{\mathbb{R}} \{\sigma_1,\sigma_2,\sigma_3\},\cr \mathbb{R}^3~\ni~&\vec{r}~=~(x^1,x^2,x^3) \cr &\quad\mapsto \quad\sigma~=~\sum_{j=1}^3 x^j\sigma_j~\in~ su(2), \cr ||\vec{r}||^2 ~=~&\det(\sigma),\end{align}\tag{4}$$ cf. the answer by Thomas Fritsch. For a relation to the Bloch sphere, see also e.g. this related Phys.SE post.

3. As a consistency check, one may show that if we transform a spinor $\psi\in \mathbb{C}^2$ with an element of $SU(2)$, then the corresponding vector $\vec{r}\in\mathbb{R}^3$ from eq. (3+4) is rotated with the corresponding element of $SO(3)$ from eq. (2).

--

$^1$ The rotation vector $\vec{\alpha}\in \mathbb{R}^3$ that OP mentions represents an element of the Lie groups (2) rather than the vector spaces (1).

Answer 2

I should in theory be able to map a vector from $U\rightarrow V$ but I'm not sure how to do this.

Let $$|\psi\rangle=\begin{pmatrix} a \\ b \end{pmatrix}$$ be a spinor in $V=\mathbb{C}^2$. Then you can construct a corresponding vector in $U=\mathbb{R}^3$ by using the Pauli matrices (see A. Steane - An introduction to spinors, page 4, equation 12):

$$\begin{align} x&=\langle\psi|\sigma_x|\psi\rangle \\ y&=\langle\psi|\sigma_y|\psi\rangle \\ z&=\langle\psi|\sigma_z|\psi\rangle \end{align}$$

or explicitly spelled out $$ \vec{r}= \begin{pmatrix} x \\ y \\ z \end{pmatrix}= \begin{pmatrix} a^*b+b^*a \\ -ia^*b+ib^*a \\ a^*a-b^*b \end{pmatrix}$$

You see, it is not a linear map (as already pointed out in @Qmechanic's answer), but it is more a square-like map. Especially, all spinors $e^{i\alpha}|\psi\rangle$ (with $\alpha\in\mathbb{R}$) result in the same vector $\vec{r}$.

When the spinor $|\psi\rangle$ is transformed by a matrix from $SU(2)$, then the corresponding vector $\vec{r}$ is transformed by a matrix from $SO(3)$.

The inverse mapping (from vector $\vec{r}\in\mathbb{R}^3$ to spinor $|\psi\rangle\in \mathbb{C}^2$) is more difficult. And as noticed above, it is not unique.
First convert the Cartesian coordinates ($x,y,z$) to spherical coordinates ($r,\theta,\phi$). From these you can then calculate the components of $|\psi\rangle$ (see A. Steane - An introduction to spinors, page 3, equation 1): $$\begin{align} a&=\sqrt{r}\cos(\theta/2)e^{i(-\alpha+\phi)/2} \\ b&=\sqrt{r}\sin(\theta/2)e^{i(-\alpha-\phi)/2} \end{align}$$ where $\alpha\in\mathbb{R}$ is any arbitrary phase.