Classical spin viewed as $SU(2)$

Question

In which sense is the configuration variable of a classical spin $SU(2)$? I can view a classical spin as a unit vector in $\mathbb{S}^2$ (2-dim. sphere), but it seems it is really given by a matrix $U$ in $SU(2)$. The Hopf map $$H:SU(2)\rightarrow \mathbb{S}^2$$ given by $$H(U)=U\sigma_3U^{\dagger}$$ whose image can be identified with an element in $\mathbb{S}^2$ gives what I imagined to be this classical spin.

Since with a magnetic field $B$ the interaction is just $H(U) \cdot B$ there would be no problem on just considering $S \in \mathbb{S}^2$ as a configuration variable, but I read the following:

A classical particle of mass $m$, with position $x$ and spin $S$ moving on a fixed external magnetic field $B$ can be described by the Lagrangian function on the tangent bundle of the configuration space $\mathbb{R} \times SU(2)$ given as

$$L=\frac{1}{2}\dot{x}^2+i\lambda Tr(\sigma_3U^{\dagger}\dot{U})+\mu Tr(H(U)\dot B).$$

So the second term is explicitly in terms of $U$.

EDIT: I got this from "Gauge symmetries and fiber bundles". Balachandran et al.

Answer 1

I'm not altogether sure what you are asking, but I suspect the following may help. To represent rotations, spins and vectors in $SU(2)$ we work as follows.

Rotations live in $SU(2)$.

Vectors (in the physicist's sense) live in the algebra $\mathfrak{su}(2)$. The position vector $(x,\,y,z)$ is:

$$X =x\,\hat{s}_x+y\,\hat{s}_y+z\,\hat{s}_z = \left(\begin{array}{cc}i z&i\,x-y\\i\,x+y&-i z\end{array}\right)$$

which is a superposition of the Pauli matrices with a factor of $i$ thrown in to put our vector in the skew-Hermitian $\mathfrak{su}(2)$.

A Rotation $\gamma\in SU(2)$ acts on a vector $X\in\mathfrak{su}(2)$ through the spinor map:

$$X\mapsto \gamma\,X\,\gamma^{-1}=\gamma\,X\,\gamma^\dagger$$

The cross product between the two vectors $X,\,Y\in\mathfrak{su}(2)$ is the Lie bracket $[X,\,Y]$. The inner product can be thought of either as the anticommutator $\{X,\,Y\}=X\,Y+Y\,X$ and is always a scale factor times the identity matrix (it is thus a "scalar") or the scale factor alone can also be found as $\mathrm{tr}(X\,Y)$ (which, for $\mathfrak{su}(2)\cong\mathrm{ad}(\mathfrak{su}(2))=\mathfrak{so}(3)$, is the same as the negative of the Killing form, since $X^T = -X$).

An angular velocity defines the time derivative of a rotation, as such it is also a member of the Lie algebra $\mathfrak{su}(2)$ and is left translated to become the time derivative of a rotation operator:

$$\mathrm{d}_\tau\,\gamma(\tau) = \gamma\,\Omega$$

where $\Omega = \gamma^{-1}\,\mathrm{d}_\tau\,\gamma(\tau)\in\mathfrak{su}(2)$ is the angular velocity. The instantaneous velocity of a constant position vector $X$ under the action of $X\mapsto \gamma(\tau)\,X\,\gamma^{-1}(\tau)$ is then $[\Omega,\,X]$. The energy of interaction between $\Omega$ and a magnetic induction $B\in\mathfrak{su}(2)$ is, by the above, the inner product $\mathrm{tr}(\Omega\,B)$ (modulo a gyromagnetic ratio)

Does this help?