0

So we can write the $SU(2)$ matrices multiplication as this.

$$\begin{bmatrix}\alpha&\beta\\-\beta^*&\alpha^*\end{bmatrix}\begin{bmatrix}z&x-iy\\x+iy&-z\end{bmatrix}\begin{bmatrix}\alpha^*&-\beta\\\beta^*&\alpha\end{bmatrix}=\begin{bmatrix}Z&X-iY\\X+iY&-Z\end{bmatrix}$$

now is this always rotation ($SO(3)$) provided that $|\alpha|^2+|\beta|^2=1$ which makes them $SU(2)$.

and therefore we can rotate Pauli vectors with the help of $SU(2)$ and we know that Pauli vectors are like 3d vectors so we are effectively rotating the 3d space with the help of $SU(2)$ instead of $SO(3)$.

edit:-

and so we then can imagine Pauli spinors present within the geometry which just rotate by $SU(2)$ operation on them.

Qmechanic
  • 220,844
abx_pradB
  • 15

1 Answers1

0

Hint: a real 3-vector $\vec r=(x,y,z)^T$ is mapped, in your complex doublet SU(2) illustration to "Pauli vectors", $$ \vec r \mapsto \vec r\cdot \vec \sigma, $$ in the fundamental of the su(2) Lie algebra, (but of course, it could be dotted to generators $\vec T$ in any representation, of any dimension.)

In your doublet representation, you may likewise map your SU(2) group matrix to the conventional Pauli doublet representation, $$ \begin{bmatrix}\alpha&\beta\\-\beta^*&\alpha^*\end{bmatrix}=\cos\theta +i\sin\theta ~\hat n \cdot \vec \sigma = e^{i\theta \hat n\cdot \vec \sigma}, $$ where you can work out the relation of $\vec \theta$ to complex unimodular $(\alpha, \beta)$.

You should then be able to check that, for $\vec R=(X,Y,Z)^T$, $$ M^{ij}=\tfrac{1}{2} \operatorname{Tr}(\sigma^i e^{i\theta \hat n\cdot \vec \sigma} \sigma^j e^{-i\theta \hat n\cdot \vec \sigma}), \leadsto \\ R^i= M^{ij} r^j, $$ possibly up to signs.

The 3-by-3 matrix M turns out to be real orthogonal, so, then in the adjoint of SU(2) and SO(3).

Cosmas Zachos
  • 67,623