- Rotation Matrix in Euclidean Space Using Spin Operator Sz
Introduction
This document explores how to represent rotation matrices in Euclidean space using the spin operator $S_z$ along the rotation axis $n_z$.
Angular Momentum Operator $L_z$
The angular momentum operator $L_z$ can be represented as:
$$L_z = \begin{pmatrix} 0 & -i & 0 \\ i & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$This operator has eigenvalues $+1$, $0$, and $-1$ with corresponding eigenvectors:
$$|+1\rangle = \frac{1}{\sqrt{2}}(1, i, 0)$$ $$|0\rangle = (0, 0, 1)$$ $$|-1\rangle = \frac{1}{\sqrt{2}}(1, -i, 0)$$
Basis Transformation
We need to transform from the standard basis ${(1,0,0), (0,1,0), (0,0,1)}$ to the basis of eigenvectors ${\frac{1}{\sqrt{2}}(1, i, 0), (0, 0, 1), \frac{1}{\sqrt{2}}(1, -i, 0)}$.
The unitary transformation matrix $U$ is:
$$U = \begin{pmatrix} \frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \\ \frac{i}{\sqrt{2}} & 0 & -\frac{i}{\sqrt{2}} \\ 0 & 1 & 0 \end{pmatrix}$$Relationship Between $S_z$ and $L_z$
The spin operator $S_z$ is the representation of $L_z$ in the transformed basis:
$$\langle a’|L_z|b’\rangle = \langle a|U^{\dagger}L_zU|b\rangle$$
Therefore: $$S_z = U^{\dagger}L_zU$$
And conversely: $$L_z = US_zU^{\dagger}$$
Rotation Matrix
The rotation matrix around the $n_z$ axis by angle $\theta$ can be expressed as:
$$R(\theta, n_z) = e^{-i\theta L_z} = e^{-i\theta US_zU^{\dagger}} = Ue^{-i\theta S_z}U^{\dagger}$$
This formula allows us to compute rotation matrices in Euclidean space using the spin operator $S_z$.