- Rotations in the Lie Algebra so(3)
1. Basis of so(3)
The Lie algebra so(3) has a basis consisting of three elements: $L_x$, $L_y$, and $L_z$.
- $L_x$: Represents rotation in the $yz$-plane (analogous to $\sigma_y$)
- $L_y$: Represents rotation in the $zx$-plane (analogous to $\sigma_y$)
- $L_z$: Represents rotation in the $xy$-plane (analogous to $\sigma_y$)
These basis elements can be explicitly written as matrices:
$$L_x = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & -i \\ 0 & i & 0 \end{pmatrix}$$$$L_y = \begin{pmatrix} 0 & 0 & i \\ 0 & 0 & 0 \\ -i & 0 & 0 \end{pmatrix}$$$$L_z = \begin{pmatrix} 0 & -i & 0 \\ i & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$For any element $X \in \text{so}(3)$, we can express it as a linear combination: $X = xL_x + yL_y + zL_z$, where $[X] = (x, y, z)$
If we define $\mathbf{n} = (n_x, n_y, n_z)$ and $\mathbf{L} = (L_x, L_y, L_z)$, then: $\mathbf{n} \cdot \mathbf{L} = n_xL_x + n_yL_y + n_zL_z \in \text{so}(3)$
2. Rotation Operators
The rotation operator can be expressed using the exponential map:
$$\begin{align*} R(\theta) &= e^{-i\theta \mathbf{n}\cdot \mathbf{L}}\\ &= 1 - i\sin\theta (\mathbf{n}\cdot\mathbf{L}) - (1-\cos\theta)(\mathbf{n}\cdot\mathbf{L})^2 \end{align*}$$Where we use the property: $(\mathbf{n}\cdot\mathbf{L})^3 = \mathbf{n}\cdot\mathbf{L}$
Explicitly, the rotation matrix $R(\theta, \hat{\mathbf{n}})$ can be written as:
$$R(\theta, \hat{\mathbf{n}}) =$$
$(\cos\theta + (1-\cos\theta)n_x^2$, $\quad -\sin\theta n_z + (1-\cos\theta)n_x n_y $, $\quad \sin\theta n_y + (1-\cos\theta)n_x n_z)$
$(\sin\theta n_z + (1-\cos\theta)n_x n_y$ ,$\quad \cos\theta + (1-\cos\theta)n_y^2 $, $\quad -\sin\theta n_x + (1-\cos\theta)n_y n_z)$
$(-\sin\theta n_y + (1-\cos\theta)n_x n_z $,$\quad \sin\theta n_x + (1-\cos\theta)n_y n_z$, $\quad \cos\theta + (1-\cos\theta)n_z^2)$
$$$$
This matrix represents a rotation by angle $\theta$ around the axis defined by the unit vector $\hat{\mathbf{n}}$.