Generators and the Lie Algebra of SO(3) - #2 Generators

 We derived the representation of SO(3) on the last post, 'Generators and the Lie Algebra of SO(3) - #1 Representation', which shown up to be the below three matrices. 

\[ R_{x}(\psi)=\begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos\phi & \sin\phi \\ 0 & -\sin\phi & \cos\phi \end{pmatrix} \quad R_{y}(\phi)=\begin{pmatrix} \cos\phi & 0 & -\sin\phi \\ 0 & 1  & 0 \\ \sin\phi & 0 & \cos\phi \end{pmatrix} \quad R_{z}(\theta)= \begin{pmatrix} \cos\theta & \sin\theta & 0 \\ -\sin\theta & \cos\theta & 0 \\ 0 & 0 & 1  \end{pmatrix} \]

And we know that multiplying all of them would represent any arbitrary element of SO(3), but we won't do that. Generators could be derived by differentiating each representations.

\[ \mathcal{R}_{x}=\frac{\partial}{\partial \psi}R_{x}(\psi)|_{\psi=0}\ \quad \textrm{,} \quad \mathcal{R}_{y}=\frac{\partial}{\partial \phi}R_{y}(\phi)|_{\phi=0} \quad \textrm{,} \quad\mathcal{R}_{z}=\frac{\partial}{\partial \theta}R_{z}(\theta)|_{\theta=0} \]

Thus we can obtain 3 generators, $\mathcal{R}_{x}$, , $\mathcal{R}_{y}$, and , $\mathcal{R}_{z}$.

\[ \mathcal{R}_{x} = \begin{pmatrix} 0&0&0\\ 0&0&1\\ 0&-1&0 \end{pmatrix} \]

\[ \mathcal{R}_{y} = \begin{pmatrix} 0&0&-1\\ 0&0&0\\ 1&0&0 \end{pmatrix} \]

\[ \mathcal{R}_{z} = \begin{pmatrix} 0&1&0\\ -1&0&0\\ 0&0&0 \end{pmatrix} \]

We can check if these generators generate each rotations by using the exponential function. And we can check the below equations, which tells us they do generate the rotations.

\[ R_{x}(\psi)=e^{\psi\mathcal{R}_{x}} \quad \textrm{,} \quad R_{y}(\phi)=e^{\phi\mathcal{R}_{y}}\quad \textrm{,} \quad R_{z}(\theta)=e^{\theta\mathcal{R}_{z}}  \]

Defining new matrices by multiplying $-i$ on each generators, we can find about commutation relations. Thus, we define new a new matrix as

\[ J_{x} \equiv -i\mathcal{R}_{x} = \begin{pmatrix} 0&0&0\\ 0&0&-i\\ 0&i&0 \end{pmatrix} \]

\[ J_{y}  \equiv -i\mathcal{R}_{y} = \begin{pmatrix} 0&0&i\\ 0&0&0\\ -i&0&0 \end{pmatrix} \]

\[ J_{z} \equiv -i\mathcal{R}_{z}= \begin{pmatrix} 0&-i&0\\ i&0&0\\ 0&0&0 \end{pmatrix} \]

Commutating each of the $J$s would show us the commutation relations.

\[ \left[ J_x , J_y \right] = \begin{pmatrix} 0&0&0\\ 0&0&-i\\ 0&i&0 \end{pmatrix}\begin{pmatrix} 0&0&i\\ 0&0&0\\ -i&0&0 \end{pmatrix} - \begin{pmatrix} 0&0&i\\ 0&0&0\\ -i&0&0 \end{pmatrix} \begin{pmatrix} 0&0&0\\ 0&0&-i\\ 0&i&0 \end{pmatrix}\]

\[ \left[ J_x , J_y \right] = \begin{pmatrix} 0&0&0\\ -1&0&0\\ 0&0&0 \end{pmatrix} - \begin{pmatrix} 0&-1&0\\ 0&0&0\\ 0&0&0 \end{pmatrix} = \begin{pmatrix} 0&1&0\\ -1&0&0\\ 0&0&0 \end{pmatrix}\]

Thus,

\[ \left[ J_x , J_y \right] = \begin{pmatrix} 0&1&0\\ -1&0&0\\ 0&0&0 \end{pmatrix}= i J_z \]

Doing this for every $J$s, we can obtain below equations.

\[ \left[ J_x , J_y \right]= i J_z \quad , \quad \left[ J_y , J_z \right]= i J_x \quad , \quad \left[ J_z , J_x \right]= i J_y \]

\[ \left[ J_y , J_x \right]= - i J_z \quad , \quad \left[ J_z , J_y \right]= - i J_x \quad , \quad \left[ J_x , J_z \right]= - i J_y \]

And it is trivial that commutation between the same $J$ would be 0, which means that $\left[ J_i , J_i \right]=  0 $. Obtaining every commutation relation, the structure constant of SO(3) can be easily expressed by using the Levi-Civita symbol.

\[ \left[ J_i , J_j \right]= i \epsilon_{ijk} J_k \]

Thus, the structure constant of SO(3) is $\epsilon_{ijk}$, and the Lie algebra is $\left[ J_i , J_j \right]= i \epsilon_{ijk} J_k $.