Generators and the Lie Algebra of SO(3) - #1 Representation

 SO(3) group is the Special Orthogonal group of 3-dimension, and we can easily understand it's property by using rotations of 3-dimension. There are many ways to represent an arbitrary rotation in $R^3$. If you ever had a Classic Mechanics or Analytical Mechanics lecture, you may have learned the 'Euler Angle' representation for rotations. Rather using the Euler angle representation, I'll introduce a slightly different but very similar representation which leads SO(3) to have 3 parameters for any arbitrary rotation. The basic notion of Euler angle is to rotate 3 times, and each rotation has an invariant axis. Thus, there are lots of Euler angles, for example, we can rotate by $(z,x,z)$, which means that the first rotation is based on the z-axis, the second rotation is based  on the rotated y-axis, and the last rotation is based on the z-axis. The type of Euler angle is differed by people or book, we can just use what is convenient to us. 

 Euler angles are great to represent random rotations, but it makes us hard to derive the generators of SO(3). Thus we need a different representation, which is very similar to the Euler method. Not every axis rotation is involved in 'Euler angles', which means that $(x,y,z)$, $(y,z,x)$, $(z,x,y)$, $(z,y,x)$, $(y,x,z)$, $(x,z,y)$ are included. Then what about the representation that is consisted with every three base rotations..........? They are called 'Tait-Bryan angles', which includes $(x,y,z)$, $(y,z,x)$, $(z,x,y)$, $(z,y,x)$, $(y,x,z)$, $(x,z,y)$. ( Some people call them both 'Euler angles' and call $(x,y,x)$, $(y,x,y)$, $(z,x,z)$, $(x,z,x)$, $(y,z,y)$, $(z,y,z)$ as 'Proper Euler angles'. We'll just call them Euler angles & Tait-Bryan angles.  )

 So, using Tait-Bryan angle representation, we could represent any element of SO(3) with 3 parameters. The rotation based on the $x$ axis with the angle of $\psi$, $y$ axis with the angle of $\phi$, and the $z$ axis with $\theta$ could be represented by a $3\times 3$ matrix, and multiplying every matrix would represent the whole rotation. Using mathematical language, we can write

\[ R_{x_{i}}(\theta_{i}) |x_{i}\rangle = |x_{i}\rangle \quad \textrm{when} \quad |x_{i}\rangle = \begin{pmatrix} \delta_{i1}x_1 \\ \vdots \\ \delta_{ii}x_i \\ \vdots  \end{pmatrix} = \begin{pmatrix} \vdots \\ x_i \\ \vdots \end{pmatrix}=x_{i}|e_{i}\rangle \]

which tells us that $x_i$based rotation doesn't affects $x_i$ rotation. And, the total representation would look like, 

\[ R(\theta,\psi,\phi)=R_{x}(\psi)R_{y}(\phi)R_{z}(\theta) \quad \textrm{and} \quad R(\theta,\psi,\phi) \in SO(3) \]

when we rotate with the sequence of $z$ , $y$, and then $x$. 

 Now, lets take a closer look on each rotations. 

 First, the rotation of $z$ axis $R_{z}(\theta)$, is the rotation which rotates $\theta$ in counter-clockwise and leaves the $z$ axis invariant, and transforms the $x$ and $y$ as $R_{z}(\theta)|x\rangle$ and $R_{z}(\theta)|y\rangle$ .  Refer to the below figure for better understanding. 

 Thus, we can derive the representation of $R_{z}(\theta)$ by comparing how the axis changed. 

\[R_{z}(\theta)|x\rangle=\begin{pmatrix} x\cos\theta \\ -x\sin\theta \\ 0 \end{pmatrix}\ \quad R_{z}(\theta)|y\rangle=\begin{pmatrix} y\sin\theta \\ y\cos\theta \\ 0 \end{pmatrix}\ \quad R_{z}(\theta)|z\rangle=\begin{pmatrix} 0 \\ 0 \\ z \end{pmatrix} \]

Which tells us that,

\[ R_{z}(\theta)|x\rangle + R_{z}(\theta)|y\rangle + R_{z}(\theta)|z\rangle =R_{z}(\theta) \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} x\cos\theta+y\sin\theta \\ -x\sin\theta+y\cos\theta \\ z \end{pmatrix}   \]

So we can assume $R_{z}(\theta)$ is,

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

 Things are same on different axes. 

 $y$ axis rotation in the same fashion : 

\[ R_{y}(\phi)|x\rangle=\begin{pmatrix} x\cos\phi \\ 0 \\ x\sin\phi \end{pmatrix}\ \quad R_{y}(\phi)|y\rangle=\begin{pmatrix} 0 \\ y \\ 0 \end{pmatrix}\ \quad R_{y}(\phi)|z\rangle=\begin{pmatrix} -z\sin\phi \\ 0 \\ z\cos\phi \end{pmatrix}\]

\[ R_{y}(\phi)|x\rangle + R_{y}(\phi)|y\rangle + R_{y}(\phi)|z\rangle = R_{y}(\phi) \begin{pmatrix} x \\ y \\ z \end{pmatrix}\ = \begin{pmatrix} x\cos\phi-z\sin\phi \\ y \\ x\sin\phi+z\cos\phi  \end{pmatrix} \]

Thus we can assume $R_{y}(\phi)$ as, 

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

 And the last rotation, $R_{x}(\psi)$, which leaved the $x$ axis invariant is derived as same as the others. 

Using mathematical notation,

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

\[ R_{x}(\psi)|x\rangle+R_{x}(\psi)|y\rangle+R_{x}(\psi)|z\rangle=R_{x}(\psi)\begin{pmatrix} x \\ y \\ z \end{pmatrix}=\begin{pmatrix} x \\   y\cos\phi+ z\sin\phi \\  -y\sin\phi+   z\cos\phi \end{pmatrix} \]

Thus, we can finally get the representation of our last rotation :

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

Deriving each representations,

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

and multiplying all of them would be our final representation of SO(3), the $R(\theta,\psi,\phi)$.

 But, for deriving the generators, we don't need to use $R(\theta,\psi,\phi)$,( and it also looks boring to multiply 3 $3 \times 3 $ matrices) but we need to think of each of it's rotations individually. It'll be discussed on later posts.