$$T(\phi)=
\begin{bmatrix}
\cos(\theta) &\sin(\theta) & 0 \\
\sin(\theta)\cos(\phi) & -\cos(\theta)\cos(\phi) & \sin(\phi)\\
\sin(\theta)\sin(\phi) & -\cos(\theta)\sin(\phi) &-\cos(\phi)
\end{bmatrix}$$
Here the matrix $T$ is parametrized by $\phi$ and $\theta$=some constant angle. Can I find out the generators of this orthogonal transformation parametrized by the angle $\phi$ ?
If my approach is wrong, how do I find the generators of this matrix and exponentiate it?
I have derived an infinitesimal transformation which leads to $$T(\delta\phi)=\bigg[I+t\delta\phi\bigg]$$
where $t$ is,
$$t=
\begin{bmatrix}
0 &0 & 0 \\
0 & 0 & 1\\
\sin(\theta) & -\cos(\theta) &0
\end{bmatrix}.$$
Here $\theta$ is some fixed angle, say $120$ degrees or $69$ degrees or anything, but it remains constant. Can I exponentiate this matrix to get $$e^{-\hat{t}\phi}$$
Is it correct? Where am I going wrong if I am completely incorrect?
Edit: if $\theta$ is a fixed constant, there is no way I could get the identity element, so what if $T$ is parametrized by both $\theta$ and $\phi$ ? I will surely get the identity element. Now how do I proceed from here?
Best Answer
Your orthogonal matrix $$R(\phi,\theta)= \begin{bmatrix} \cos(\theta) &\sin(\theta) & 0 \\ \sin(\theta)\cos(\phi) & -\cos(\theta)\cos(\phi) & \sin(\phi)\\ \sin(\theta)\sin(\phi) & -\cos(\theta)\sin(\phi) &-\cos(\phi) \end{bmatrix}$$ must have antisymmetric generators.
To find them, you must expand around the origin, $\phi=\pi, \theta=0$. To allay confusion, define $\Phi\equiv = \pi-\phi$, so the origin is at $\Phi=\theta=0$. $$R(\Phi,\theta)= \begin{bmatrix} \cos(\theta) &\sin(\theta) & 0 \\ - \sin(\theta)\cos(\Phi) & \cos(\theta)\cos(\Phi) & \sin(\Phi)\\ \sin(\theta)\sin(\Phi) & -\cos(\theta)\sin(\Phi) &\cos(\Phi) \end{bmatrix}$$
Evaluate $R(\delta\Phi, 0)=\bigg[I+t\delta\Phi\bigg]$, so $$t= \begin{bmatrix} 0 &0 & 0 \\ 0 & 0 & 1\\ 0 & - 1 &0 \end{bmatrix}.$$
Can you also evaluate $R(0,\delta \theta)$?
Note added as per comments.
The above rotation matrix R then, in the conventions of WP, is but $$ e^{-\Phi L_x} e^{-\theta L_z} , $$ which you might choose to compose by BCH, $$\exp (-\Phi L_x -\theta L_z+ \Phi \theta [L_x,L_z]/2+... ), $$ or the Gibbs finite rotation formula, etc. if you were so inclined. For orthogonal axes like yours, Gibbs' formula all but collapses: the effective axis of rotation is just parallel to $\hat z \tan (\theta/2) +\hat x \tan (\Phi/2) +\hat y \tan (\theta/2) \tan (\Phi/2)$ ! (Can you see that this is precisely the invariant vector of R?)
In any case, the limiting procedure at the origin yielding the generators from your finite rotation matrix has sacrificed information: convince yourself that several different rotation matrices may share this identical behavior at the origin, of course--think of reversing the order of the two factors above; so you should not expect to reconstitute this specific rotation matrix from the tangent space behavior at the origin, in general. (Here you already factored your finite rotations in advance. What Lie's 3rd theorem guarantees is essentially Euler's theorem: the two component rotations will combine to a single rotation about a new axis.)