# Rodrigues' rotation formula from SO(3) comutator properties

• I
• sergiokapone

#### sergiokapone

Is any way to get Rodrigues' rotation formula from matrix exponential

e^{i\phi (\star\vec{n}) } = e^{i\phi (\vec{n}\cdot\hat{\vec{S}}) } = \hat{I} + (\star\vec{n})\sin\phi + (\star\vec{n})^2( 1 - \cos\phi ).

using SO(3) groups comutators properties ONLY like
$$\hat S_k\hat S_j-\hat S_j\hat S_k=i\varepsilon_{kjl}\hat S_l,\qquad \hat S^2=2\hat I ?$$
where ##\vec{n} = (n_x,n_y,n_z)^{\top}##, ##\vec{n}^2 = 1##, and
\label{}
(\vec{n}\cdot\hat{\vec{S}}) = \star\vec{n} =
\begin{pmatrix}
0 & -n_z & n_y \\
n_z & 0 & -n_x \\
-n_y & n_x & 0 \\
\end{pmatrix},

\begin{multline}\label{}
\hat{\vec{S}} =
\begin{pmatrix}
0 & 0 & 0 \\
0 & 0 & i \\
0 & -i & 0 \\
\end{pmatrix}
\vec{e}_x
+
\begin{pmatrix}
0 & 0 & -i \\
0 & 0 & 0 \\
i & 0 & 0 \\
\end{pmatrix}
\vec{e}_y
+
\begin{pmatrix}
0 & i & 0 \\
-i & 0 & 0 \\
0 & 0 & 0 \\
\end{pmatrix}
\vec{e}_z =
\hat{S}_x \vec{e}_x +
\hat{S}_y \vec{e}_y +
\hat{S}_z \vec{e}_z
.
\end{multline}

References: https://en.wikipedia.org/wiki/Axis–angle_representation#Exponential_map_from_so(3)_to_SO(3)

Last edited:
The problem you will run into is that you will be expanding a commutator action [when you exponentiate] and that's not associative. [You want to work with an associative product when expanding an exponential.]

For a purely algebraic derivation you can work it through most easily (imnsho) utilizing quaternion algebra. You have to be aware that in that algebra you must act adjointly (left action yields the spinor representation).

I suggest working with quaternions in scalar+3-vector format. The quaternion product then manifests as:
$$(a+\mathbf{u})(b+\mathbf{v}) = ab+a\mathbf{v}+b\mathbf{u} -\mathbf{u}\bullet\mathbf{v}+\mathbf{u}\times\mathbf{v}$$ where ##\times## is the cross product and ##\bullet## the dot product.

$$R_{\theta,\mathbf{u}} \mathbf{v} = e^{\theta\mathbf{u}/2} \mathbf{v}e^{-\theta\mathbf{u}/2}$$
where ##\mathbf{u}## is a unit vector in the direction about which you rotate ( right hand rule). The exponential of a pure quaternion is then a version of the Euler formula as ##\mathbf{u}^2 = -1##:
$$e^{\theta\mathbf{u}/2} = \cos(\theta/2)+\mathbf{u}\sin(\theta/2)$$ Work out the adjoint action of this on ##\mathbf{v}## and play with your vector product identities and half-angle trig. identities and the Rodigues' formula will pop right out.

sergiokapone
There is an analog of the binomial theorem for noncommutative associative algebras. It is quite useful when the commutator assumes a special form. However, it’s not strictly necessary in this case - for so(3), it’s simple enough just to use the characteristic polynomial ##\mathbf{[x]_\times^3}+\mathbf{|x\|^2[x]_\times}=0## to compute the exponential.

MisterX and jambaugh
suremarc