# Computing the exponential of the operator $H(\textbf{x}) = \alpha \textbf{n} \times\textbf{x}$

Define the vector operator: $$H(\textbf{x}) \equiv \alpha \textbf{n} \times\textbf{x}$$ For unit vector $\textbf{n}$ and some constant $\alpha$. We define further the operator: $$G \equiv I + H + \frac{H^2}{2!} + \frac{H^3}{3!} + ...$$ Where powers of $H$ represent iterations and $I$ is defined as the identity such that $I(\textbf{x}) = \textbf{x}$

We are to prove $$G\textbf{x} = \textbf{x} +(\textbf{n}\times\textbf{x})\sin{\alpha} + \textbf{n}\times(\textbf{n}\times\textbf{x})(1-\cos{\alpha})$$

My attempt at the proof involved noting: $$G = e^{H}$$ And thus $$G\textbf{x} = e^H\textbf{x}$$

But I can't seem to see where to go from here, or even if the last step is a legitimate one. Many thanks in advance.

• What does $\displaystyle{\left(\,{\bf n} \times {\bf x}\,\right)^{\, 2}}$ mean ?. – Felix Marin Jan 1 '15 at 21:16

I don’t quite know whether the $e^H$ approach takes us anywhere useful, because I’m not entirely sure what the exponential of a vector operator means. Anywhere, here's one method that seems to work:

$\newcommand{\myvec}[1]{\mathbf{#1}} \newcommand{\cross}{\times} \newcommand{\nn}{\myvec{n}} \newcommand{\xx}{\myvec{x}}$ First, a couple of useful identities. Lagrange's formula, or triple product expansion: $$\myvec{a} \cross (\myvec{b} \cross \myvec{c}) = \myvec{b}\left(\myvec{a} \cdot \myvec{c}\right) - \myvec{c}\left(\myvec{a} \cdot \myvec{b}\right)$$ and the power series for sine and cosine: \begin{align*} \cos \theta &= 1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \cdots &= \sum_{k=0}^\infty (-1)^k \frac{\theta^{2k}}{(2k)!} \\[6pt] \sin\theta &= x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots &= \sum_{k=0}^\infty (-1)^k\frac{\theta^{2k+1}}{(2k+1)!} \end{align*} So here's what I'd do:

1. Expand $I\myvec{x}$, $H\myvec{x}$, $H^2\myvec{x}$, for a few small terms, to see if I could spot a pattern. Here I've done it up to $H^4\xx$:

\begin{align*}I\xx &= \xx \\H\xx &= \alpha\left(\nn \cross \xx\right) \\H^2\xx &= \alpha\left(\nn \cross \alpha\left(\nn \cross \xx\right)\right) \\&= \alpha^2 \left(\nn \cross \left(\nn \cross \xx\right)\right) \\H^3\xx &= -\alpha^3 \left(\nn \cross \xx\right) \\H^4\xx &= -\alpha^4 \left(\nn \cross \left(\nn \cross \xx\right)\right)\end{align*}

2. Identify a general formula for $H^m\xx$, and prove that it works.

I'm omitting the proof, but here's the general formula I came up with:$$H^m\xx =\begin{cases}\xx & \text{if m=0,} \\\alpha^m (-1)^{(m-1)/2}\left(\nn \cross \xx\right) & \text{if m odd,} \\\alpha^m (-1)^{m/2-1}\left(\nn \cross \left(\nn \cross \xx\right)\right) & \text{if m even.}\end{cases}$$You can prove this by induction.

3. Now that we have a general formula for $H^m\xx$, we substitute this into the definition of $G$ and rearrange terms.

We can now write $G$ as (being careful with the signs)

$$G \equiv\xx+\underbrace{\left(\nn \cross \xx\right) \sum_{m=0}^\infty (-1)^{(m-1)/2}\frac{\alpha^{2m+1}}{(2m+1)!}}_{\text{odd terms}}-\underbrace{\left(\nn \cross \left(\nn \cross \xx\right)\right) \sum_{m=1}^\infty (-1)^{m/2}\frac{\alpha^{2m}}{(2m!)}}_{\text{even terms}}$$

Looking at the identities above, we can simplify this expression as

$$G \equiv \xx+\left(\nn \cross \xx\right) \sin\alpha - \left(\nn \cross \left(\nn \cross \xx\right)\right) (\cos\alpha-1)$$

which is the desired expression.

• I see! Many thanks for your help! Greatly appreciated – user211337 Jan 1 '15 at 11:19
• The exponential of a square matrix (at least in finite dimensions) is fairly well defined: en.m.wikipedia.org/wiki/Matrix_exponential – DanZimm Jan 1 '15 at 19:02
• @DanZimm: Cheers for the link. I hadn’t quite made the connection between the operator and a square matrix; I’m obviously getting a little rusty! :) – alexwlchan Jan 1 '15 at 19:09

$H x = \alpha \, n \times x$ gives the matrix: $$H = \alpha \left( \begin{matrix} 0 & -n_3 & n_2 \\ n_3 & 0 & -n_1 \\ -n_2 & n_1 & 0 \end{matrix} \right) = \alpha N$$ Then for your proof $$G = I + \sin(\alpha) N + (1-\cos\alpha) N^2$$ needs to be shown.

So we calculate $N^2$ and $N^3$: \begin{align} N^2 &= \left( \begin{matrix} 0 & -n_3 & n_2 \\ n_3 & 0 & -n_1 \\ -n_2 & n_1 & 0 \end{matrix} \right) \left( \begin{matrix} 0 & -n_3 & n_2 \\ n_3 & 0 & -n_1 \\ -n_2 & n_1 & 0 \end{matrix} \right) \\ &= \left( \begin{matrix} -(n_2^2+n_3^2) & n_1 n_2 & n_1 n_3 \\ n_1 n_2 & -(n_1^2 + n_3^2) & n_2 n_3 \\ n_1 n_3 & n_2 n_3 & -(n_1^2 + n_2^2) \end{matrix} \right) \\ &= \left( \begin{matrix} n_1^2-1 & n_1 n_2 & n_1 n_3 \\ n_1 n_2 & n_2^2 - 1 & n_2 n_3 \\ n_1 n_3 & n_2 n_3 & n_3^2 - 1 \end{matrix} \right) \end{align} where we used $1 = n_1^2 + n_2^2 + n_3^2$ for the unit vector $n$ and then \begin{align} N^3 &= \left( \begin{matrix} n_1^2-1 & n_1 n_2 & n_1 n_3 \\ n_1 n_2 & n_2^2 - 1 & n_2 n_3 \\ n_1 n_3 & n_2 n_3 & n_3^2 - 1 \end{matrix} \right) \left( \begin{matrix} 0 & -n_3 & n_2 \\ n_3 & 0 & -n_1 \\ -n_2 & n_1 & 0 \end{matrix} \right) \\ &= \left( \begin{matrix} 0 & n_3 & -n_2 \\ -n_3 & 0 & n_1 \\ n_2 & -n_1 & 0 \end{matrix} \right) = -N \end{align} This means all $N^k$ can be expressed with $I$, $N$ and $N^2$: $$N^0 = I \quad N^{2k} = (-1)^k N^2 \quad N^{2k+1} = (-1)^k N$$ for $k > 0$ and we can calculate \begin{align} G = e^H &= \sum_{k=0}^\infty \frac{1}{k!} H^k \\ &= \sum_{k=0}^\infty \frac{1}{(2k)!} H^{2k} + \sum_{k=0}^\infty \frac{1}{(2k+1)!} H^{2k+1} \\ &= \sum_{k=0}^\infty \frac{1}{(2k)!} \alpha^{2k} N^{2k} + \sum_{k=0}^\infty \frac{1}{(2k+1)!} \alpha^{2k+1} N^{2k+1} \\ &= I + \sum_{k=1}^\infty \frac{(-1)^k}{(2k)!} \alpha^{2k} N^2 + \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!} \alpha^{2k+1} N \\ &= I + (1-\cos \alpha) N^2 + \sin(\alpha) N \end{align}

• A very ingenious approach to the problem! Many thanks for your help! – user211337 Jan 1 '15 at 11:19