# Equivalent Rotation using Baker-Campbell-Hausdorff relation

Is there a way in which one can use the BCH relation to find the equivalent angle and the axis for two rotations? I am aware that one can do it in a precise way using Euler Angles but I was wondering whether we can use just the algebra of the rotation group to perform the same computation?

-
If you like this question you may also enjoy reading this answer. –  Qmechanic Sep 4 '13 at 14:27

I) The Baker-Campbell-Hausdorff formula for the 3-dimensional rotations can indeed be summed up. Here we will just state the result in the notation of Ref. 1.

Three-dimensional rotations are described by the Lie group $SO(3)$. The corresponding Lie algebra $so(3)$ is

$$\tag{1} [L_i, L_j] ~=~ i\sum_{k=1}^3\epsilon_{ijk} L_k, \qquad i,j,k~\in~ \{1,2,3\}, \quad \epsilon_{123}~=~1, \quad i^2=-1.$$

In adjoint representation, the three Lie algebra generators $iL_k\in{\rm Mat}_{3\times 3}(\mathbb{R})$, $k\in\{1,2,3\}$, are $3\times 3$ real antisymmetric matrices,

$$\tag{2} i(L_i)_{jk} ~=~ \epsilon_{ijk} .$$

II) A rotation matrix

$$\tag{3} R(\vec{\alpha})~\in~ SO(3)~ \subseteq ~{\rm Mat}_{3\times 3}(\mathbb{R})$$

can be specified by a rotation axis and an angle. Here we will use a 3-vector

$$\tag{4} \vec{\alpha}~=~\alpha \vec{n}_{\alpha}~\in~ \mathbb{R}^3,$$

where $\vec{n}_{\alpha}\in\mathbb{R}^3$ is a unit vector parallel to the rotation axis, $|\vec{n}_{\alpha}|=1$ ; and $\alpha\in \mathbb{R}$ (without an arrow on top) is the angle of rotation.

Define for later convenience

$$\tag{5} c_{\alpha} ~:=~\cos(\alpha) ~\in~ \mathbb{R},$$ $$\tag{6} \vec{s}_{\alpha}~:=~\sin(\alpha)\vec{n}_{\alpha} ~\in~ \mathbb{R}^3,$$ $$\tag{7} \vec{t}_{\alpha}~:=~\tan(\alpha)\vec{n}_{\alpha} ~\in~ \mathbb{R}^3.$$

The formula for the rotation matrix in terms of $\vec{\alpha}$ reads

$$\tag{8} R(\vec{\alpha}) ~=~e^{i \vec{\alpha}\cdot \vec{L}}~=~ {1}_{3\times 3} - (1-c_{\alpha}) (\vec{n}_{\alpha}\cdot \vec{L})^2 + i\vec{s}_{\alpha}\cdot \vec{L}.$$

III) The composition of two rotations is again a rotation

$$\tag{9} R(\vec{\gamma})~=~R(\vec{\alpha})R(\vec{\beta}).$$

The "addition formula" for the corresponding $3$-vectors reads

$$\tag{10} \vec{t}_{\gamma} ~=~\frac{\vec{t}_{\alpha}+\vec{t}_{\beta}-\vec{t}_{\alpha}\times\vec{t}_{\beta} }{1-\vec{t}_{\alpha}\cdot \vec{t}_{\beta}}.$$

The derivation of eq. (10) simplifies if one uses the fact that $SU(2)\cong U(1,\mathbb{H})$ is the double cover of $SO(3)$.

References:

1. G 't Hooft, Introduction to Lie Groups in Physics, lecture notes, chapter 3. The pdf file is available here.
2. S. Weigert, J. Phys. A30 (1997) 8739, arXiv:quant-ph/9710024.
-
Thank you! Is \tag{} documented somewhere? It looks lovely :) –  Claudius Dec 9 '12 at 17:24
Another reference: K. Engoe, On the BCH-formula in so(3), 2001. –  Qmechanic Sep 4 '13 at 11:33

As $\mathrm{SO}(3)$ is a connected group, $\exp(\mathsf{L}(\mathrm{SO}(3))) = \mathrm{SO}(3)$ and hence this should – in theory – work. Let us work in the fundamental representation of $\mathrm{SO}(3)$, that is orthogonal 3x3 matrices.

Assume you have a rotation $B$ acting first and a second rotation $A$, the resulting rotation is then given by $AB \equiv C \in \mathrm{SO}(3)$. Furthermore, we can express $A$, $B$ and $C$ by $\exp(a)$, $\exp(b)$ and $\exp(c)$ for $a,b,c \in \mathsf{L}(\mathrm{SO}(3))$. We then have¹

$$\exp(a) \exp(b) = AB = C = \exp(c) = \exp\left(a + b + \frac{1}{2}[a,b] + \frac{1}{12} [ a, [a,b]] - \frac{1}{12}[b,[a,b]]+ \ldots\right)\quad.$$

Now, the problem with verifying this by an example is that these commutators are rather ugly. I shall do two examples:

## First example: Two rotations about the $x$ axis

Take $A$ to rotate about $(1,0,0)$ by $\theta$ and $B$ to rotate about the same axis by $\phi$. We then have

$$A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos(\theta) & -\sin(\theta) \\ 0 & \sin(\theta) & \cos(\theta) \end{pmatrix}$$

and similarly for $B$. The associated $a$ is then simply:

$$a = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & -\theta \\ 0 & \theta & 0 \end{pmatrix}\quad$$

and again similarly for $b$ with $\theta \to \phi$. You can check easily that $\exp(a)$ gives you indeed $A$. Now since $a$ and $b$ commute, we have $[a,b] = 0$ and hence $c = a + b$ - which is

$$c = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & -\theta -\phi \\ 0 & \theta + \phi & 0 \end{pmatrix}\quad.$$

This very likely illuminates better than $AB$ that two rotations about the same axis are equivalent to one rotation by the sum of the angles. You can again check that $\exp(c)$ gives you $C$.

## Second Example: One rotation about $y$, a second about $x$.

This one is more difficult, as we will have to calculate annoying commutators. The result presented here will hence only be approximate, not exact.

Take

$$A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos(\theta) & -\sin(\theta) \\ 0 & \sin(\theta) & \cos(\theta) \end{pmatrix} \qquad B = \begin{pmatrix} \cos(\phi) & 0 & \sin(\phi) \\ 0 & 1 & 0 \\ -\sin(\phi) & 0 & \cos(\phi) \end{pmatrix} \quad .$$

You can calculate that

$$AB = C = \begin{pmatrix} \cos(\phi) & 0 & \sin(\phi) \\ \sin(\theta) \sin(\phi) & \cos(\theta) & -\sin(\theta) \cos(\phi) \\ \sin(\phi)\cos(\theta) & \sin(\theta) & \cos(\phi)\cos(\theta) \end{pmatrix} \quad .$$

Similarly to the above, we have

$$a = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & -\theta \\ 0 & \theta & 0 \end{pmatrix} \qquad b = \begin{pmatrix} 0 & 0 & \phi \\ 0 & 0 & 0 \\ -\phi & 0 & 0 \end{pmatrix} \quad.$$

Now the tricky part is to calculate

$$c = a + b + \frac{1}{2} [ a,b] + \frac{1}{12} [ a, [a,b]] - \frac{1}{12} [b,[a,b]] + \ldots$$

to such a precision that $\exp(c)$ gives remotely sensible results. At this, I mostly failed, but here's what I got:

$$\frac{1}{2} [ a,b] = \frac{1}{2} \begin{pmatrix} 0 & -\theta\phi & 0 \\ \theta\phi & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}\quad,$$

which looks an awful lot like the element of the Lie algebra basis corresponding to a rotation about the $z$ axis, but unfortunately doesn’t fit in at all (something linear in either $a$ or $b$ would have been nice…). I then went on to calculate $[a,[a,b]]$ and $[b,[a,b]]$ and arrived at

$$c \approx \begin{pmatrix} 0 & -\frac{1}{2}\theta\phi & \phi - \frac{1}{12} \theta^2 \phi \\ \frac{1}{2} \theta \phi & 0 & -\theta + \frac{1}{12} \theta\phi^2 \\ -\phi + \frac{1}{12} \theta^2 \phi & \theta - \frac{1}{12} \theta \phi^2 & 0 \end{pmatrix} \quad .$$

The nice thing here is that this is still an antisymmetric matrix and hence (can be) in $\mathsf{L}(\mathrm{SO}(3))$. In order to now compare this to anything, we have to approximate $C$. Recall the expression from above. As a first approximation, I will set $\cos(x) = 1 - \frac{1}{2}x^2$, $\sin(x) = x - \frac{1}{6} x^3$. I then get

$$C \approx \begin{pmatrix} 1 - \frac{\phi^2}{2} & 0 & \phi - \frac{\phi^3}{6} \\ \left(\theta - \frac{\theta^3}{6}\right) \left(\phi - \frac{\phi^3}{6}\right) & 1 - \frac{\theta^2}{2} & -\left(1-\frac{\phi^2}{2}\right)\left(\theta - \frac{\theta^3}{6}\right) \\ -\left(1-\frac{\theta^2}{2}\right)\left(\phi-\frac{\phi^3}{6}\right) & \theta - \frac{\theta^3}{6} & \left(1 - \frac{\theta^2}{2}\right)\left(\phi - \frac{\phi^3}{6}\right) \end{pmatrix} \quad ,$$

expanding out the brackets and throwing away anything of order four, I arrive at

$$C \approx \begin{pmatrix} 1 - \frac{\phi^2}{2} & 0 & \phi - \frac{\phi^3}{6} \\ \theta\phi & 1 - \frac{\theta^2}{2} & -\theta + \frac{\phi^2\theta}{2} \\ -\phi + \frac{\theta^2\phi}{2} & \theta - \frac{\theta^3}{6} & \phi - \frac{\theta^2\phi}{2} \end{pmatrix}\quad.$$

This expression should be roughly equal to

$$1_3 + c + \frac{1}{2} c^2 + \frac{1}{6} c^3 \quad,$$

which is the expansion of $\exp(c)$. After again throwing away everything of order four, we arrive at

$$\exp(c) \approx \begin{pmatrix} 1-\frac{\phi^2}{2} & 0 & \phi - \frac{\phi^3}{6} \\ \theta\phi & 1 - \frac{\theta^2}{2} & -\theta+\frac{\theta^3}{6} +\frac{\theta\phi^2}{2}\\ -\phi +\frac{\theta^2 \phi}{2} & \theta - \frac{\theta^3}{6} & 1 - \frac{\theta^2}{2} - \frac{\phi^2}{2} \end{pmatrix} \quad .$$

The remaining ‘wrong’ terms here most probably cancel with higher orders of $c$, but I have to admit I am slightly too lazy for that.

## Conclusion

The main problem with the BCH formula is really that, in general, $[a,b] \neq 0$ and you hence most often not even get an exact expression for $c$ – from which one could most likely deduce angle and axis of rotation without evaluating that pesky exponential. Without an exact expression for $c$, however, all is lost, as non-exact expressions merely rely on the fact that for infinitesimal angles of rotation, all rotations commute.

I would love to hear other opinions, though, especially regarding the ‘theoretical’ part what one could do with $c$, if it was known exactly.

-
This is right; the BCH formula gives the right answer, but the expansion doesn't terminate, so it's only so useful. The easiest way I can think of to quickly extract axis and angle is to use quaternions. –  Muphrid Dec 5 '12 at 21:09