Equivalent Rotation using Baker-Campbell-Hausdorff relation

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

$$ exp(a) exp(b) = AB = C = exp(c) = expleft(a + b + frac{1}{2}[a,b] + frac{1}{12} [ a, [a,b]] - frac{1}{12}[b,[a,b]]+ ldotsright)quad.$$

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:

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

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

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

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

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

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

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

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.