In Group Theory by Morton Hamermesh, he states on page 32:

For a body of the finite extension, a molecule or the macroscopic form of a mineral, only the first two symmetry types [rotation, reflection] are possible. In fact, all transformations of the symmetry group of a finite body must leave at least one point of the body fixed. In other words, all axes of rotation and all planes of reflection must intersect in (at least) one point. Clearly, successive rotations about nonintersecting axes or reflections in nonintersecting planes will result in the introduction of translation and a continual shift of the body.

I have been trying to justify how this is true but have been unable so far, I am only able to say that:

  1. Classically:

Suppose we perform two rotations about the non-intersecting axis, then any general motion can be represented as a rotation followed by translation, in our case, if there is no translation, then it has to be pure rotation, that means there should be a certain axis that transforms to itself and once such axis is found, then clearly there is no translation and only rotation.

So trying to show such an axis does not exist if two rotations happen along non-intersecting axes will be sufficient to show that there must be translation, now I am not able to prove this, I can say that the second rotation leaves only the points along 2nd rotational axis unchanged, now these points have already been moved once during 1st rotation, hence both the rotational axes are certainly not invariant under the action, but what if there is a certain line that comes to itself after two rotations such that shifts in both rotations cancel each other, I am not able to show

how to show this cannot be possible??

  1. quantum mechanically

here I am totally clueless, I can only translate the action into matrices as: $$T_r(\vec{r})D(\hat{n_2},\theta_2)T_r(-\vec{r})D(\hat{n_1},\theta_1) = D(\hat{n_3},\theta_3)$$ Where $T_r(\vec{r})$ translates the wavefunction by $\vec{r}$ and $D(\hat{n_2},\theta_2)$ rotates the wavefunction by angle $\theta$ along axis $\hat{n}$. I think this is correct because first $D(\hat{n_1},\theta_1)$ rotates the system around a certain axis by a certain angle, then we translate the system such that the new axis of rotation intersects the origin, now we rotate the second time and then we shift back the system to undo the effect of translation and all this should amount to a single rotation around some axis

Please help!!!


1 Answer 1


The issue is that the composition of two rotations about non-intersecting axes from a translation iff the axes are parallel and the rotations' angles are opposite. In general, their composition will lead to a third rotation or a screw rotation with a different axis and angle.

Mathematically, you can write the rotations in their affine form: $r_1 : x\to R_1(x-y_1)+y_1$ and $r_2 : x\to R_2(x-y_2)+y_2$, with $R_1,R_2$ rotation matrices. Unless the above condition is satisfied, which translates to $R_2 = R_1^{-1}$, the composition $r_1r_2$ is a rotation when it has a fixed point or else a screw rotation .

What the author essentially means is that the orbits of the group generated by two such rotations are unbounded. Hence, it cannot represent the symmetry group of a finite body. This would be easy to prove if the generated symmetry group contains a translation or a screw rotation. Indeed, by iterating one of such transformations you get an unbounded subset of the orbit.

However, there is no reason for this to happen. You have a translation in the group iff the $R_1,R_2$ don’t form a free basis. Intuitively, this conditions means that there is no multiplicative relation between the matrices (like $R_1R_2=1$). I guess it was this scenario that the author had in mind, which greatly simplifies the proof.

The existence of a screw rotation isn’t guaranteed either. Take the case when the axes are parallel, the group consists only of rotations whose axis are parallel to the two original generators.

The last case is more tricky since the group consists only of rotations. Therefore, only an aperiodic composition of its elements can potentially send a point to infinity, and there is no easy way to control mathematically this divergence.

To illustrate the difficulty, consider the previously mentioned example, which reduces the problem to a planar one. This way, I can represent the matrices by simple complex numbers: $r_1: z\to a_1z+b_1$ and $r_2: z\to a_2 z+b_2$ with $|a_1|=|a_2| = 1$ (rotations) $b_1\neq b_2$ (different centers) and $1,\arg a_1,\arg a_2$ rationally independent (free group). The goal is by a judicious combination of $r_1,r_2,r_1^{-1}, r_2^{-1}$ to send any $z$ (say the origin for simplicity) to $\infty$. Even in this simple case, it isn’t so obvious. I think that you can use the density of $\langle a_1,a_2\rangle$ which essentially takes you as close as you want to a non-free scenario, but the full mathematical formalism seems tedious.

There is something I don't understand though, why are you distinguishing the quantum and classical case? This is irrelevant, you are studying the symmetry group of a finite body, whether you apply it in a classical model or a quantum model through a representation shouldn’t matter.

Hope this helps and tell me if you need more details.

  • $\begingroup$ It is very heavy for me to understand, I am trying to slowly understand your answer! $\endgroup$
    – Kutsit
    Jun 17, 2022 at 9:09

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