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In the book I'm reading there is a question,

Prove that given a rotation $R(\phi \hat{n})$ and a reflection $\sigma$, they commute iff $\sigma \perp \hat{n}$ and $\phi=\pi$. (Binary rotation normal to the reflection plane.)

The answer in the back of the book is,

Take $R(\phi z)$. For $\phi=\pi$, the eigenvectors, with eigenvalues in brackets, are: $\vec{z}(+1), \vec{x}(-1), \vec{y}(-1)$. If $\sigma$ is the $\vec{x}\vec{y}$ plane its eigenvectors are $\vec{x}(+1), \vec{y}(+1), \vec{z}(-1)$ and coincide with those of $R(\pi\vec{z})$.

My question:

I thought an eigenvector of a symmetry operation is one that after the symmetry operation $g\vec{r}$ leaves r changed only by the eigenvalues +1 or -1. In my mind, this means that the eigenvectors of a reflection operator $\sigma$ would be all vectors on the reflection plane and normal to the plane. Not just the two particular vectors that span the plane.

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The given statement is false: If $\sigma\perp\hat n$, the rotation angle may be anything. Furthermore there is an extra case: If the rotation axis $\hat n$ lies in the reflection plane $\sigma$ and $\phi=\pi$ then the rotation and the reflection do also commute. Furthermore the "proof in the book" only proves the easy half of the (wrong) statement. – Christian Blatter Feb 15 '11 at 11:55
Well I'm glad I didn't understand the question so I could find out this wasn't a true statement anyway! I drew out a picture and quickly verified what you said is correct. Thank you. – coderdave Feb 15 '11 at 20:46
up vote 1 down vote accepted

Eigenvectors are not unique in general and can be with respect to any eigenvalue. Generally what one does instead is pick a basis for a given eigenspace (the space of eigenvectors with respect to a particular eigenvalue) in the cases where such a basis exists. In this case the eigenspace associated to eigenvalue $1$ is the entire reflection plane, and any basis of this works.

However, in the generic case eigenspaces have dimension $1$ and this issue doesn't occur.

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Let me rephrase your answer a bit to see if I have an understanding of what you said. Would it be correct to say that two symmetry operations can commute if they share the same eigenspace? Also, for a space of N it would require N eigenvalues, correct? – coderdave Feb 15 '11 at 20:44
No; that condition is necessary (and they must share all of the same eigenspaces, plural, although not with the same eigenvalues) but not sufficient in general. – Qiaochu Yuan Feb 15 '11 at 20:46
Ok, so the sufficient statement is that two symmetry operations can commute iff they share the same eigenvectors, correct? I'm a bit confused about the difference in the two statements. I'll have to think about it for a bit. – coderdave Feb 15 '11 at 20:51
That is true if both of them have a basis of eigenvectors. – Qiaochu Yuan Feb 15 '11 at 21:04
Ok, thank you for your patience in explaining this to me as this is very new material for me. Anyway, it seems you can pick multiple eigenvector basis on the reflection plane because multiple basis can span the plane. Would it be correct to revise the statement to 'Two symmetry operations can commute iff there is a common eigenvector basis between them'. – coderdave Feb 15 '11 at 21:16

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