# Why does the C6z operator have 6 eigenvectors in this paper?

In this paper, I couldn't understand this line in the 3rd section (i.e. III. SYMMETRY ARGUMENTS):

In the absence of SOC, the eigenvalues of $$C_{6z}$$ rotation operator are $$e^{i2\pi n/6}$$ where $$n = 0$$ to $$5$$ and the corresponding eigenstates are denoted by $$\psi_1 , \psi_2 , \psi_3, \psi_4 , \psi_5 , \text{and} \psi_6$$, respectively.

But, as we know, the rotation matrix for an angle $$\phi$$ rotation about z-axis in 3-D cartesian space is given by:

$$\begin{pmatrix} \cos\phi & -\sin\phi & 0\\ \sin\phi & \cos\phi & 0\\ 0 & 0 & 1 \end{pmatrix},$$ with eigenvalues $$\lambda_1 = 1 , \lambda_2=e^{i\phi} , \lambda_3 =e^{-i\phi}$$ and their corresponding eigenvectors $$v_1 = (0,0,1) , v_2 = (i,1,0), v_3=(-i,1,0)$$. I don't know what I am missing to correctly interpret and find the mathematical logic for the eigenvalues of the $$C_{6z}$$ rotation operator, and what I am confused about is that I am getting 3 eigenvalues but in the paper, it is 6.

• On which Hilbert space does this (representation of?) rotation act on...? It seems to be six-dimensional... Commented Jan 26 at 11:22
• Thats what I am confused about. When going through the Matrix representation for the C6z operator, I am getting 3 eigenvalues but in the paper, it is 6 . In the paper it is coming from the C6v point Group from group theory. Commented Jan 26 at 11:30

The paper appears to have an error.

When a $$C_6$$ operator acts on an atom's (x,y,z) coordinates, the operator's matrix representation is the 3x3 matrix that you correctly provided in your question post. The rotation matrix is a 3x3 matrix whether it is representing a $$C_{3}$$ operator, $$C_{6}$$ operator, $$C_{1402}$$ operator, or any other $$C_n$$ operator with $$n$$ being a positive integer. In the paper to which you referred, $$C_{6z}$$ is the specific $$C_{6}$$ operator for a rotation around the z-axis, meaning that it is the $$R_z(\theta)$$ operator in which $$\theta$$ is 360/6 = 60 degrees.

In Fig 2a of the arXiv version of the paper, what they call the "eigenstates" of the $$C_{6z}$$ operator, are labeled as $$\psi_1,\psi_2,\ldots,\psi_6$$, even though these just appear to be positions in (x,y,z)-space that are obtained by applying the C6 operator. For example, if a nucleus resides at (x,y,z) = (1,2,3), there are 6 different locations to which the $$C_{6z}$$ operator can transform the coordinates of the nucleus, one of these being the original coordinates: (1,2,3). This is very different from eigenvectors: what eigenvectors would be in this situation, are (x,y,z) coordinates at which the $$C_{6z}$$ operator does not change the coordinates. For example, you suggested the vector (x,y,z) = (0,0,1), which after any rotation around the z-axis, remains to be (0,0,1).

If the authors were using the terminology correctly, then in Fig 2a we would expect for $$\psi_1,\psi_2,\ldots,\psi_6$$ to all be at the same location, because they would represent locations at which the operator does not change the coordinates!

A rotation operator can have 6 eigenvalues and 6 eigenvectors if it is an operator that is acting on two sets of (x,y,z) coordinates (for example if it is acting on the positions of two nuclei, rather than just one), but even then the rotation operator would usually be given by the standard 3x3 matrix that you presented in your post, and it would just be applied to the two sets of (x,y,z) coordinates independently (the 6x6 matrix would be a block diagonal matrix composed of two 3x3 rotation matrices).

• Are the eigenvalue correct ? i.e, $e^{i2\pi n/6}$ because the rotation operator is a 3x3 matrix and it will only have 3 eigenvalues (degenerate or non-degenerate depends). Commented Jan 26 at 15:14
• @user192399 the eigenvalues can be found in WolframAlpha or Symbolab or with any mathematics package that can do eigenvalues (although if the package doesn't have symbolic computation capabilities, you you would need to plug values of "n" into the expression in your comment, in order to match them with the result given by the package). If you want to ask what the eigenvalues of the matrix are, it is different from "Thats what I am confused about. When going through the Matrix representation for the C6z operator, I am getting 3 eigenvalues but in the paper, it is 6." Please post a new question. Commented Jan 26 at 17:53
• What about spin? If it is an operator representing rotation, it could be an operator on e.g. $L^2\otimes \mathbb C ^2$. But I haven't checked the paper... Commented Jan 26 at 18:49
• @Jakob if you want to ask a question about what it would look like with spin, it would be best to post a new question, so that one of our 7500+ users can write an answer if I don't know it, and so that the answer can actually fit, because comments only allow 600 characters and we are actually not supposed to use them for new questions. However even if we consider spin, Fig 2a of the paper makes it appear that they think $M\vec{v} = \lambda \vec{w}$ makes $\vec{w}$ an eigenvector, so $\vec{w_1},\vec{w_2},\ldots ,\vec{w_6}$ are being labeled as eigenvectors even though they are just "generated". Commented Jan 26 at 20:04