Why does (2z²-x²-y², x²-y²) have -1 trace in the character table of Td for C3 rotation?

Question

I'm trying to understand the irreducible rep for $(2z^2-x^2-y^2, x^2-y^2)$ in the $T_\mathrm{d}$ point group. Specifically, for $C_3$ rotations, a trace of $-1$ is listed in the character table for this group. For any matrix, one can rotate the coordinates (basis for the matrix) - i.e. an orthogonal transformation, yielding a matrix with different entries, but the trace will remain the same. So in the case of $120^\circ$ rotations around an axis through a vertex and opposite face of the tetrahedron, one can rotate the coordinates so that the $z$-axis aligns with rotation axis making the calculation of the trace easier.

Any rotation in this basis will leave $2z^2-x^2-y^2$ unchanged, ie. a $1$ on the diagonal. However, the projection of a rotated $x^2-y^2$ back to its own basis will yield $(\cos 2\theta)(x^2-y^2)$. The trace will then be $1 + \cos 2\theta$. Can anyone help me where I went wrong?

Answer 1

I think I know now.

Choose rotation axis that makes equal angles with the $x$-, $y$-, and $z$-axes so that the $C_3$ rotation essentially permutes the axes. Without loss of generality, I could choose a $C_3$ rotation that permutes $x$ to $z$, $z$ to $y$, and $y$ to $x$. Rotation in either direction will give the same trace with this $C_3$ rotation. With this rotation (permutation), $(2z^2-x^2-y^2,x^2-y^2)$ becomes $(2y^2-z^2-x^2,z^2-x^2)$. Multiplication of basis function vector a by $2\times 2$ matrix in this representation yields

$$\begin{pmatrix} -1/2 & -3/2\\ 1/2 & -1/2 \end{pmatrix} \begin{pmatrix} 2z^2-x^2-y^2 \\ x^2-y^2 \end{pmatrix} = \begin{pmatrix} 2y^2-z^2-x^2\\ z^2-x^2 \end{pmatrix} $$

The trace of this matrix is $-1$ as in the character table. Choosing any axis via orthogonal transformation of this matrix yields same trace.

The tricky thing about this example was that I needed to use a rotation in a three-dimensional vector space and project it into a rotation in a two-dimensional vector space.

Answer 2

This is answer number #2:

Choosing the $z$-axis as the rotation axis just made things more complicated in the end, that is, in trying to simplify the calculation of the trace in this example.

Other examples of pairs of quadratic basis functions commonly used in point groups such as $D_{n\mathrm{h}}$ includes: $(x^2-y^2,xy)$.

For this, one can replace $x$ and $y$ with $(x\cos\theta - y\sin\theta)$ and $(x\sin\theta + y\cos\theta)$ respectively, which represents a rotation around the $z$-axis by an angle $\theta$. After using some trigonometric identities and algebra, one finds that the matrix for this representation is

$$\begin{pmatrix} \cos 2\theta & -2 \sin 2\theta\\ (\sin 2\theta)/2 & \cos2\theta \end{pmatrix}$$

yielding a trace of $2 \cos2\theta$ for $C_n$ rotations about the $z$-axis. So, in this example of finding the character for representations associated with $(x^2-y^2,xy)$ for $C_n$ rotations, using the $z$-axis turns out to be useful because the the result of the operation could be expressed as a linear combination of your basis functions which you can not always do.

Answer 3

The following is a solution alternative to OP's own. As OP correctly observed, any $C_3$ rotation belonging to the group $T_d$ is similar to the $C_3$ rotation about $\hat{\textbf{n}} = (\hat{\textbf{x}}+\hat{\textbf{y}}+\hat{\textbf{z}})/\sqrt{3}$, the axis making equal angles with all three coordinate axes (of course, with appropriate choice of the coordinate system). This rotation switches the coordinate variables such that $x\to y$, $y\to z$, and $z\to x$.

Notice that $C_3$ rotations are also elements of $\mathrm{SO}(3)$, the group of three-dimensional rotations. The five $d$ orbitals furnish an irreducible representations of $\mathrm{SO}(3)$ corresponding to $l=2$ (angular momentum quantum number). If we take the five $d$ orbitals to be the ones with definite magnetic quantum number $m$ ($-2\le m\le 2$) and the axis of rotation to be $\hat{\textbf{z}}$, the rotation by $2\pi/3$ radians (the angle of rotation for $C_{3}$) is equivalent to multiplying the phase factor $e^{-2\pi i m/3}$ to each $d$ orbital. Therefore, the character of the corresponding matrix is given by \begin{equation} \chi(C_{3}(\hat{\textbf{z}});l=2) = \sum_{m=-2}^{2}e^{-2\pi i m/3} = -1. \end{equation} Because rotation about any axis is similar to the rotation about the $z$ axis with the same rotation angle, we have \begin{equation} \chi(C_{3}(\hat{\textbf{n}});l=2) = -1. \end{equation}

Next, we know that in the group $T_{d}$, the $l=2$ representation formed by the five $d$ orbitals are no longer irreducible: it further splits into the $e_{g}$ and $t_{2g}$ representations. Hence, \begin{equation} \chi(C_{3}(\hat{\textbf{n}});l=2) = \chi(C_{3}(\hat{\textbf{n}});e_{g}) + \chi(C_{3}(\hat{\textbf{n}});t_{2g}). \end{equation} Observe that $C_{3}(\hat{\textbf{n}})$ simply switches between $t_{2g}$ orbitals such that $xy \to yz$, $yz \to zx$, and $zx \to xy$. Therefore, the diagonal entries of the corresponding matrix are zero, meaning that $\chi(C_{3}(\hat{\textbf{n}});t_{2g}) =0$. It then follows that \begin{equation} \chi(C_{3}(\hat{\textbf{n}});e_{g}) = \chi(C_{3}(\hat{\textbf{n}});l=2) - \chi(C_{3}(\hat{\textbf{n}});t_{2g}) = -1. \end{equation}