Deriving the irreps of the d-orbitals under C3 and S6 operations in octahedral symmetry

Contextual background: I am trying to reproduce the symmetries in the graph on [this](https://en.wikipedia.org/wiki/Ligand_field_theory) page (MO diagram for $\ce{ML6}$ complex). I have reproduced the metal s and p orbital representations as well as the linear combination of representations for the $\ce{L6}$ fragment. All that remain are the metal d orbitals.


So I know the five d-orbitals collectively reduce to the $\mathrm{e\_g}$ + $\mathrm{t\_{2g}}$ representation. I put my five d-orbitals at the origin and went through the operations for the reducible representation. Everything else adds up, except I can't find that $-1$ in the $C\_3$ and $S\_6$ operations.


### Character table


$$\small\begin{array}{c|cccccccccc|cc}\hline
O\_\mathrm{h} & E & 8C\_3 & 6C\_2 & 6C\_4 & \begin{aligned}3C\_2 \\ \scriptsize=C\_4^2\end{aligned} & i & 6S\_4 & 8S\_6 & 3\sigma\_\mathrm{h} & 6\sigma\_\mathrm{d} & & \\ \hline
\mathrm{A\_{1g}} & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & & x^2+y^2+z^2 \\
\mathrm{A\_{2g}} & 1 & 1 & -1 & -1 & 1 & 1 & -1 & 1 & 1 & -1 & & \\
\mathrm{E\_g} & 2 & -1 & 0 & 0 & 2 & 2 & 0 & -1 & 2 & 0 & & \begin{aligned}(2z^2-x^2-y^2,\\ x^2-y^2)\,\,\,\,\,\, \end{aligned} \\
\mathrm{T\_{1g}} & 3 & 0 & -1 & 1 & -1 & 3 & 1 & 0 & -1 & -1 & (R\_x,R\_y,R\_z) & \\
\mathrm{T\_{2g}} & 3 & 0 & 1 & -1 & -1 & 3 & -1 & 0 & -1 & 1 & & (xy,xz,yz) \\
\mathrm{A\_{1u}} & 1 & 1 & 1 & 1 & 1 & -1 & -1 & -1 & -1 & -1 & & \\
\mathrm{A\_{2u}} & 1 & 1 & -1 & -1 & 1 & -1 & 1 & -1 & -1 & 1 & & \\
\mathrm{E\_u} & 2 & -1 & 0 & 0 & 2 & -2 & 0 & 1 & -2 & 0 & & \\
\mathrm{T\_{1u}} & 3 & 0 & -1 & 1 & -1 & -3 & -1 & 0 & 1 & 1 & (x,y,z) & \\
\mathrm{T\_{2u}} & 3 & 0 & 1 & -1 & -1 & -3 & 1 & 0 & 1 & -1 & & \\ \hline
\end{array}$$


### Work


$$\begin{array}{c|cccccccccc} \hline
O\_\mathrm{h} & E & 8C\_3 & 6C\_2 & 6C\_4 & 3C\_2 & i & 6S\_4 & 8S\_6 & 3\sigma\_\mathrm{h} & 6\sigma\_\mathrm{d} & & \\ \hline
\mathrm{E\_g} & 2 & -1 & 0 & 0 & 2 & 2 & 0 & -1 & 2 & 0 \\
\mathrm{T\_{2g}} & 3 & 0 & 1 & -1 & -1 & 3 & -1 & 0 & -1 & 1 \\ \hline
\Gamma\_{\text{d-orbitals}} & 5 & ? & 1 & -1 & 1 & 5 & -1 & ? & 1 & 1 \\ \hline
\end{array}$$


As can be seen, everything matches up but the $C\_3$ and $S\_6$ operations. I tried using [Molecule Viewer](http://molecule-viewer.com/) and rotating the d-orbitals, but I still don't see the $-1$. They all look like zero to me.


I would expect the $\mathrm d\_{x^2-y^2}$ orbital to correspond to $E\_g$ from the character table, but I only see it rotate into a "$\mathrm d\_{y^2-z^2}$" orbital, with the orbitals along the y and z axes instead of x and y axes. The $\mathrm d\_{z^2}$ orbital rotates into a "$\mathrm d\_{x^2}$" orbital, lying along the x axis instead of the z axis.


Anyone know which d-orbital rotates into $-1$ for a $C\_3$ operation? I assume a $-1$ for $S\_6$ is the same orbital, but if not then does anyone which that one is as well?


Thank you




The $C\_3$ operation rotates $x \rightarrow y$, $y \rightarrow z$ and $z \rightarrow x$.
Hence,
$$\phi\_1=2z^2-(x^2+y^2)$$ 
rotates into 
$$-(1/2)\phi\_1 +(3/2)\phi\_2$$ and
$$\phi\_2=x^2-y^2$$  

rotates into 
$$-(1/2)\phi\_1 -(1/2)\phi\_2$$
so, in the basis
 $( \phi\_1, \phi\_2 )$ the matrix that represents $C\_3$ is


$$\left [\begin{matrix} -1/2 & -1/2 \\ 3/2 & -1/2 \end{matrix}\right]$$


Denoting this 2 by 2 matrix by *M* what we have is
$C\_3(\phi\_1 \phi\_2) = (\phi\_1 \phi\_2)M$.
The trace of this matrix is $-1$. 


This is what you find in the character table.


Since $C\_3^3=I$ (identity) its eigenvalues are the cubic roots of unity:
$1, \lambda$ and $ \lambda^\*$
where $\lambda = \exp(2\pi i/3)=-1/2+i\sqrt 3/2$,
and $\lambda^\*$ is the complex conjugate of $\lambda $.


It is easy to find eigenvectors with the eigenvalue 1 [e.g., an s orbital or x+y+z].


You can find a linear combination of $\phi\_1$ and $\phi\_2$ that will be an eigenvector of $C\_3$ with the eigenvalue $\lambda$ (transform into itself multiplied by $\lambda$)
and another linear combination that will have eigenvalue $\lambda^\*$. Now, the matrix representing $C\_3$ will be diagonal with $ \lambda$ and $\lambda^\*$ along the diagonal. The trace remains $-1$.
Something similar should work for $S\_6$.


 3

From reading what you wrote I think you don't entirely understand how degenerate irreps work. Firstly, it's incorrect to say "$\mathrm{d}\_{x^2-y^2}$ corresponds to $\mathrm{E\_g}$". The $\mathrm{d}\_{x^2-y^2}$ orbital by itself *does not transform as anything*. It only makes sense to say that the two orbitals *transform together* as $\mathrm{E\_g}$. But anyway, let's go from the start.




---


For non-degenerate irreps, the behaviour of an orbital is that the symmetry operation transforms it into some multiple of itself:


$$\hat{R}\psi = \chi^R\psi \tag{1}$$


where $\chi^R$ is the character under the symmetry operation $\hat{R}$. For example if $\psi$ is an s-orbital, then all the symmetry operations leave it untouched: $\hat{R}\psi = \psi$. Therefore $\chi^R = 1$ for all symmetry operations, and $\psi$ transforms as the totally symmetric irrep ($\mathrm{A\_{1g}}$ in this case).




---


Taking this logic and applying it to degenerate irreps is very dangerous. When two orbitals $(\psi\_1, \psi\_2)$ transform *together* as a degenerate irrep, what happens is that


$$\begin{align}
\hat{R}\pmatrix{\psi\_1 & \psi\_2} &= \pmatrix{\psi\_1 & \psi\_2}\pmatrix{\Gamma\_{11} & \Gamma\_{12} \\ \Gamma\_{21} & \Gamma\_{22}} \\[8pt]
&= \pmatrix{\Gamma\_{11}\psi\_1 + \Gamma\_{21}\psi\_2 & \Gamma\_{12}\psi\_1 + \Gamma\_{22}\psi\_2} \tag{2}
\end{align}$$


where $\Gamma = \pmatrix{\Gamma\_{11} & \Gamma\_{12} \\ \Gamma\_{21} & \Gamma\_{22}}$ is some matrix that you have to find out by investigating the effect of $\hat{R}$ on the two functions.


$\Gamma$ is called a "matrix representation" of the symmetry operation. The *character* under this symmetry operation (which is what is listed in the character table) is the trace of this matrix $\Gamma$, i.e.


$$\chi^R = \Gamma\_{11} + \Gamma\_{22}. \tag{3}$$


(Notice that in the one-dimensional case the matrix $\Gamma$ simply becomes a number and the character is simply that number itself, which regenerates equation $(1)$.)


Now let's say $\psi\_1 = \mathrm{d}\_{x^2-y^2}$, $\psi\_2 = \mathrm{d}\_{z^2}$, and $\hat{R} = \hat{C}\_3$. The issue is that you seem to expect that the rotation $C\_3$ to have a nicely well-defined effect on $\psi\_1$:


$$\hat{R}\psi\_1 = k\_1\psi\_1 \tag{4}$$


But for a degenerate irrep this is simply not true. From equation $(2)$ one can see that 


$$\hat{R}\psi\_1 = \Gamma\_{11}\psi\_1 + \Gamma\_{21}\psi\_2 \tag{5}$$


Furthermore there is no guarantee that any of these two numbers $\Gamma\_{11}$ or $\Gamma\_{21}$ are related to the number in the character table. The only requirement is that $\Gamma\_{11} + \Gamma\_{22}$ (the latter of which we haven't even started talking about!) is equal to the character under the $C\_3$ operation, i.e. $-1$. So now you see why asking "which d-orbital rotates into $−1$ for a $C\_3$ operation?" doesn't make much sense. *Neither* of those two d-orbitals do.


Hopefully with this description you are in a better position to understand the other answer, which is essentially telling you that $\Gamma\_{11} = -1/2$ and $\Gamma\_{22} = -1/2$.


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