I have another SALC question. @Andrew and @user5713492 did a phenomenal job clarifying my previous confusion with Methane T2 SALC production, so I attempted to apply this to a new point group, and have hit another wall.

Utilizing the full projection operator, I can perform the following analysis to get one of the e1 SALC orbitals in C5v (assuming a purely sigma framework through 1s orbitals as basis on the ligands):

The e1 matrix representation can be generated from {x,y}, resulting in:

$$\begin{array}{cccccc}E&\sigma_{v}&C_{5}&C_{5}'\\ \begin{bmatrix}1&0\\0&1\\\end{bmatrix}& \begin{bmatrix}-1&0\\0&1\\\end{bmatrix}& \begin{bmatrix}Cos(72)&-Sin(72)\\Sin(72)&Cos(72)\\\end{bmatrix}& \begin{bmatrix}Cos(-72)&-Sin(-72)\\Sin(-72)&Cos(-72)\\\end{bmatrix} \end{array}$$

$$ \begin{array}{cccccc}C_{5}^2&C_{5}'^2\\ \begin{bmatrix}Cos(144)&-Sin(144)\\Sin(144)&Cos(144)\\\end{bmatrix}& \begin{bmatrix}Cos(-144)&-Sin(-144)\\Sin(-144)&Cos(-144)\\\end{bmatrix} \end{array} $$

$$ \begin{array}{cccccc}\sigma_{v}'&\sigma_{v}''\\ \begin{bmatrix}-Cos(-144)&Sin(-144)\\Sin(-144)&Cos(-144)\\\end{bmatrix}& \begin{bmatrix}-Cos(72)&Sin(72)\\Sin(72)&Cos(72)\\\end{bmatrix} \end{array} $$

$$ \begin{array}{cccccc}\sigma_{v}'''&\sigma_{v}''''\\ \begin{bmatrix}-Cos(-72)&Sin(-72)\\Sin(-72)&Cos(-72)\\\end{bmatrix}&\begin{bmatrix}-Cos(144)&Sin(144)\\Sin(144)&Cos(144)\\\end{bmatrix} \end{array} $$

Using the labeling scheme below (ignoring the $\psi$1 1s orbital at the top of the pentagonal pyramid) I can also generate the transformation series of each of these basis functions. Let's do $\psi$2 first:

enter image description here

$$ \begin{array}{cccccc}E&\sigma_{v}&C_{5}&C_{5}'&C_{5}^2&C_{5}'^2&\sigma_{v}'&\sigma_{v}''&\sigma_{v}'''&\sigma_{v}''''\\ \begin{bmatrix}\psi2\\\end{bmatrix}&\begin{bmatrix}\psi2\\\end{bmatrix}&\begin{bmatrix}\psi3\\\end{bmatrix}&\begin{bmatrix}\psi6\\\end{bmatrix}&\begin{bmatrix}\psi4\\\end{bmatrix}&\begin{bmatrix}\psi5\\\end{bmatrix}&\begin{bmatrix}\psi3\\\end{bmatrix}&\begin{bmatrix}\psi4\\\end{bmatrix}&\begin{bmatrix}\psi5\\\end{bmatrix}&\begin{bmatrix}\psi6\\\end{bmatrix} \end{array} $$

Now, when I choose to project using the matrix elements (1,1) and the basis set $\psi$2, I get the following:

$$ SALC_{e1}'\propto C(72)(\psi3-\psi4)+C(-72)(\psi6-\psi5)+C(144)(\psi4-\psi6)C(-144)(\psi5-\psi3) $$

$$ \propto \psi3+\psi6-\psi4-\psi5 $$ enter image description here

Where C($\theta$) and S($\theta$) are cosine and sine of theta respectively.

This SALC is obviously transforming as $y$ in our chosen coordinate frame (+y up the page, -y down the page, +x to the right and -x to left).

My confusion starts when I take any other combination of the $\psi$ projections and multiply by any other (row,column) combination of the matrix elements. Intuition tells me that I should either get the complementary SALC which transforms as x, or get the y-like SALC again. This is not true.

For instance, if I choose $\psi3$ as my starting point, and do (row,column) = (1,2), I get:

$$ \begin{array}{cccccc}E&\sigma_{v}&C_{5}&C_{5}'&C_{5}^2&C_{5}'^2&\sigma_{v}'&\sigma_{v}''&\sigma_{v}'''&\sigma_{v}''''\\ \begin{bmatrix}\psi3\\\end{bmatrix}&\begin{bmatrix}\psi6\\\end{bmatrix}&\begin{bmatrix}\psi4\\\end{bmatrix}&\begin{bmatrix}\psi2\\\end{bmatrix}&\begin{bmatrix}\psi5\\\end{bmatrix}&\begin{bmatrix}\psi6\\\end{bmatrix}&\begin{bmatrix}\psi2\\\end{bmatrix}&\begin{bmatrix}\psi3\\\end{bmatrix}&\begin{bmatrix}\psi4\\\end{bmatrix}&\begin{bmatrix}\psi5\\\end{bmatrix} \end{array} $$


$$ SALC_{e1}'\propto S(72)(\psi2+\psi3-2\psi4)+S(144)(\psi6-\psi2) $$

$$ \propto a\psi2+b\psi3-2\psi4+c\psi6 $$


enter image description here

Which does not, at least at first glance, appear to satisfy our requirements of orthogonality, at least at position $\psi$5.

Can anyone indicate what I've done wrong here? Or show where I strayed off the correct path? Because I am quite confused. I would like to stick to the full projection method if at all possible, as I understand that that method should work, meaning I've just messed up somewhere.



You are doing the right thing, but I think it was Bones who said that Evil usually triumphs unless Good is very, very careful. And you just weren't being careful enough. You seem to be going wrong with your geometric interpretation of your vertical reflection matrices. Not surprising since you don't offer an obvious physical description of which is which. I am going to say that $\sigma_{vj}$ is the reflection through the plane containing $\psi_j$. Then it should be obvious that $$D^{(E_1)}(\sigma_{v2})=\begin{bmatrix}-1&0\\0&1\end{bmatrix}$$ To get to $\sigma_{v3}$, we employ a similarity transformation that moves $\psi_3$ into where $\psi_2$ was, then reflects with $\sigma_{v2}$, and then moves $\psi_3$ back from its position where $\psi_2$ was back to where it started, so it will be a reflection that ends up leaving $\psi_3$ invariant: $$\begin{align}D^{(E_1)}(\sigma_{v3})&=D^{(E_1)}(C_5)D^{(E_1)}(\sigma_{v2})D^{(E_1)}(C_5^4)\\ &=\begin{bmatrix}c_1&-s_1\\s_1&c_1\end{bmatrix}\begin{bmatrix}-1&0\\0&1\end{bmatrix}\begin{bmatrix}c_4&-s_4\\s_4&c_4\end{bmatrix}\\ &=\begin{bmatrix}-c_1&-s_1\\-s_1&c_1\end{bmatrix}\begin{bmatrix}c_4&-s_4\\s_4&c_4\end{bmatrix}=\begin{bmatrix}-c_3&s_3\\s_3&c_3\end{bmatrix}\end{align}$$ Where we have adopted the notation that $c_j=\cos(2\pi j/5)$ and $s_j=\sin(2\pi j/5)$. This is your $\sigma_v^{\prime}$ matrix, but you have $\sigma_v^{\prime}$ swapping $\psi_2$ and $\psi_3$, which is what $\sigma_{v5}$ does. Just as a check, note that a vector in the direction of $\psi_3$ should be an eigenvector of $D^{(E_1)}(\sigma_{v3})$ with eigenvalue $1$: $$\begin{bmatrix}-c_3&s_3\\s_3&c_3\end{bmatrix}\begin{bmatrix}-s_1\\c_1\end{bmatrix}=\begin{bmatrix}s_4\\c_4\end{bmatrix}=\begin{bmatrix}-s_1\\c_1\end{bmatrix}$$ So it works! I offer a table as in my previous post: $$\begin{array}{ccccc}E&C_5&C_5^2&C_5^3&C_5^4\\ \begin{bmatrix}1&0\\0&1\end{bmatrix}&\begin{bmatrix}c_1&-s_1\\s_1&c_1\end{bmatrix}&\begin{bmatrix}c_2&-s_2\\s_2&c_2\end{bmatrix}&\begin{bmatrix}c_3&-s_3\\s_3&c_3\end{bmatrix}&\begin{bmatrix}c_4&-s_4\\s_4&c_4\end{bmatrix}\\ \psi_2&\psi_3&\psi_4&\psi_5&\psi_6\\ \psi_3&\psi_4&\psi_5&\psi_6&\psi_2\\ \psi_4&\psi_5&\psi_6&\psi_2&\psi_3\\ \psi_5&\psi_6&\psi_2&\psi_3&\psi_4\\ \psi_6&\psi_2&\psi_3&\psi_4&\psi_5\\ \hline\end{array}$$ $$\begin{array}{ccccc}\sigma_{v2}&\sigma_{v3}&\sigma_{v4}&\sigma_{v5}&\sigma_{v6}\\ \begin{bmatrix}-1&0\\0&1\end{bmatrix}&\begin{bmatrix}-c_3&s_3\\s_3&c_3\end{bmatrix}&\begin{bmatrix}-c_1&s_1\\s_1&c_1\end{bmatrix}&\begin{bmatrix}-c_4&s_4\\s_4&c_4\end{bmatrix}&\begin{bmatrix}-c_2&s_2\\s_2&c_2\end{bmatrix}\\ \psi_2&\psi_4&\psi_6&\psi_3&\psi_5\\ \psi_6&\psi_3&\psi_5&\psi_2&\psi_4\\ \psi_5&\psi_2&\psi_4&\psi_6&\psi_3\\ \psi_4&\psi_6&\psi_3&\psi_5&\psi_2\\ \psi_3&\psi_5&\psi_2&\psi_4&\psi_6\\ \end{array}$$ Using this, we can see that the $(1,1)$ projector annihilates $\psi_2$. You knew something was going wrong here because the $(j,k)$ projector with $k=1$ has to spit out something that transforms as $x$, not $y$. Operating with the same projector on $\psi_3$ I get $$(1-c_3)\psi_3+(c_1-c_2)\psi_4+(c_2-c_1)\psi_5+(c_3-1)\psi_6=-2s_1\left(-s_1\psi_3+s_3\psi_4-s_3\psi_5+s_1\psi_6\right)$$ And this is $x$-like in that $\sigma_{v2}$ turns it into its additive inverse. In this case it seems easiest to get the $y$-like partner by operating on $\psi_3$ with the $(1,2)$ projection/transfer operator and the result I get eventually is $$-2s_1\left(\psi_2+c_1\psi_3+c_3\psi_4+c_3\psi_5+c_1\psi_6\right)$$ You may check for yourself that these transform like an $\{x,y\}$ pair.

  • $\begingroup$ Thank you! And I especially appreciate the color commentary with the Star Trek reference. I will double check for myself. $\endgroup$ – Yajibromine Feb 12 at 0:03

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