Chemical Forums
Chemistry Forums for Students => Inorganic Chemistry Forum => Topic started by: nonscience_guy on February 13, 2020, 03:17:08 PM
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Hey!
I am a college student working on an app to help chemistry students get the solution to a point group. Similar to the webpage, http://gernot-katzers-spice-pages.com/character_tables/index.html .
I am currently having trouble get the solution for certain group points such as: c3, c4, c5, and c6.
When I go about inputting [24, 0] for c3 in my app it gives me the value of 8A + 16E, but the correct answer is 8A + 8E.
I use the given table in http://gernot-katzers-spice-pages.com/character_tables/C3.html , to calculate the answer. This is how I go about solving it.
= coefficient * table value * input
A = 1 * 1 * 24 + 2 * 1 * 0 = 24 / h = 24 / 3 = 8 A
E = 1 * 2 * 24 + 2 * -1 * 0 = 48 /h = 48 / 3 = 16 E
I was wondering if anyone here could possibly show me a step by step solution on how to get the correct answer?
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Read the footnotes on the webpage you linked to. These groups are a little unusual:
"The single “E” representation is reducible but almost behaves like a true irreducible representation. Its norm, however, is twice the group order. Therefore, is has been marked with an asterisk in the table. This is essential when trying to decompose a reducible representation into “irreducible” ones using the familiar projection formula."
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Could you please elaborate a bit more?
Could you show a step by step solution using the projection formula?
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The two dimensional representation in the C3 character table is actually reducible, but you must use complex numbers. Because complex numbers are generally undesirable in chemistry applications, the two dimensional representation is usually treated as basically irreducible and often expressed by combining the complex numbers with their complex conjugates to form a "real valued" representation.
The C3 character table is rigorously expressed as:
[tex]
\begin{array}{c|ccc}
C_4&\hat{E}&\hat{C_3}&\hat{C_3}^2 \\ \hline
A&1&1&1\\
E&\lbrace\begin{array}{l}1\\1\end{array}&\begin{array}{l}\epsilon\\ \epsilon^*\end{array}&\begin{array}{l}\epsilon^*\\\epsilon\end{array}\rbrace \\ \end{array}
[/tex]
Where ε = exp(2πi/3).
Because these representations are actually reducible, you run into some issues when trying to use the various formulas and rules for manipulating character tables, including the reduction formula, that are intended for use with irreducible representations. You can go through and derive the results long hand, but if you replace h by 2h in the reduction formula it will give you the right answer when working with these pseudo-reducible 2-dimensional representations.