MM 219: Lecture 3 - Point Groups & Symmetry

Recap: $S_6$ Operations in Staggered Ethane

In the previous lecture, we established the presence of a six-fold roto-reflection axis ($S_6$) in the staggered conformation of ethane. Let's analyze what happens when we apply this operation successively.

Successive $S_6$ Operations:

$S_6^1 = S_6$
$S_6^2 = (C_6\sigma_h)(C_6\sigma_h) = C_6^2 = \mathbf{C_3}$
$S_6^3 = (C_6\sigma_h)^3 = C_6^3\sigma_h = C_2\sigma_h = \mathbf{S_2} = \mathbf{i}$
$S_6^4 = (C_6\sigma_h)^4 = C_6^4 = \mathbf{C_3^2}$
$S_6^5 = C_6^5\sigma_h$ (Unique operation)
$S_6^6 = (C_6\sigma_h)^6 = C_6^6 = \mathbf{E}$ (Identity)

Drag to Rotate

Mathematical Proof: $S_2 = i$

Let's verify this using transformation matrices. We know that $S_2 = C_2 \sigma_h$. Assuming the $C_2$ rotation axis is along the Z-axis, and $\sigma_h$ is reflection across the XY plane:

Rotation ($C_2^z$)

\[ \begin{bmatrix} \cos 180^\circ & -\sin 180^\circ & 0 \\ \sin 180^\circ & \cos 180^\circ & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \]

$\times$

Reflection ($\sigma_h^{xy}$)

\[ \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{bmatrix} \]

Result: Inversion ($i$)

\[ \begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{bmatrix} = \begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{bmatrix} = \mathbf{i} \]

Group Multiplication Table ($S_6$ Subgroup)

Let's construct the multiplication table for the cyclic subgroup generated by the $S_6$ operation on staggered ethane. Click on any cell with a ? to dynamically compute the result. The animation applies the Row Operation first, followed by the Column Operation to compute the sequence, and compares it with the direct mathematical Result Operation. All rotations are evaluated strictly as Counter-Clockwise (CCW).

2nd \ 1st	$E$	$S_6$	$C_3$	$i$	$C_3^2$	$S_6^5$

Sequence (Row → Col)

Direct Mathematical Result

Current Calculation

Select a cell to begin

PG

Point Groups

Definition

The group of all possible symmetry operations that can be performed on an object, which leaves at least one point in that object unmoved.

To classify as a mathematical group, the set of symmetry operations must satisfy four fundamental rules:

Closure

The product of any two elements is an element in the group.

Inverse

The inverse of every element is an element in the group.

Identity

Presence of an identity element ($E$).

Associativity

$(AB)C = A(BC)$ holds true for all operations.

Derivation of Basic Point Groups

Proper Rotations Only

Group $C_n$

Involves only proper rotations around a single axis.

$C_n = \{E, C_n, C_n^2, \dots, C_n^{n-1}\}$

Example: $C_3$ Order = 3

	$E$	$C_3$	$C_3^2$
$E$	$E$	$C_3$	$C_3^2$
$C_3$	$C_3$	$C_3^2$	$E$
$C_3^2$	$C_3^2$	$E$	$C_3$

Reflections Only

Group $C_s$

Involves only reflection operations ($s$ stands for spiegel, mirror in German).

$C_s = \{E, \sigma\}$

Order = 2

	$E$	$\sigma$
$E$	$E$	$\sigma$
$\sigma$	$\sigma$	$E$

Inversion Only

Group $C_i$

Involves only inversion through a center of symmetry.

$C_i = \{E, i\}$

Order = 2

	$E$	$i$
$E$	$E$	$i$
$i$	$i$	$E$

Improper Rotations ($S_n$) (continued..)

Let's recall some key rules regarding point groups that involve only improper rotations ($S_n$).

When $n$ is Even

$C_{n/2}$ must necessarily be present.
Neither $C_n$ axis nor $\sigma_h$ necessarily exist as standalone symmetry operations (e.g., Staggered Ethane has $C_3$, but $C_6$ and $\sigma_h$ are not standalone elements of its group).
Order of the group: $\mathbf{O(S_n) = n}$

When $n$ is Odd

$C_n$ is the highest order rotation axis present.
Both $C_n$ and $\sigma_h$ exist as independent operations.
Order of the group: $\mathbf{O(S_n) = 2n}$
These are often represented as $\mathbf{C_{nh}}$ (explained later).

When is Inversion ($i$) present?

We know $S_2 = i$. Inversion is present when there exists an element of the type $C_2^m \sigma_h^m$ where $m$ is odd, because:

$C_2^m \sigma_h^m = C_2 \sigma_h = S_2 = i$

Example: $S_6$ Group

$n=6$ (even, not a multiple of 4).
$S_6^3 = C_6^3 \sigma_h^3 = C_2 \sigma_h = i$
$\rightarrow i$ IS present.

Example: $S_8$ Group

$n=8$ (even, multiple of 4).
$S_8^4 = C_8^4 \sigma_h^4 = C_2 \cdot E = C_2$
We never reach a state where $m$ is odd for both $C$ and $\sigma$.
$\rightarrow i$ IS NOT present.

$C_{nh}$

Point Groups $C_{nh}$

Constructed by combining a proper rotation axis ($C_n$) with a reflection plane perpendicular to it ($\sigma_h$).

Construction Requirements:

Should contain Identity ($E$).
Should contain all rotations: $C_n, C_n^2, \dots, C_n^{n-1}$.
Should contain the horizontal plane: $\sigma_h$.
Should contain all possible products of the above elements to satisfy closure.

Analyzing Products:

1. $C_n \times \sigma_h$
We defined $S_n = C_n\sigma_h$.

2. What is $C_n^2 \times \sigma_h$?

$\begin{bmatrix} \cos 2\theta & -\sin 2\theta & 0 \\ \sin 2\theta & \cos 2\theta & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{bmatrix} = \begin{bmatrix} \cos 2\theta & -\sin 2\theta & 0 \\ \sin 2\theta & \cos 2\theta & 0 \\ 0 & 0 & -1 \end{bmatrix}$

This has the form of an $S_n$ matrix. Let's call it $S_n'$. It is an improper rotation.

Generally: The product of any rotation ($C_n^k$) and $\sigma_h$ yields an improper rotation ($S_n^k$, denoted here as $S_n, S_n', S_n''\dots$).

Final Composition of $C_{nh}$

$C_{nh} = \{E, C_n, C_n^2, \dots, C_n^{n-1}, \sigma_h, S_n, S_n', \dots, S_n^{(n-2)'}\}$

Order: $O(C_{nh}) = 2n$

Practice Assessment

Did It Sink In? - 3

Test your mastery of Successive Operations, Roto-reflections, and Point Group Axioms.

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