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Chemistry LibreTexts

1.1: Symmetry Elements and Operations

  • Page ID
    221669
  • Consider the symmetry properties of an object (e.g. atoms of a molecule, set of orbitals, vibrations). The collection of objects is commonly referred to as a basis set

    • classify objects of the basis set into symmetry operations
    • symmetry operations form a group
    • group mathematically defined and manipulated by group theory

    A symmetry operation moves an object into an indistinguishable orientation

    A symmetry element is a point, line or plane about which a symmetry operation is performed

    There are five symmetry elements, which will be defined relative to point with coordinate (x1, y1, z1) :

    1. identity, E

    \[ E\left(x_{1}, y_{1}, z_{1}\right)=\left(x_{1}, y_{1}, z_{1}\right) \]

    2. plane of reflection, σ

    3. inversion, i

    \[\mathrm{i}\left(\mathrm{x}_{1}, \mathrm{y}_{1}, \mathrm{z}_{1}\right)=\left(-\mathrm{x}_{1},-\mathrm{y}_{1},-\mathrm{z}_{1}\right)\]

    4. proper rotation axis, Cn \( \text { (where }\left.\theta=\frac{2 \pi}{n}\right) \)

    convention is a clockwise rotation of the point

    \[ \mathrm{C}_{2}(\mathrm{z})\left(\mathrm{x}_{1}, \mathrm{y}_{1}, \mathrm{z}_{1}\right)=\left(-\mathrm{x}_{1},-\mathrm{y}_{1}, \mathrm{z}_{1}\right) \]

    5. improper rotation axis, Sn

    two step operation: Cn followed by σ through plane ⊥ to Cn

    \[\mathrm{S}_{4}(\mathrm{z})\left(\mathrm{x}_{1}, \mathrm{y}_{1}, \mathrm{z}_{1}\right)=\sigma(\mathrm{xy}) \mathrm{C}_{4}(\mathrm{z})\left(\mathrm{x}_{1}, \mathrm{y}_{1}, \mathrm{z}_{1}\right)=\sigma(\mathrm{xy})\left(\mathrm{y}_{1},-\mathrm{x}_{1}, \mathrm{z}_{1}\right)=\left(\mathrm{y}_{1},-\mathrm{x}_{1}-\mathrm{z}_{1}\right) \]

    Note: rotation of pt is clockwise; Corollary is that axes rotate counterclockwise relative to fixed point

    In the example above, we took the direct product of two operators:

    \( \sigma_{\mathrm{h}} \cdot \mathrm{C}_{\mathrm{n}}=\mathrm{S}_{\mathrm{n}} \)

    clipboard_e7bc5cd244d43be307fe114bd54a923ee.png

    Horizontal mirror plane (normal to Cn)

    \(\text { for } n \text { even }: S_{n}^{n}=C_{n}^{n} \cdot \sigma_{h}^{n}=E \cdot E=E \)

    \(\begin{aligned}
    \text { for } n \text { odd: } & S_{n}^{n}=C_{n}^{n} \cdot \sigma_{h}^{n}=E \cdot \sigma_{h}=\sigma_{h} \\
    & S_{n}^{2 n}=C_{n}^{2 n} \cdot \sigma_{h}^{2 n}=E \cdot E=\sigma_{h}
    \end{aligned} \)

    \( \text { for } \mathrm{m} \text { even: } \mathrm{S}_{\mathrm{n}}^{\mathrm{m}}=\mathrm{C}_{\mathrm{n}}^{\mathrm{m}} \cdot \sigma_{\mathrm{h}}^{\mathrm{m}}=\mathrm{C}_{\mathrm{n}}^{\mathrm{m}} \)

    \( \text { for } m \text { odd: } \quad S_{n}^{m}=C_{n}^{m} \cdot \sigma_{h}^{m}=C_{n}^{m} \cdot \sigma_{h}=S_{n}^{m} \)

    Symmetry operations may be represented as matrices. Consider the vector \( \overline{\mathbf{v}} \)

    1. identity: \( E\begin{bmatrix}
    x_1 \\
    y_1 \\
    z_1
    \end{bmatrix}
    =
    \begin{bmatrix}
    & & \\
    & ? & \\
    & &
    \end{bmatrix}
    \begin{bmatrix}
    x_1 \\
    y_1 \\
    z_1
    \end{bmatrix}
    =
    \begin{bmatrix}
    x_1 \\
    y_1 \\
    z_1
    \end{bmatrix}\)

    matrix satisfying this condition is:

    \( \left[\begin{array}{lll}
    1 & 0 & 0 \\
    0 & 1 & 0 \\
    0 & 0 & 1
    \end{array}\right] \)

    \(\therefore \mathrm{E}=\left[\begin{array}{lll}
    1 & 0 & 0 \\
    0 & 1 & 0 \\
    0 & 0 & 1
    \end{array}\right] \)... E is always the unit matrix

    2. reflection: \( \sigma(\mathrm{xy})\left[\begin{array}{l}
    \mathrm{x}_{1} \\
    \mathrm{y}_{1} \\
    \mathrm{z}_{1}
    \end{array}\right]=\left[\begin{array}{r}
    \mathrm{x}_{1} \\
    \mathrm{y}_{1} \\
    -\mathrm{z}_{1}
    \end{array}\right] \quad \therefore \sigma(\mathrm{xy})=\left[\begin{array}{rrr}
    1 & 0 & 0 \\
    0 & 1 & 0 \\
    0 & 0 & -1
    \end{array}\right] \)

    similarly \( \text { similarly } \sigma(\mathrm{xz})=\left[\begin{array}{rrr}
    1 & 0 & 0 \\
    0 & -1 & 0 \\
    0 & 0 & 1
    \end{array}\right] \text { and } \sigma(\mathrm{yz})=\left[\begin{array}{rrr}
    -1 & 0 & 0 \\
    0 & 1 & 0 \\
    0 & 0 & 1
    \end{array}\right] \)

    3. inversion: \( \mathrm{i}\left[\begin{array}{l}
    \mathrm{x}_{1} \\
    \mathrm{y}_{1} \\
    \mathrm{z}_{1}
    \end{array}\right]=\left[\begin{array}{l}
    -\mathrm{x}_{1} \\
    -\mathrm{y}_{1} \\
    -\mathrm{z}_{1}
    \end{array}\right] \quad \therefore \quad \mathrm{i}=\left[\begin{array}{rrr}
    -1 & 0 & 0 \\
    0 & -1 & 0 \\
    0 & 0 & -1
    \end{array}\right] \)

    4. proper rotation axis:

    because of convention, φ, and hence zi, is not transformed under Cn(θ) ∴ projection into xy plane need only be considered… i.e., rotation of vector v(xi,yi) through θ

    \( x_{1}=\bar{v} \cos \alpha \)

    \( y_{1}=\vec{v} \sin \alpha \)

    \( {C}_{n}(\theta) \)

    \( \mathrm{x}_{2}=\overline{\mathrm{v}} \cos [-(\theta-\alpha)]=\overline{\mathrm{v}} \cos (\theta-\alpha) \)

    \( y_{2}=\vec{v} \sin [-(\theta-\alpha)]=-\bar{v} \sin (\theta-\alpha) \)

    using identity relations:

    \(x_{2}=\vec{v} \cos (\theta-\alpha)=\vec{v} \cos \theta \cos \alpha+\vec{v} \sin \theta \sin \alpha=x_{1} \cos \theta+y_{1} \sin \theta \)

    \( y_{2}=-\bar{v} \sin (\theta-\alpha)=-[\bar{v} \sin \theta \cos \alpha-\bar{v} \cos \theta \sin \alpha]=-x_{1} \sin \theta+y_{1} \cos \theta \)

    Reformulating in terms of matrix representation:

    \( \mathrm{C}_{\mathrm{n}}(\theta)\left[\begin{array}{l}
    \mathrm{x}_{1} \\
    \mathrm{y}_{1} \\
    \mathrm{z}_{1}
    \end{array}\right]=\left[\begin{array}{r}
    \mathrm{x}_{1} \cos \theta+\mathrm{y}_{1} \sin \theta \\
    -\mathrm{x}_{1} \sin \theta+\mathrm{y}_{1} \cos \theta \\
    \mathrm {z}_{1}
    \end{array}\right] \)

    \( \therefore \mathrm{C}_{\mathrm{n}}(\theta)=\left[\begin{array}{rrr}
    \cos \theta & \sin \theta & 0 \\
    -\sin \theta & \cos \theta & 0 \\
    0 & 0 & 1
    \end{array}\right] \quad \text { where } \theta=\frac{2 \pi}{\mathrm{n}} \)

    Note… the rotation above is clockwise, as discussed by HB (pg 39). Cotton on pg. 73 solves for the counterclockwise rotation… and presents the clockwise result derived above. To be consistent with HB (and math classes) we will rotate clockwise as the convention.

    The above matrix representation is completely general for any rotation θ…

    Example: \(C_{3}, \theta=\frac{2 \pi}{n} \)

    \( C_{3}=\left[\begin{array}{ccc}
    \cos \frac{2 \pi}{3} & \sin \frac{2 \pi}{3} & 0 \\
    -\sin \frac{2 \pi}{3} & \cos \frac{2 \pi}{3} & 0 \\
    0 & 0 & 1
    \end{array}\right]=\left[\begin{array}{ccc}
    -\frac{1}{2} & \frac{\sqrt{3}}{2} & 0 \\
    -\frac{\sqrt{3}}{2} & -\frac{1}{2} & 0 \\
    0 & 0 & 1
    \end{array}\right] \)

    5. improper rotation axis :

    σ h ⋅ Cn(θ) = Sn(θ)

    \( \left[\begin{array}{rrr}
    1 & 0 & 0 \\
    0 & 1 & 0 \\
    0 & 0 & -1
    \end{array}\right] \cdot\left[\begin{array}{rrr}
    \cos \theta & \sin \theta & 0 \\
    -\sin \theta & \cos \theta & 0 \\
    0 & 0 & 1
    \end{array}\right]=\left[\begin{array}{rrr}
    \cos \theta & \sin \theta & 0 \\
    -\sin \theta & \cos \theta & 0 \\
    0 & 0 & -1
    \end{array}\right] \)

    Like operators themselves, matrix operations may be manipulated with simple matrix algebra…above direct product yields matrix representation for Sn.

    Another example:

    \( \left[\begin{array}{rrr}
    1 & 0 & 0 \\
    0 & 1 & 0 \\
    0 & 0 & -1
    \end{array}\right] \cdot\left[\begin{array}{rrr}
    -1 & 0 & 0 \\
    0 & -1 & 0 \\
    0 & 0 & 1
    \end{array}\right]=\left[\begin{array}{rrr}
    -1 & 0 & 0 \\
    0 & -1 & 0 \\
    0 & 0 & -1
    \end{array}\right] \)

    σxy (≡ σh ) ⋅ C2 (z) = i

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