2.5: Rotations in a Lattice

In a single molecule, the elements of every rotational symmetry intersect a common point called the origin. And, these operations are represented by matrices using Cartesian coordinates as a basis. In a crystal, on the other hand, a motif repeats periodically throughout real space, so that symmetry elements intersect at many different points. In addition, the unit cell vectors of the lattice serve as the basis of a coordinate system, but these vectors need not be all mutually orthogonal. Therefore, it is important to learn how to express rotations by matrices using a lattice as the basis as well as when rotation axes do not intersect the designated origin.

In 3-d space rotations are represented by 3×3 orthogonal matrices, but they can act either actively or passively. Active rotations move the object while keeping the coordinate system fixed, whereas passive rotations move the coordinate system and keep the object fixed. Now, an object in a crystal is a set of points u specified by fractional coordinates \(x,y,z\) with respect to the unit cell lattice vectors a, b, c. Therefore, using matrix notation, every point \(\mathbf{u}\) in the crystal can be expressed as:

\[\boldsymbol{u} = x\boldsymbol{a} + y\boldsymbol{b} + z\boldsymbol{c} = \begin{pmatrix} \boldsymbol{a} & \boldsymbol{b} & \boldsymbol{c} \\ \end{pmatrix}\begin{pmatrix} x \\ y \\ z \\ \end{pmatrix} = \begin{pmatrix} x & y & z \\ \end{pmatrix}\begin{pmatrix} \boldsymbol{a} \\ \boldsymbol{b} \\ \boldsymbol{c} \\ \end{pmatrix}. \nonumber \]

The second part of this equation shows two different ways to represent the coordinates and basis vectors using matrices. These different representations are important for distinguishing active and passive rotations. Any rotation \(R_{\text{Axis}}\) transforms the point \(\boldsymbol{u}\) to the point \(R_{\text{Axis}}\boldsymbol{u}\) as follows:

\[ R_{\text{Axis}}\boldsymbol{u} = \begin{pmatrix} \boldsymbol{a} & \boldsymbol{b} & \boldsymbol{c} \\ \end{pmatrix}R_{\text{Axis}}\begin{pmatrix} x \\ y \\ z \\ \end{pmatrix} = \begin{pmatrix} \boldsymbol{a} & \boldsymbol{b} & \boldsymbol{c} \\ \end{pmatrix}\begin{pmatrix} R_{11} & R_{12} & R_{13} \\ R_{21} & R_{22} & R_{23} \\ R_{31} & R_{32} & R_{33} \\ \end{pmatrix}\begin{pmatrix} x \\ y \\ z \\ \end{pmatrix} \nonumber \]

\[ = \left( R_{11}x + R_{12}y + R_{13}z \right)\boldsymbol{a} + \left( R_{21}x + R_{22}y + R_{23}z \right)\boldsymbol{b} + \left( R_{31}x + R_{32}y + R_{33}z \right)\boldsymbol{c}. \nonumber \]

The matrix for \(R_{\text{Axis}}\) is the active mode representative because the matrix transforms the coordinates. To determine the passive mode representative, rearrange the second equation to factor out the coordinates \(x,y,z\):

\[R_{\text{Axis}}\boldsymbol{u} = x\left( R_{11}\boldsymbol{a} + R_{21}\boldsymbol{b} + R_{31}\boldsymbol{c} \right) + y\left( R_{12}\boldsymbol{a} + R_{22}\boldsymbol{b} + R_{32}\boldsymbol{c} \right) + z\left( R_{13}\boldsymbol{a} + R_{23}\boldsymbol{b} + R_{33}\boldsymbol{c} \right) \nonumber \]

\[ = \begin{pmatrix} x & y & z \\ \end{pmatrix}\begin{pmatrix} R_{11} & R_{21} & R_{31} \\ R_{12} & R_{22} & R_{32} \\ R_{13} & R_{23} & R_{33} \\ \end{pmatrix}\begin{pmatrix} \boldsymbol{a} \\ \boldsymbol{b} \\ \boldsymbol{c} \\ \end{pmatrix} = \begin{pmatrix} x & y & z \\ \end{pmatrix}R_{\text{Axis}}^{T}\begin{pmatrix} \boldsymbol{a} \\ \boldsymbol{b} \\ \boldsymbol{c} \\ \end{pmatrix}, \nonumber \]

in which the matrix \(R_{\text{Axis}}^{T}\) transforms the lattice. Therefore, if the matrix representative for the active mode rotation is \(R_{\text{Axis}}\), then the transpose² \(R_{\text{Axis}}^{T}\) is the matrix representative for the passive mode rotation. Since rotation matrices are orthogonal matrices, then the transpose is also the inverse. As a result, an active counterclockwise (ccw) rotation is identical to a passive clockwise (cw) rotation by the same rotation angle and axis.

According to the compatibility between rotational and translational symmetry, the allowed rotations for a lattice lead to rotation matrices with only integer matrix elements. Rotation matrices for the passive mode can be derived using geometrical relationships among the unit cell lattice vectors. The corresponding matrices for the active mode arise by taking the transposes. For example, consider a tetragonal lattice with the fourfold rotation axis directed parallel to \(\mathbf{c}\), i.e., \(C_{4\boldsymbol{c}} = 4_{\boldsymbol{c}}\). The matrix for \(4_{\boldsymbol{c}}\) can be determined by determining how this operation affects each lattice vector: \(4_{\boldsymbol{c}}\boldsymbol{a} = \boldsymbol{b}\), \(4_{\boldsymbol{c}}\boldsymbol{b} = 4_{\boldsymbol{c}}\left( 4_{\boldsymbol{c}}\boldsymbol{a} \right) = 4_{\boldsymbol{c}}^{2}\boldsymbol{a} = 2_{\boldsymbol{c}}\boldsymbol{a} = -\boldsymbol{a}\), and \(4_{\boldsymbol{c}}\boldsymbol{c} = \boldsymbol{c}\). From these results, we can construct the passive mode matrix for \(4_{\boldsymbol{c}}\):

\[4_{\boldsymbol{c}}\begin{pmatrix} \boldsymbol{a} \\ \boldsymbol{b} \\ \boldsymbol{c} \\ \end{pmatrix} = \begin{pmatrix} \boldsymbol{b} \\ -\boldsymbol{a} \\ \boldsymbol{c} \\ \end{pmatrix} = \begin{pmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix}\begin{pmatrix} \boldsymbol{a} \\ \boldsymbol{b} \\ \boldsymbol{c} \\ \end{pmatrix} \nonumber\]

so that

\[4_{\boldsymbol{c}}\text{(Passive)} = \begin{pmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix}. \nonumber \]

Using the transpose relationship between the active and passive mode rotation matrices, the active mode matrix for \(4_{\boldsymbol{c}}\) is:

\[4_{\boldsymbol{c}}\text{(Active)} = \left\lbrack 4_{\boldsymbol{c}}\text{(Passive)} \right\rbrack^{T} = \small{\begin{pmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix}^{T} = \begin{pmatrix} 0 & - 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix}} \nonumber \]

, so that

\[4_{\boldsymbol{c}}\small{\begin{pmatrix} x \\ y \\ z \\ \end{pmatrix} = \begin{pmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix}\begin{pmatrix} x \\ y \\ z \\ \end{pmatrix} = \begin{pmatrix} -y \\ x \\ z \\ \end{pmatrix}} \nonumber \]

The figures below illustrate the difference between the active and passive modes for the 90º rotation associated with a tetragonal lattice.

Figure 2.21a: Three lattices in a line: (Middle) square molecule at lattice point of tetragonal lattice.(Right) Active 90 degree ccw rotation of the molecule while lattice remains fixed.(Left) Passive 90 degree cw rotation of lattice while the molecule remains fixed.

The origin point of the lattice is designated by the red square, which symbolizes tetragonal symmetry through each lattice point. Also, the square planar molecule identifies a “pattern” that must be replicated at each lattice point (only one molecule is illustrated). One of the ligands is highlighted in blue and corresponds to the point (vector) \(\boldsymbol{u}\) relative to the origin. The inverse relationship between the active and passive mode rotations is evident, and the description of the transformed point \(4_{\boldsymbol{c}}\boldsymbol{u}\) is identical for both cases:

Active Mode:

\[4_{\boldsymbol{c}}\boldsymbol{u} = \begin{pmatrix} \boldsymbol{a} & \boldsymbol{b} & \boldsymbol{c} \\ \end{pmatrix}\begin{pmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix}\begin{pmatrix} x \\ y \\ z \\ \end{pmatrix} = \begin{pmatrix} \boldsymbol{a} & \boldsymbol{b} & \boldsymbol{c} \\ \end{pmatrix}\begin{pmatrix} -y \\ x \\ z \\ \end{pmatrix} = - y\boldsymbol{a} + x\boldsymbol{b} + z\boldsymbol{c} \nonumber \]

Passive Mode:

 \[4_{\boldsymbol{c}}\boldsymbol{u} =  \begin{pmatrix} x & y & z \\ \end{pmatrix}\begin{pmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix}\begin{pmatrix} \boldsymbol{a} \\ \boldsymbol{b} \\ \boldsymbol{c} \\ \end{pmatrix} = \begin{pmatrix} x & y & z \\ \end{pmatrix}\begin{pmatrix} \boldsymbol{b} \\ - \boldsymbol{a} \\ \boldsymbol{c} \\ \end{pmatrix} = - y\boldsymbol{a} + x\boldsymbol{b} + z\boldsymbol{c}. \nonumber \]

Rotations with axes away from the origin \(R_{\text{Axis}}(\boldsymbol{\tau})\)

In every lattice, there are points \(\boldsymbol{\tau}\) other than lattice points [mnp] where rotational symmetry exists. So, where are these points and how are the corresponding rotations expressed as matrices? Since rotation matrices with respect to a designated origin are known, then the general procedure using the active mode involves the following three steps:

1. Translate the entire structure from point \(\boldsymbol{\tau}\) to the origin 0, which is denoted by \(T(-\boldsymbol{\tau})\);
2. Perform the rotation \(R_{\text{Axis}}\) because its axis now intersects the origin, i.e., \(R_{\text{Axis}}(\boldsymbol{0})\);
3. Translate the entire structure from the origin 0 back to point \(\boldsymbol{\tau}\), denoted as \(T(+\boldsymbol{\tau})\).

As an example, consider a fourfold rotation about the unit cell center of the tetragonal lattice:

Figure 2.21b: Two ways for 90-degree ccw rotation of four square molecules (structure) about center of square unit cell. (Upper left) Starting point; (Lower left) Displacement of structure from unit cell center to unit cell edge; (Lower right) 90-degree ccw rotation of the structure about the origin; (Upper right) Displacement of structure from unit cell edge back to unit cell center.One atom of one square molecule is highlighted (blue) to follow the movements. 

By following the motion of the position \(\boldsymbol{u}\), notice that the single fourfold ccw rotation about the axis at the center of the unit cell \(\boldsymbol{\tau}\), symbolized by \(4_{\boldsymbol{c}}(\boldsymbol{\tau})\), involves a rotation about the origin followed by a lattice translation. The same outcome arises by following the three-step procedure. This result is written according to the equation

\[4_{\boldsymbol{c}}(\boldsymbol{\tau})\ \boldsymbol{u} = T(+\boldsymbol{\tau})\ 4_{\boldsymbol{c}}(\boldsymbol{0})\ T(-\boldsymbol{\tau})\ \boldsymbol{u} \nonumber \]

The operations of the three-step procedure are written right-to-left because of their sequential effect on the position \(\boldsymbol{u}\). Since a translation by \(-\boldsymbol{\tau}\) is the inverse of a translation by \(+\boldsymbol{\tau}\), then the general expression for a rotation \(R_{\text{Axis}}(\boldsymbol{\tau})\) is

\[R_{\text{Axis}}(\boldsymbol{\tau}) = T(+\boldsymbol{\tau})\ R_{\text{Axis}}(\boldsymbol{0})\ T(-\boldsymbol{\tau}) = T(\boldsymbol{\tau})\ R_{\text{Axis}}(\boldsymbol{0})\ T^{-1}(\boldsymbol{\tau}). \nonumber \]

Therefore, \(R_{\text{Axis}}(\boldsymbol{\tau})\) and \(R_{\text{Axis}}(\boldsymbol{0})\) are related to each other by a similarity transformation in Euclidean space. However, the two operations are not related by a similarity transformation of the lattice unless \(\boldsymbol{\tau}\) is a lattice vector. Now, since \(R_{\text{Axis}}(\boldsymbol{\tau})\) is the product of three operations, two of which are translations, how are these operations represented by matrices?

The Augmented Matrix Representation

Any rotation \(R_{\text{Axis}}\) with its axis intersecting the origin can be written as the 3×3 orthogonal matrix. However, a 3×3 matrix cannot properly represent a displacement \(T(\boldsymbol{\tau})\). To accomplish this, we must use a 4×4 matrix called an augmented matrix. Because a rotation \(R_{\text{Axis}}(\boldsymbol{\tau})\) with its axis intersecting the point \(\boldsymbol{\tau}\) is identical to the same rotation about the origin \(R_{\text{Axis}}\) followed by a lattice translation \(\boldsymbol{T}_{mnp}(\boldsymbol{\tau})\), which depends on the location of \(\boldsymbol{\tau}\), the augmented matrix for \(R_{\text{Axis}}(\boldsymbol{\tau})\) is

\[R_{\text{Axis}}(\boldsymbol{\tau}) =\begin{pmatrix} R_{\text{Axis}} & \boldsymbol{T}_{mnp}(\boldsymbol{\tau}) \\ 0 & 1 \\ \end{pmatrix}=\small{\begin{pmatrix} \begin{matrix} R_{11} & R_{12} \\ R_{21} & R_{22} \\ \end{matrix} & \begin{matrix} R_{13} & m(\boldsymbol{\tau}) \\ R_{23} & n(\boldsymbol{\tau}) \\ \end{matrix} \\ \begin{matrix} R_{31} & R_{32} \\ 0 & 0 \\ \end{matrix} & \begin{matrix} R_{33} & p(\boldsymbol{\tau}) \\ 0 & 1 \\ \end{matrix} \\ \end{pmatrix}}. \nonumber \]

This matrix transforms a point \(\boldsymbol{u}\) to \(R_{\text{Axis}}(\boldsymbol{\tau})\ \boldsymbol{u}\) as follows:

\[R_{\text{Axis}}(\boldsymbol{\tau})\boldsymbol{u}= \begin{pmatrix} \begin{matrix} R_{11} & R_{12} \\ R_{21} & R_{22} \\ \end{matrix} & \begin{matrix} R_{13} & m(\boldsymbol{\tau}) \\ R_{23} & n(\boldsymbol{\tau}) \\ \end{matrix} \\ \begin{matrix} R_{31} & R_{32} \\ 0 & 0 \\ \end{matrix} & \begin{matrix} R_{33} & p(\boldsymbol{\tau}) \\ 0 & 1 \\ \end{matrix} \\ \end{pmatrix}\begin{pmatrix} x \\ y \\ z \\ 1 \\ \end{pmatrix}\\ = \begin{pmatrix} \begin{matrix} R_{11}x + R_{12}y + R_{13}z + m(\boldsymbol{\tau}) \\ R_{21}x + R_{22}y + R_{23}z + n(\boldsymbol{\tau}) \\ \end{matrix} \\ \begin{matrix} R_{31}x + R_{32}y + R_{33}z + p(\boldsymbol{\tau}) \\ 1 \\ \end{matrix} \\ \end{pmatrix} \\ = R_{\text{Axis}}\boldsymbol{u} + \boldsymbol{T}_{mnp}(\boldsymbol{\tau}) \nonumber \]

To use an augmented matrix, the position \(\boldsymbol{u}\) must be expressed as a 4×1 column matrix with the 4^th component being 1. The augmented matrices for a pure translation by vector \(\boldsymbol{\tau}\), \(T(\boldsymbol{\tau})\), and a pure rotation with respect to the origin \(R_{\text{Axis}}(\boldsymbol{0})\) are

Translation \(T(\boldsymbol{\tau})\):

\[T(\boldsymbol{\tau})\boldsymbol{u} = \begin{pmatrix} 1 & \boldsymbol{\tau} \\ 0 & 1 \\ \end{pmatrix}\begin{pmatrix} \boldsymbol{u} \\ 1 \\ \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 & \tau_{1} \\ 0 & 1 & 0 & \tau_{2} \\ 0 & 0 & 1 & \tau_{3} \\ 0 & 0 & 0 & 1 \\ \end{pmatrix}\begin{pmatrix} x \\ y \\ z \\ 1 \\ \end{pmatrix} = \begin{pmatrix} x + \tau_{1} \\ y + \tau_{2} \\ z + \tau_{3} \\ 1 \\ \end{pmatrix} \nonumber \]

\[T(\boldsymbol{\tau})\boldsymbol{u} = \boldsymbol{u} + \boldsymbol{\tau} \nonumber \]

Rotation \(R_{\text{Axis}}(\boldsymbol{0})\):

\[R_{\text{Axis}}(\boldsymbol{0})\boldsymbol{u} = \begin{pmatrix} R_{\text{Axis}} & \boldsymbol{0} \\ 0 & 1 \\ \end{pmatrix}\begin{pmatrix} \boldsymbol{u} \\ 1 \\ \end{pmatrix} = \small{\begin{pmatrix} R_{11} & R_{12} & R_{13} & 0 \\ R_{21} & R_{22} & R_{23} & 0 \\ R_{31} & R_{32} & R_{33} & 0 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix}\begin{pmatrix} x \\ y \\ z \\ 1 \\ \end{pmatrix} = \begin{pmatrix} R_{11}x + R_{12}y + R_{13}z \\ R_{21}x + R_{22}y + R_{23}z \\ R_{31}x + R_{32}y + R_{33}z \\ 1 \\ \end{pmatrix}} \nonumber \]

\[ = R_{\text{Axis}}\boldsymbol{u} + \boldsymbol{\tau}. \nonumber \]

Using augmented matrices, the 90º ccw rotation about the c-axis through the point \(\boldsymbol{\tau} =\) \(\small{\begin{pmatrix} ½ \\ ½ \\ 0 \\ \end{pmatrix}}\) of a tetragonal lattice is:

\[\begin{alignat*}{1} 4_{\boldsymbol{c}}( \boldsymbol{\tau} ) &= T( + \boldsymbol{\tau} )\ 4_{\boldsymbol{c}}\ T( - \boldsymbol{\tau} ) \\ &= \begin{pmatrix} 1 & 0 & 0 & ½ \\ 0 & 1 & 0 & ½ \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix}\begin{pmatrix} 0 & - 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix}\begin{pmatrix} 1 & 0 & 0 & - ½ \\ 0 & 1 & 0 & - ½ \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix} \\ &= \begin{pmatrix} 0 & - 1 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix}\\ \end{alignat*} \nonumber \]

so that

\[4_{\boldsymbol{c}}( \boldsymbol{\tau} )\boldsymbol{u =}\begin{pmatrix} 0 & - 1 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix}\begin{pmatrix} x \\ y \\ z \\ 1 \\ \end{pmatrix} = \begin{pmatrix} - y + 1 \\ - x + 0 \\ - z + 0 \\ 1 \\ \end{pmatrix} = 4_{\boldsymbol{c}}\boldsymbol{u} + \boldsymbol{a} \nonumber \]

Therefore, this operation is a 90º ccw rotation about the c-axis through the origin followed by a lattice translation of one step along the a-direction. We can arrive at the same result by considering the successive operations:

\[4_{\boldsymbol{c}}(\boldsymbol{\tau})\boldsymbol{u}=T(+\boldsymbol{\tau})\ 4_{\boldsymbol{c}}\ T(-\boldsymbol{\tau}) \boldsymbol{u} = T(+\boldsymbol{\tau})\ 4_{\boldsymbol{c}}( \boldsymbol{u}-\boldsymbol{\tau}) = T(+\boldsymbol{\tau})( 4_{\boldsymbol{c}}\boldsymbol{u}-4_{\boldsymbol{c}}\boldsymbol{\tau}) = 4_{\boldsymbol{c}}\boldsymbol{u}+(\boldsymbol{\tau}-4_{\boldsymbol{c}}\boldsymbol{\tau}) \nonumber \]

\[\boldsymbol{\tau}\boldsymbol{-}4_{\boldsymbol{c}}\boldsymbol{\tau} =\) \(\small{\begin{pmatrix} ½ \\ ½ \\ 0 \\ \end{pmatrix} - \begin{pmatrix} 0 & - 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix}\begin{pmatrix} ½ \\ ½ \\ 0 \\ \end{pmatrix} = \begin{pmatrix} ½ \\ ½ \\ 0 \\ \end{pmatrix} - \begin{pmatrix} - ½ \\ ½ \\ 0 \\ \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ 0 \\ \end{pmatrix}}\) \(= \boldsymbol{a} \nonumber \].

Seitz Notation

The augmented matrix for the operation \(R_{\text{Axis}}( \boldsymbol{\tau})\) can be written concisely using the Seitz notation:

\[R_{\text{Axis}}\left( \boldsymbol{\tau} \right) \equiv \left( R_{\text{Axis}} \middle| \boldsymbol{\ }\boldsymbol{T}_{mnp}\left( \boldsymbol{\tau} \right) \right) = \begin{pmatrix} R_{\text{Axis}} & \boldsymbol{T}_{mnp}\left( \boldsymbol{\tau} \right) \\ 0 & 1 \\ \end{pmatrix} \nonumber \]

so that

\[R_{\text{Axis}}\left( \boldsymbol{\tau} \right)\boldsymbol{u} \equiv \left( R_{\text{Axis}} \middle| \boldsymbol{\ }\boldsymbol{T}_{mnp}\left( \boldsymbol{\tau} \right) \right)\boldsymbol{u} = R_{\text{Axis}}\boldsymbol{u} + \boldsymbol{T}_{mnp}\left( \boldsymbol{\tau} \right). \nonumber \]

Example \(\PageIndex{1}\)

Use the Seitz notation and augmented matrices for the operation \(4_{\boldsymbol{c}}\left( \boldsymbol{\tau} \right)\) at \(\boldsymbol{\tau} =\) \(\small{\begin{pmatrix} ½ \\ ½ \\ 0 \\ \end{pmatrix}}\) of a tetragonal lattice to follow its progression four consecutive times.

Solution

We determined that \(4_{\boldsymbol{c}}\left( \boldsymbol{\tau} \right) = \left( 4_{\boldsymbol{c}} \middle| \boldsymbol{a} \right)\). Therefore,

\(\left( 4_{\boldsymbol{c}} \middle| \boldsymbol{a} \right)\boldsymbol{u} = 4_{\boldsymbol{c}}\boldsymbol{u} + \boldsymbol{a = u'}\)

\(\left( 4_{\boldsymbol{c}} \middle| \boldsymbol{a} \right)\boldsymbol{u} =\) \(\small{\begin{pmatrix} 0 & - 1 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix}\begin{pmatrix} x \\ y \\ z \\ 1 \\ \end{pmatrix} = \begin{pmatrix} - y + 1 \\ - x + 0 \\ - z + 0 \\ 1 \\ \end{pmatrix}}\)

\(\left( 4_{\boldsymbol{c}} \middle| \boldsymbol{a} \right)^{2}\boldsymbol{u} = \left( 4_{\boldsymbol{c}} \middle| \boldsymbol{a} \right)\left\lbrack 4_{\boldsymbol{c}}\boldsymbol{u} + \boldsymbol{a} \right\rbrack = 4_{\boldsymbol{c}}\left\lbrack 4_{\boldsymbol{c}}\boldsymbol{u} + \boldsymbol{a} \right\rbrack + \boldsymbol{a} = 4_{\boldsymbol{c}}^{2}\boldsymbol{u} + 4_{\boldsymbol{c}}\boldsymbol{a} + \boldsymbol{a} = 2_{\boldsymbol{c}}\boldsymbol{u} + \boldsymbol{b} + \boldsymbol{a}\)

\(\left( 4_{\boldsymbol{c}} \middle| \boldsymbol{a} \right)^{2}\boldsymbol{u} = \left( 2_{\boldsymbol{c}} \middle| \boldsymbol{a} + \boldsymbol{b} \right)\boldsymbol{u = u''}\)

\(\left( 4_{\boldsymbol{c}} \middle| \boldsymbol{a} \right)^{2}\boldsymbol{u} =\) \(\small{\begin{pmatrix} 0 & - 1 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix}^{2}\begin{pmatrix} x \\ y \\ z \\ 1 \\ \end{pmatrix} = \begin{pmatrix} - 1 & 0 & 0 & 1 \\ 0 & - 1 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix}\begin{pmatrix} x \\ y \\ z \\ 1 \\ \end{pmatrix} = \begin{pmatrix} - x + 1 \\ - y + 1 \\ - z + 0 \\ 1 \\ \end{pmatrix}}\)

\(\left( 4_{\boldsymbol{c}} \middle| \boldsymbol{a} \right)^{3}\boldsymbol{u} = \left( 4_{\boldsymbol{c}} \middle| \boldsymbol{a} \right)\left\lbrack 2_{\boldsymbol{c}}\boldsymbol{u} + \boldsymbol{a} + \boldsymbol{b} \right\rbrack = 4_{\boldsymbol{c}}^{3}\boldsymbol{u} + 4_{\boldsymbol{c}}\boldsymbol{a} + 4_{\boldsymbol{c}}\boldsymbol{b} + \boldsymbol{a} = 4_{\boldsymbol{c}}^{3}\boldsymbol{u} + \boldsymbol{b - a} + \boldsymbol{a} = 4_{\boldsymbol{c}}^{3}\boldsymbol{u} + \boldsymbol{b}\)

\[\left( 4_{\boldsymbol{c}} \middle| \boldsymbol{a} \right)^{3}\boldsymbol{u} = \left( 4_{\boldsymbol{c}}^{3} \middle| \boldsymbol{b} \right)\boldsymbol{u = u'''} \nonumber \]

\[\left( 4_{\boldsymbol{c}} \middle| \boldsymbol{a} \right)^{2}\boldsymbol{u} =\) \(\small{\begin{pmatrix} 0 & - 1 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix}^{3}\begin{pmatrix} x \\ y \\ z \\ 1 \\ \end{pmatrix} = \begin{pmatrix} 0 & 1 & 0 & 0 \\ - 1 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix}\begin{pmatrix} x \\ y \\ z \\ 1 \\ \end{pmatrix} = \begin{pmatrix} - y + 0 \\ - x + 1 \\ - z + 0 \\ 1 \\ \end{pmatrix}} \nonumber \]

\[\left( 4_{\boldsymbol{c}} \middle| \boldsymbol{a} \right)^{4}\boldsymbol{u} = \left( 4_{\boldsymbol{c}} \middle| \boldsymbol{a} \right)\left\lbrack 4_{\boldsymbol{c}}^{3}\boldsymbol{u} + \boldsymbol{b} \right\rbrack = 4_{\boldsymbol{c}}^{4}\boldsymbol{u} + 4_{\boldsymbol{c}}\boldsymbol{b} + \boldsymbol{a} = \boldsymbol{u - a} + \boldsymbol{a} = \left( 1 \middle| \boldsymbol{0} \right)\boldsymbol{u} = \boldsymbol{u} \nonumber \]

\[\left( 4_{\boldsymbol{c}} \middle| \boldsymbol{a} \right)^{2}\mathbf{u} =\) \(\small{\begin{pmatrix} 0 & - 1 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix}^{4}\begin{pmatrix} x \\ y \\ z \\ 1 \\ \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix}\begin{pmatrix} x \\ y \\ z \\ 1 \\ \end{pmatrix} = \begin{pmatrix} x \\ y \\ z \\ 1 \\ \end{pmatrix}} \nonumber \]

Figure 2.22: \(\mathbf{u} = \left( 4_{\mathbf{c}} \middle| \mathbf{a} \right)^{4}\mathbf{u}\)

Figure 2.23: \(\mathbf{u}^{\prime} = \left( 4_{\mathbf{c}} \middle| \mathbf{a} \right)\mathbf{u}\)

Figure 2.24: \(\mathbf{u}^{\prime\prime} = \left( 4_{\mathbf{c}} \middle| \mathbf{a} \right)^{2}\mathbf{u}\)

Figure 2.25: \(\mathbf{u}^{\prime\prime\prime} = \left( 4_{\mathbf{c}} \middle| \mathbf{a} \right)^{3}\mathbf{u}\)

Rotations in 2-d Lattices

The allowed rotational symmetries for lattices in 2-dimensions give five crystal systems described by their point groups at lattice points. For each unit cell, there are sites other than the corners that have the same point group:

System	Oblique	Rectangular	Tetragonal	Trigonal	Hexagonal
Unit Cell	Fig 2.26	Fig 2.27	Fig 2.28	Fig 2.29	Fig 2.30
	a ≠ b; γ ≠ 90°	a ≠ b; γ = 90°	a = b; γ = 90°	a = b; γ = 120°	a = b; γ = 120°
Lattice Symmetry	\mathcal{C}_{2} = 2	\mathcal{C}_{2v} = 2mm	\mathcal{C}_{4v} = 4mm	\mathcal{C}_{3v} = 3m	\mathcal{C}_{6v} = 6mm
Additional Symmetry Points	2: \small{\begin{pmatrix} ½ \\ 0 \\ \end{pmatrix},\begin{pmatrix} 0 \\ ½ \\ \end{pmatrix},\begin{pmatrix} ½ \\ ½ \\ \end{pmatrix}}	2mm: \small{\begin{pmatrix} ½ \\ 0 \\ \end{pmatrix},\begin{pmatrix} 0 \\ ½ \\ \end{pmatrix},\begin{pmatrix} ½ \\ ½ \\ \end{pmatrix}}	4mm: \small{\begin{pmatrix} ½ \\ ½ \\ \end{pmatrix}} 2mm: \small{\begin{pmatrix} ½ \\ 0 \\ \end{pmatrix},\begin{pmatrix} 0 \\ ½ \\ \end{pmatrix}}	3m: \small{\begin{pmatrix} ⅓ \\ ⅔ \\ \end{pmatrix},\begin{pmatrix} ⅔ \\ ⅓ \\ \end{pmatrix}}	3m: \small{\begin{pmatrix} ⅓ \\ ⅔ \\ \end{pmatrix},\begin{pmatrix} ⅔ \\ ⅓ \\ \end{pmatrix}} 2: \small{\begin{pmatrix} ½ \\ 0 \\ \end{pmatrix},\begin{pmatrix} 0 \\ ½ \\ \end{pmatrix},\begin{pmatrix} ½ \\ ½ \\ \end{pmatrix}}

Every lattice has inversion, which is the 2-fold rotation in 2-d. Therefore, the oblique system is the lowest lattice symmetry and there are no restrictions on the independent repeating lengths and the angle between them, i.e., the unit cell constants. The rectangular system includes orthogonal reflections, which requires the unit cell angle to be 90º. Tetragonal, trigonal, and hexagonal systems include higher order rotations as well as reflections. Now, since the combination of a 3-fold and 2-fold rotation yields a 6-fold rotation, the lattices (and unit cells) for the trigonal and hexagonal systems are identical.

The illustrated unit cells also point out rotational symmetries at points other than the corners. In particular, additional 2-fold axes occur at edge midpoints and cell centers for the oblique, rectangular, tetragonal, and hexagonal cells. There is a 4-fold axis at the center of the tetragonal cell and two 3-fold axes in the hexagonal and trigonal cells. To determine the fractional coordinates of these n-fold rotational symmetries for a given 2-d lattice, we need to identify all possible positions \(\mathbf{\tau}\) that can occur for rotations \(n_{\boldsymbol{c}}(\boldsymbol{\tau})\) in the unit cell.

Example \(\PageIndex{1}\)

Verify the locations of the 3-fold axes in the 2-d trigonal system.

Solution

We must identify all points \(\boldsymbol{\tau}\) such that \(3_{\boldsymbol{c}}(\boldsymbol{\tau}) = T(+\boldsymbol{\tau}) \cdot 3_{\boldsymbol{c}} \cdot T(-\boldsymbol{\tau})\) is a 3-fold rotation followed by a lattice translation. Using the lattice basis, \(3_{c}\boldsymbol{a} = \boldsymbol{b}\) and \(3_{c}\boldsymbol{b} = -\boldsymbol{a} - \boldsymbol{b}\), so

\[3_{\boldsymbol{c}}(\text{Passive}) = \small{\begin{pmatrix} 0 & 1 & 0 \\ - 1 & - 1 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix}}\) and \(3_{\boldsymbol{c}}(\text{Active}) = 3_{\boldsymbol{c}}^{T}(\text{Passive}) = \small{\begin{pmatrix} 0 & - 1 & 0 \\ 1 & - 1 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix}}. \nonumber \]

For a 2-d lattice, the coordinates of \(\boldsymbol{\tau}\) are unknown in the plane, \(\tau_{1},\tau_{2},0\). Then, using augmented matrices

\[3_{\boldsymbol{c}}(\boldsymbol{\tau}) = T(+\boldsymbol{\tau}) \cdot 3_{\boldsymbol{c}}(\boldsymbol{0}) \cdot T(-\boldsymbol{\tau}) \nonumber \]

\[\ \ \ \ \ \ \ = \small{\begin{pmatrix} 1 & 0 & 0 & \tau_{1} \\ 0 & 1 & 0 & \tau_{2} \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix}\begin{pmatrix} 0 & - 1 & 0 & 0 \\ 1 & - 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix}\begin{pmatrix} 1 & 0 & 0 & {- \tau}_{1} \\ 0 & 1 & 0 & {- \tau}_{2} \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix} = \ \begin{pmatrix} 0 & - 1 & 0 & \tau_{1} + \tau_{2} \\ 1 & - 1 & 0 & {- \tau}_{1} + {2\tau}_{2} \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix}} \nonumber \]

The only allowed translations for \(3_{\boldsymbol{c}}(\boldsymbol{\tau})\) are lattice translations \(\boldsymbol{T}_{mn} = m\boldsymbol{a} + n\boldsymbol{b}\) for integers \(m\) and \(n\). Therefore, \(\tau_{1} + \tau_{2} = m\) and \(- \tau_{1} + 2\tau_{2} = n\). There are a few mathematical strategies to solve this system of equations. By combining the two equations, we conclude that \(3\tau_{2} =\) integer, so that there are three possible solutions for \(\left| \tau_{2} \right| < 1:\ \tau_{2} = 0,⅓,\ ⅔\). Therefore, in one unit cell, \(3_{\boldsymbol{c}}(\boldsymbol{\tau})\) axes occur at \(\boldsymbol{\tau} =\small{\begin{pmatrix} 0 \\ 0 \\ 0 \\ \end{pmatrix},\ \begin{pmatrix} ⅔ \\ ⅓ \\ 0 \\ \end{pmatrix},\ \begin{pmatrix} ⅓ \\ ⅔ \\ 0 \\ \end{pmatrix}}\).

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1. Two groups are isomorphous if and only if they have the same multiplication table.↩
2. The transpose M^T of a square matrix M is determined by switching the row and column indices of each matrix element of M: M_ij^T = M_ji.↩