Crystals are classified by their symmetries. Every crystal can be associated with one of the 32 point groups. A point group is the collection of symmetries (rotations, reflections, inversion) that a crystal has where at least one point in the crystal does not move during the transformation. Point group symmetries can be represented by $3\times 3$ matrices. Rotational symmetries are described by the following matrices where the main symmetry axis has been taken to be the $z$-axis.
\begin{equation} C_1=\left[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}\right]\hspace{1cm} C_2=\left[\begin{matrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{matrix}\right]\hspace{1cm} C_3=\left[\begin{matrix} -\frac{1}{2} & -\frac{\sqrt{3}}{2} & 0 \\ \frac{\sqrt{3}}{2} & -\frac{1}{2} & 0 \\ 0 & 0 & 1 \end{matrix}\right]\hspace{1cm} C_4=\left[\begin{matrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{matrix}\right]\hspace{1cm} C_6=\left[\begin{matrix} \frac{1}{2} & -\frac{\sqrt{3}}{2} & 0 \\ \frac{\sqrt{3}}{2} & \frac{1}{2} & 0 \\ 0 & 0 & 1 \end{matrix}\right]. \end{equation}All of the symmetries of a point group can be generated from at most three generating matrices. Multiplying the generating matrices with themselves and with each other results in other symmetries of that point group. The resulting matrices are multiplied with themselves and with the generating matrices until no new symmetries appear. For instance, $C_4$ is the generating matrix of the point group 4. All of the elements of point group 4 are given below.
\begin{equation} C_4=\left[\begin{matrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{matrix}\right]\hspace{1cm} C_4^2=C_2=\left[\begin{matrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{matrix}\right]\hspace{1cm} C_4^3=C_4^{-1}=\left[\begin{matrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{matrix}\right]\hspace{1cm} C_4^4=C_1=E=\left[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}\right]. \end{equation}Multiplying any of these matrices with any other, results in one of these four matrices. These are the four elements of point group 4.
Matrix multiplication |
Using the generating matrix of point group 3,
\begin{equation} C_{6h}=\left[\begin{matrix} \frac{1}{2} & -\frac{\sqrt{3}}{2} & 0 \\ \frac{\sqrt{3}}{2} & \frac{1}{2} & 0 \\ 0 & 0 & -1 \end{matrix}\right]. \end{equation}find all of the elements in this group.
All of the elements of a point group can be determined from the generating matrices of that group. The generating matrices of C2h are,
\begin{equation} C_2=\left[\begin{matrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{matrix}\right]\hspace{1.5cm}\sigma_h=\left[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{matrix}\right]. \end{equation}What are all the elements of this point group?
Every equilibrium property of a crystal can be expressed as a derivative of an energy. A system where all of the extrinsic variables are held fixed goes to the minimum of the internal energy. Extrinsic variables are those that scale with the size of the system like the number $N$, the volume $V$, the entropy $S$, the magnetization $\vec{M}$, and the electric polarization $\vec{P}$. For a solid, the internal energy is a sum of the electron contribution and a phonon contribution.
\[ \begin{equation} u=\int\limits_{-\infty}^{\infty}ED_e(E)f_{FD}(E)dE+\int\limits_{-\infty}^{\infty}ED_{ph}(E)f_{BE}(E)dE \end{equation} \]Here $E$ is integrated over all energies, $D_e$ is the electron density of states, $D_{ph}$ is the phonon density of states, $f_{FD}$ is the Fermi-Dirac function and $f_{BE}$ is the Bose-Einstein function. Often in an experiment it is easier to hold the temperature constant than the entropy. If the temperature is held constant then the system will go to a minimum of the Helmholtz free energy $F$ which is constructed from the internal energy $U$ by a Legendre transformation,
\[ \begin{equation} F=U-TS. \end{equation} \]Further Legendre transformations are typically made to construct a Gibbs free energy $G$ where the independent variables are temperature $T$, electric field $\vec{E}$, magnetic field $\vec{H}$, and the stress tensor $\sigma$. The conjugate variables to these independent variables are,
\[ \begin{equation} S = -\frac{\partial G}{\partial T}\qquad P_i = -\frac{\partial G}{\partial E_i}\qquad M_i = -\frac{\partial G}{\partial H_i}\qquad\epsilon_{ij} = -\frac{\partial G}{\partial \sigma_{ij}} \end{equation} \]Here $\epsilon$ is the strain tensor and Einstein notation is used for the tensors. Taking further derivatives of these quantities results in many measurable quantities such as:
| Stiffness tensor | $s_{ijkl} = -\frac{\partial\epsilon_{ij}}{\partial\sigma_{kl}}=-\frac{\partial^2 G}{\partial \sigma_{ij}\partial\sigma_{kl}}$ |
| Thermal expansion | $\alpha_{ij} = \frac{\partial \epsilon_{ij}}{\partial T}= -\frac{\partial^2 G}{\partial \sigma_{ij}\partial T}$ |
| Piezoelectric effect | $d_{ijk} = \frac{\partial P_i}{\partial \sigma_{jk}}=-\frac{\partial^2 G}{\partial E_i \partial \sigma_{jk}}$ |
| Electric susceptibility | $\chi^E_{ij} = \frac{\partial P_{i}}{\partial E_j} =-\frac{\partial^2 G}{\partial E_{i}\partial E_j}$ |
| Magnetic susceptibility | $\chi^M_{ij} = \frac{\partial M_{i}}{\partial H_j}=-\frac{\partial^2 G}{\partial H_{i}\partial H_j}$ |
| Pyroelectric effect | $\pi_{i} =\frac{\partial P_{i}}{\partial T}= -\frac{\partial^2 G}{\partial E_{i}\partial T}$ |
It is also possible to further Taylor expand the physical quantities to include nonlinear effects such as,
\[ \begin{equation} P_i=\chi^{(1)}_{ij}E_j+\chi^{(2)}_{ijk}E_jE_k+\cdots \end{equation} \]Here $\chi^(2)_{ijk}$ is the nonlinear electric susceptibility that is used in nonlinear optics. The strain can also be expanded to higher order in the electric field,
\[ \begin{equation} \epsilon_{ij}=d_{ijk}E_k+Q_{ijkl}E_kE_l+\cdots. \end{equation} \]Here $d_{ijk}$ are the piezoelectric coefficients and $Q_{ijkl}$ are the electrostriction coefficients.
Tensor quantities like the electric susceptibility must be unchanged by the application of a symmetry element of the point group. This restricts the form that tensors can have for every crystal class. Some of these restrictions are listed in the table of crystal classes. If a tensor is unchanged by the generating matrices of a group, it will be unchanged by all elements of the group. It is only necessary to determine how the tensor transforms under the operation of the generating matrices. All odd-rank tensors are zero if inversion is an element of the point group. Thus, a crystal does not exhibit pyroelectricity, piezoelectricity, piezomagnetism, or a nonlinear optical response if there is inversion in the point group. Symmetry also requires all second-rank tensors for cubic crystals to be proportional to the identity matrix.
The form of tensors that describe nonequilibrium properties such as the electrical conductivity or the thermal conductivity are also restricted by the symmetry of the crystal.
| Electrical conductivity | $\sigma_{ij} = \frac{\partial j^e_{i}}{\partial E_{j}}$ | $\vec{j}$ is the current density and $\vec{E}$ is the electric field. | |
| Thermal conductivity | $\kappa_{ij} = -\frac{\partial j^{th}_{i}}{\partial (\nabla T)_j}$ | $\vec{j}^{th}$ is the heat current and $T$ is the temperature. |
Pyroelectricity describes how the electric polarization changes as the temperature changes. The pyroelectric coefficients form a rank 1 tensor $\pi_i$,
\[ \begin{equation} \pi_i=\frac{\partial P_i}{\partial T}. \end{equation} \]The generating matrix of the point group m is,
\begin{equation} \sigma_h=\left[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{matrix}\right]. \end{equation}(a) Give the independent tensor elements for this point group.
(b) How could the pyroelectric coefficient be calculated from the microscopic quantum states of the crystal?
The piezoelectric effect is described by a rank 3 tensor which relates stress with the electrostatic polarization.
\[ \begin{equation} d_{ijk}=\frac{\partial P_i}{\sigma_{jk}}. \end{equation} \](a) Write down the tensor equations for the piezoelectric effect and for the inverse piezoelectric effect.
(b) Explain how you would determine the independent components of the piezoelectric tensor for the tetragonal point group 4.
Determine the point group (International notation) for NaCl, β-Sn, and Graphite. The flowchart on this site can help you.
Graphite (the material in pencils) has a layered structure where the carbon atoms are strongly bonded in two-dimensional planes but the planes bond weakly with each other. It has a hexagonal Bravais lattice. The electrical conductivity is not the same parallel to the layers and perpendicular to the layers. The electrical conductivity is defined as,
$$\vec{j}=\sigma\vec{E},$$where $\vec{j}$ is the current density, $\vec{E}$ is the electric field, and $\sigma$ is the conductivity. What is the form of the electrical conductivity tensor for graphite?
The thermal expansion coefficients $\alpha_{ij}$ describe how the strain changes as the temperture changes,
\[ \begin{equation} \alpha_{ij}=\frac{\partial \epsilon_{ij}}{\partial T}. \end{equation} \]For the mineral calcite has a trigonal crystal class and there are only the three nonzero components to the thermal expansion tensor, $\alpha_{11} = \alpha_{22} = -5.6\times10^{-6} [\text{K}^{-1}]$ and $\alpha_{33} = 25\times10^{-6} [\text{K}^{-1}]$. Calculate the expansion of a calcite sphere with diameter $R$ for a temperature increase of 5 Kelvin.
Tensor quantities must be unchanged by the application of a symmetry element of the point group. This restricts the form that tensors can have for every crystal class. Some of these restrictions are listed in the table of crystal classes. All odd-rank tensors are zero if inversion is an element of the point group. Thus, a crystal does not exhibit pyroelectricity, piezoelectricity, piezomagnetism, or a nonlinear optical response if there is inversion in the point group. Symmetry also requires all second-rank tensors for cubic crystals to be proportional to the identity matrix.
(a) What rank tensor is the thermal conductivity? What form does this tensor have for gold (fcc)?
(b) The accelerometer in a phone typically consists of a cantilever beam of silicon that bends when the phone accelerates. The conductivity of the silicon changes as a function of the stress. What rank tensor is the piezoconductivity?
(c) How do rank-2 tensors transform under inversion?
(d) What matrix rotates a cubic crystal by 120° around the [111] direction?
(a) Explain why the magnetic susceptibility, $\chi_{ij}^M =\frac{\partial M_i}{\partial H_j}$, must be a symmetric tensor.
(b) Why don't the electrons contribute to the thermodynamic properties of diamond? Why do you need to calculate the electronic band structure anyway when you calculate the thermodynamic properties of diamond?
(c) Why is the change in the magnetization when you change the electric field equal to the change in electric polarization when you change the magnetic field?
$$\frac{\partial M_i}{\partial E_j}=\frac{\partial P_j}{\partial H_i}$$