Solid State Physics
An introduction to solid state physics. Logistics Heat Capacity of Solids Einstein Model of Heat Capacity of Solids Debye Model of Heat Capacity of Solids Electrical Properties (Drude Model) Constant E&M Force Thermal Conductivity Sommerfield Model Fermi-Dirac Statistics Sommerfield Model (cont.) LCAO (Linear Combination of Atomic Orbitals)/ Tight Binding Theory Chemistry Review Shell Theory Ionic Bonds Covalent Bonds LCAO Basics Vibrations 2 Types of Springs Crystal Structure Reciprocal Space Logistics Location: 307 Pupin Time: 2:40-3:55 PM Tuesdays and Thursdays Textbook: The Oxford Solid State Basics Grading Scheme 30% problem sets (5 in total. 1 per 2 weeks) Due on Saturdays at noon 20% midterm Closed book 20% research project presentation 30% final exam Heat Capacity of Solids Heat capacity is defined as $C = \frac{\partial Q}{\partial T}$ For solids, heat capacity at constant pressure and at constant volume are essentially the same, since the atoms are so closely packed For a gas, we have that $\frac{C_{v}}{N} = \frac{3}{2} k_{b}$ For (most) solids, we know that $\frac{C}{N} = 3k_{b}$. Called the law of Dulong-Petit For these derive the above (like Boltzmann did), pretend that we have N atoms trapped in a box. Assume that each atom is trapped within a harmonic potential well of the form $U = \frac{k|\vec{r}|^{2}}{2}$ This arises due to electrostatic forces from neighboring atoms From the equipartition theorem, we have that for every quadratic degree of freedom in the Hamiltonian adds $\frac{k_{b}}{2}$ to the heat capacity For a monoatomic gas, we have 3 quadratic degrees of freedom from the 3 momentum components (position is not a d.o.f. since the particles don’t interact with each other) For solids, we have 6 degrees of freedom, which implies that $\frac{C}{N} = 3k_{b}$ As the tempurature starts to get lower, the tempurature starts to fall of rapidly Einstein Model of Heat Capacity of Solids The main idea is that Einstein used the same model of the N particles in a box, but changed the energy model of the confining potential Instead of $E = \frac{kx^{2}}{2}$, we utilize $E = \hbar \omega(n+\frac{1}{2})$ The original derivation omitted the $\frac{\hbar \omega}{2}$ (ie. the zero point energy), since the Schrodinger equation hadn’t been invented yet $\omega$ is called the Einstein frequency, and is a free parameter of the system Recall the partition function is $Z = \Sigma_{n} exp(-\beta E_{n})$ where $\beta = \frac{1}{kT}$ $<E> = -\frac{1}{Z}\frac{\partial Z}{\partial \beta}$ Plug in energy of quantum harmonic oscillator, turn the crack, and then take derivative w.r.t. temperature yields the specific heat Can define Einstein temperature as $\frac{\hbar\omega}{k_{b}}$, which defines the “freeze out” temperature of the system Problem with this is that at lower potentials, the head capacity goes as $T^{3}$ experimentally, while Einstein is exponential decays Debye Model of Heat Capacity of Solids Basic idea: individual atomic displacements are coupled, which creates a collective excitation/mode which acts like a sound wave The particles which arise from quantizing sound waves are called phonons acts like bosons For sound waves, the dispersion relationship is $\omega = v_{s}|k|$ The average energy is now $<E> =3\Sigma_{k}\hbar \omega(k)(n+\frac{1}{2})$ Namely, now the frequency depends on the wavevector (ie. the dispersion relationship) Can replace sum with integral: $\Sigma = \int 4\pi k^{2} dk$ where k is the magnitude of the wavenumber This comes from assuming periodic boundary conditions, deducing that we have 1 state per every $\frac{(2\pi)^{3}}{L_{x}L_{y}L_{z}} = \frac{(2\pi)^{3}}{V}$ volume in k-space, then converting to spherical coordinates and integrating out solid angle $<E> = 3 \frac{V}{(2\pi)^{3}}4\pi \int_{0}^{\infty} \omega^{2} (\frac{1}{v_{s}})^{3}\hbar\omega(n+\frac{1}{2}) = \int_{0}^{\infty} d\omega g(\omega) \hbar\omega(n+\frac{1}{2})$ Made change of variables from k to $\omega$ $g(\omega) = \frac{12\pi \omega^{2}}{(2\pi)^{3}v_{s}^{3}}V = N\frac{9\omega^{2}}{\omega_{d}^{2}}$ is the density of state Can think of this as the number of modes that exist in the frequency band of $\omega$ and $\omega+d\omega$ $\omega_{d}^{3} = 6\pi^{2}\frac{N}{V}v_{s}^{3}$ Do the tedious integral. The following is useful: $\int_{0}^{\infty} dx \frac{x^{3}}{e^{x}-1} = \frac{\pi^{4}}{15}$ Taking T derivative yields the experimentally observed low tempurature power-law scaling $C_{v} = Nk_{b}(\frac{k_{b}}{\hbar \omega})^{3} \frac{12\pi^{4}}{15} T^{3}$ $k_{b}T_{D} = \hbar \omega_{d}$ Currently, the model blows up at higher tempuratures. Problem is that by integrating to infinity, we are assuming an infinite number of modes. But we are limited by the number of modes present in the system The physical assumption is that $\int_{0}^{\omega_{d}} g(\omega) d\omega = 3N$ ie. there exists a cut-off frequency $\omega_{d}$ that includes all possible modes of the system The integral becomes intractable (ie. must be numerically solved) Problems with model cutoff is a bit ad. hoc. the dispersion relationship is a bit different than sound waves not an exact match The heat capacity of metals follows a different pattern: $\alpha T^{3}+\gamma T$ Electrical Properties (Drude Model) Apply classical kinetic theory to electrons in materials Assumptions Having a scattering time $\tau$ between collisions implies that the probability of a scatter occuring in some dt is $\frac{dt}{\tau}$ After scattering, the averaged momentum of the ensemble is 0 (ie. $<\vec{p}> = 0$) Between scattering events, electrons feel E&M forces since they are charged Consider an electron with some $p(t)$ at time t. We want to know p(t_dt) $(1-\frac{dt}{\tau})(p(t)+Fdt)+\frac{dt}{\tau}*0 = p(t)+Fdt-p(t)\frac{dt}{\tau}\rightarrow \frac{d\vec{p}}{dt} = \vec{F}-\frac{\vec{p}}{\tau}$ Constant E&M Force Let $\vec{F} = q(\vec{E}+\vec{v}\times\vec{B})$ Define current density $\vec{j} = ne\vec{v} \rightarrow \vec{v} = \frac{-\vec{j}}{ne}$ $\vec{E} = \frac{1}{ne}\vec{j}\times \vec{B}+\frac{m}{ne^2\tau}\vec{j}$ First term is an $E_{\perp}$ (Hall field) and second term is $E_{\parallel}$, where we take parallel to be in the flow of the current If B=0, then $\vec{j} = \frac{ne^{2}\tau}{m}\vec{E} = \sigma \vec{E}$ where $\sigma$ is the conductivity (ie. $\sigma = \frac{1}{\rho}$ where $\rho$ is resistivity) $E_{Hall} = =R_{H} \vec{B}\times \vec{j}$ $R_{H} = \frac{-1}{ne}$ Thermal Conductivity $j_{q} = R \nabla T$ is the heat current density For monoatomic gases, $R = \frac{1}{3}nc_{v}<v>\lambda$ $v = \sqrt{\frac{8k_{b}T}{m\pi}}$ Assuming that R of monoatomic gases works for solids, replace v with the appropriate expression Taking the ratio of thermal conductivity to electrical conductivity (called the Lorenz number), we get something on te order of $10^{8}$ called the Wiedemann-Franz law Sommerfield Model Assumptions ...