Solid State Physics Notes

Phonons

Crystal Vibration

Single Atom

Assume the elastic response of crystal is the linear function of force. So, the force is proportional to the displacement difference of two planes: \begin{equation} F_s = C(u_{s+1}-u_s)+C(u_{s-1}-u_s) \end{equation}

Equation of Motion

Thus the equation of motion is \begin{equation} M\frac{d^2 u_s}{dt^2}=C(u_{s+1}+u_{s-1}-2u_s) \end{equation} where \(M\) is the mass of an atom and \(C\) is the force constant between nearest planes. We require the solutions with time relation \(e^{-i\omega t}\), so \begin{equation} -M\omega^2 u_s=C(u_{s+1}+u_{s-1}-2u_s) \end{equation} has the solutions in following form: \begin{equation} u_{s \pm 1} = u e^{isKa} e^{\pm iKa} \end{equation} where \(a\) is the distance between planes.

Dispersion Relation

From previous formulas we can obtain the dispersion relation: \begin{equation} \color{red} \omega^2 = (2C/M)(1-\cos Ka) \end{equation} \begin{equation} \omega = \sqrt{4C/M} |\sin \frac{1}{2}Ka| \end{equation}

Group Velocity

The group velocity is the velocity of the wave package, \begin{equation} v_g = \frac{w\omega}{dK} =\nabla_{\vec K} \omega(\vec K) \end{equation}

Two Atoms in Basis.

Equation of Motion

Similiarly, we have: \begin{equation} M_1\frac{d^2u_s}{dt^2} = C(v_s+v_{s-1}-2u_s) \end{equation} \begin{equation} M_2\frac{d^2v_s}{dt^2} = C(u_{s+1}+u_s-2v_s) \end{equation} where \(u,v\) are the displacements of different planes.

Dispersion Relation

To make this soluable, we require: \begin{equation} \color{red} M_1 M_2 \omega^4 -2C(M_1+M_2)\omega^2+2C^2(1-\cos Ka) = 0 \end{equation} from which we can solve \(\omega^2\) strictly.

Heat Capacity of Phonons

Planck Distribution

The occupation number of phonons with wavevector \(K\) and polarization mode \(p\) under equilibrium is: \begin{equation} \langle n \rangle = \frac{1}{e^{\hbar\omega /\tau}-1} \end{equation}

State Density

1-Dimension

Considering the periodic boundary condition \(u(sa) = u(sa+L)\), for walking wave solutions, the allowed values of \(K\) are: \begin{equation} K=0, \pm\frac{2\pi}{L},\pm\frac{4\pi}{L},\cdots ,\frac{N\pi}{L} \end{equation} So the number of states in elementary \(K\) is \(L/2\pi\). The number of states at \(\omega\) can be written as: \begin{equation} D(\omega)d\omega = \frac{L}{\pi}\frac{dK}{d\omega} d\omega = \frac{L}{\pi}\frac{d\omega}{d\omega /dK} \end{equation}

3-Dimension

In 3-dimension, for each polarization and each branch, there is only one \(\vec K\) value in an elementary volume \((2\pi /L)^3\). So the modes number with wavevector smaller than \(K\) is \begin{equation} N = \frac{V}{(2\pi)^3} \frac{4\pi K^3}{3} \end{equation} for each polarization. Thus the state density: \begin{equation} D(\omega) = \frac{dN}{d\omega}= \frac{VK^2}{2\pi^2}\frac{dK}{d\omega} \end{equation} for each polarization.

Debye Model

State Density

In Debye approximation, we assume phonon speed stays the same: \begin{equation} \omega = vK. \end{equation} Thus the state density is \begin{equation} \color{red} D(\omega) = \frac{V\omega^2}{2\pi^2 v^3}. \end{equation} And the cutoff frequency \(\omega_D\) and wavevector \(K_D\) are: \begin{equation} \omega_D^3=\frac{6\pi^2v^3N}{V} \end{equation} \begin{equation} K_D = (6\pi^2 v^3 N/V)^{1/3}. \end{equation}

Internal Energy

Assuming the phonon velocity is independent of polarization states, we can multiple a factor of 3: \begin{equation} U = 3\int d\omega D(\omega) \langle n \rangle \hbar\omega = \frac{3Vk_BT^4}{2\pi^2 v^3 \hbar^3}\int_0^{x_D} dx\frac{x^3}{e^x-1} \end{equation} where \(x = \hbar\omega / k_B T\) and \(x_D = \hbar\omega_D /k_B T = \theta /T\). We define the Debye temperature: \begin{equation} \theta = \frac{\hbar v}{k_B}\cdot (\frac{6\pi^2 N}{V})^{1/3} \end{equation} Thus the phonon energy can be written as: \begin{equation} U = 9Nk_B T (\frac{T}{\theta})^3 \int_0^{x_D} dx\frac{x^3}{e^x -1} \end{equation}

Heat Capacity

The heat capacity is the differential to the energy: \begin{equation} \color{red} C_V = 9Nk_B (\frac{T}{\theta})^3 \int_0^{x_D}\frac{x^4 e^x}{(e^x -1)^2} \end{equation}

\(T^3\) Law

In very low temperature (\(T< \theta /50\)), the integral approximate: \begin{equation} \int_0^{\infty}dx \frac{x^3}{e^x-1} = \frac{\pi^4}{15} \end{equation} is a constant. So: \begin{equation} C_V \approx 234Nk_B (\frac{T}{\theta})^3 \end{equation} which is the Debye \(T^3\) law.

Einstein Model

In Einstein model, we assume the frequencies are all \(\omega_0\). Thus the state density is \(D(\omega) = N\delta(\omega - \omega_0)\). And the internal energy in 3 demension: \begin{equation} U = 3 \frac{N\hbar\omega}{e^{\hbar\omega/\tau} -1} \end{equation} The heat capacity of this is: \begin{equation} C_V = 3Nk_B (\frac{\hbar\omega}{\tau})^2 \frac{e^{\hbar\omega /\tau}}{(e^{\hbar\omega /\tau}-1)^2} \end{equation} with the high temperature limit value \(3Nk_B\), which is called the Dulong-Petit Value.

Free Electron Fermi Gas

We consider the electrons inside metals as free gas that follows the Pauli’s principle.

1-Dimensional Energy Level

Using the quantum mechanics we can obtain the eigen values of infinite square well: \begin{equation} \epsilon_n = \frac{\hbar^2}{2m}(\frac{n\pi}{L})^2 \end{equation} The Fermi energy is defined as the highest filled energy level of the ground state of a \(N\) electrons system, which can be determined by \(2n_F = N\): \begin{equation} \epsilon_F = \frac{\hbar^2}{2m}(\frac{N\pi}{2L})^2 \end{equation}

Fermi-Dirac Distribution

From Fermi-Dirac distribution we can obtain the occupation probability of the orbit with energy \(\epsilon\) is: \begin{equation} \color{red} f(\epsilon) = \frac{1}{e^{(\epsilon - \mu)/k_B T}+1} \end{equation} where \(\mu\) is a function of temperature. At absolute zero: \(\mu = \epsilon_F\); at any temperature: \(f(\mu) = \frac{1}{2}\).

Fermi Gas in 3-Dimension

The solution to the 3-D Schrodinger equation is the standing wave: \begin{equation} \psi_n(\vec r) = A\sin (\pi n_x x/L)\sin (\pi n_y y/L)\sin(\pi n_z z/L) \end{equation} To meet the periodic boundary condition, the wave function must have the form of plane waves: \begin{equation} \psi_k(\vec r) = e^{i\vec k \cdot \vec r} \end{equation} where the wavevector \(k_{x,y,z} = 0, \pm \frac{2\pi}{L}, \pm \frac{4\pi}{L},\cdots\).

Fermi Sphere

Under ground state, the occupied orbit can be presented as a sphere in \(k\)-space. The radius meets: \begin{equation} \epsilon_F = \frac{\hbar^2}{2m}k_F^2 \end{equation} The orbit number inside this sphere is: \begin{equation} N = 2\cdot \frac{4\pi k_f^3 /3}{(2\pi/L)^3} = \frac{V}{3\pi^2}k_F^3 \end{equation} where the factor \(2\) is the degeneracy due to spin. Thus: \begin{equation} \color{red} k_F = (\frac{3\pi^2 N}{V})^{1/3} \end{equation} only depends on particle concentration.

State Density

The orbit number in energy unit interval is \begin{equation} \color{red} D(\epsilon) = \frac{dN}{d\epsilon}=\frac{V}{2\pi^2}\cdot (\frac{2m}{\hbar^2})^{3/2} = \frac{3N}{2\epsilon} \end{equation}

Heat Capacity of Electron Gas

When the temperature increases, only electrons with energy near the Fermi energy level in the range \(k_B T\) will be thermal excited. From this we know in \(NT/T_F\) electrons, each has the internal energy of \(k_BT\), so the total energy is about \begin{equation} U\approx (NT/T_F)k_BT \end{equation} And the electron heat capacity: \begin{equation} C_{el} \approx Nk_B(T/T_F) \end{equation} which is proportional to the temperature.

Energy Bands

Nearly Free Electron Model

From the Bragg conditon, the Bragg Reflection occurs at: \begin{equation} k = \pm\frac{1}{2}G = \pm n\pi /a \end{equation} The wave function is a standing wave that has two different form: \begin{equation} \psi_{+} = 2\cos(\pi x/a) \end{equation} \begin{equation} \psi_{-} = 2i\sin(\pi x/a) \end{equation}

Energy Gap

The electron probability density functions of these two standing waves are: \begin{equation} \rho_{+} ~ \cos^2 \frac{\pi x}{a} \end{equation} \begin{equation} \rho_{-} ~ \sin^2 frac{\pi x}{a} \end{equation} If we write the periodic potential of lattice as \begin{equation} U(x)=U\cos^2 \frac{\pi x}{a}, \end{equation} the energy gap is \begin{equation} E_g = 2\int dx U\cos(2\pi x/a)(\cos^2 \pi x/a - \sin^2 \pi x/a) = U \end{equation} We can see that the energy gap equals to the Fourier component of lattice potential.

Bloch Function

The Bloch Theorem says that for Schrodinger equation that has periodic potential, the solution must have following form: \begin{equation} \psi_{\vec k}(\vec r) = u_{\vec k}(\vec r)e^{i\vec k \cdot \vec r} \end{equation} where \(u_{\vec k}(\vec r)\) has the period of lattice.

Kronig-Penney Model

In Kronig-Penney model we assume the potential as periodic square wells.
In the region \(0<x<a\), \(U=0\), we have the solution: \begin{equation} \psi(x) = Ae^{iKx}+Be^{-iKx}, K=\frac{sqrt{2m\epsilon}}{\hbar} \end{equation} In the region \(-b<x<0\), \(U=U_0\), we have: \begin{equation} \psi(x) = Ce^{lx}+De^{-lx}, l=\frac{\sqrt{2m(U_0 - \epsilon)}}{\hbar} \end{equation} And the Bloch condition: \(\psi(a<x<a+b) = \psi(-b<x<0)e^{ik(a+b)}\) to determine the wavevector \(k\). Using the boundary condition at \(x=0\) and \(x=a\), we obtain the equation: \begin{equation} [(l^2-K^2)/2lK]\sinh (lb)\sin(Ka)+\cosh(lb)\cos(Ka) = \cos [k(a+b)] \end{equation}

\(\delta\)-Potential Limit

To simplify this, we use the limit \(b \to 0, U_0\to\infty\), thus \(l>>K,lb<<1\). So the equation is simplified to: \begin{equation} \frac{P}{Ka}\sin(Ka)+\cos(Ka) = \cos(ka) \end{equation} from which we can know the relation between \(k\) and \(K\) or energy \(\epsilon\).

Wave Equation of Electrons in Periodic Potential

The periodic potential of lattice can be expanded to a Fourier series: \begin{equation} U(x) = \sum_{G} U_{G}e^{iGx} = 2\sum_{G>0}U_G \cos Gx \end{equation} The wave equation of electrons is (Schrodinger equation): \begin{equation} (\frac{p^2}{2m} + \sum_G u_G e^{iGx})\psi(x) = \epsilon\psi(x) \end{equation} The wave function can be represented by a Fourier series as well: \begin{equation} \psi = \sum_k C_k e^{ikx} \end{equation}

Central Equation

Substitute the function into the equation to obtain the central equation: \begin{equation} (\lambda_k -\epsilon)C_k +\sum_G U_G C(k-G) = 0 \end{equation} where \(\lambda_k = \hbar^2 k^2/2m\)