PHY.F20 Molecular and Solid State Physics
16.08.2016

First name
Last name
Matrikelnr.

Problem 1

The starting point for the quantum description of an atom with $Z$ protons and $Z$ electrons is the many-electron Hamiltonian,

\begin{equation} H_{\text{me}}= - \sum \limits_{i=1}^Z \frac{\hbar^2}{2m_e}\nabla_i^2 - \sum \limits_{i=1}^Z \frac{Ze^2}{4\pi\epsilon_0 | \vec{r}_i |} + \sum \limits_{i\lt j}\frac{e^2}{4\pi\epsilon_0 |\vec{r}_i - \vec{r}_j | }. \end{equation}

An approximate solution of the many-electron Hamiltonian for an atom can be found by neglecting the electron-electron interactions. The resulting Hamiltonian is called the reduced Hamiltonian. The reduced Hamiltonian is the sum of $Z$ identical atomic orbital Hamiltonians.

\begin{equation} H_{\text{red}}= \sum \limits_{i=1}^Z H_{\text{ao}}. \end{equation}

(a) What is the atomic orbital Hamiltonian $H_{\text{ao}}$ for lithium $Z =3$?


$H_{\text{ao}}^{\text{Li}}=$


(b) What are the eigenfunction solutions to the atomic orbital Hamiltonian for lithium?




(c) What are the eigenfunction solutions to the reduced Hamiltonian for lithium?




(d) What is the ground state many-electron wavefunction for lithium?




(e) How would you calculate the energy of the ground state many-electron wavefunction for lithium including the electron-electron interactions?




Problem 2

$\Delta k_x$

$\Delta k_y$

$\Delta k_z$

Intensity

0

0

0

1

-1.5E10

0

0

0.229

0

-1.5E10

0

0.229

0

0

-1.5E10

0.229

0

0

1.5E10

0.229

0

1.5E10

0

0.229

1.5E10

0

0

0.229

-1.5E10

-1.5E10

0

0.667

-1.5E10

0

-1.5E10

0.667

-1.5E10

0

1.5E10

0.667

-1.5E10

1.5E10

0

0.667

0

-1.5E10

-1.5E10

0.667

0

-1.5E10

1.5E10

0.667

0

1.5E10

-1.5E10

0.667

0

1.5E10

1.5E10

0.667

1.5E10

-1.5E10

0

0.667

1.5E10

0

-1.5E10

0.667

1.5E10

0

1.5E10

0.667

1.5E10

1.5E10

0

0.667

-1.5E10

-1.5E10

-1.5E10

0.192

-1.5E10

-1.5E10

1.5E10

0.192

-1.5E10

1.5E10

-1.5E10

0.192

1.5E10

1.5E10

1.5E10

0.192

1.5E10

-1.5E10

-1.5E10

0.192

1.5E10

-1.5E10

1.5E10

0.192

1.5E10

1.5E10

-1.5E10

0.192

1.5E10

1.5E10

1.5E10

0.192

The table to the right shows the intensity of diffraction peaks that were measured in an x-ray diffraction experiment on a single crystal. The three components of the scattering vector $\Delta k$ are given in units of [1/m]. The intensities have been normalized to their largest value.

(a) What are the primitive lattice vectors of this crystal in real space?




(b) What Bravais lattice does this crystal have?




(c) How can you estimate the number of atoms in the basis? How many atoms are there in the basis of this crystal?




(d) How could you determine the point group of this crystal?




(e) How could you determine how the atoms are arranged in the basis of this crystal?

Problem 3
The effective densities of states and the band gaps of several semiconductors are given in the following table.

$N_c$

$N_v$

$E_g$

 Si (300 K)

2.78 × 1025 m-3

9.84 × 1024 m-3

1.12 eV

 Ge (300 K)

1.04 × 1025 m-3

6.0 × 1024 m-3

0.66 eV

 GaAs (300 K) 

 4.45 × 1023 m-3 

7.72 × 1024 m-3

1.424 eV

 4H-SiC (300 K) 

 1.7 × 1025 m-3 

2.5 × 1025 m-3

3.3 eV

Silicon is doped with boron (an acceptor) at 1017 1/cm³.

(a) In silicon, the mobility of electrons is 1500 cm²/Vs and the mobility of holes is 450 cm²/Vs. What is the conductivity of this doped silicon at 300 °C? (This is in the extrinsic regime.)




(b) What is the chemical potential of this doped silicon at 300 °C?




(c) What are the minority carriers? What is their concentration at 300 °C?




(d) What is the approximate temperature dependence of the majority carriers and the minority carriers in the extinsic regime?




(e) How could you calculate the electronic contribution to the specific heat for this doped silicon? How would it compare to the phonon contribution to the specific heat?

Problem 4

Tight binding is a method to calculate the electronic band structure of a crystal. The tight-binding wavefunction is,

\begin{equation} \psi_{\vec{k}}\left(\vec{r}\right)=\frac{1}{\sqrt{N}}\sum\limits_{h,j,l}e^{i\left(h\vec{k}\cdot\vec{a}_1 + j\vec{k}\cdot\vec{a}_2 + l\vec{k}\cdot\vec{a}_3\right)} \psi_{\text{unit cell}}\left(\vec{r}-h\vec{a}_1-j\vec{a}_2-l\vec{a}_3\right). \end{equation}

Here $N$ is the number of unit cells in the crystal; $h$, $j$, and $l$ are integers that are used to label all the unit cells in the crystal; $\vec{k}$ is a wave vector; $\vec{a}_1$, $\vec{a}_2$, and $\vec{a}_3$ are primitive lattice vectors in real space, and $\psi_{\text{unit cell}}(\vec{r})$ is a trial wavefunction that is constructed from the valence orbitals $\phi_{i}$ of all of the atoms in a primitive unit cell,

\begin{equation} \psi_{\text{unit cell}}\left(\vec{r}\right)=\sum\limits_{i}c_{i}\phi_{i}\left(\vec{r}-\vec{r}_i\right). \end{equation}

Copper has an fcc Bravais lattice and the valence orbitals are the 3d and 4s atomic orbitals.

(a) What is $\psi_{\text{unit cell}}$ for copper? How many coefficients $c_i$ are there?




(b) How do you determine the coefficients $c_i$ for copper?




(c) If the coefficients $c_i$ are known, how can you construct the dispersion relation for copper?




(d) If the dispersion relation for copper is known, how can you construct the density of states?




(e) If the density of states for copper is known, how can you calculate the chemical potential? Describe how this calculation depends on the number of valence orbitals that are included in the tightbinding calculation.




Quantity

Symbol

Value

Units

electron charge

e

1.60217733 × 10-19

C

speed of light 

c

2.99792458 × 108

 m/s

Planck's constant

h

6.6260755 × 10-34

J s 

reduced Planck's constant

$\hbar$

1.05457266 × 10-34

J s

Boltzmann's constant

 kB

1.380658 × 10-23

J/K

electron mass

me

9.1093897 × 10-31

kg 

Stefan-Boltzmann constant

σ

5.67051 × 10-8

W m-2 K-4

Bohr radius

a0

0.529177249 × 10-10

m

atomic mass constant

mu

1.6605402 × 10-27

kg

permeability of vacuum

μ0

4π × 10-7

N A-2

permittivity of vacuum

ε0

8.854187817 × 10-12

F m-1

Avogado's constant

NA

6.0221367 × 1023

mol-1