PHY.K02 Molecular and Solid State Physics
07.10.2022

Name Matrikelnr.

Every subproblem (a), (b), (c), etc. is worth 3 points.

Problem 1

(a) Write down the total Hamiltonian for a helium atom ($Z=2$).

(b) What is the reduced Hamiltonian for helium? Why do we use the reduced Hamiltonian instead of the total Hamiltonian?

(c) Write down a two-electron wavefunction that solves the reduced Hamiltonian for helium. You can write this wavefunction in terms of the atomic orbitals.

(d) What is the exchange energy? How is it calculated?

Solution

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(a) The total Hamiltonian includes the kinetic energy terms for both electrons, the two electron-nucleus Coulomb terms, and the electron-electron interaction.

$$H_{\text{total}}^{\text{He}}= - \frac{\hbar^2}{2m}\nabla_1^2 - \frac{\hbar^2}{2m}\nabla_2^2 - \frac{2e^2}{4\pi\epsilon_0 r_1} - \frac{2e^2}{4\pi\epsilon_0 r_2} + \frac{e^2}{4\pi\epsilon_0 |\vec{r}_1-\vec{r}_2| }.$$

(b) In the reduced Hamiltonian we neglect the electron-electron term. This makes the Hamiltonan a sum of single-electron molecular-orbital Hamiltonians. The Hamiltonian can be solved by the separation of variables and the solutions are products of the atomic orbitals.

$$H_{\text{red}}^{\text{He}}= - \frac{\hbar^2}{2m}\nabla_1^2 - \frac{\hbar^2}{2m}\nabla_2^2 - \frac{2e^2}{4\pi\epsilon_0 r_1} - \frac{2e^2}{4\pi\epsilon_0 r_2} .$$

(c) Any product of atomic orbitals for $Z=2$ will solve the reduced Hamiltonian. For instance,
$\phi_{1s}^{Z=2}(\vec{r}_1)\phi_{1s}^{Z=2}(\vec{r}_2)$ is a solution. There is an additional condition that the wavefunction should be antisymmetric. An antisymmetric wave function can be constructed usings a Slater determinant.

(d) The exchange energy is half the difference of a two electron state with a symmetric spin component and an antisymmetric spin component. To calculate it you start with a product state of two atomic orbital and construct the four Slater determinants for the four spin possiblilities $\uparrow\uparrow$, $\downarrow\downarrow$, $\downarrow\uparrow$, and $\uparrow\downarrow$. After a transformation to $\uparrow\uparrow$, $\downarrow\downarrow$, $\downarrow\uparrow + \uparrow\downarrow$, and $\downarrow\uparrow - \uparrow\downarrow$, you evaluate the energy of these four states including the electron-electron interaction term. Three of these states (the triplet) will have the same energy and the singlet $\downarrow\uparrow - \uparrow\downarrow$ will have another energy. The difference between these two energies is twice the exchange energy.

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Problem 2

The conventional unit cell of a CsCl crystal is shown below.

(a) What is the Bravais lattice of CsCl?

(b) What are the primitive reciprocal lattice vectors of CsCl?

(d) How many optical branches are there in the phonon dispersion relation of CsCl?

(e) Express the phonon contribution to the specific heat as an integral including the phonon density of states.

Solution

Problem 3

(a) In a diffraction experiment, x-rays with wavevector $\vec{k}$ are directed at a crystal and x-rays with a wavevector $\vec{k}'$ are diffracted from the crystal. The wavelength of the incoming waves $\vec{k}$ is set by the x-ray source. What determines the wavelength of the diffracted waves $\vec{k}'$?

(b) How can the reciprocal lattice vectors be determined by measuring $\vec{k}'$?

(c) Suppose one hundred reciprocal lattice vectors have been measured. How could you use this data to determine something about the crystal structure?

(d) If you also recorded the intensity of each diffraction peak associated with the one hundred reciprocal lattice vectors, what could you determine with this information?

(e) What is the Ewald sphere?

Solution

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(a) For elastic scattering, $|\vec{k}|= |\vec{k}'|$ so the wavelength of the diffracted waves is the same as that of the incident waves.
(b) Knowing $\vec{k}$ and $\vec{k}'$, the reciprocal lattice vectors can be determined from the Laue condition $\vec{k}'-\vec{k}=\vec{G}$.
(c) One would try to fit the data to one set of primitive reciprocal lattice vectors.

$$\vec{G}_{hkl} = h\vec{b}_1+ k\vec{b}_2+l\vec{b}_3.$$

The primitive reciprocal lattice vectors could then be used to construct the real space lattice vectors. This would tell us the Bravais lattice and the volume of the primitive unit cell.

(d) The intensities of the peaks can be used to determine how the atoms are arranged in the basis of the crystal. Typically the basis is guessed and the structure factors are calculated. The squares of the structure factors are then compared to the measured intensities. If all the structure factors match the measured intensities, that indicates that the correct crystal structure has been found.

(e) The Ewald sphere is a graphical construction that shows which reciprocal lattice vectors will have diffraction peaks. The sphere has a radius of $2|\vec{k}|$ in reciprocal space.

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Problem 4

A tight-binding calculation is performed for copper. Copper has 29 electrons in total with an electron filling of 1s²2s²2p⁶3s²3p⁶3d¹⁰4s¹. The tight binding calculation only includes the five 3d atomic orbitals and the 4s atomic orbital.

(a) Write down the trial wave function that is used in this calculation.

(b) How many bands will be calculated?

(c) How would you calculate the chemical potential? Hint: the electron density you use to find the chemical potential depends on the number of atomic orbitals in the tight-binding wave function.

(d) Why do you need to calculate the chemical potential to determine if this is a metal, a semiconductor, or an insulator?

Solution

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(a) The tight binding wave function repeats a unit cell wave function at every unit cell of the crystal. Each unit cell is given a phase factor with an exponential factor so that the wavefunction commutes with the translation operator.

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

In this case the unit cell wave function would be a linear combination of 3d and 4s atomic orbitals.

$$\psi_{\text{unit cell}}(\vec{r}) = c_1\phi^{\text{Cu}}_{3dxy}+c_2\phi^{\text{Cu}}_{3dyz}+c_3\phi^{\text{Cu}}_{3dxz}+c_4\phi^{\text{Cu}}_{3dz^2}+c_5\phi^{\text{Cu}}_{3dx^2-y^2}+c_6\phi^{\text{Cu}}_{4s}.$$

(b) 6 bands.

(c) From the band structure, the density of states $D(E)$ would have to be constructed. The condition for the electron density is,

$$n = \int\limits_{-\infty}^{\infty}D(E)f_{FD}(E)dE.$$

Here $f_{FD}(E)$ is the Fermi-Dirac function which contains the chemical potential. This equation is typically solved numerically for the chemical potential. Since only the 3d and 4s states were included in the calculation, the electron density that should be used is $n=$ 11 electrons/unit cell.

(d) If the chemical potential cuts through a band, the material is a metal. If the chemical potential falls in a band gap, the material is a semiconductor if the band gap is smaller than 3 eV and an insulator if the band gap is larger than 3 eV.

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PHY.K02 Molecular and Solid State Physics07.10.2022

PHY.K02 Molecular and Solid State Physics
07.10.2022