from matplotlib import pyplot as plt 
import numpy as np
from math import pi
  

$$\$\$\$\$\$\$\$\$$$

Solutions for lecture 11 exercises

Exercise 1: Bloch's model

Subquestion 1

It must obey the crystal symmetry, such as the translational symmetry of the lattice described by the lattice vectors ${\mathbf{a}}_1,{\mathbf{a}}_2$ and ${\mathbf{a}}_3$.

Subquestion 2

From the periodicity of the wavefunction with the real space lattice vectors, it follows that the kinetic part of the Hamiltonian, here denoted $\hat{K}$, will commute with the translation operator ${\hat{T}}_{\alpha ,\beta ,\gamma }$. \begin{align} [\hat{T}{\alpha,\beta,\gamma},\hat{H}]&= \left(V(\mathbf{r}-\alpha \mathbf{a}_1-\beta \mathbf{a}_2 - \gamma \mathbf{a}_3)-V(\mathbf{r})\right)\hat{T}=0\ \end{align} Where we used that the lattice potential is periodic with integer lattice constants.

Subquestion 3

\begin{align} \hat{T}_{\alpha,\beta,\gamma}u_n(\mathbf{r})e^{i\mathbf{k}\cdot\mathbf{r}}&=e^{-i\mathbf{k}\cdot (\alpha \mathbf{a}_1+\beta \mathbf{a}_2 + \gamma \mathbf{a}_3)} u_n(\mathbf{r})e^{i\mathbf{k}\cdot \mathbf{r}}\ \end{align} The eigenfunctions of $\hat{H}$ must also be Bloch waves.

Subquestion 4

\begin{equation} \nabla^2\psi_n(\mathbf{r})=e^{i\mathbf{k}\cdot\mathbf{r}}\left[ \nabla^2 u_n(\mathbf{r})-k^2u_n(\mathbf{r})+2i\mathbf{k}\nabla u_n(\mathbf{r}) \right]. \end{equation} Once this is explicitly written in the Schr. eqn, the complex exponentials cancel out.

Subquestion 5

$u_n(\mathbf{r})$ becomes a normalization constant that is independent of position. Hence, the momentum operators return zero, and the only term that remains is ${\hslash }^2{k}^2/2m$ (which is indeed the free electron dispersion).

Exercise 2: The Central Equation in 1D

Subquestion 1

All $k_n$ that differ by an integer multiple of $2\pi /a$ from $k_0$ have the exact same wavefunction.

Subquestion 2

\begin{equation} \phi_n(x)=\frac{1}{\sqrt{\Omega}} \exp\left[i \left(k_0+\frac{2\pi n}{a}\right)x \right] \end{equation} $$E_n=\frac{{\hslash }^2}{2m}{(k_0+\frac{2\pi n}{a})}^2$$

Subquestion 3

def dispersions(N = 5):

    x0 = np.linspace(-N*2*pi,N*2*pi,2*N+1)
    x = np.tile(np.linspace(-N*pi,N*pi,500),(2*N+1,1))

    y = []
    for i, offset in enumerate(x0):
        y.append((x[i]-offset)**2)

    plt.figure(figsize=(5,5))
    plt.axvspan(-pi, pi, alpha=0.2, color='red')

    for i in range(2,9):
        plt.plot(x[i],y[i])
        plt.axvline(1+x0[i],0,1,color='k')    

    plt.text(1.2-2*pi,37,'$k_{-1}$',fontsize=16)
    plt.text(1.2,37,'$k_0$',fontsize=16)
    plt.text(1.2+2*pi,37,'$k_1$',fontsize=16)
    plt.text(-2,59,'1st BZ',fontsize=19)

    plt.axhline(0.8,0,1,linestyle='dashed')
    plt.axhline(28,0,1,linestyle='dashed')
    plt.axhline(53,0,1,linestyle='dashed')

    plt.xlim([-3*pi,3*pi])
    plt.ylim([0,70])
    plt.xlabel('$k$',fontsize=19)
    plt.ylabel('$E$',fontsize=19)
    plt.xticks((-2*pi, -pi, 0 , pi,2*pi),('$-2\pi/a$','$-\pi/a$','$0$','$\pi/a$','$2\pi/a$'),fontsize=15)
    plt.yticks((1,27,54),('$E_0$','$E_1$','$E_{-1}$'))

dispersions(5)

Subquestion 4

First the kinetic term, $$⟨{\varphi }_m|\hat{K}|{\varphi }_n⟩=C_m\frac{{\hslash }^2{k}_m^2}{2m}$$ And the potential term, $$\begin{array}{r}⟨{\varphi }_m|V(x)|\psi ⟩=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{C}_n{\int }_0^{a}V(x)e^{-i\frac{2\pi }{a}(m-n)x}dx\end{array}$$ then relabel indices and combine both expressions to find the final answer and expression for ${\epsilon }_m$ (which is the free electron dispersion).

Subquestion 5

From the expression for the energy, it is clear that the difference with respect to the free electron model is given by the Fourier component $V_{m-n}$, describing the coupling between two states $m$ and $n$. The question becomes: when does this term contribute significantly? For that we look at two orthogonal states ${\varphi }_n$ and ${\varphi }_m$, and construct the Hamiltonian in the basis (${\varphi }_n$,${\varphi }_m$),
$$\hat{H}=\left(\begin{array}{ccc}{\displaystyle \frac{{\hslash }^2{k}_n^2}{2m}}& V_{n-m}{V}_{m-n}& {\displaystyle \frac{{\hslash }^2{k}_m^2}{2m}}\end{array}\right)$$ The eigenvalues of this fellow are $$E=\frac{{\hslash }^2(k_n^2+k_m^2)}{4m}\pm \sqrt{\frac{{\hslash }^4}{16m^2}(k_n^2-k_m^2{)}^2+|V_{n-m}{|}^2}.$$ This clearly displays that only if $|k_n|\approx |k_m|$, the band structure will be affected (given that the potential is weak, and therefore small). This nicely demonstrates how an avoided crossing arises.

Exercise 3: The Tight Binding Model versus the Nearly Free Electron Model

Subquestion 1

We construct the Hamiltonian (note that we have exactly one delta-peak per unit cell of the lattice), \begin{equation} \hat{H}= $$\left(\begin{array}{ccc}⟨{\psi }_1|\hat{H}|{\psi }_1⟩& ⟨{\psi }_1|\hat{H}|{\psi }_2⟩⟨{\psi }_2|\hat{H}|{\psi }_1⟩& ⟨{\psi }_2|\hat{H}|{\psi }_2⟩\end{array}\right)$$ \end{equation} The bottom band means we have to pick the lowest energy band, i.e. the dispersion with the lowest eigenvalues, which is \begin{equation} E_-(k) = -\frac{\lambda}{a}+\frac{\hbar^2}{4m}\left[k^2+\left(k-\frac{2\pi}{a}\right)^2 \right]-\sqrt{\left(\frac{\hbar^2}{4m}\left[k^2-\left(k-\frac{2\pi}{a}\right)^2 \right]\right)^2 + \left(\frac{\lambda}{a}\right)^2} \end{equation}

Subquestion 2

See the lecture notes!

Subquestion 3

We split the Hamiltonian into two parts $H=H_n+H_{\bar{n}}$, where $H_n$ describes a particle in a single delta-function potential well, and $H_{\hat{n}}$ is the perturbation by the other delta functions: $$\begin{array}{rlr}{H}_n=& \frac{-{\hslash }^2}{2m}\frac{{\partial }^2}{\partial x^2}-\lambda \delta (x-na)H_{\bar{n}}=& -\lambda \sum _{m\ne n}\delta (x-ma)\end{array}$$ such that $H_n|n⟩={ϵ}_0|n⟩=-{\hslash }^2{\kappa }^2/2m|n⟩$ with $\kappa =m\lambda /{\hslash }^2$. We can now calculate $$⟨n|H|n⟩={ϵ}_0+⟨n|H_{\bar{n}}|n⟩$$ Note that the last term represents the change in energy of the wavefunction $|n⟩$ that is centered at the $n$-th delta function caused by the presence of the other delta functions. This term yields $$⟨n|H_{\bar{n}}|n⟩=-\kappa \lambda \sum _{m\ne 0}\int e^{-2\kappa |x|}\delta (x-ma)=-\kappa \lambda \sum _{m\ne 0}{e}^{-2\kappa |ma|}=-2\kappa \lambda (\frac{1}{1-e^{-2\kappa a}}-1)$$ Note that the result should not depend on $n$, so we chose $n=0$ for convenience.

Similarly, we can calculate $$⟨n-1|H|n⟩={ϵ}_0⟨n-1|n⟩+⟨n-1|H_{\bar{n}}|n⟩$$ where $⟨n-1|n⟩$ is the overlap between two neighbouring wavefunctions: $$⟨n-1|n⟩=2\kappa {\int }_0^{\mathrm{\infty }}{e}^{-\kappa |x-a/2|}{e}^{-\kappa |x+a/2|}=e^{-\kappa a}(1+\kappa a)$$ and $$\begin{array}{rlr}⟨n-1|H_{\bar{n}}|n⟩=& -\kappa \lambda \sum _{m\ne 0}\int e^{-\kappa |x-a|}\delta (x-ma)e^{-\kappa |x|}=& -\kappa \lambda \sum _{m\ne 0}{e}^{-\kappa a|m-1|}{e}^{-\kappa a|m|}=-\kappa \lambda (e^{ka}+e^{-ka})\sum _{m=1}^{m=\mathrm{\infty }}{e}^{-2\kappa am}\end{array}$$

In the limit $\kappa a\gg 1$ (i.e., where the distance between the delta functions is large compared to the width of the isolated orbitals), the onsite energy becomes $$⟨n|H|n⟩={ϵ}_0(1-4e^{-2\kappa a})$$ and the hopping becomes $$⟨n-1|H|n⟩={ϵ}_0(\kappa a-2)e^{-\kappa a}\approx {ϵ}_0\kappa a{e}^{-\kappa a}$$

Subquestion 4

..	Lower Band minimum	Lower Band Width
TB model	${\epsilon }_0-2t$	$4t$
NFE model	$E_{-}(0)$	$E_{-}(\pi /2)-E_{-}(0)$

Subquestion 5

Notice which approximations were made! For large $\lambda a$, the tight binding is more accurate, while for small $\lambda a$, the nearly free electron model is more accurate. The transition point for the regimes lies around $\lambda a\approx {\hslash }^2/m$.