Assignment 3: the reciprocal lattice, scattering, and semiconductors

The third assignment can be found here

Exercise 1 - The reciprocal lattice

Following the normal conventions, let us denote $\mathbf{a}_i$ and $\mathbf{b}_i$ as the real-space and reciprocal space lattice vectors.

1. A construction of lattice vectors can be achieved using the relation $$\mathbf{b}_i = 2\pi \frac{\mathbf{a}_j \times \mathbf{a}_k}{\mathbf{a}_1 \cdot (\mathbf{a}_2 \times \mathbf{a}_3)}$$ Explicitly compute $\mathbf{a}_1 \cdot \mathbf{b}_1$, $\mathbf{a}_2 \cdot \mathbf{b}_1$, and $\mathbf{a}_3 \cdot \mathbf{b}_1$. Do these computations accord with the definition of the reciprocal lattice?

   The computation of these vectors is aided by knowledge of properties of the scalar triple product. In calculating the product:

   $$\mathbf{a}_i \cdot \mathbf{b}_1 = 2\pi \frac{\mathbf{a}_i \cdot (\mathbf{a}_2 \times \mathbf{a}_3)}{\mathbf{a}_1 \cdot (\mathbf{a}_2 \times \mathbf{a}_3)}$$

   it should be immediately obvious that $\mathbf{a}_2 \times \mathbf{a}_3$ is orthogonal to both $\mathbf{a}_2$ and $\mathbf{a}_3$, so the dot product between $\mathbf{a}_2$ and $\mathbf{a}_3$ will be zero. In the case of $\mathbf{a}_1$, we then have identical vector products on both numerator and denominator and therefore the product evaluates to $2\pi$.

   The construction of the reciprocal lattice is baser around the identity

   $$e^{i\mathbf{G}\cdot\mathbf{R}} = 1$$

   for any lattice point $\mathbf{R}$ and any reciprocal lattice point $\mathbf{G}$. That relation only holds when

   $$\mathbf{a}_i \cdot \mathbf{b}_j = 2\pi \delta_{ij}$$

   and hence our relations above are looking good.

2. The volume of a primitive unit cell with lattice vectors $\mathbf{a}_i$ is given by $V = \left|\mathbf{a}_1 \cdot (\mathbf{a}_2 \times \mathbf{a}_3) \right|$. Find the volume of the corresponding primitive unit cell in reciprocal space.

   The volume in reciprocal space $V'$ should be computed using the same method as provided:

   $$\begin{align} V' & = \left|\mathbf{b}_1 \cdot (\mathbf{b}_2 \times \mathbf{b}_3) \right| \\ & = 2\pi \left| \frac{(\mathbf{a}_2 \times \mathbf{a}_3)}{\mathbf{a}_1 \cdot (\mathbf{a}_2 \times \mathbf{a}_3)} \cdot (\mathbf{b}_2 \times \mathbf{b}_3) \right| \\ & = \frac{2\pi}{V}\left| (\mathbf{a}_2 \times \mathbf{a}_3) \cdot (\mathbf{b}_2 \times \mathbf{b}_3) \right| \\ & = \frac{2\pi}{V}\left| (\mathbf{a}_2 \cdot \mathbf{b}_2)(\mathbf{a}_3 \cdot \mathbf{b}_3) - ( \mathbf{a}_2 \cdot \mathbf{b}_3)( \mathbf{a}_3 \cdot \mathbf{b}_2) \right| \\ & = \frac{(2\pi)^3}{V} \end{align}$$

3. Show that the general direction $[hkl]$ in a cubic crystal is normal to the planes with Miller indices $(hkl)$.

   Consider a (real-space) lattice with basis vectors $\mathbf{a}$, $\mathbf{b}$, and $\mathbf{c}$ which are assumed orthogonal (cubic crystal). Let the lengths of these vectors be $a$, $b$, and $c$ respectively. The plane $(hkl)$ will have intercepts with the axes at (a/h, 0, 0), (0, b/k, 0), (0, 0, c/l), and from these 3 points, one can construct a vector normal to the plane (which defines the plane) by taking the cross products between any two vectors between the points above. For example, consider

   $$\begin{align} \mathbf{n} & = (\mathbf{a}/h - \mathbf{b}/k) \times (\mathbf{a}/h - \mathbf{c}/k) \\ & = \frac{abc}{hkl} \left( \frac{h}{a^2}\mathbf{a} + \frac{k}{b^2}\mathbf{b} + \frac{l}{c^2}\mathbf{c} \right) \end{align}$$

   and this is only parallel to the vector $[hkl]$ in the case of $a = b = c$.

4. Is the above statement true for an orthorhombic crystal? Justify your response.

   No, it is not true, see the question above!

5. Show that the distance between two adjacent Miller planes $(hkl)$ of any lattice is $d=2\pi/|\mathbf{G}_{\textrm{min}}|$, where $\mathbf{G}_{\textrm{min}}$ is the shortest reciprocal lattice vector perpendicular to these Miller planes.

   The unit vector normal to the plane can be computed via

   $$\hat{\mathbf{n}} = \frac{\mathbf{G}}{|\mathbf{G}|}.$$

   Let us consider a very simple case in which we have the miller planes $(h00)$. For lattice planes, there is always a plane intersecting the zero lattice point $(0,0,0)$. As such, the distance from this plane to the closest next one is given by

   $$d = \hat{\mathbf{n}} \cdot \frac{\mathbf{a_1}}{h} = \frac{2\pi}{|\mathbf{G}|}$$

6. Find the family of Miller planes of the BCC lattice that has the highest density of lattice points. It may of useful to think about the density of lattice points per unit area on a Miller plane which is given by $\rho=d/V$.

   Since $\rho=d/V$, to maximise $\rho$ me must either must maximize $d$ or minimise $V$, the latter of which is fixed. Therefore, to maximise $d$, we minimize must $|\mathbf{G}|$ and thus the smallest possible reciprocal lattice vectors are the {110} family of planes.

Exercise 2 - Scattering

1. What is the origin of the Laue condition? That is, why is the amplitude of a scattered wave zero if $\mathbf{k'} - \mathbf{k} \ne \mathbf{G}$?

   If $\mathbf{k'} - \mathbf{k} \ne \mathbf{G}$, then the argument of the exponent has a phase factor dependent on the real-space lattice points. Because we sum over each of these lattice points, each argument has a different phase. Summing over all these phases results in an average amplitude of 0, resulting in no intensity peaks.

2. Consider a two-dimensional crystal with a rectangular lattice and lattice vectors $\mathbf{a}_1 = (0.468, 0) \mathrm{nm}$ and $\mathbf{a}_2 = (0, 0.342) \mathrm{nm}$ (so that $\mathbf{a}_1$ points along $x$-axis and $\mathbf{a}_2$ points along $y$-axis)

   1. Sketch the reciprocal lattice of this crystal

   

   1. Consider an X-ray diffraction experiment performed on this crystal using monochromatic X-rays with wavelength $0.166\mathrm{nm}$. Assuming elastic scattering, find the magnitude of the wave vectors of the incident and reflected X-rays

   With elastic scattering, we have $|k| = |k'| = 2\pi\lambda = 37.9 \mathrm{nm^{-1}}$

   1. On your sketch of the reciprocal lattice, draw the "scattering triangle" corresponding to the diffraction from (210) planes. Explicitly, use the Laue condition $\Delta \mathbf{k} = \mathbf{G}$ for constructive interference of diffracted X-rays

   

Exercise 3 - Structure determination

A diffraction experiment with an unknown crystalline powder sample was performed using a tungsten X-ray tube. Tungsten has $K_{\alpha}$ emission lines $K_{\alpha_1} = 59 318.8\mathrm{eV}$ and $K_{\alpha_2} = 57 981.9\mathrm{eV}$, and the ratio of intensities of the emissions lines is $\alpha_2/\alpha_1 \approx 0.115$.

1. Explain how X-ray tubes produce X-rays, and describe how one would go about performing a powder diffraction experiment.

   This is a more detailed response than I would expect, but for a complete picutre: in an X-ray tube, electrons are emitted from the filament and accelerated towards the anode, and two forms of radiation are produced from the interaction between the electrons and the anode. The first kind is Bremsstrahlung radiation, which will be emitted as the electron is decelerated by the atomic nuclei of the anode. The emission spectrum is continuous, with the approximate shape described by Kramers' Law:

   $$I(\lambda)d\lambda = K \left(\frac{\lambda}{\lambda_{\textrm{min}}}-1\right)\frac{1}{\lambda^2}d\lambda$$

   where the cutoff wavelength, $\lambda_{\textrm{min}}$ is given by the Duane-Hunt law:

   $$\lambda_{\textrm{min}} = \frac{hc}{eV}.$$

   The key point relating to the distribution is that with an increased acceleration voltage, the Bremsstrahlung radiation intensity increases and shifts towards higher frequencies.

   The second kind of radiation is the characteristic X-ray emission, which occurs when a high-lying electron decays to a (vacant) lower energy level, resulting in the emission of a photon. The bombardment of the anode by energetic electrons results in the emission of inner-shell electrons from the atoms in the anode. Such vacancies are filled by the decay of outer-shell electrons and due to the unique energy structure of each element, each atom emits photons of distinct frequencies, completely analogous to atomic optical spectra.

   The combination of these effects means that the expected emission spectrum would expect a smooth, continuous spectrum punctuated by emissions of characteristic X-rays.

   Powder diffraction experiments are performed by placing a crystalline sample in powdered form in an X-ray beamline and collecting the scattered radiation from the sample. The diffraction pattern will by design form rings rather than spots, as every possible crystal orientation is assured by using a powder rather than a monocrystalline sample.

2. A plot of the data measured from the experiment when $K_{\alpha_1}$ emission was used is shown below:

   1. Following the recipe discussed in class, produce a table with columns of angle, plane separation, ratio of the square of first plane separation to plane separation, $N = h^2 + k^2 + l^2$, ${hkl}$, and $a$ (assuming some kind of cubic lattice)

      This should be relatively straightforward, given the example we did in class was trickier and the code used to do the analysis was freely distributed. The peaks as found from the data are shown below:

      

      Peak finding code

      peaks, _ = find_peaks(data['Intensity'])
      
      fig, ax = plt.subplots()
      
      ax.plot(data['Angle'], data['Intensity']/maxamp, label = r'W $K_{\alpha_1}$', color = 'C0')
      ax.plot(data['Angle'][peaks], data['Intensity'][peaks]/maxamp, "x", color = 'C1', label = 'Peaks')
      ax.set_xlabel(r'Angle [2$\theta$]')
      ax.set_xlim((19,82))
      ax.set_ylabel('Intensity [arb]')
      ax.set_title('X-ray scattering of unknown material');
      plt.legend()
      
      loc = matplotlib.ticker.MultipleLocator(1)
      ax.xaxis.set_minor_locator(loc)
      ax.tick_params(which='minor')
      
      if True:
      plt.savefig('A-6-peaks.svg', facecolor='white', transparent=False, bbox_inches='tight')
      
      plt.show()
      

      from which the following data table can be constructed:

      Peak number	$2\theta$	$d~[\mathrm{pm}]$	$d_a^2/d^2$	$N$	${hkl}$	$a~[\mathrm{pm}]$
      1	20.461	58.84	1.00	1	[1, 0, 0]	58.84
      2	29.086	41.61	1.99	2	[1, 1, 0]	58.84
      3	35.828	33.97	2.99	3	[1, 1, 1]	58.84
      4	41.609	29.42	3.99	4	[2, 0, 0]	58.84
      5	46.788	26.32	4.99	5	[2, 1, 0]	58.84
      6	51.567	24.02	5.99	6	[2, 1, 1]	58.84
      7	60.302	20.80	7.99	8	[2, 2, 0]	58.84
      8	64.380	19.61	8.99	9	[2, 2, 1]	58.84
      9	68.327	18.61	9.99	10	[3, 0, 0]	58.84
      10	72.164	17.74	10.99	11	[3, 1, 0]	58.84
      11	75.920	16.98	11.99	12	[3, 1, 1]	58.84
      12	79.617	16.32	12.99	13	[2, 2, 2]	58.84

      The difficulty of finding the multiplier to make all values of $d_a^2/d^2$ an integer is not present in this example, so the process should be straightforward to find $a=58.84 \mathrm{pm}$

   2. Use the table above to determine the lattice structure of the crystal

      The lattice is a simple cubic lattice, which can be inferred by the calculated values of N above and the selection rule for simple cubic lattices ($N$ is all integers excluding $7, 15, 23, \ldots$), and this is directly visible from the diffraction data: peak 7 is clearly missing.

3. The basis of the lattice is given by $X = [0,0,0]$ and $Y = [1/2, 1/2, \beta], [1/2, 1/2, (1 - \beta)] , [1/2, \beta, 1/2], [(1 - \beta), 1/2, 1/2], [1/2, 1/2, \beta],$ and $[(1 - \beta), 1/2, 1/2]$ where $X$ and $Y$ are different atomic species, and $\beta \approx 0.2$.

   1. Draw the unit cell for the crystal using your lattice and the basis specified above.

      Below is the crystal structure (many unit cells) and can be well described as a simple cubic with a neat prism inside. The material is $\textrm{LaB}_6$, which is the most gorgeous shade of pink, and is a neat material for making hot cathode emitters (electron sources).

      

   2. Explain how the intensity of the peaks could be used to determine $\beta$, and obtain an expression for the ratio of the first two diffraction peaks. Note: You do not need to solve this equation for $\beta$, just arrive at something that could be used to calculate $\beta$.

      The intensity of the peaks is related to the square (well, modulus squared) of the structure factor $S$. As we have been using it in class, the structure factor is

      $$S_{(hkl)} = \sum_\alpha f_\alpha e^{i\mathbf{G}\cdot\mathbf{R}_\alpha}$$

      which in reality means that one must compute

      $$S_{(hkl)} = \sum_\alpha f_\alpha e^{2\pi i(hkl)\cdot[xyz]_\alpha}$$

      This is not so difficult, but can be a bit tedious depending on the basis of the unit cell. Given we are looking at the intensities of the first two peaks, this means we will need to look at $S_{(100)}$:

      $$\begin{aligned} S_{(100)} & = f_\textrm{X} + f_\textrm{Y}\left( e^{i\pi} + e^{i\pi} + e^{2\pi i \beta} + e^{2\pi i (1-\beta)} + e^{i\pi} + e^{i\pi} \right) \\ & = f_\textrm{X} + f_\textrm{Y}\left( e^{2\pi i \beta} + e^{-2\pi i \beta} e^{2\pi i} - 4 \right) \\ & = f_\textrm{X} + f_\textrm{Y}\left( 2\cos(2\pi\beta) - 4 \right) \\ & = f_\textrm{X} - 2 f_\textrm{Y}\left( 1 + \sin^2(\pi\beta) \right) \\ \end{aligned}$$

      and $S_{(110)}$:

      $$\begin{aligned} S_{(110)} & = f_\textrm{X} + f_\textrm{Y}\left( e^{2\pi i} + e^{2\pi i} + 2\left( e^{2\pi i \beta}e^{i\pi} + e^{2\pi i (1-\beta)}e^{i\pi} \right) \right) \\ & = f_\textrm{X} + f_\textrm{Y}\left( 2 - 2\left( e^{2\pi i \beta} + e^{-2\pi i \beta} e^{2\pi i} \right) \right) \\ & = f_\textrm{X} + f_\textrm{Y}\left( 2 - \cos(2\pi\beta) \right) \\ & = f_\textrm{X} + f_\textrm{Y}\left( 1 - 2\sin^2(\pi\beta) \right) \\ \end{aligned}$$

      One could then look at the ratio $I_{(100)}/I_{(110)} = |S_{(100)}|^2/|S_{(110)}|^2$ and try and solve for $\beta$, but that will be a mess, so one should solve it numerically. The real value is $\beta = 0.1981$, although I am not sure exactly what the ratio of $I_{(100)}/I_{(110)}$ would return.

4. Imagine the experiment was altered such that both $K_{\alpha_1}$ and $K_{\alpha_2}$ emission lines were present. How would this alter the data as recorded above? Would you expect that one could still uniquely determine the crystal structure of the sample?

   Shown below is the spectrum with both emission lines:

   

   With two emission lines, one would expect to have essentially to diffraction patterns superimposed, but as there is sufficient energy separation between the two emission lines, one can resolve the two peaks (well, in simulated data). The question explicitly states that the ratio of intensities for emission from the different transitions is $\alpha_2/\alpha_1 \approx 0.115$, meaning the $K_{\alpha_2}$ peaks will be noticeably smaller. The main point is that there will be two distinct diffraction patterns, and provided one can uniquely distinguish from which emission line the peak comes, one would actually be able to better determine the crystal structure, as one effectively has double the number of peaks. This method is used more broadly, that is, not using monochromatic X-rays but rather polychromatic X-rays in order to get more information per unit diffraction, but one must know well the illuminating radiation. This is usually called "pink" beam illumination.

5. Now imagine that the experiment were altered such that only $K_{\alpha_1}$ radiation were used, but a monocrystalline sample were used. What would be the difference in the recorded diffraction pattern?

   If one were to use a monocrystalline sample, the diffraction pattern would have distinct Bragg spots rather than rings, but then we need to care much more about crystal orientation (it also usually very difficult to get large monocrystalline samples). The image below shows diffraction from monocrystalline silicon:

   

6. Unfortunately, the beautiful single crystal was dropped before it could be used, resulting in a sample that is neither amorphous nor monocrystalline, rather something between the two. How would this alter the appearance of the diffraction pattern?

   With a sample between monocrystalline and amorphous, there are extended regions of order and one can imagine either the Bragg spots blurring out around the optical axis (around $\theta$ in polar coordinates), or the rings "sharpening up", that is still ring-like but with more intensity at the Bragg spots. The image below shows both powder and grain-oriented diffraction from aluminium:

   

Exercise 4 - The nearly-free electron model

Consider an electron in a weak periodic potential in one dimension $V(X) = V(x+a)$. It is natural to write the potential as $$V(x) = \sum_G e^{iGx} V_G$$ where the sum is over the reciprocal lattice $G = 2\pi n/a$ and $V_G* = V_{-G}$ assures the potential $V(x)$ is real.

1. Explain why for k near to a Brillouin zone boundary (such as $k$ near $\pi/a$) the electron wavefunction should be taken to be
   $$\psi = A e^{ikx} + B e^{i(k+G)x}$$ where $G$ is a reciprocal lattice vector such that $|k|$ is close to $|k+G|$.

   A periodic lattice can only scatter a wave by a reciprocal lattice vector (Bragg diffraction). In the nearly free electron picture, the scattering perturbation is weak, so that we can treat the scattered wave in perturbation theory. In this case, there is an energy denominator which suppresses mixing of $k$-vectors which have greatly different unperturbed energies. Thus, the only mixing that can occur is between two states with similar energies that are separated by a reciprocal lattice vector. Degenerate perturbation theory tells us that we should first diagonalize within the degenerate space spanned by only these two eigenstates.

2. We have seen that with the above wavefunction, the energy (that is, the eigenvalues) at this wavevector are given by $$E = \frac{\hbar^2 k^2}{2m} + V_0 \pm |V_G|$$ where $G$ is chosen such that $|k| = |k+G|$.

   1. Give a qualitative explanation of why these two states are separated in energy by $2|V_G|$

      If we consider only the $V_{2n\pi/a}$ and $V_{-2n\pi/a}$ Fourier modes of the potential then we have $V = 2 V_{2n\pi/a} \cos(2n\pi r/a)$. Assuming $V_{2n\pi/a} > 0$, then the higher energy state is the $\psi = \cos(2n\pi r/a)$ which puts the maximum amplitude of the wavefunction exactly at the maxima of the potential. Similarly, the lower energy wavefunction is the $\sin(2n\pi r/a)$ which has the minimum amplitude of the wavefunction at the maximum of the potential. In the case of $V_{2n\pi /a} < 0$ the sin is the higher energy wavefunction.

   2. Provide a sketch or plot of the energy as a function of $k$ in both the extended and reduced zone schemes. Note that one need not compute $E$ for all $k$, emphasis should be on the general features of the energy spectrum.

      These plots below are actually calculated by solving the eigenvalue problem as outlined in the next question for all $k$ (where the perturbation is valid, that is where the term in $\delta k$ to second order is small compared $|V_{2n\pi/a}|^2$).

      In the reduced zone scheme, one should have something like:

      

      Reduced-zone scheme code

      # Use colors from the default color cycle
      default_colors = plt.rcParams['axes.prop_cycle'].by_key()['color']
      blue, orange, *_ = default_colors
      
      def energy(k, V=1):
          k = (k + np.pi) % (2*np.pi) - np.pi
          k_vals = k + 2*np.pi * np.arange(-1, 2)
          h = np.diag(k_vals**2) + V * (1 - np.identity(3))
          return np.linalg.eigvalsh(h)
      
      energy = np.vectorize(energy, signature="(),()->(m)")
      
      fig, ax = plt.subplots(1, 1)
      
      momenta = np.linspace(-np.pi, np.pi, 400)
      energies = energy(momenta, 0)
      max_en = 41
      energies[energies > max_en] = np.nan
      ax.plot(momenta, energies, c=blue, label = 'Free electron')
      energies = energy(momenta, 2)
      max_en = 41
      energies[energies > max_en] = np.nan
      ax.plot(momenta, energies, c=orange, label = 'Nearly free electron')
      
      ax.set_xlabel("$ka$")
      ax.set_ylabel("$E$")
      ax.set_ylim(-.5, max_en + 5)
      ax.set_xticks(np.pi * np.arange(-1, 2))
      ax.set_xticklabels(r"$-\pi$ $0$ $\pi$".split())
      
      
      def legend_without_duplicate_labels(ax):
          handles, labels = ax.get_legend_handles_labels()
          unique = [(h, l) for i, (h, l) in enumerate(zip(handles, labels)) if l not in labels[:i]]
          ax.legend(*zip(*unique))
      
      legend_without_duplicate_labels(ax)
      
      if True:
          plt.savefig('A-6-reduced.svg', facecolor='white', transparent=False, bbox_inches='tight')
      
      plt.show()
      

      Whereas in the extended zone scheme, one should have:

      

      Extended-zone scheme code

      fig, ax = plt.subplots(1, 1)
      
      momenta = np.linspace(-3*np.pi, 3*np.pi, 400)
      energies = energy(momenta, 0)
      max_en = 41
      energies[energies > max_en] = np.nan
      energies[~((abs(momenta) // np.pi).reshape(-1, 1) == np.arange(3).reshape(1, -1))] = np.nan
      ax.plot(momenta, energies, c=blue, label = 'Free electron')
      energies = energy(momenta, 2)
      max_en = 41
      energies[energies > max_en] = np.nan
      energies[~((abs(momenta) // np.pi).reshape(-1, 1) == np.arange(3).reshape(1, -1))] = np.nan
      ax.plot(momenta, energies, c=orange, label = 'Nearly free electron')
      
      ax.set_xlabel("$ka$")
      ax.set_ylabel("$E$")
      ax.set_ylim(-.5, max_en + 5)
      ax.set_xticks(np.pi * np.arange(-3, 4))
      ax.set_xticklabels(fr"${i}\pi$".replace("1", "") if i else "$0$" for i in range(-3, 4))
      
      legend_without_duplicate_labels(ax)
      
      if True:
          plt.savefig('A-6-extended.svg', facecolor='white', transparent=False, bbox_inches='tight')
      
      plt.show()
      

3. Let us look at the case where $k$ is not at the Brillouin zone boundary, but rather close to the boundary. Following the same method as used to achieve the above result, show that at the point $k = n\pi/a + \delta k$ the energy to second order in $\delta k$ is given by $$E_{\pm}=\frac{\hbar^{2}(n \pi / a)^{2}}{2 m}+V_{0} \pm\left|V_{2 n \pi / a}\right|+\frac{\hbar^{2}(\delta k)^{2}}{2 m}\left(1 \pm \frac{\hbar^{2}(n \pi / a)^{2}}{m\left|V_{2 n \pi / a}\right|}\right)$$

   Technically the wavefunction should have a normalisation

   $$\psi = (A e^{ikx} + B e^{i(k+G)x})/\sqrt{L}$$

   but it is rarely the case that one is actually probing for normalisation compliance, and in any case, to maintain normalization we can insist that $|A|^2 + |B|^2 = 1$. Taking $k$ and $k+G$ both on a Brillouin zone boundary we have $k = n\pi/a$ and $k+G = -n\pi/a$, where here we have chosen the $n^{\textrm{th}}$ zone boundary, and we must have $G = -2n\pi/a$ the reciprocal lattice vector. The Hamiltonian $H$ in question is the usual Kinetic term plus $V(x)$. From here, one can either use the variational method or diagonalise the Hamiltonian the degenerate space, the latter of which was done in class and is done here.

   We must compute the matrix elements:

   $$\begin{aligned} \langle k|H| k\rangle &=\hbar^{2}(\delta k+n \pi / a)^{2} /(2 m)+V_{0} \\ \langle k+G|H| k+G\rangle &=\hbar^{2}(\delta k-n \pi / a)^{2} /(2 m)+V_{0} \\ \langle k|H| k+G\rangle &=V_{2 n \pi / a} \\ \langle k+G|H| k\rangle &=V_{-2 n \pi / a} \end{aligned}$$

   which means that me must diagonalise the matrix

   $$\left(\begin{array}{cc} \hbar^{2}(\delta k+n \pi / a)^{2} /(2 m)+V_{0} & V_{2 n \pi / a} \\ V_{-2 n \pi / a} & \hbar^{2}(\delta k-n \pi / a)^{2} /(2 m)+V_{0} \end{array}\right).$$

   Performing the diagonalisation, one finds

   $$E_{\pm}=\frac{\hbar^{2}\left[(\delta k)^{2}+(n \pi / a)^{2}\right]}{2 m}+V_{0} \pm \sqrt{\left[\frac{\hbar^{2} 2(\delta k) n \pi / a}{2 m}\right]^{2}+\left|V_{2 n \pi / a}\right|^{2}}$$

   and the square root should be expanded and the common terms in $\delta k$ collected to obtain the result

   $$E_{\pm}=\frac{\hbar^{2}(n \pi / a)^{2}}{2 m}+V_{0} \pm\left|V_{2 n \pi / a}\right|+\frac{\hbar^{2}(\delta k)^{2}}{2 m}\left(1 \pm \frac{\hbar^{2}(n \pi / a)^{2}}{m\left|V_{2 n \pi / a}\right|}\right)$$

   as required.

4. Calculate the effective mass of an electron at this wavevector

   The effective mass can be obtained from

   $$\frac{1}{2m^*} = \frac{1}{2m}\left(1 \pm \frac{\hbar^{2}(n \pi / a)^{2}}{m\left|V_{2 n \pi / a}\right|}\right)$$

   or equivalently

   $$m^{*}=\left|\frac{m}{1 \pm \frac{\hbar^{2}(n \pi / a)^{2}}{m\left|V_{2 n \pi / a}\right|}}\right|$$

Exercise 5 - Semiconductors: holes

1. In the context of semiconductor physics, what is meant by a hole and why is it useful?

   A hole is the absence of an electron in an otherwise filled valence band. This is useful since instead of describing the dynamics of all the (many) electrons in the band, it is equivalent to describe the dynamics of just the (few) holes.

2. An electron near the top of the valence band in a semiconductor has energy $$E = -10^{-37}|k|^2$$ where $E$ is in Joules, and $k$ is in $\mathrm{m^{-1}}$. An electron is removed from a state $k = 2 \times 10^8 \mathrm{m^{-1}} \hat{x}$, where $\hat{x}$ is the unit vector in the $x-$direction. For a hole, calculate (including the sign)

   1. the effective mass

      Effective mass $\hbar^2 k^2/(2m^∗) = 10^{-37} k^2$. So $m^∗ = 5 \times 10^{-32} \mathrm{kg}$ or $.05$ the mass of the electron. This mass is positive in the usual convention.

   2. the energy

      The energy is $E = 10^{-37} k^2 = 4\times 10^{-21} \mathrm{J}$, or about $0.025\mathrm{eV}$. This energy is positive (it takes energy to "push" the hole down into the fermi sea, like pushing a balloon under water).

   3. the velocity

      Getting the velocity (and momentum) right are tricky. First, note that the velocity of an eigenstate is the same whether or not the state is filled with an electron. It is always true that the velocity of an electron in a state is $\nabla_k E_k/\hbar$ where $E_k$ is the electron energy. Thus the hole velocity here is negative $v = -\hbar k/m^∗ = -3.8 \times 10^5 \mathrm{m/s}$ (i.e. the velocity is in the negative $\hat{x}$) direction.

   4. the momentum

      For momentum, since a filled band carries no (crystal) momentum, and for electrons crystal momentum is always $\hbar k$, the removal of an electron leaves the band with net momentum $-\hbar k$ which we assign as the momentum of the hole. Thus we obtain hole momentum $-\hbar k = -2.1 \times 10^{-26} \mathrm{kg ~ m/s}$ which is also in the negative \hat{x} direction (this matches well to the intuition that $p = mv$ with a positive effective mass for holes).

3. If there is a density $p = 10^5 \mathrm{m^{-3}}$ of such holes all having almost exactly this same momentum, calculate the current density and its sign.

   With $p$ the density of such holes, the total current density is $p e v = -6 \times 10^{-9} \mathrm{A/m^2}$ also in the negative \hat{x} direction (noting that the charge of the hole is positive).

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Last update: October 30, 2023

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