### Solution

A1  ?? Substitute $\Psi(x) = e^{ikx/a} u(x/a)$ into the Schrödinger equation and get the equation
$u'' + 2ik u'+\left[ \frac{2ma^2}{\hbar^2} \left( E- U(x) \right) - k^2 \right] u = 0$
on $u(y)$.

$\Psi' = \frac{ik}{a} e^{ikx/a} u + e^{ikx/a} u', \quad \Psi'' = -\frac{k^2}{a^2} e^{ikx/a}u + \frac{2ik}{a}e^{ikx/a}u' + e^{ikx/a} u''$
and substitute the results into the Schrödinger's equation.

A2  ?? Consider a linear differential equation with constant coefficients $\alpha$ and $\beta$:
$u'' + 2i\alpha u' +(\beta-\alpha^2) u = 0.$
Find the values of $\lambda_{1,2}$ for which the function $u_{1,2}(y) = e^{\lambda_{1,2} y}$ is a solution to this equation. Check that for any constants $A$ and $B$, the function $u(y) = A e^{\lambda_1 y} + B e^{\lambda_2 y}$ is also a solution to this equation.

Substituting the $u=e^{\lambda y}$ into the equation, we have
$e^{\lambda y} \left( \lambda^2 + 2i \alpha \lambda + \beta - \alpha^2 \right) = 0.$
So, let's solve the quadratic equation:
$\mathcal{D} = -4 \alpha^2 - 4 \beta + 4 \alpha^2 = -4 \beta, \quad \lambda_{1,2} = \frac{-2i \alpha \pm 2 \sqrt{-\beta}}{2}$

Answer: $\lambda_1 = -i\alpha + \sqrt{-\beta}, \quad \lambda_2 = -i\alpha - \sqrt{-\beta}$

A3  ?? Write a system of equations for the coefficients $A_1$, $A_2$, $B_1$ and $B_2$ corresponding to the requirements on the functions $u_I(x)$ and $u_II(x)$, in matrix form
$M \begin{pmatrix} A_1 \\ A_2 \\ B_1 \\ B_2 \end{pmatrix} = \begin{pmatrix} M_{11} & M_{12} & M_{13} & M_{14} \\ M_{21} & M_{22} & M_{23} & M_{24} \\ M_{31} & M_{32} & M_{33} & M_{34} \\ M_{41} & M_{42} & M_{43} & M_{44} \end{pmatrix} \begin{pmatrix} A_1 \\ A_2 \\ B_1 \\ B_2 \end{pmatrix} = 0.$
Write the coefficients of the matrix $M_{ij}$ by $\lambda_{11}$, $\lambda_{12}$, $\lambda_{21}$, $\lambda_{22}$, $a$, and $b$.

There are required relations for the constants:
$\begin{cases} A_1 e^{-\lambda_{11}b/a} + B_1 e^{-\lambda_{12}b/a} = A_2 e^{\lambda_{21}(1-b/a)} + B_2 e^{\lambda_{22}(1-b/a)}\\ A_1 + B_1 = A_2 + B_2 \\ \lambda_{11} A_1 + \lambda_2 B_1 = \lambda_{21} A_2 + \lambda_{22} B_2\end{cases}$
The 4th equation could be simplified to
$u'_I(-b/a) = u_{II}'(1-b/a) \quad \Rightarrow \quad A_1 \lambda_{11} e^{-\lambda_{11}b/a} + B_1 \lambda_{12} e^{-\lambda_{12}b/a} = A_2 \lambda_{21} e^{\lambda_{21}(1-b/a)} + B_2 \lambda_{22} e^{\lambda_{22}(1-b/a)}$

Answer: $M=\begin{pmatrix} 1 & -1& 1 & -1\\ \lambda_{11} & -\lambda_{21} & \lambda_{12} & -\lambda_{22} \\ e^{-\lambda_{11}b/a} & -e^{\lambda_{21}(1-b/a)} & e^{-\lambda_{12}b/a} & -e^{\lambda_{22}(1-b/a)} \\ \lambda_{11}e^{-\lambda_{11}b/a} & -\lambda_{21}e^{\lambda_{21}(1-b/a)} & \lambda_{12}e^{-\lambda_{12}b/a} & -\lambda_{22}e^{\lambda_{22}(1-b/a)} \\ \end{pmatrix}$

A5  ?? Show that the de Broglie wave $\Psi(x) = e ^{ikx/a}$ has the energy $E=\frac{\hbar^2 k^2}{2m a^2}$.

Answer: To find the energy we can just substitute the de Broglie wave to the Schödinger's equation.
$\frac{\hbar^2}{2m} \cdot \frac{k^2}{a^2} e^{ikx/a} = E e^{ikx/a}$

A6  ?? Express $E_F$ in terms of $L$, $N$, $m$, and $\hbar$.

To find the $E_F$ we have to consider the "box" with $N$ occupied levels. As $k= \pi n a /L$ and we have two different spin directions, $k_\text{max}=\pi N a / 2L$. Thus, the energy of the highest occupied level
$E_F = \frac{\hbar^2 k_\text{max}^2}{2ma^2} = \frac{\pi^2 \hbar^2 N^2}{8 m L^2}$

Answer: $E_F = \frac{\pi^2 \hbar^2 N^2}{8m L^2}$

A7  ?? Use the program energy.py to obtain the dependence of the band gap $E_g$, where the Fermi level is located, on the parameter $k_0$ at $b/a=0.1$. Consider $a=0.5~\text{nm}$, electron mass $m=9.1 \cdot 10^{-31}~\text{kg}$, Planck constant $\hbar=1.05 \cdot 10^{-34}~\text{J} \cdot \text{s}$, elementary charge $e=1.60 \cdot 10^{-19}~\text{C}$.
Fill in the table.txt table and use the program bandgap.py to draw a graph of this relationship. In table.txt table, the separator between columns is tab.

In the program we work with ondimensionalized energy $\varepsilon$. To find the energy $E$ that corresponds to it we have to use such an equation
$E = \frac{\varepsilon b}{a} U_0 = \varepsilon \frac{k_0^2 \hbar^2}{2ma^2}$

A8  ?? What $k_0$ corresponds to the forbidden zone of silicon $E_{\rm Si} = 1.12~\text{eV}$? Do all further calculations for this $k_0$.

Answer: $k_0=2.0$

A9  ?? Use the program energy.py to find the values of $E_c-E_F$, $E_F-E_v$, $\alpha$, and $\beta$. Check for yourself that $E_c - E_v = E_{\rm Si}$. Also find the values of $(E_c - E_F)/k_\text{B}T$ and $(E_F - E_v)/k_\text{B}T$ for $T=300~\text{K}$.

Answer: $E_c = E_F + 0.71~\text{eV}, \quad \alpha = 1.99~\text{eV}, \quad \frac{E_c - E_F}{k_\text{B}T}=27$
$E_v = E_F - 0.48~\text{eV}, \quad \beta = 1.68~\text{eV}, \quad \frac{E_F - E_v}{k_\text{B}T}=18$

A10  ?? Show that the total number $p$ of NOT-filled states in the valence band is
$p = \frac{L}{a} e^{\frac{E_v - E_F}{k_\text{B}T}} \sqrt{\frac{k_B T}{\pi\beta}}.$
This number $p$ is the number of holes involved in the transfer processes.

Show that $p \cdot n$ does not depend on the position of the Fermi level $E_F$. Find the value of $n_i= \sqrt{p \cdot n} \cdot a / L$ for silicon at temperature $T=300~\text{K}$.

$\begin{split} p &= \sum_{E \leq E_v} \left[ g(\Delta E, k) - F(g(\Delta E,k),E) \right] = \sum_{E \leq E_v} \left[ g_v (\Delta E,k) - \frac{g_v (\Delta E,k)}{e^{\frac{E-E_F}{k_\text{B}T}}+1} \right] \simeq \sum_{E \leq E_v} g_v e^{\frac{E-E_F}{k_\text{B}T}} \simeq\\&\simeq \int\limits_{-\infty}^{E_v}\frac{L}{\pi a \alpha k} e^{\frac{E - E_F}{k_\text{B}T}} dE = \frac{2L}{\pi a} e^{\frac{E_v- E_F}{k_\text{B}T}} \int\limits_0^{\infty} e^{-\frac{\beta k^2}{k_BT}}dk = \frac{L}{a} e^{\frac{E_v - E_F}{k_\text{B}T}} \sqrt{\frac{k_B T}{\pi \beta}}. \end{split}$

Answer: $pn = \frac{L^2}{a^2} e^{-\frac{E_g}{k_\text{B}T}} \frac{k_\text{B} T}{\pi\sqrt{\alpha \beta}}$

$n_i = \sqrt{pn} \cdot a/L = 2.7 \cdot 10^{-11}$

A11  ?? What should the Fermi level $E_F$ be so that the crystal is electrically neutral, i.e., the number of holes $p$ is equal to the number of electrons $n$?

Answer: $E_F = \frac{E_v + E_c}{2} - \frac{k_\text{B}T}{2} \ln \sqrt{\frac{\beta}{\alpha}}$

A12  ?? What fraction $c$ of silicon $\rm Si$ atoms must be replaced by aluminum $\rm Al$ to increase the ratio of holes to electrons $p/n$ to $q=10^6$ with respect to pure silicon? Consider the temperature to be $300~\text{K}$.

In the pure silicon $\rm Si$ we have $p=n$, so in the doped silicon there is $p = qn$. Changing atoms only shift the Fermi level, so $\sqrt{pn}=n_i$ and
$p =n_i \sqrt{q}, \quad n = \frac{n_i}{\sqrt{q}}.$

Now, let's satisfy the electronutrality condition:
$p - \frac{cL}{a} = n$
$c = \left( \sqrt{q} + \frac{1}{\sqrt{q}} \right) \cdot 2.6 \cdot 10^{-11} \simeq 2.6 \cdot 10^{-8}.$
Answer: $c = \left( \sqrt{q} + \frac{1}{\sqrt{q}} \right) \cdot 2.6 \cdot 10^{-11} \simeq 2.6 \cdot 10^{-8}.$