Question 3.7: Using the method outlined in Section 3.4, write a computer p......

Using the method outlined in Section 3.4, write a computer program to solve the Schrödinger wave equation for the first four eigenvalues and eigenstates of an electron with effective mass m^{*}_{e} =0.07\times m_{0} confined to a rectangular potential well of width L = 10 nm bounded by infinite barrier potential energy.

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We would like to use the method outlined in Section 3.4 to numerically solve the onedimensional Schrödinger wave equation for an electron of effective mass m^{*}_{e} =0.07\times m_{0} in a rectangular potential well of width L = 10 nm bounded by infinite barrier energy. Our solution is a MATLAB computer program which may be found on this book’s website and which consists of two parts. The first part deals with input parameters such
as the length L, effective electron mass, the number of discretization points, N, and the plotting routine. It is important to choose a high enough value of N so that the wave function does not vary too much between adjacent discretization points and so that the three-point finite-difference approximation used in Eq. (3.41) is reasonably accurate.

\frac{d^{2} }{dx} \psi (x_{j} )=\frac{\psi (x_{j  –  1} )  –  2\psi (x_{j} )  +  \psi (x_{j  +  1} )}{h^{2}_{0} } (3.41)

The second part of the computer program solves the discretized Hamiltonian matrix (Eq. (3.45)) and is a function, solve_schM, called from the main routine, Chapt3Exercise7. The diagonal matrix element given by Eq. (3.43) has potential values V_{j} =0. The adjacent off-diagonal matrix elements are given by Eq. (3.44). In this particular exercise the first four energy eigenvalues are E_{0} = 0.0537  eV, E_{1} = 0.2149  eV, E_{2} = 0.4834  eV, and E_{3} = 0.8592  eV. The eigenfunctions generated by the program and plotted in the figure are not normalized.

  d_{j} =\frac{\hbar ^{2} }{m\hbar ^{2}_{0} } +V_{j}               (3.43)

u_{j} =\frac{\hbar ^{2} }{2m\hbar ^{2}_{0} }                             (3.44)

(H-E1)\psi                                                                         (3.45)

=  \left [ \begin{matrix} (d_{1}-E )   & -u_{2} & 0 & 0 & \cdot & \cdot \\ -u_{2} & (d_{2}-E ) & -u_{3} & 0 & \cdot & \cdot \\ 0 & -u_{3} & (d_{3}-E ) & -u_{4} & \cdot & \cdot \\ \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot & \cdot & -u_{N-1} & (d_{N-1}-E ) \end{matrix} \right ] \left [ \begin{matrix} \psi _{1} \\ \psi _{2} \\ \psi _{3} \\ \cdot \\ \cdot \\ \psi _{N-1} \end{matrix} \right ] =0
Screenshot 2023-02-15 194313

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