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Simulation on the Nanoscale ENGM03
Part I Energy eigenvalues and eigenvectors in a finite potential well
In this lab you will numerically evaluate SchrÃ¶dingerâ??s equation using the built-in
MatLab function â??eigâ?? to yield the eigenvalues and eigenvectors associated with a
finite quantum well.
Finite Potential Well â?? eig
From your notes and Lab 3, we determined that the position dependent onedimensional form of SchrÃ¶dingerâ??s equation may be expressed as:
â?2 ð?? 1 ð??ð??(ð?§)
ð??0 ð??(ð?§) = ð¸ð??(ð?§)
2 ð??ð?§ ð??(ð?§) ð??ð?§
Where â? is Plankâ??s constant, ð?§ is the spatial position along the barrier-well structure,
ð??(ð?§) is the effective mass, ð??0 is the relative barrier height with respect to the well
bottom (eV), ð¸ and ð??(ð?§) are the eigenenergy (discrete eigenvalues) and
corresponding eigenvector (wavefunction) in the barrier-well-barrier system.
Using the finite difference technique (described in lectures), equation 1 can be
spatially discretised into the following form:
â?? ð??ð?? = ð¸ð??ð??
Where the index ð?? refers to the specific spatial location across the barrier-wellbarrier structure; in this case each consecutive integer value of ð?? is Î?ð?§ =
â?« (1 Ã? 10â??10m) and ð´ = â?? 2Î?ð?§ 2ð??. Equation 2 is now in tri-diagonal format, and can
be written in matrix form as:
ð??ð??â??1ð??â??1 ] [ð??ð??â??1 ]
where the ð??ð??ð?? â?²ð? refer to the coefficients of equation 2 at the ð?? ð?¡â?? mesh position, here
ð?? refers to the number of mesh points across a user defined potential (see above
We may express the coefficient ð?? Ã? ð?? matrix in equation 3 as a 3 Ã? ð?? matrix (we just
use the three leading diagonals of the coefficient matrix, as all other entries are
zero). In this form, SchrÃ¶dingerâ??s equation must be entered as the input to the
MATLAB function eig (see help eig), the function will then return the eigenvalues and
eigenvectors associated with the barrier-well system.
(a) Using the following parameters, develop a script file that constructs the 3 Ã? ð??
coefficient matrix, making sure you correctly assign the correct spatial values of
effective mass and potential at each discretised spatial location. The barriers either
side of the well are identical.
(i) hbar â?? Plankâ??s reduced constant = 6.636e-34/(2ð??)[Js]
(ii) q â?? Charge on an electron = 1.6e-19[C]
(iii) mfe â?? Mass of a free electron = 9.11e-31[kg]
(iii) delta_z â?? Mesh spacing = 1e-10[m]
(iv) bh â?? barrier height [eV] (see part (b))
(v) ww â?? well width (see part (b))
(vi) bm â?? effective mass in barrier = 0.1
(vii) wm â?? effective mass in well = 0.063
(b) Evaluate your 3 Ã? ð?? coefficient matrix via the eig function for the following
barrier-well-barrier parameters. For each situation plot the wavefunctions as a
function of spatial position and energy on top of the barrier-well-barrier structure.
(i) bh = 1eV and ww = 130Ã?.
(ii) bh = depth 2eV and ww = 130Ã?.
(iIi) bh = depth 1eV and ww = 150Ã?.
Part II Electron reflection and transmission across a multi-quantum barrier
To determine the reflection and transmission probabilities of an incident electron
through a quantum structure, SchrÃ¶dingerâ??s equation and its derivative must be
evaluated in each material layer. By matching corresponding solutions in adjacent
material regions, a transfer matrix can be derived and expressions for both
probabilities determined from the ratios of particular matrix elements. To illustrate
this concept, expressions for the electron transmission and reflection from a single
perfectly flat interface are deduced below. SchrÃ¶dingerâ??s equation for electrons in a
particular semiconductor layer may be written as
â??2 ð?? ð?? 2 ð?? = 0
here ï¹~ and k refer to the electronâ??s wavefunction and wave number respectively,
ð??(ð¸) = â??
(ð¸ â?? ð??0 )
here Ä§ is the reduced Planckâ??s constant and ð??â?? , ð??0 and ð¸ refer to the electron effective
mass, potential and incident energy respectively. Typically, solutions to SchrÃ¶dingerâ??s
equation are written as a superposition of two oppositely traveling plane waves. In the
one-dimensional limit, where the x-direction is both the device growth and electron
propagation directions, the solution may be expressed as
ð?? = ð´ð?? ð??ð??ð?§ ðµð?? â??ð??ð??ð?§
To ensure charge continuity the electron wavefunction (equation 5) and its first
derivative divided by the appropriate effective mass must be continuous across each
material boundary; for example, when propagating from material 1 to material 2 the
continuity relations may be expressed as
ð??1 = ð??2
Equations 6 and 7 are solved to relate the unknown amplitude coefficients of equation
3 and its derivative in the two adjacent material regions of the semiconductor to yield
[ 1 ] = ð??1â??1 ð??2 [ 2 ] = ð??12 [ 2 ]
Where the elements of the transfer matrix, ð??12 are calculated to be
ð??11 = (1 ) ð?? â??ð??(ð??1â??ð??2)ð?§
ð??12 = (1 â?? ) ð?? â??ð??(ð??1ð??2)ð?§
ð??21 = (1 â?? ) ð?? â??ð??(ð??1ð??2)ð?§
ð??22 = (1 ) ð?? â??ð??(ð??1â??ð??2)ð?§
Where ð??ð?? , is the wavenumber in the jth material layer respectively. The reflection and
transmission probabilities across the interface can be calculated for the entire
incident electron energy spectrum by the following ratios of the transfer matrix
This formalism can be easily expanded to an arbitrary number of interfaces, by
calculating the matrix product of all the transfer matrices describing transfer across a
single heterojunction, i.e.
[ 1 ] = â? ð??ð??ð?? [ ð? ]
where the unknown amplitudes of the wavefunction for all layers can be obtained with
knowledge of the electron wavefunction in the extreme regions (see figure below).
(a) Using the parameters from Part I, develop a script file that evaluates
equations 9-11 over the energy interval ð¸ = [0 ð??0 ]ð??ð??. Use at least 100,000
energy values in this period, i.e. E = linspace(0, ð??0,1e5)â??;
(b) Calculate and plot the reflection and transmission probabilites for the
following quantum structures.
(i) Well â?? Barrier â?? Well; where the barrier height = 1eV and barrier width =
(ii) Well â?? Barrier â?? Well; where the barrier height = 1eV and barrier width =
(iii) Well â?? Barrier â?? Wellâ?? Barrier â?? Well; where the barrier height = 1eV and
the barrier widths = 40Ã? and the middle well width is 50 Ã?. Comment on the
features of the transmission probability.
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