Lecture 08

# Lecture 08 - EEE 434/591Quantum Mechanics L8:1 Institute...

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EEE 434 Quantum Mechanics http://www.eas.asu.edu/~ferry/EEE434.htm L8:1 EEE 434/591—Quantum Mechanics David K. Ferry Regents’ Professor Arizona State University Institute for Advanced Study Princeton, New Jersey

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EEE 434 Quantum Mechanics http://www.eas.asu.edu/~ferry/EEE434.htm L8:2 Numerical Solutions to the Schrödinger Equation There are times when the potential and/or the boundary conditions are such that a straight-forward analytical solution of the Schrödinger equation is not easy (or even possible in some cases). This is certainly the case with the Airy functions, but is even more the case when one wants to couple the SE with e.g. the Poisson equation for the potential. Poisson’s Eqn. Schrödinger Eqn. V(x) n(x)
EEE 434 Quantum Mechanics http://www.eas.asu.edu/~ferry/EEE434.htm L8:3 Thus, we want to develop a numerical approach to solve first the time-independent form of the equation The first step is to chose a grid (we illustrate this in 1D, but it can be easily extended to 2D and 3D) x 0 x 1 x i+1 x i x i+2 x i+3 x i-1 x i-2 x i-3 x N a Boundary conditions are applied at the end points.

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EEE 434 Quantum Mechanics http://www.eas.asu.edu/~ferry/EEE434.htm L8:4 Let us first expand the second derivative term using the Taylor series expansion
EEE 434 Quantum Mechanics http://www.eas.asu.edu/~ferry/EEE434.htm L8:5 This leads to Now, we define the discrete values of the wave function as And we define the hopping energy 2 2 ma 2 ψ ( x i + a ) + ( x i a ) 2 ( x i ) [ ] + V ( x i ) ( x i ) = E ( x i )

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EEE 434 Quantum Mechanics http://www.eas.asu.edu/~ferry/EEE434.htm L8:6 This now gives us the general equation We have a difference at x 1 and x N-1 : The values at x 0 and x N are the boundary values that are applied, and we can write the general matrix equation t ψ i + 1 + i 1 2 i ( ) + V i i = E i E 2 t V i ( ) i + t i + 1 + t i 1 = 0 E 2 t V 1 ( ) 1 + t 2 = t 0 E 2 t V N 1 ( ) N 1 + t N 2 = t N
EEE 434 Quantum Mechanics http://www.eas.asu.edu/~ferry/EEE434.htm L8:7 The main matrix is square with dimension N-1 . The length of the device is given by Na = L . If the boundary conditions are that ψ = 0 at these boundaries, then we must find the eigenvalues (roots) of the determinant of the square matrix S . Let us consider the infinite square well as an example.

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EEE 434 Quantum Mechanics http://www.eas.asu.edu/~ferry/EEE434.htm L8:8 Boundary conditions that ψ 0,N =0: (infinite barrier at edges) ( E 2 t ) t ... 0 0
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## This note was uploaded on 10/03/2010 for the course EEE 434 taught by Professor Roedel during the Fall '08 term at ASU.

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Lecture 08 - EEE 434/591Quantum Mechanics L8:1 Institute...

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