schrodinger project

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School of Engineering

Department of Electrical Engineering

EEGR 215

Electronic Materials and Devices

Project 1

Schrodingers Project

02/20/2012

1)


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The general stationary state solution for the Schrdinger equation for the

stationary wave is given by

(x) = A eikx +A eikx -----------(1) with V(x) = 0 12

This equation represents the superposition of two waves traveling in the +x

direction with amplitude of A1 and other traveling in -x direction with amplitude of

A2

Also, the one dimensional Schrdinger equation is given by

---------------------- (2)

For V(x)=0 since it corresponds to the free particle,


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and

Now, rewriting the eqn 1 using the Eulers relation,

(x) = A1 (cos kx + i sin kx) + A2{cos(kx) + isin(kx)}

= ( A1 + A2 ) cos kx + i( A1 A2 ) sin kx

For x = 0,

(0) = A1 + A2or, A2= - A1

Hence,(x) = 2iA1sin kx = C sin kx-------------- (3)

k can be any number, to satisfy the boundary condition

kL = n

so,

Hence, for V (0) equation (2) becomes


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Using the De-broglies wave length,

Now, substituting the value of k in eqn 3

where,n=1,2,3.......

The normalization condition for the particle to exist between x = - and x = +

must be equal to 1, i.e


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Now, plotting the graph for (x) for the three wave functions and energy levels,

Using the equations, 4 and 5

The data points are obtained on MSExcel, We know,

h= 6.67E-34Js

m= 9.31E-31kg

L= 10000000000m

The values for x was obtained from 0-10 in an interval of pi/6 which is shown


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below, and the corresponding (x) was obtained for n=1 to 3

Similarly, the Energy was obtained for n= 1 to 3

nE

. 1 2.35815E-56J

. 2 9.43261E-56J

. 3 2.12234E-55J

The Graphs were obtained, which are listed below,


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2) The distance of the well was extended to 1000A so the interval for x was also

extended this time at the interval of 60. The data points and plots for the three

energy levels are shown below:


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3)

Given,Potential height (U0) = 2.4 eV


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Energy (E) = 2.1 eVThe probability of the tunneling is given by,

where

Therefore, by comparing the probabilities, the probability of finding the particle at


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a distance of 12A is more than 48A.

Plotting the Probability Curve,The data points were first obtained in Excel andthen the graphs were plotted as shown below:


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Observing the decaying probability curve, it is clear that the probability is

decreasing exponentially as the length of the tunnel, or width L increases. Also,

the probability of finding the particle at the barrier is very high, from the graph.

Considering the particles kinetic energy and the width L there is a probability of

finding the particle at the right of the barrier after the tunneling phenomena. At

such case the probability T has to be proportional to the square of the ratio of the

amplitudes of sinusoidal wave functions on the two sides of the barrier.

4) Given,


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The probability of finding the particle are as follows:

(a) 0


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Hence, the probability is 23.8%

(c) 0


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Now, dividing both sides of equation 1 by

Now, we know that, for the one dimension Schrdingers equation

Since, the particle has three different component of momentum given by px, py

and pz

Hence, the kinetic energy is given by,


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For, n=3 and a= 10 the 3d graph obtained in Matlab is shown below:


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schrodinger project

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