Tuesday, November 27, 2012

Gaussian Wave Packets

Purpose: It is to visualize and understand wave packets and see how the uncertainty principle relates to their construction. This will also prove as a visualization of the Schrodinger equation and probability distribution functions by superimposing various waves that will cause a localization of a particle and display its wave property.

For this experiment we will use VPython in order to show first create a Gaussian distribution, then  create a sine function, a superpositioons of various sine functions

To simulate a Gaussian function the following was code used:



from pylab import *
center=5
sigma=1
coeff=1/sqrt(2*pi)*sigma
gauss_list=[]

for x in arange (0,10,0.1):
    gauss=coeff*exp(-(x-center)**2/(2.*sigma**2))
    gauss_list.append(gauss)
     
plot(gauss_list)
show()


----------------------------------------------------
To create a sine function the following was used:


from pylab import *
center=5
sigma=1
coeff=1/sqrt(2*pi)*sigma
omega=1
sine_list=[]

for x in arange (0,10,0.1):
    sine=coeff*sin(omega*x)
    sine_list.append(sine)
     
plot(sine_list)
show()
------------------------------------------

To create a superposition of the sine functions:


from pylab import *
center=5
sigma=1
coeff=1/sqrt(2*pi)*sigma
sine_list=[]
A=5
omega=0.5
supaa = []

for i in range (1,4):
    x=[]
    sine_list=[]
    for t in arange (-2*pi,2*pi,0.01):

        sine=A*sin(i*omega*t)
        sine_list.append(sine)
        x.append(t)
    plot(x,sine_list)
    supaa.append(sine_list)

superposition= zeros(len(sine_list))
for function in supaa:
    #print function                 
    for i in range(len(function)):
        superposition[i] += function[i]
        #print superposition
plot(x,superposition)
show()
-------------------------------------------------------

For a Gaussian we have:

from pylab import *  #Need this for plotting functions
center = 3  #Define the center of the guassian
sigma = 1.0   #Set the standard deviation to 1
coeff = 1 / ((sqrt(2* pi))*sigma)  #This the normalization coefficient

#Define Constants
w = 1   #Set the frequency coefficient
gauss_list=[]
A=gauss_list
Fourier_Series = []  #Initialize the list of sine functions
#Calculate the harmonics of the sine functions
for x in arange(1,50):
    gauss=coeff*exp(-(x-center)**2/(2.*sigma**2))
    gauss_list.append(gauss)
    
for i in range(1,50):
    x = [] #This will let us plot the value from -pi to pi
    sine_function = [] #This contains the sine function

    for t in arange(-3.14,3.14,0.01):  #Loop from -3.14 to 3.14 by 0.1
        sine_f=gauss_list[i-1]*sin(i*w*t)
        sine_function.append(sine_f)  #Add the calculated value to the list of values
        x.append(t)
    Fourier_Series.append(sine_function)

superposition = zeros(len(sine_function)) #set as zeros of length equal to the sine

for function in Fourier_Series:
    for i in range(len(function)):
        superposition[i]+= function[i]

plot(x,superposition)

show()


Questions:Using the integral in \psi(x)=\int_{0}^{\infty}B(k) \:{\rm cos}\: kx \: dk, determine the wave function \psi \left( {x} \right) for a function B\left( k \right) given by 
  B\left( k \right) = \left\{ {\begin{array}{*{20}c}     0 & {k < 0}  \\     {1/k_0 {\rm{,}}} & {0 \le k \le k_0 }  \\     {0,} & {k > k_0 }  \\  \end{array}} \right.   This represents an equal combination of all wave numbers between 0 and k_0. Thus \psi \left( x \right) represents a particle with average wave number k_0 /2, with a total spread or uncertainty in wave number of k_0. We will call this spread the width w_{\rm k} of B\left( k \right), so w_{\rm k} = k_0.

a. Graph B\left( k \right) versus k for the case k_0 = 2\pi /L, where L is a length.

This is a straight line with a height of 2*pi/L


b. Graph \psi \left( {x} \right) versus k for the case k_0 = 2\pi /L, where L is a length.



c. Locate the two points closest to this maximum (one on each side of it) where \psi \left( x \right) = 0, and define the distance along the x-axis between these two points as w_{\rm x}, the width of \psi \left( x \right). What is the value of w_{\rm x} if k_0 = 2\pi /L?
This is 1.00 L
d.Repeat part C for the case k_0 = \pi /L.


e. Repeat part D for the case k_0 = \pi /L.

f. The momentum p is equal to hk/2\pi, so the width of B in momentum is w_{\rm p} = hw_k /2\pi. Calculate the product w_{\rm p} w_{\rm x} for the case k_0 = 2\pi /L. 
This is h
g. Calculate the product w_{\rm p} w_{\rm x} for the case k_0 = \pi /L.
This is also h.
h. Discuss your results in light of the Heisenberg uncertainty principle.
The packets follow the uncertainty principle. As we have greater values of k(a wider range of k) which relate the uncertainty in the momentun increases. therefore the positions should be more precise. 
This is seen from using various harmonics in the function and you see a large amplitude in the center that quickyly dies off as you leave the center. This corresponds to a more
 localized particle (just as stated by the uncertainty principle. They all also have the same uncertainty which agrees with the principle.



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