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.



Friday, October 26, 2012

CD Diffraction

Objective: CD's have very small grooves on them which make it possible to store music on them. Measure the distances between the groves on a CD by using a laser and diffraction.

Equipment:
  • Laser
  • Compact Disk
  • Meter Stick
  • A Screen with a hole in the middle.
Procedure: Arrange the  laser and screen so that the laser points through the hole in the screen and hits the CD with nearly normal incidence. (we used a white board to help mount the CD.


   Move the CD around as needed in order to see a diffraction pattern coming from the CD to the screen. Make it so that the first order maxima will appear on the screen on either side of the hole the  laser is coming through. (the zero order maxima strikes the hole so it is not seen. Once this happens the distance between the screen and the CD is measured, L, and the distance from the hole to the first order maxima is measured, x, (this part was done by marking on the screen the location of the maxima and measuring the length and dividing by 2). Also note the wavelength of the laser.
                      λ=633 nm              L=3.8 ± 0.02 cm          2x= 2.5 ± 0.1 cm

Using these calculations it is seen that our error is high from the standard manufacturer's standard value of 1600 nm. This error is accredited to the the fact that our experiment was not stable. We couldn't get the diffraction to show on our screen so for the experiment we held the CD (off the table) in order to get the diffraction to show on the screen. As a result our measurements were not very accurate because the CD was moving around slightly. But on the low side of our uncertainty we came up with a 9.31% deviation from the standard value.

Relativity in Length Active Physics

Again using Active Physics, concepts from relativity are explored. This time Length contraction is explored.
http://media.pearsoncmg.com/bc/aw_young_physics_11/pt2a/Media/ModernPhysics/1702RelOfLength/Main.html
 In this experiment light is going in the same direction as motion.

The time measurement of a round trip for the ray in the lights frame is independent of weather the light clock is moving relative to the earth or not.

The round trip time interval as measured on a stationary point on the earth will be longer than the frame of reference of the time clock.


  • As mentioned in the previous lab, time dilation in one frame is the product of the Lorentz factor and the proper time. This is consistant with what is observed here.
The light clock shrinks in the second frame. If it didn't and the lengths in both frames were the same, then the the the dime dilation relation couldn't hold true.

If the length of the light clock were 1000m (propper length) to find the length in the second frame we can use the relation, the length in any frame is given by the proper length divided by the Lorentz Factor. If the Lorentz factor were 1.3 the the length of the clock in the second frame would be(1000)m/1.3=769m

Relativity in Active Physics

Using Acive Physics from the following website, I explored the effects or time dilation near the speed of light by varying the value of the gamma value (essentially changing velocity).
http://media.pearsoncmg.com/bc/aw_young_physics_11/pt2a/Media/ModernPhysics/1701RelOfTime/Main.html


The length traveled by the light is longer in the second frame of reference (the one not moving with the light).

Because the speed of light is constant in all frames of reference the time it takes in the second frame of reference to complete a cycle is longer than the time in the initial frame. Specifically, according to the simulation, if the mirrors are moving at gamma=1.4 the time difference is 2.73 µs.

In the frame of reference of the time clock, the time required to complete a round trip is independent of weather the mirrors are moving or not.















The difference in light pulse travel time between the earth's timers and the light clock's timers will decrease as the time clock's speed slows down and becomes closer to that of earth's.

The equation for this effect (time dilation) is Δt = γΔtproper where the proper time is the time in the frame of the light clock. for γ = 1.2 the time seen in the second frame should be 8.00 µs which agrees with the experiment when changing γ to 1.2.

If the time for the observer in earth's frame is 7.45 µs then using the same equation we get gamma to be about 1.12. When the program is used and 1.12 is used then the timing on the earth's frame is consistent with this.


Saturday, October 13, 2012

Measuring a Human Hair

Objective: To measure the thickness of a human hair using a laser and micrometer.

Equipment:
  • Laser
  • Meter Stick
  • Note Card 
  • Hair
  • Micrometer
Procedure: Punch a hole in the note card and secure the hair in between the hole as to split the hole into two. Make sure it is taut. Make the note card parallel to a vertical surface (like a wall or board) and measure the length between the surface and the note card, L. 


     Mount the laser so it shines over the hair and an interference pattern on the vertical surface. The wavelength of the laser is given to be 633nm. Measure the distance between maximum on the  diffraction pattern (y) and record the order of the maximum (m).




It was easier to mark the maximum and and measure the marks afterwards.

        The equation used to determine the diameter was:    
                                     d=λLm
                                           y







The measurement obtained from this  was compared to the measurement obtained from measuring a hair with a micrometer.

The Length of separation was 1.00 ± 0.01m
     Sample 1: y= 3.1 ± 0.1cm
                    m=6
                      Diameter = 0.000123 ± 0.000005 m
     Sample 2: y= 3.3 ± 0.1cm
                    m=4
                      Diameter = 0.0000767 ± 0.0000032 m
     Sample 3: y= 1.2 ± 0.2cm
                    m=1
                      Diameter = 0.0000528 ± 0.00001.11 m
     The diameter measured was on the correct order of magnitude as expected of the diameter of hair. Using the micrometer was slightly difficult but it gave similar values. For sample 1 the micrometer gave a value of 0.0002 m. This is a error of 48%. The  micrometer is less accurate because the minimum increment of measurement is a tenth of a millimeter  The laser method can better measure small lengths because it can more accurately measure the parameters needed and obtain much smaller values.

Wednesday, October 3, 2012

Lenses

Objective: To observe characteristics of a converging lens when the object is placed on one side of the lens and the real, inverted image is placed on the other side of the lens.

Equipment:
  • socket lamp with V-shaped filament
  • Large converging lens
  • masking tape
  • Lens Holder
  • piece of cardboard (or other flat surface)
  • Track for lens
  • Meter stick
Procedure: 
     The focal length was recorded by taking a source that was infinitely far (the sun) and arranging it around to find a point where the image was focused. A meter stick was used to find the distance between the lens and the focused image. This was 0.0485 ±0.0030 m.

     The following was set up by placing the lens into the lens holder and creating a track for the lens (using a meter stick and some stands). 
     The length of the arm of the image from the circle to the end of the line was take as the object image (9.2±2.0 cm). We place the image about 1.5 focal lengths away and used the cardboard to focus the resulting image. The image height and distance from the lens was recorded. From this the magnification could be found by dividing the image height by the object height. (If the lens was rotated the image remained the same). The image was always inverted.
This was repeated for various focal lengths.
Object Distance Image Distance  Image Height Object Height. Magnification
1 25 6 2.1 12 0.175 ± 0.02
2 20 5.4 2.9 12 0.242 ± 0.02
3 15 6.4 3.8 12 0.317 ± 0.02
4 10 8.6 7.3 12 0.608 ± 0.03
5 7.5 7.5 13 12 1.083 ± 0.04
All of these values were measured in cm. The object distance and image height values were plus or minus 0.2cm while the image distance was about plus or minus 1.5cm.

Data Analysis: If the object distance was less then one focal point the object height was too great to record. A graph of image distance vs object distance was made and showed a nonlinear relationship. But when we graphed the inverse of negative object distance vs the inverse of image distance we got a somewhat linear relation (within experimental error arising from difficulties in finding the exact image distance). Do to it being an outlyer the fifth data point was removed from the calculation of the plot.
The y intercept was 0.2208 which represents the inverse of image distance as the object distance reaches infinity. This is the inverse of the focus! The relationship between Inverse object distance (x) and inverse of image distance (y) is given in the equation in the above image

Friday, September 28, 2012

Concave and Convex Mirrors

Objective: Explore the images formed by convex and concave mirrors.

Equipment:
  • Convex mirror
  • Concave mirror
  • An object
  • Ruler
Procedure:
     convex mirror- A marker was placed in front of the convex mirror.The image appears smaller than the actual object but the object is upright. The image seems further from the mirror then the actual object.        With a ruler placed normal to the mirror's center  we held the mirror a distance d0=0.50m. The height of the marker (our object) was h0=0.12m. Measuring (on the mirror) the size of the image of the marker gave us hi=0.067m when the object is moved closer the image appears bigger and it gets smaller as it moves away.

     concave mirrors- the object was now placed in front of a convex mirror. The object appeared inverted and larger. When the object moved closer it appears upright, and still magnified. The image also appears closer to the mirror then the actual object. Using the ruler  we again placed it 0.50m from the center and we observed an inverted height of 0.21m.
Analysis: This phenomenon can be explained by using a diagram of light rays being reflected off the pen.
 For the convex mirror you see that the point where these rays intersect is where the top of the marker's image appears and this agrees with our observation
For the concave mirror we see the focus and center of the sphere is outside of the mirror. Using the rays it can be seen that the object appears to be inverted which also agrees with our observation.
From this it can be concluded that the image size and orientation depend on the focus or center of the spherical mirror.