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Spiral.. Physics
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Modern Physics
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Matter Waves
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Copyright © 2003
Paul D’Alessandris
Spiral Physics
Rochester, NY 14623
All rights reserved. No part of this book may be reproduced or transmitted in any
form or by any means, electronic or mechanical, including photocopying, recording,
or any information storage and retrieval system without permission in writing from
the author.
This project was supported, in part, by the National Science Foundation. Opinions expressed are those of
the author and not necessarily those of the Foundation.
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Matter Waves
Treating Particles as Waves
Inspired by the dual nature of light, in 1923 Louis DeBroglie postulated, in his PhD thesis,
that material particles also have both a particle-like and a wave-like nature. He conjectured
that the frequency and wavelength of a “particle” are related to its energy and momentum in
the same way as the frequency and wavelength of light are related to its energy and
momentum, namely
E  hf
h
p

After the experimental verification of this prediction, DeBroglie was awarded the Nobel Prize
in 1929.
Bragg Diffraction
You should immediately ask, “How was the wave-like nature of matter experimentally
verified?” If matter has a wave-like nature, it should exhibit interference in a manner
completely analogous to the interference of light. Thus, when passing through a regular array
of slits, or reflecting from a regular array of atoms, an interference pattern should form. In
1927 Clinton Davisson and David Germer tested this hypothesis by directing a beam of
electrons at a crystal of nickel.
Incoming waves reflecting from the first crystal plane will interfere with waves reflecting
from the second (and subsequent) crystal planes forming an interference pattern. This
interference, termed Bragg diffraction, had been initially investigated using x-rays.
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For constructive interference, the path length difference between the two reflected beams
must differ by an integer multiple of a complete wavelength. From the diagram above, the
wave reflecting from the second crystal plane travels an additional distance of 2dsin. Thus,
the relation for constructive interference is:
2d sin   n
where

d is the distance between adjacent crystal planes, termed the lattice spacing,

 is the angle, measured from the crystal face, at which constructive interference
occurs,

and is the wavelength of the disturbance.
A beam of electrons is accelerated through a potential difference of 54 V and is incident
on a nickel crystal. The primary interference maximum is detected at 65o from the crystal
face. What is the lattice spacing of the crystal?
If a beam of electrons is accelerated through a potential difference of 54 V, it gains a kinetic
energy of 54 eV. This results in a momentum of
Etotal  ( pc) 2  (mc 2 ) 2
2
pc  Etotal  (mc 2 ) 2
2
pc  (54  511000) 2  (511000) 2
pc  7.43keV
and, by DeBroglie’s relation, a wavelength of
hc
pc
1240eVnm

7430eV
  0.167 nm

Inserting this result into the Bragg relation results in
2d sin   n
2d sin 65  (1)(0.167nm)
d  0.092nm
This value agrees with the known lattice spacing of nickel.
The presence of distinct interference maxima validates the idea that matter has a wave-like
nature, and the agreement in lattice spacing illustrates that DeBroglie’s relationship between
the momentum and wavelength of matter is correct. For their experimental validation of
DeBroglie’s relation, Davisson (but not poor Mr. Germer) was awarded the Nobel Prize in
1937.
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The Double Slit with Matter
A beam of very cold neutrons with kinetic energy 5.0 x 10-6 eV is directed toward a
double slit foil with slit separation 1 m. What is the angular separation between
adjacent interference maxima?
In addition to Bragg diffraction, the wave-like nature of matter can be demonstrated in the
same experimental manner as the wave-like nature of light was first demonstrated, by passing
the matter wave through a pair of adjacent slits. You should remember the result for the
location of interference maxima in a double slit experiment, but nonetheless I’ll remind you:
d sin   n
where

d is the distance between adjacent slits,

 is the angle at which constructive interference occurs,

and is the wavelength of the disturbance.
The kinetic energy of the neutrons is so small we can use classical physics to determine the
momentum. Remembering the classical relationship between kinetic energy and momentum
1
(mv) 2 p 2 ( pc) 2
KE  mv 2 


2
2m
2m 2mc 2
leads to
pc  2( KE )mc 2
pc  2(5 x10 6 )(939.6 x10 6 )
pc  96.9eV
and, by DeBroglie’s relation, a wavelength of
hc
pc
1240eVnm

96.9eV
  12.8nm

Inserting this result into the double slit relation results in
d sin   n
(1000nm) sin   (1)(12.8nm)
  0.730
Thus, adjacent maxima are separated by 0.73 degrees.
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Thermal Wavelength
How “cold” is a beam of very cold neutrons with kinetic energy 5.0 x 10-6 eV?
You may have been confused when I referred to the neutron beam in the previous example as
being “very cold”. However, physicists routinely talk about temperature, mass and energy
using the same language. An ideal (non-interacting) gas of particles at an equilibrium
temperature will have a range of kinetic energies. You may recall from your study of the ideal
gas that:
KE mean 
3
kT
2
where

KEmean is the mean kinetic energy of a particle in the sample,

k is Boltzmann's constant,

and Tis the temperature of the sample, in Kelvin.
Technically, we shouldn’t talk about the temperature of a mono-energetic beam, since by
definition a temperature implies a range of energies. However, let’s be sloppy and assume the
energy of the beam corresponds to the mean kinetic energy of a (hypothetical) sample. Then:
KE mean 
3
kT
2
2 KE mean
3k
2(5 x10 6 eV )
T
3(8.617 x10 5 eV / K )
T  0.039 K
T
So the neutron beam really is pretty cold!
Note that if we wanted to find the DeBroglie wavelength corresponding to this mean kinetic
energy, we would find (assuming non-relativistic speeds)
KE 
( pc) 2
2mc 2
pc  2( KE )mc 2
3
pc  2( KT )mc 2
2
pc  3mc 2 kT
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and thus


hc
pc
hc
3mc 2 kT
This is the DeBroglie wavelength corresponding to the mean kinetic energy of a gas at
temperature, T. However, a more useful value would be the mean wavelength of all of the
particles in the gas. The mean wavelength is not equal to the wavelength of the mean energy.
Calculating this mean wavelength, termed the thermal DeBroglie wavelength is a bit beyond
our skills at this point, but it is the same as the result above but with a different numerical
factor in the denominator:
thermal 
hc
2mc 2 kT
For an ideal gas sample at a known temperature, we can quickly determine the average
wavelength of the particles comprising the sample.
One important use for this relationship is to determine when the gas sample is no longer ideal.
If the mean wavelength becomes comparable to the separation between the particles in the
gas, this means that the waves begin to overlap and the particles begin to interact. When these
waves begin to overlap, it becomes impossible (even in principle) to think of each of the
particles as a separate entity. When this occurs, some really cool stuff starts to happen…
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A Plausibility Argument for the Heisenberg Uncertainty Principle
Imagine a wave passing through a small slit in an opaque barrier. As the wave passes through
the slit, it will form the diffraction pattern shown below.
Remember that the location of the first minima of the pattern is given by
a sin   
From the geometry of the situation,
tan  
y
D
If the detecting screen is far from the opening,
sin   tan
so
a sin   
a tan   
y
a( )  
D
D
y
a
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Now, consider the “wave” to be a “particle”. The time to traverse the distance from slit to
screen is given by
t
D
vx
while during this time interval the particle also travels a distance in the y-direction given by:
y  vyt
Combining these relations yields
y  vy (
D
)
vx
Combining this “particle” expression with the “wave” expression above gives:
D
a

 vy (
D
)
vx
vya
vx
Substituting the DeBroglie relation results in,
vya
h

mv x
vx
h  mv y a
Notice that the term mvy is the uncertainty in the y-momentum (  p y ) of the particle, since the
particle is just as likely to move in the +y or the –y-direction with this momentum. Also, a is
twice the uncertainty in the y-position (  y ) of the particle, since the particle has a range of
possible positions of +a/2 to –a/2.
Therefore, our expression can be written as
(2 y )( p y )  h
( y )( p y ) 
h
2
Thus, the uncertainty in the y-position of the particle is inversely proportional to the
uncertainty in the y-momentum. Neither of these quantities can be determined precisely,
because the act of restricting one of these parameters automatically has a compensating effect
on the other parameter, i.e., making the hole smaller spreads out the pattern, and the only way
to make the pattern smaller is to increase the size of the hole!
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A more careful analysis (for circular openings rather than slits) shows that the minimum
uncertainty in the product of position and momentum can be reduced by a factor of 2,
resulting in:
1 h
2 2
h
( y )( p y ) 
2
( y )( p y ) 
where the symbol ħ is defined to be Planck’s constant divided by 2.
The Spatial Form of the Uncertainty Principle
The electrons in atoms are confined to a region of space approximately 10 -10 m across.
What is the minimum uncertainty in the velocity of atomic electrons?
The wave-like nature of matter forces certain restrictions on the precision with which a
particle can be located in space and time. These restrictions are known as the Heisenberg
Uncertainty Principle. (The word “uncertainty” is a poor choice. It is not that we are uncertain
of the speed and location of the particle at a specific time; rather it is that the “particle” does
not have a definite speed and location! The wave-like nature of the “particle” forces it to be
spread out in space and time, analogous to the spreading of classical waves in space.)
The spatial form of Heisenberg’s Uncertainty Principle is
( x )( px ) 
h
2
where

x is the uncertainty, or variation, in the particle’s position,

px is the uncertainty, or variation, in the particle’s momentum in the same direction,

and ħ is Planck’s constant divided by 2.
With the center of the atom designated as the origin, the position of the electron can be
represented as
x  (0  0.5) x10 10 m
thus
 x  0.05nm
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Using the uncertainty principle results in
( x )( px )  h / 2
p 
x
h
2( x )
m vx 
h
2( x )
v 
h
2m( x )
v 
hc
c
2mc 2 ( x )
x
x
Just as “hc” will pop up in numerous equations throughout this course, the constant “hc” is
also quite common and has an equally friendly value, 197.4 eV nm. Thus,
v 
x
197.4eVnm
c
2(511000eV )(0.05nm)
 v  3.86 x10 3 c
x
 v  1.2 x10 6 m / s
x
Thus the velocity of an atomic electron has an inherent, irreducible uncertainty of about a
million meters per second! If anyone tells you they know how fast an atomic electron is
moving to a greater precision than a million meters per second, you know what to tell them…
The Temporal Form of the Uncertainty Principle
Empty space can never be completely empty. Particles can spontaneously “pop” into
existence and then disappear. Imagine a proton and antiproton spontaneously created
from the vacuum with kinetic energy 1.0 MeV each. For how long can these particles
exist and how far could they travel in this time?
An analogous argument to the one that led to the spatial form of the uncertainty principle can
be made that leads to the temporal form of Heisenberg’s Uncertainty Principle:
( E )( t )  h / 2
where

E is the uncertainty, or variation, in the particle’s total energy,

and t is the time interval over which the energy was measured.
This form of the uncertainty principle implies that the precise value of the energy of a particle
or system can never be known, since the time interval over which the value is measured
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inversely affects the precision of the measurement. This even applies to a region of space in
which the energy is, nominally, zero.
In this example, the energy of a certain region of empty space, naively thought to be equal to
zero, spontaneously fluctuates by an amount equal to the total energy of the two created
particles. This variation can only last for
( E )( t )  h / 2
t 
t 
h
2 E
0.658 x10 15 eV  s
2[2(938MeV  1.0 MeV )]
 t  1.8 x10 25 s
The particle’s speed is given by
KE  (  1)mc 2
1.0  (  1)(938)
  1.001066
v  0.046c
In this incredibly short amount of time the particles will be able to travel
d  vt
d  (0.046c)(1.8 x10  25 )
d  2.4 x10 18 m
This distance is approximately one-thousandth the width of a single proton. Although this is
an incredibly short distance, our modern understanding of the nature of forces and the
evolution and fate of the universe depend on the affects of these virtual particles.
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The Meaning of the Uncertainty Principle
There is much confusion regarding the meaning of the Uncertainty Principle. In fact, the
uncertainty principle is really just a statement about waves and how simple waves can be
combined to form wave packets. A wave packet is a localized wave disturbance.
Unlike simple sine and cosine representations of waves, such as:
( x, t )  A sin(kx  t )
which extend at equal amplitude to ±∞, a wave packet has amplitude that is larger in one
region of space than another. Mathematically, wave packets are formed by adding together
appropriately chosen simple sine and cosine waves.
For example,
note that the wave packet (C) has larger amplitude in some regions of space than in other
regions. We can say that the wave packet C is localized in space. If we want to more narrowly
localize C in space we will need to add together a larger range of different wavelength waves.
Thus, the spatial localization of C is inversely related to the range of wavelengths used to
construct C. Since a larger range of wavelengths corresponds to a larger range of momenta
(by DeBroglie’s hypothesis), spatial localization comes at the price of an increased range of
momenta. This is all the uncertainty principle says! The width of spatial localization is
inversely proportional to the range in momenta. This is true of all waves and is not special, in
any way, to matter waves.
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Additionally, since the wave packet is mathematically
comprised of many waves with different momenta,
these constituent waves all move through space at
different rates. The speed of each constituent wave is
referred to as the phase velocity. The speed of the
wave packet is the group velocity. Since the different
constituent waves have different phase velocities, this
leads, over time, to a change in the overall shape and
extent of the wave packet. This change in shape of the
wave packet over time is termed dispersion and is
illustrated at left (time increases as you scan from top
to bottom).
This spreading of the wave packet is natural and is
completely analogous to, for example, the spreading
of water waves on a pond. In the case of a matter
wave, however, this spreading is interpreted as the
increasing uncertainty as to the location of the
“particle” that the wave represents. The wave
(actually the square of the wave) represents the
probability of finding the particle at a certain location
in space when a measurement is performed.
Before the measurement is made, however, the
“particle” must be thought of as existing at all of the
locations where the probability is non-zero.
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A Problem with DeBroglie’s Hypothesis?
Aside from the “minor” issue of trying to understand what it means for a particle to have a
frequency and a wavelength, DeBroglie’s hypothesis also leads to a more technical issue
regarding how the frequency and wavelength of the particle are related to its velocity.
Hopefully you remember that the standard relationship between a wave’s velocity, frequency
and wavelength is
v  f
Substituting DeBroglie’s relationships yields:
 E  h 
v    
 h  p 
E
v
p
v
( pc) 2  (mc 2 ) 2
p
v  c2 
(mc 2 ) 2
p2
This should bother you. Why? Because the second term under the radical is obviously positive
and if added to c2 seems to require that the velocity of the “matter wave” is greater than the
speed of light! To resolve this apparent contradiction, we have to be much more careful in
how we conceptualize the wave-representation of a particle.
The key is to distinguish between the phase velocity and group velocity as defined above. The
wave packet, and hence the group velocity, is what represents the particle. The constituent
waves that add together to form the packet are not, individually, physically meaningful. The
simple relation between frequency and wavelength mentioned above is actually only correct
for these individual constituent waves. Hence, it holds for the phase velocity only:
v phase  f
What we need to worry about is not the value of these phase velocities, but the value of the
group velocity. If the group velocity is greater than c, we’ve got some serious problems.
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So how do we define the group velocity? Well, let’s first review how we define the phase
velocity. The phase velocity is the speed at which a particular point on the wave moves
through space. This particular point has a constant phase, so let’s set the phase in the standard
definition of a wave,
( x, t )  A sin(kx  t )
equal to a constant. For simplicity we’ll call that constant zero.
(kx  t )  0
kx  t
x 

t k
v phase 

k
If you recall the definitions of angular velocity () and wave number (k) from your study of
waves, this reduces to:
v phase 
2f
2
( )

v phase  f
Ok, so that was easy. Now what do we change to get an expression for group velocity?
In a wave packet, each constituent wave has a constant phase velocity but the packet as a
whole experiences dispersion and changes its shape with time. Because of this continually
changing shape, let’s try to define the group velocity not as the ratio of the angular frequency
to the wave number but rather as the derivative of the angular frequency with respect to wave
number:
v phase 

and
k
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v group 

k
Before we tackle this derivative, let’s express DeBroglie’s hypothesis in terms of  and k:
E  hf  h(
p
h



)  h
2
h
 hk
 2 


 k 
Thus,

k
E
( )
v group  h
p
( )
h
E
v group 
p
v group 
This means that the velocity of the wave packet is the rate at which the particle’s energy
changes with respect to its momentum.
Using the relativistic relationship between energy and momentum yields
v group 

( p 2c 2  m2c 4 )
p
1

1
v group  ( p 2 c 2  m 2 c 4 ) 2 (2 pc 2 )
2
pc 2
v group 
p 2c 2  m2c 4
v group 
mvc 2
E

mvc 2
v group 
mc 2
v group  v
Thus the group velocity of the wave packet is exactly the same as the velocity of the particle
the packet is intended to represent. Even though each of the constituent waves travels with a
phase velocity greater than c, they combine to form a wave packet that moves slower than
light. Isn’t mathematics amazing?
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Matter Waves
Activities
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The following objects move at the speeds indicated.
A
B
C
D
E
F
a baseball at 90 mph
a person running at 10 mph
a car at 60 mph
a beam of red light
an electron at 0.01 c
a proton at 0.01 c
a. Rank these objects on the basis of their momentum.
Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest
_____ The ranking cannot be determined based on the information provided.
Explain the reason for your ranking:
b. Rank these objects on the basis of their wavelength.
Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest
_____ The ranking cannot be determined based on the information provided.
Explain the reason for your ranking:
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I. If all matter (electrons, protons, tables, people, ...) has “wave-like” properties, why do tables and people
seem so “particle-like”? Three students ponder this question.
Xue:
Tables and people appear particle-like because they are built from smaller wave-like entities
like electrons and protons. As you put more and more of these waves together, the resulting
object appears more and more particle-like because of the interference between the individual
waves.
Mario: Objects that have mass, especially a lot of mass like tables and people, have a huge amount of
energy by E = mc2. This large amount of energy corresponds to a ridiculously high frequency,
by E = hf. Therefore, tables are wave-like, but the fields that comprise them are oscillating at
much too high a frequency to ever be detectable.
Robert: Waves and particles are just two categories that humans invented as they were trying to figure
out nature. Basically, nothing is purely a wave or a particle, everything is a little of both.
Some things are more wave-like and others are more particle-like.
Which, if any, of the students are correct? For each incorrect response, provide a short explanation why it is
incorrect. If no one is correct, provide a correct answer below.
II. Three students debate whether quantum mechanical indeterminism guarantees free will.
Raheem: Since the determinism of Newton’s Laws implies that we can’t have free will, clearly the
indeterminism that quantum mechanics creates in the universe allows us to choose our future
path, since even the paths of individual electrons are fundamentally unpredictable.
Alek:
The paths of electrons are more than unpredictable, they are actually random (within the
limits imposed by the interference pattern). If you apply quantum mechanical indeterminism
to human behavior you don’t get free will, you get a random future that we can’t control. How
is that any better than a perfectly determined future we can’t control?
Mizuno: The interference pattern is a good analogy for human behavior. There are places the electron
can’t land (just like there are things that we aren’t free to do, like fly), but within the places
where the pattern says the electron can land, the path of the electron is unpredictable, just like
human behavior.
Which, if any, of the students are correct? For each incorrect response, provide a short explanation why it is
incorrect. If no one is correct, provide a correct answer below.
20
After discussing numerous variations of the double-slit experiment, conducted with individual “particles”,
many students are confused about the implications of quantum mechanics. Their confusion can often be
divided into one of four main categories:
A
B
C
D
E
Wave-Particle Duality
Indeterminism
Non-Locality
Wavefunction Collapse
Measurement Problem
Each of these categories describes some aspect of quantum mechanics that differs from our preconceived
view of how the world works. For each statement in the conversation below, select the category into which
the student confusion best fits.
________ 1. If you shoot a bunch of electrons, which you know act as particles, at the two slits you get an
interference pattern on the screen, but particles can’t exhibit interference.
________ 2. What’s weirder, however, is if you shoot only one electron at the two slits, it still hits the
screen at a location that implies interference. But what’s it interfering with? Does this mean it
went through both slits? How can one electron go through two slits?
________ 3. Moreover, there’s no way to predict in advance exactly where that single electron will strike
the screen.
________ 4. If you put detectors behind each slit and try to detect the electron going through both slits at
once, you never see it in two places at once. Only one or the other detector goes off.
________ 5. And you still can’t predict which detector will go off when. There doesn’t seem to be any
pattern to which detector “sees” the each electron
________ 6. After the detector goes off, the electron seems to travel as a pure particle from that point on.
________ 7. I think it’s even stranger when there is only one detector behind one of the slits. When that
one detector “goes off”, the electron acts like a particle from that point on.
________ 8. But, somehow, even when the detector does not go off, the electron acts like a pure particle
and there’s no interference pattern on the screen.
________ 9. That’s not half as weird as what happens when the “detector” doesn’t make some sort of
record that can be viewed by an observer. Somehow, this doesn’t seem to “count” as a
detector and the electron still creates an interference pattern.
________ 10. Or when the detector is only put in place after the electron has passed the slits! If the
electron is already past the slits, how can you detect it back by the slits? Yet somehow this
can still sometimes cause the electron to act like a pure particle.
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No experiment has ever been conducted that contradicts any aspect of quantum mechanics. However,
beyond simply making correct mathematical predictions (for a very wide range of phenomenon), many
scientists also want to know what quantum mechanics means, what it implies about the nature of the
physical world. Several interpretations of the meaning of quantum mechanics have been proposed.
A
B
C
D
E
F
Strong Copenhagen Interpretation
Weak Copenhagen Interpretation
Many Worlds Interpretation
Decoherence Interpretation
Consciousness Interpretation
Hidden Variables Interpretation
For each statement below, select the interpretation that the student is most closely espousing.
________ 1. Although quantum mechanics works, in that it gives the correct answer for any experimental
test we’ve ever tried, something is missing from the theory. Nature just can’t be as weird as
quantum mechanics implies.
________ 2. It’s not quantum mechanics that’s weird, it’s the questions people worry about. It only seems
weird because people try to talk about things before they observe them or about things they
can’t observe. Who knows whether the electron goes through both slits when nobody is
looking? Or whether Schrödinger’s cat is alive or dead before we look? Those are
meaningless questions. Science is about observation. If you start to worry about things that
you can’t observe, you’re not doing science any more.
________ 3. You’re right, quantum mechanics is not strange at all. The electron is never in two places at
the same time and that damn cat is never dead and alive. The universe branches every time a
system is given a choice of paths and both paths are followed. The cat is dead in one branch
of the universe and alive in another. There is no indeterminism present at all.
________ 4. What are you smoking? There is only one physical reality. You just have to accept that
certain concepts have meaning in the macroscopic world and have no meaning in the
microscopic world. The concept of location in space just does not exist for an electron, so
trying to figure out where and electron is is basically like asking it a meaningless question.
That’s why the electron “answers” in such a confusing manner. Quantum mechanics clearly
implies that on the scale of the electron, there is no such thing as position.
________ 5. You are almost correct. The electron is not anywhere until we measure it to be somewhere. It
is our act of measurement that creates the electrons location. We, the observers, are
responsible in many ways for the existence of the macroscopic world.
________ 6. The electron existing at a specific position rather than at several positions at once has nothing
to do with the presence of an actual observer. It’s simply a natural process by which the
electron’s wave function gets “inter-meshed” with the wavefunction of its macroscopic
environment. It’s this interaction with the macroscopic world that would cause, for example,
Schrödinger’s cat to very rapidly become either alive or dead.
________ 7. So it “very rapidly” becomes either dead or alive? But what is it before this very rapid process
takes place? You haven’t really addressed the central problem. The bottom line is we can
never know what state the cat is in before we observe it (even if it is only in this state for a
nanosecond), so to argue about it is silly.
________ 8. It’s “silly” to try to figure out how the universe works? Your point of view undermines all of
science! Quantum mechanics is very clear on this point, the cat does not have the
characteristic of being alive or dead. Only in the system of “cat plus observer” does that
characteristic have meaning.
22
What is the wavelength of
a. a 1.0 kg baseball traveling at 90 mph?
b. a proton accelerated through a potential difference of 1000 V?
c. an electron at 0.99c?
d. What is the mean wavelength of an oxygen molecule at room temperature (20° C)?
Mathematical Analysis
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In order to have a wavelength of 10-10 m (about the size of an atom),
a. through what potential difference must protons be accelerated?
b. through what potential difference must electrons be accelerated?
c. In order to have a mean wavelength of 10 -10 m, at what temperature should a neutron gas be held?
Mathematical Analysis
24
In order to “see” something, you must bounce waves (or particles) off of it. For this to work the wavelength
of the incident waves must be no larger than the size of the object you are trying to see. Through what
potential difference must electrons in an electron microscope be accelerated in order to see
a. a virus (diameter 12 nm)?
b. an atom (diameter 0.12 nm)?
c. a proton (diameter 1.2 fm)?
Mathematical Analysis
25
The wavelength of particles can be altered by either accelerating a beam of particles or heating a gas of
particles.
a. What accelerating voltage is necessary to produce a wavelength equal to the thermal DeBroglie
wavelength of room temperature (293 K) electrons?
b. What temperature is necessary to produce a thermal DeBroglie wavelength equal to the wavelength of
electrons accelerated through a potential difference of 100 V?
c. Do the results above also hold for protons?
Mathematical Analysis
26
One mole of air at standard pressure and temperature (273 K) occupies about 22.5 liters.
a. What is the average distance between gas molecules?
b. What is the thermal DeBroglie wavelength of a nitrogen molecule in the above sample?
c. Can we ignore the interactions between molecules in the air sample described? Explain.
Mathematical Analysis
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The photosphere (outer surface) of the sun is at 5800 K and has a density of about 2.0 x 10-4kg/m3.
a. Assuming the photosphere consists of equal numbers of protons and electrons, what is the average
distance between particles in the photosphere?
b. What is the thermal DeBroglie wavelength of an electron in the photosphere?
c. Can we ignore the interactions between electrons and protons in the photosphere? Explain.
Mathematical Analysis
28
A certain crystal has rows of atoms separated by 0.423 nm. A beam of electrons is accelerated through a
potential difference of 250 V and is incident on the crystal.
a. What is the wavelength of the electrons?
b. At what angles from the crystal face will the interference maxima be found?
c. If the beam is replaced with a beam of protons, will the interference maxima be resolvable? Assume
that if the maxima are separated by less than 1° the individual maxima cannot be resolved.
Mathematical Analysis
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A beam of electrons is used to explore a crystal with known lattice spacing of 0.165 nm.
a. If the accelerating potential is 200 V, at what angles from the crystal face will the interference maxima
be found?
b. Below what accelerating potential will only one interference maximum be present?
c. Above what accelerating potential will four interference maxima be present?
Mathematical Analysis
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In addition to electrons, crystal structure can be explored using neutron diffraction. Consider a source of
thermal neutrons at 293 K.
a. What is the thermal DeBroglie wavelength of these neutrons?
b. Explain why no interference pattern will be present if this beam of thermal neutrons is directed at a
crystalline solid.
c. Imagine selecting only those neutrons traveling with the mean kinetic energy. If these neutrons are
directed at the solid, interference maxima occur at 24.1° and 54.8° from the crystalline face. What is
the lattice spacing of the crystal?
Mathematical Analysis
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A beam of 100 eV electrons is incident on a metal foil with slits 100 nm apart.
a. What will be the approximate distance between interference maxima on a detector 1.0 m from the
slits?
b. To increase the fringe spacing by a factor of ten, should the accelerating potential be increased or
decreased? Explain.
c. What new value of accelerating potential is necessary for the new fringe spacing? Is this feasible?
Mathematical Analysis
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A gas of helium atoms is held at 20 K.
a. What is the mean kinetic energy of the atoms in the sample?
b. If atoms with this mean kinetic energy are selected and directed at a double slit with slit spacing 0.8
m, at what angles do interference maxima appear? Are these maxima easily resolvable?
Mathematical Analysis
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The largest “particles” ever directed through a double slit apparatus were buckyballs, molecules formed
from 60 carbon atoms. The buckyballs were directed with a speed of 200 m/s at a grating with slit
separation 100 nm. The detector was placed 1.25 m from the slits. What was the separation between
adjacent maxima at the detector? (A very special detecting mechanism was needed to resolve this very
small separation.)
Mathematical Analysis
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A bullet (m = 50 g) and an electron are both measured with a speed of 300 m/s, with a precision of 0.01%.
a. What is the minimum uncertainty in the bullet’s location?
b. What is the minimum uncertainty in the electron’s location?
c. For which object does the Heisenberg uncertainty principle have a larger affect on our knowledge of
the object’s location?
Mathematical Analysis
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An electron is localized to within ±0.001 nm.
a. What is the minimum uncertainty in the electron’s velocity?
b. Based on this uncertainty in velocity, how long will it take the electron’s uncertainty in position to
grow to ±1.0 m?
c. Repeat the above calculation for a proton.
Mathematical Analysis
36
Consider a (hypothetical) electron trapped inside an atomic nucleus of radius 5 x 10 -15 m.
a. What is the minimum uncertainty in the electron’s momentum?
b. Assuming the electron has a momentum equal to this minimum value, what is the electron’s kinetic
energy?
c. In nuclear beta decay, electrons are emitted from the nucleus with kinetic energies of typically no more
than 1 or 2 MeV. Based on you answer in (b), can the electrons emitted during beta decay have been
“trapped” inside of the nucleus before they were emitted?
Mathematical Analysis
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My 6 kg computer is located somewhere in my 5 m wide office. At least I think it is. After how long should I
not be so sure that’s where it is?
Mathematical Analysis
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At different times during class, I may spontaneously disappear and then reappear shortly thereafter. I’m
about 82 kg.
a. How long can I be gone?
b. You notice my presence primarily via light reflecting off my body. To notice me missing, these rays
would have to not reflect. How far can light travel during the time period I’m gone?
Mathematical Analysis
39
Empty space can never be completely empty. Particles can spontaneously “pop” into existence, “dance”
around for a period of time, and then disappear. These virtual particles have huge repercussions on the
structure and fate of the universe. Imagine an electron and positron popping into existence. If each particle
receives 100 keV of kinetic energy, determine the maximum distance each particle can travel before
disappearing.
Mathematical Analysis
40
The force holding the atomic nucleus together can be modeled as the exchange of virtual pions between the
protons and neutrons in the nucleus. In this model, a proton can spontaneously “emit” a pion and the pion
can travel a short distance until it is “absorbed” by either another proton or neutron. The distance the pion
can travel gives an estimate of the range of the nuclear force.
a. Imagine a charged pion with kinetic energy of 100 keV spontaneously emitted by a proton. How far
can this pion travel before being absorbed?
b. As the kinetic energy of the pion approaches zero, what happens to the range of the pion?
Mathematical Analysis
41
The electric force between charged particles can be modeled as the exchange of virtual photons between
the particles. In this model, a charged particle can spontaneously “emit” a photon and the photon can
travel a distance until it is “absorbed” by another charged particle. The distance the photon can travel
gives an estimate of the range of the electric force.
a. Imagine a 10 eV photon spontaneously emitted by a charged particle. How far can this photon travel
before being absorbed?
b. As the energy of the photon approaches zero, what happens to the range of the photon?
Mathematical Analysis
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