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Physics 741 – Graduate Quantum Mechanics I
Everyone Pick Up:
•Syllabus
•Two homework passes
Eric Carlson
“Eric”
“Professor Carlson”
Olin 306
Office Hours always
758-4994 (o) 407-6528 (c)
[email protected]
Materials
•Quantum Mechanics, by
Eric Carlson (free on the
web)
•Calculator
•Pencils or pens, paper
http://users.wfu.edu/ecarlson/quantum
8/26
Dr. Carlson’s Approximate Schedule
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PHY 741
PHY 215
office hour
office hour
PHY 741
PHY 215
office hour
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PHY 741
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research
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Free food
colloquium
research
faculty
fellow
•I will try to be in my office Tues. Thurs. 12-1
•When in doubt, call/email first
Reading Assignments / The Text
http://users.wfu.edu/ecarlson/quantum
• I use my own textbook
• Downloadable for free from the web
• Assignments posted on the web
• Will be updated as needed
• Readings every day
Day
Today
Friday
Monday
ASSIGNMENTS
Read
Homework
1A, 1B
none
1C, 1D, 1E 1.1
2A, 2B, 2C 1.2, 1.3, 1.4
Homework
http://users.wfu.edu/ecarlson/quantum
•
•
•
•
•
Almost all the problems in the textbook
About two problems due every day
Homework is due at 12:00 on day
Late homework penalty 20% per day
Two homework passes per semester
• Working with other student is allowed
• Seek my help when stuck
• You should understand anything you turn in
Day
Today
Friday
Monday
ASSIGNMENTS
Read
Homework
1A, 1B
none
1C, 1D, 1E 1.1
2A, 2B, 2C 1.2, 1.3, 1.4
Attendance and Tests
Attendance
•I do not grade on attendance
•Attendance is expected
•Class participation is expected
•I take attendance every day
Tests
•Midterm will be about two hours
long, approximately on Oct. 15
•Final exam will be about three
hours long, 2:00 PM on Dec. 9
Grades, Pandemic Plans
Percentage
Breakdown:
Homework 50%
Midterm
20%
Final
30%
Grade Assigned
94% A
77% C+
90% A73% C
87% B+
70% C83% B
<70% F
80% B-
Pandemic Plans
•If there is a catastrophic closing of
the university, we will attempt to
continue the class:
•Some curving
possible
Emergency contacts:
Web page
email
Cell: 336-407-6528
1. Introduction
1A. Quantum Mechanics is Weird
Photons and Quantum Mechanics
E  hf  
•Light come in packets with energy
•This equation is bizarre
–It implies both that EM energy is in waves (f) and is particles (E)
•A plane wave looks something like
E  r, t   E0 cos  kx  t  or E r, t   E0 sin  kx  t 
•Mathematically simpler (and crucial for
E  r, t   E0eikxit
quantum) to combine them into complex waves:
•Light moves at the speed of light c:   ck
• A formula from relativity for massless particles: E  cp
•Put these formulas together:
p k
cp  E    ck
Uncertainty Principle for Photons
•Waves are not generally localized in space
p k
–They have a spread in position x
•You can make them somewhat localized by combining different wave
numbers k
–Now they have a spread in k, k
•There is a precise inequality relating these two quantities
xk  12
–Proved formally in a later chapter
•If we multiply this equation by , we get the uncertainty principle
xp  12
•In quantum mechanics, photons can’t have both a definite position and a
definite momentum
It’s All Done with Mirrors
•Ordinary mirrors reflect all of the light that impinges on them
mirror
•Half-silvered mirrors have a very thin coating of metal
–They reflect only half the light and transmit the other half
half-mirror
•What happens if we shine photons on a half silvered mirror?
Photons and half-silvered mirror
Let’s send photons through a half-mirror
•The photon gets split into two equal pieces
•Each detector sees 50% of the original photons
•Even if we send photons in one at a time
•Never in both detectors
•If you send in a wave the other way, the same thing happens
•There’s a “phase difference”, but since we square the
amplitude, the probabilities are the same
•50% in each detector
50%
A 50%
Detectors
B
50%
50%
Interferometry
Now use two mirrors and two half-mirrors
•We can reconstruct the original waves
•The photon gets split into two equal pieces
•The two halves of the photons are recombined by the
second half-mirror
•Always goes to detector A
•Even one photon at a time
•If you send in a wave the other way, the
photon is still split in half
•The “phase difference” lets it
remember which way it was going
A 0% •Always in detector B
100%
Interferometry requires that we
carefully position the mirrors
100% B 0%
Non-Interferometry
How does the photon remember which way it was going?
•Replace one mirror with a detector
•The photon gets split into two equal pieces
•Half of them go to detector C
•The other half gets split in half again
•Detectors A and B each see 25%
•Even if you do it one photon at a time
•The “memory” of which way it was going is in
both halves
A
C
25%
•Depending on which experiment you do,
photons sometimes act like particles and
sometimes act like waves
50%
B
25%
Can We Have Our Cake and Eat it Too?
•When you do interference, you can tell the photon went both ways
•For other experiments, you can measure which way it went
The plan:
•Can we do both?
•Do experiment in space (no friction, etc.)
•Carefully measure momentum of mirror before
you send one photon in
•Check photon goes to detector A
•Remeasure momentum and determine the path
pbefore
A 100%
pafter
B
0%
p
0

 k
if upper path
if lower path
The problem
•If you measure the mirror’s initial momentum
accurately, you have small p, and big x
•Poor positioning of mirror ruins the interference
Conclusion
•Because Quantum applies to photons, it must apply to mirrors as well
•It seems inevitable that it applies to other things (like electrons)
•We will apply it to non-relativistic systems, because these are easier to
understand than relativistic
– We will eventually do electromagnetism / photons, but this is
harder
1B. Schrödinger’s Equation
Deriving the free Schrödinger Equation in 1D
•For electromagnetic field, we had waves like: E  r, t   E0eikxit
•For non-relativistic particles, we rename the wave function:   r, t   Neikxit
•We assume two other relationships still apply E   p  k
•We note that:



i
  k
i   
x
t
•Multiplying by , we have:




p  i
E i
i
  E  i
  p
x
t
t
x
p2
•Classical relationship between the energy E and momentum p:
E
2
2m
p
•Multiply by  on the right:
E 

2
2m

1 
 
i

•Make the substitutions above
 i
 
t
2m 
x 
Schrödinger’s Equation in 1D
2



2
p


i
E i
i
  x, t   
  x, t 
2
x
t
t
2m x
•What if we have forces?

•If the forces conserve energy, they can be
F  x, t    V  x, t 
written as the derivative of a potential
x
•This contributes a new term to the energy
p2
E
 V  x, t 
2m
p2
E   x, t  
  x, t   V  x, t    x, t 
2m
•Make the same substitutions as before
2

2
i
  x, t   
  x, t   V  x , t    x , t 
2
t
2m x
Schrödinger’s Equation in 3D
E i

t

p  i
x
•What if we are in 3D?
•Momentum and potential must be generalized:
•Redo the calculation as before
p  i 
F  r, t   V  r, t 
p2
E
 V  r, t 
2m
p2
E   r, t  
  r, t   V  r, t    r, t 
2m
•Make the new substitutions:
2

i
  r, t   
 2   r, t   V  r, t    r, t 
t
2m
Other Things to Consider:
•Particles may have spin
•Some forces (such as magnetism) are not conservative forces
•There may be multiple particles
•The number of particles may actually be indefinite
•Relativity
•We will deal with all of these later
1C. The Meaning of the Wave Function
Probability Density
•The wave function, in essence, describes where the particle is
– When it’s zero, the particle isn’t there
– When it’s large, the particle more likely is there
•In electromagnetism, the energy density is proportional to |E(r,t)|2
•Since the probability of finding a photon is proportional to the energy
density, it makes sense to make a similar conclusion for wave functions
Probability density (1D)    x, t     x, t   *  x, t    x, t 
2
Probability density (3D)    r, t     r, t   *  r, t    r, t 
2
Using Probability Density, and Normalization
•The probability density must be integrated to find
the probability that a particle is in a certain range
  x, t     x, t 
  r, t     r, t 
P  a  x  b      x, t  dx
b
2
a
What are the units of
(x,t)? Of (r,t)?
P  r  V      r, t  d r
2
3
V
•The probability that a particle is somewhere must be exactly 1

1     r,t  d 3r
2
1     x, t  dx
2

•Don’t forget how to do 3D integrals in spherical coordinates!
 f r  d r  
3

0

2
0
0
r dr  sin  d  d f  r , ,  
2
2
2
Sample Problem
A particle in a three-dimensional
infinite square well has ground state
wave function as given by
 N sin  r R  r r  R ,
  r , ,    
0
r  R.

What is the normalization constant
N? What is the probability that the
particle is at r < ½R?
P r  R  N
1
2
2

1R
2
0

1    d 3r
2
 N
 N
2


R
0
2 R
0

sin  r R  dr
2

 4 N
2
 4 N
2 1
2
R
0

1
1
sin 2  r R  dr
R  2 R N
2
d cos 
 
2
0
d
2

1
N
2 R
> integrate(sin(Pi*r/R)^2,r=0..R);
> integrate(sin(Pi*r/R)^2,r=0..R/2);
2
r dr  sin  d  d sin  r R  r 
0
0
2
2
r dr  sin  d  d sin  r R  r 
0
0
2
4 12 R 2
4 1

sin

r
R
dr

R



2 R 0
2 R 4
2
P  r  12 R   12
Some comments
2

i
  r, t   
 2   r, t   V  r, t    r, t 
t
2m
1     r,t  d 3r
2
•Schrödinger’s equation is first order in time
•(r,t = 0) determines (r,t) at all times
•Phase change in (r,t) does not affect anything
– (r,t)  ei (r,t) is effectively equivalent
– We will treat these as different in principle but experimentally
indistinguishable
•Normalization condition must be satisfied at all times
– Can be shown that if true at t = 0, Schrödinger’s equation makes sure
it is always true
– Will be proven later
•We still need to discuss how measurement changes (r,t)
– Chapter 4
1D. Fourier Transform of the Wave Function
The Fourier Transform In 1D
•The Fourier transform of a function is given by*
•You can also Fourier transform in reverse
•If the function is normalized, so is its Fourier
transform    k  2 dk     x  2 dx  1




 k   

  x  



dx
  x  eikx
2
dk
  k  e ikx
2
  x   Neik0 x
•Suppose we have a plane wave:
•It’s momentum is k0
  k   N 2   k  k0 
•It’s Fourier transform is
•You would expect, if the Fourier transform is concentrated near k0, then its
momentum would be around k0
•This suggests the Fourier transform tells you P p  k ~  k 2


 
something about what the momentum is
*The factors of 2 differ based on convention. I like this one.
The Fourier Transform and Momentum
 dk
 ikx
dx
 ikx

x


k
e
  
 
 k   
  x e

2
2
•The wave function (x) tells you the probability of finding the
particle’s position x in a certain range
x2
2
P  x1  x  x2      x  dx

x1
•The Fourier Transform tells you the probability of finding the
particle’s momentum p in a certain range
P  k1  p  k2      k  dk
k2
2
k1
•Neither of these two representation is really more fundamental than the other
•We can similarly do Fourier transforms in 3D
 k   
d 3r
 2 
 r  e
3/2
 ik r
 r   
d 3k
 2 
ik r

k
e


3/2
Expectation Values
•In any situation where you have a list of probabilities of a particular
outcome, the expectation value is the average of what you expect to get
•If there is a continuous distribution of possibilities, this becomes an integral
a   a P a
a
a   a   a  da
•For example, we can measure the average value of the position x or the position
squared x2
2
2
2
2
x   x   x  dx , x   x   x  dx
•Using the Fourier transform, we can similarly compute the momentum or its
square
2
2
2
2
p   k   k  dk , p    k    k  dk
•There are in fact ways of finding these without doing the Fourier transform
(later chapter)
Uncertainty
•The uncertainty of a quantity is the root mean square average of how far it
deviates from its average
a   a P a
 a 
 a  a  P a
2
2
a
a
•There is an easier way to calculate this:
 a   a 2  a 2
2
2
2

a

a

a
P
a

a

2
aa

a
  
   
 P a
2
2
a
a
  a 2 P  a   2a  aP  a   a 2  P  a   a2  2a a  a 2 1
a
a
•Uncertainty in
position are given by:
•These satisfy:
 x  p  
1
2
a
 x 
2
 p 

  x 2   x  dx 
2

2



2


k   k  dk 
2
2

x   x  dx
2




2
k   k  dk
2

2
Understanding from Uncertainty
 x  p   12
•We can often gain a qualitative understanding of what is going on by
considering the uncertainty principle
•What stabilizes the Hydrogen atom?
•It has potential and kinetic energy:
1 2 ke e 2
E
p 
•Classically, it wants:
2m
r
– p = 0 to minimize the kinetic energy
– r = 0 to minimize the potential energy
•But this violates the uncertainty principle!
•If we specify the position too well, the momentum will be large
•If we specify the momentum too well, the position will become uncertain
•A compromise is the best solution
Sample Problem
Classically, the Hydrogen atom has
 x  p   12
energy given by
•Let the uncertainty in the position be x  a
1 2 ke e 2
E
p 
•By the uncertainty principle, the momen2m
r
p 
tum must have at least uncertainty
2a
•Assume
the
position
(r)
is
about
x
from
Using the uncertainty principle,
estimate the ground state energy of the ideal position (r = 0), and the r  a
momentum (p) is about p from
the hydrogen atom
p
the ideal momentum (p = 0)
2a
2
2
•Substitute into the energy formula
ke e
•The minimum is at some finite value of a: E  8ma 2  a
2
2
ke e 2
dE

 2
0
3
a
4ma
a
da
4mke e2 •Off by factor of 4
•Substitute back into energy formula
•Mostly because we ignored that
2
2
2
2 4
it was a 3D problem (factor of 3)
 4mke e 2 


4
mk
e

2
k
2
e
ee m
E

  ke e 
 
2
2
2
•Partly because it’s an estimate
8m 


