Transcript quantumwaves

•Particles are actually waves
p  h
•Two de Broglie relations
•They have all the properties (and equations) that described
waves before, together with some new ones
•Rewrite the de Broglie relations in terms of k and :
p  h   hk 2  k  p
E  hf  h 2    E
•We can also find a dispersion relation for these waves:
•From this, we can find the group and phase velocity:
d
k
p
vg 


dk m
m
vp 

k 1

 2 vg
k
2m
E  hf
k  2
1 
f  
T 2
2
p
E  mv 
2m
2 2
2
k
k


2m
2m
1
2
“velocity” is group
velocity, not phase velocity
2
p k
E 
Quantum Uncertainty
xk 
1
2
Two uncertainty relations for any type of wave
1
1

t



x   k   2
2
•Multiply by h-bar
•Use quantum wave relations
1
1

t






x

p

2
2
Uncertainty in Position/Momentum
•Classical physics:
t E  12
•Particle is defined by position x and velocity v = (dx/dt)
•Or, if you prefer, exact x and exact p
•Quantum physics:
•Particle is defined by wave function (x)
•It cannot have a definite position and momentum – they tend to both be
uncertain
Uncertainty in Time/Energy
•Things that last eternally can have definite energy
•Things that last a brief time have uncertain energy
•Measurements of energy will have a spread in value
Time/Energy Uncertainty
The 0-meson has a rest mass of 134.977 MeV/c2 and lasts an
average of 8.410-17 s. What is the spread in energies of a 0meson at rest due to its finite lifespan?
t E 
E
t  8.4 1017 s
1
2
15
6.582 10 eV  s

 39 eV
17
2 t
2  8.4 10 s 
E
E
E - mc2 (eV)
Consequences of Position Uncertainty
•Position and momentum of a particle cannot Harmonic Oscillator: xp  0.500
be simultaneously specified or determined
One-dimensional box: xp  0.568
1
•We can often estimate one xp 
2
Spherical 3D box: xp  0.557
quantity if we know the
other by treating this inequality as if
Hydrogen Atom: xp  0.577
it were an equality
•Especially when the energy is being minimized
L
Energy of Particle in a Box
?
•By Carlson’s rule, the
position is uncertain by about x = ¼L.
2
•By the uncertainty principle, momentum is uncertain by
p 

2 x L
•We’d like the momentum to be zero, but we can’t
•This causes the particle to have some energy, called Zero Point Energy
p
p 
2

E
 2

2m
mL
2m
2
2
2
These methods give only
estimates of the answer
E

2
2
2
2mL
Solving Uncertainty/Energy Problems
A particle of mass m lies above an impenetrable
barrier in a gravitational field with acceleration g.
What is the minimum energy of the particle?
m
p2
E
 mgy
2m
y
1. Write an expression for the energy: potential and kinetic
p0
2. Figure out what momentum (normally 0) and what position would have
y0
the “ideal” lowest kinetic and potential energies
y  a and
3. Let x = a and p = /2a

p

2
a
4. Assume the momentum and position differ
from ideal by about x = a and p = /2a
E  2 8ma 2  mga
5. Set derivative of energy function
equal to zero, solve for a
0   2 4ma3  mg
6. Substitute a in to determine minimum energy
 4m 2 g 
E
 2 
8m 

2
2/3
1/3
 2 
 mg 
2 
 4m g 
3
  2 2 mg
4

2 1/3
1/3


a
2 
 4m g 
2
Uncertainty and the Hydrogen Atom
•There are two types of energy associated with the
xp  12
Hydrogen atom
2
2
p 2 ke2
ke
•Potential Energy E 



2
2m r
•Kinetic Energy
8ma
a
•Classically, these two energies are at a minimum when:
•The electron is at rest, p = 0
•The electron is at the origin, r = 0 (x = 0)
•In this case, the energy of the hydrogen atom E = –
•Quantum mechanically, you can’t control both position and momentum
•We will place the electron near the origin, r = 0, but
r x  a
there is an uncertainty x associated with it
•We will place the electron nearly at rest, p = 0, but
p

there is an uncertainty p associated with it
2x 2a
•You have to compromise between these two choices to minimize E
•Let x = a, and let p be as small as possible by the uncertainty principle
•Plug these into the formula for energy
Uncertainty and the Hydrogen Atom
We need to minimize the energy
•Take derivative and set it to zero
•Substitute result back in
2
2
dE
ke
0

 2
3
da
4ma
a
a
2
ke 2
E

2
8ma
a
2
4mke2
2
2
2 4
 4mke 

 mk 2 e4
4
mke
2
2mk
e
E

ke

2

4






2
2

2
2
8m 



2
2
Correct answer is 4 times smaller than this. Why so far off?
•It’s just an estimate
2 4
2
mk
e
•There is momentum in three dimensions
E
•This makes answer 3 times smaller
3 2
•Then answer we get is 4/3 of correct answer
2 4
mk e
E
2
2
The Wave Function
•We’ve been talking about waves, which require wave functions
•We’d better give it a name:
•It would have more arguments in more dimensions
•It is a complex function; it has both a real and imaginary parts
•What does it mean?
•This turns out to be very hard
•Analogy: Electromagnetic Waves
•Described by electric and magnetic fields
•What is an electric field?
•It is defined in terms of a potential for something
American Heritage Science Dictionary
“The distribution in space of the strength and
direction of forces that would be exerted on an
electric charge at any point in that space”.
  x, t 
E  x, t 
B  x, t 
Dr. Carlson, PHY 114
“Electric Field is
potential for force at a
distance on a charge”.
The Wave Function: what does it mean?
  x, t 
•We are talking about one particle – but it is not at
one location in space
•If we measured its position, where would we
be likely to find it?
The Wave Function is also called the probability amplitude
•Clearly, where the wave function is small (or zero), you wouldn’t expect to find the
particle
•Where it’s negative or imaginary, wouldn’t expect to have negative or imaginary
probability
•We’d better make darn sure that the probability is always positive
2
•For electric fields, the energy density is proportional to the field squared
u E
•If working with complex waves, take amplitude first
•How about we make probability density proportional to wave function magnitude
squared:
2
  x, t 
Sample
Problem
1
  x, t  
a
 i1t
  x  i2t
 2 x  
sin 

e sin  a   e
a





A wave in the region 0 < x < a has the wave function above.
What is the probability density at all locations x at all times t?
  x, t 
2
1

a
 i1t
  x  i2t
 2 x  
sin 

e sin  a   e


 a 

1

a
 i1t
  x  i2t
 2 x  
e sin  a   e sin  a  





Helpful Identity:
ei  e i  2 cos 
1  2 x 
  x   2 x  i1t i2t i2t i1t 
2  2 x 
 sin 
e

  sin 
  sin 
 sin 
 e

a
 a 
 a 
 a   a 

Sample
Problem
1
  x, t  
a
 i1t
  x  i2t
 2 x  
sin 

e sin  a   e
a





A wave in the region 0 < x < a has the wave function above.
What is the probability density at all locations x at all times t?
  x, t 
2
 2 x 

2  2 x 
sin  a   sin  a 

1







a
  x   2 x 
 2sin 
 sin 
 cos 1  2  t  
 a   a 


What Does Probability Density Mean?
•The probability density (in 1D) has units of m-1
•In a small region of size dx, the probability of finding the particle is there is given
by ||2dx.
b
2
•To find probability over a larger
P  a  x  b     x, t  dx
a
region, you have to integrate it

2
Normalization: The probability that the particle
1
  x, t  dx

is somewhere must be 1
•If we integrate over all x, we must get 1
… in the region
•In some cases, the problem implies that we restrict to some region 0 < x < a …





  x, t  dx     x, t  dx
2
a
2
0
1  2 x 

  x   2 x 
2  2 x 
  sin 
  sin 
  2sin 
 sin 
 cos t  
a 0
 a 
 a 
 a   a 

1a a

    0 cos  t     1
a2 2

a
At t = 0, the wave function is given by the expression below.
(a) What is the normalization constant N?
(b) What is the probability that the particle is at x > ½a?
Sample
Problem
 N  a  x  for 0  x  a

  x, 0    N  a  x  for  a  x  0

0
elsewhere


1     x,0  dx  N

N
2
2

0
a
2

2
a
2

2
2
a

2
ax

x
dx

a

2
ax

x

 
 dx
2
2
a
0
0
a
2
2
2
2

1 3
1 3
 N  a x  ax  3 x    a x  ax  3 x  

a
0

2
3
3
3
3
1 3
1 3 

 N  a  a  3 a    a  a  3 a   23 N 2 a 3
2

a  x  dx    a  x  dx

a
0
0
3
N
2a 3
Sample
Problem
At t = 0, the wave function is given by the expression below.
(a) What is the normalization constant N?
(b) What is the probability that the particle is at x > ½a?
 N  a  x  for 0  x  a

  x, 0    N  a  x  for  a  x  0

0
elsewhere

P  x  a 
1
2


a /2
  x, 0  dx  N
2
3
N
2a 3
a
2
 a  x
2
dx  N
2
 a x  ax
2
a /2
 N 2  a3  a3  13 a3    12 a3  14 a3  241 a3 
3 3
1 1 1 1 3 1
1
 3 a 1  1       

2a
3 2 4 24  2 24 16

2
 13 x
3

a
a /2
Nx
  x  2 2
x a
Sample Problem
At t = 0, the wave function is given by the expression above.
(a) What is the most likely / least likely places to find the particle?
(b) What is the normalization constant N?
(c) What is the probability that the particle is at 0 < x < a?
•Least likely when function vanishes, at x = 0
•Most likely when function is largest positive or negative
2
2
x

a

 N  Nx  2 x 
d
0

dx
Normalization:
•Let x = atan 
x

1


  N a
2
 2


a

 2
N 2 x 2 dx
x  a
2

2 2



2
x
2
a

2 2
N a tan  a sec  d
2
2
a
2
2
2
tan   a
2
sin  d   N a     sin  2   
2
2
2

2 2
N  a2  x2 
x  a
2
1
2
1
4

2 2
 2
2
N
  N 2 2a
2a

Sample Problem
Nx
  x  2 2
x a
At t = 0, the wave function is given by the expression above.
(a) What is the most likely / least likely places to find the particle?
(b) What is the normalization constant N?
(c) What is the probability that the particle is at 0 < x < a?
a
P 0  x  a  
0
N 2 x 2 dx
x
2
a

2 2
N
 4
N2 1
2  1
1



sin
2

  0    

2
4
a
  8 4
2a

Quantum Wave Equations You Need:
  x, t   Aei kx t 
p k h 
E    hf
xp 
1
2
t E 
1
2
P  a  x  b      x  dx
b
a

1     x  dx

End of material
for Test 2
2
2