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Given the Uncertainty Principle, how do you write an equation of motion
for a particle?
•First, remember that a particle is only a particle sort of, and a wave sort of,
and it’s not quite like anything you’ve encountered in classical physics.
We need to use Fourier’s Theorem to represent
the particle as the superposition of many waves.
wavefunction
of the
electron
adding varying amounts of
an infinite number of waves sinusoidal expression
for harmonics

( x,0)   a(k )e dk
ikx

amplitude of wave with
wavenumber k=2p/l
•We saw a hint of probabilistic behavior in the double slit experiment.
Maybe that is a clue about how to describe the motion of a “particle” or
“wavicle” or whatever.
We can’t write a deterministic equation of motion as in
Newtonian Mechanics, however, we know that a large number of
events will behave in a statistically predictable way.
probability for an
electron to be found
between x and x+dx
(x,t)
b
2
dx
P   ( x, t ) dx
2
a
Assuming this particle exists, at any given time it must be
somewhere. This requirement can be expressed mathematically as:
If you search from
hither to yon



( x, t ) dx  1
2
you will find your particle
once, not twice (that would be
two particles) but once.
Conceptual definition
When two or more wave moving through the same region of space,
waves will superimpose and produce a well defined combined effect.
Mathematical definition
For a linear homogeneous ordinary differential equation,
if
and
are solutions, then so is
.
When two waves of equal amplitude and frequency
but opposite directions of travel superimpose, you
get a standing wave- a wave that appears not to
move-its nodes and anti-nodes stay in the same
place. This happens when traveling waves on a
guitar string get to the end of the string and are
reflected back.
( x, t )  Aei ( kx t )  A{cos( kx  t )  i sin( kx  t )}
Euler' s equation
Trigonomet ric functions :
ei  cos   i sin 
ei  e i
cos  
2
ei  e i
sin  
2i
e i  cos   i sin 
ii  1
d u
du
e  eu
dx
dx
conjugate (a  ib )  (a  ib )
if z  a  ib, then :
z * z  (a  ib )( a  ib )
 a i b
2
2 2
 a b
2
2
Graphical representation of
a complex number z as a
point in the complex plane.
The horizontal and vertical
Cartesian components give
the real and imaginary parts
of z respectively.
Note that we can construct a wavefunction only if the
momentum is not precisely defined. A plane wave is
unrealistic since it is not normalizable.



e
i ( kx t )
2

dx   1 dx  

You can’t get around the uncertainty principle!
Alright, we think we might have an acceptable wavefunction. Let’s give it a whirl… If
we think we know what our wavefunction looks like now, how do we propogate it
through time and space?
The Schrodinger Equation:
•Describes the time evolution of your wavefunction.
•Takes the place of Newton’s laws and conserves energy of the system.
•Since “particles” aren’t particles but wavicles, it won’t give us a precise position
of an individual particle, but due to the statistical nature of things, it will precisely
describe the distribution of a large number of particles.
If you know the position of a particle at time t=0, and you constructed a localized wave
packet, superimposing waves of different momenta, then the wave will disperse
because, by definition, waves of different momenta travel at different speeds.
the position
of the real
part of the
wave…
the
probability
density
expression for
kinetic energy
p2
KE 
;
2m
p  k
the potential
dU
dx
i.e., for a classical spring :
F 
1 2
kx
2
dU
d 1

F 
   kx2   kx
dx
dx  2

U
kinetic plus potential
energy gives the total
energy
Remember our guitar
string? We had the
boundary condition
that the ends of the
string were fixed.
Quantum mechanical versionthe particle is confined by an
infinite potential on either side.
The boundary condition- the
probability of finding the particle
outside of the box is ZERO!
 (0, t )  0 at x  0
 ( L, t )  0 at x  L
For the quantum mechanical case:
For the guitar string:
Assume a general case:
1( x, t )   2( x, t )
 A1e i ( kx t )  A2 e i ( kx t )
 A1e i (
2 mE x /   Et /  )
 A2 ei (
2 mE x /   Et /  )
at x  0
A1e i (  Et /  )  A2 e i (  Et /  )  0  A1  A2
at
xL
 A1e i (
2 mE L /   Et /  )

 Ae iEt /  e i
 A2 ei (
2 mE L / 
 ei
2 mE L /   Et /  )
2 mE L / 
 2iA sin( 2mE L /  )  0
 2mE L /   np
n 2p 2  2
 En 
2mL2
n  1, 2 ,3...
 0
0
Note that while =0 solves the problem mathematically, it does not
satisfy the conditions for the quantum mechanical wavefunction,
because we know that at any given time, the particle must be found
SOMEWHERE. N=0 is not a solution.
Appealing to the uncertainty principle can give us a clue.
px  h
h
p 
L
2

p 
h2
E

2m
2mL2
The wavefunction can be written as the sum of two parts.
ei ( kx t ) eikx e it
  ( x, t )   ( x) (t )
Note that:
 (t ) * (t )  e it eit e0  1
|  ( x, t ) |2 |  ( x) (t ) |2   ( x)
2
For a “stationary state” the probability of finding a particle is static!
mevr  n n  1, 2, 3...
  h / 2p
an integer number of wavelengths
fits into the circular orbit
nl  2pr
where
h
l
p
l is the de Broglie wavelength