Download 2.2 Schrödinger`s wave equation

2.2 Schrödinger’s wave equation
Slides: Video 2.2.1 Schrödinger wave
equation introduction
Text reference: Quantum Mechanics
for Scientists and Engineers
Section Chapter 2 introduction
Schrödinger’s wave equation
Quantum mechanics for scientists and engineers
David Miller
2.2 Schrödinger’s wave equation
Slides: Video 2.2.2 From de Broglie to
Schrödinger
Text reference: Quantum Mechanics
for Scientists and Engineers
Sections 2.1 – 2.2
Schrödinger’s wave equation
From de Broglie to Schrödinger
Quantum mechanics for scientists and engineers
David Miller
Electrons as waves
de Broglie’s hypothesis is that the electron
wavelength  is given by
h

p
where p is the electron momentum and
h is Planck’s constant
h  6.62606957 1034 J s
Now we want to use this to help construct a
wave equation
A Helmholtz wave equation
If we are considering only waves of one
wavelength  for the moment
i.e., monochromatic waves
we can choose a Helmholtz wave equation
d 2
2
2
k



with k 
2
dz

which we know works for simple waves
with solutions like
sin(kz), cos(kz), and exp(ikz)
(and sin(–kz), cos(–kz), and exp(–ikz))
A Helmholtz wave equation
In three dimensions, we can write this as
2
2
2






 2  2  2  2   k 2
x
y
z
which has solutions like
sin(k  r), cos(k  r), and exp(ik  r)
(and sin(-k  r), cos(-k  r), and exp(-ik  r))
where k and r are vectors
From Helmholtz to Schrödinger
With de Broglie’s hypothesis   h / p
and the definition k  2 / 
then k  2 p / h  p / 
where we have defined   h / 2
so k 2  p 2 /  2
Hence we can rewrite our Helmholtz equation
p2
2
  2

or
  2 2  p 2
From Helmholtz to Schrödinger
If we are thinking of an electron, we can
divide both sides by its mass mo to obtain
2 2
p2

 

2mo
2mo
But we know from classical mechanics that
p2
 kinetic energy of electron
2mo
and in general
Total energy (E )=Kinetic energy + Potential energy (V  r  )
From Helmholtz to Schrödinger
So Kinetic energy = p 2 / 2mo
= Total energy (E ) - Potential energy (V  r  )
2 2
p2
 

Hence our Helmholtz equation 
2mo
2mo
2 2
    E  V  r  
becomes the Schrödinger equation 
2mo
or equivalently
 2 2

  V  r    E

 2mo

Schrödinger’s time-independent equation
We can postulate a Schrödinger equation for
any particle of mass m
 2 2

  V  r    E

 2m

Formally, this is the
time-independent Schrödinger equation
Probability densities
Born’s postulate is that
the probability P  r  of finding an electron
near any specific point r in space
2
is proportional to the modulus squared   r 
of the wave amplitude   r 
2
  r  can therefore be viewed as a
“probability density”
with   r  called a “probability amplitude”
or a “quantum mechanical amplitude”
2.2 Schrödinger’s wave equation
Slides: Video 2.2.4 Diffraction by two
slits
Text reference: Quantum Mechanics
for Scientists and Engineers
Section 2.3 (first part)
Schrödinger’s wave equation
Diffraction by two slits
Quantum mechanics for scientists and engineers
David Miller
Young’s slits
An opaque mask has two slits cut in it, a distance s apart
s
Young’s slits
We shine a plane wave on the mask from the left
s
Young’s slits
What will be the pattern on a screen at a large distance zo?
?
s
zo
Young’s slits
The slits as point sources give an interference pattern
Young’s slits
The distance from the upper source to point x
2
on the screen is
2
x

s
/
2

z

 o
xs/2
zo
s
2
 x  s / 2
2
 z  zo 1   x  s / 2  / zo2
2
2
o
 zo   x  s / 2  / 2 zo
2
 zo  x 2 / 2 zo  s 2 / 8 zo  sx / 2 zo
x
Young’s slits
The distance from the lower source to point x
on the screen is  x  s / 2 2  z 2
o
 zo   x  s / 2  / 2 zo
x
2
s
2
 zo  x 2 / 2 zo  s 2 / 8 zo  sx / 2 zo
zo
xs/2
Young’s slits
For large zo the waves are approximately uniformly “bright”
i.e., using exponential waves for convenience
2
2
2


 s  x   exp ik  x  s / 2   zo  exp ik  x  s / 2   zo2 




Using our approximate formulas for the distances gives
 s  x   exp  i exp ik  sx / 2 zo    exp  ik  sx / 2 zo  
where   k  zo  x 2 / 2 zo  s 2 / 8 zo 
Young’s slits
Now exp  i   exp  i   2cos  

 sx
so  s  x   exp  i  exp  ik
 2 zo



sx  
  exp  ik

2
z
o 


 sx 
  sx 
 exp  i  cos  k
  exp  i  cos 

2
z
z

o 

 o
so the “intensity” of the beam
2
1
2
 s  x   cos  sx /  zo   1  cos  2 sx /  zo  
2
Young’s slits
The interference fringes are spaced by d s  zo / s
ds
s
zo
Young’s slits
This allows us to measure small wavelengths   d s s / zo
ds
s
zo
2.2 Schrödinger’s wave equation
Slides: Video 2.2.6 Interpreting
diffraction by two slits
Text reference: Quantum Mechanics
for Scientists and Engineers
Section 2.3 (second part)
Schrödinger’s wave equation
Interpreting diffraction by two slits
Quantum mechanics for scientists and engineers
David Miller
Young’s slits
If the upper slit is blocked – no interference pattern
Young’s slits
If the lower slit is blocked – no interference pattern