Schrödinger Representation – Schrödinger Equation
Time dependent Schrödinger Equation
i
 ( x , y, z , t )
 H ( x , y , z , t ) ( x , y , z , t )
t
Developed through analogy to Maxwell’s equations and knowledge of
the Bohr model of the H atom.
H classical
Hamiltonian
Q.M.
p2

V
2m
Sum of kinetic energy and potential energy.
kinetic potential
energy energy
 2  2
H
 V ((x))
2
2m  x
2 2
H
  V ( x, y, z )
2m
2m
one dimension
three dimensions
recall
p  i

x
2
2
2
 


2
2
 x  y  z2
2
The potential, V, makes one problem different form another H atom, harmonic oscillator.
Copyright – Michael D. Fayer, 2009
Getting the Time Independent Schrödinger Equation
( x , y, z , t )
wavefunction

i
 ( x , y , z , t )  H ( x , y , z , t ) ( x , y , z , t )
t
If the energy is independent of time
H ( x, y, z )
Try solution
( x , y , z , t )   ( x , y , z )F ( t )
product of spatial function and time function
Then

i
 ( x , y , z )F ( t )  H ( x , y , z ) ( x , y , z )F ( t )
t

i   ( x, y, z )
F ( t )  F ( t ) H ( x , y , z ) ( x , y , z )
t
independent of t
divide through by
  F
independent of x, y, z
Copyright – Michael D. Fayer, 2009
dF ( t )
dt  H ( x , y , z )  ( x , y , z )
F (t )
 ( x, y, z )
i
depends only
on t
depends only
on x, y, z
Can only be true for any x,
x yy, zz, t if both sides equal a constant
constant.
Changing t on the left doesn’t change the value on the right.
Changing x,
x y,
y z on right doesn’t change value on left.
left
Equal constant
dF
dt  E  H 
F

i
Copyright – Michael D. Fayer, 2009
dF
dt  E  H 
F

i
Both sides equal a constant, E.
H ( x , y , z ) ( x , y , z )  E  ( x , y , z )
Energy eigenvalue problem – time independent Schrödinger Equation
H is energy operator.
Operate
p
on  g
get  back times a number.
’s are energy eigenkets; eigenfunctions; wavefunctions.
E
Energy Eigenvalues
Observable values of energy
Copyright – Michael D. Fayer, 2009
Time Dependent Equation (H time independent)
dF ( t )
dt  E
F (t )
ii

i
dF ( t )
 E F (t )
dt
dF ( t )
i
  E dt .
F (t )

ln F  
i Et
C

Integrate both sides
Take initial condition at t = 0, F = 1, then C = 0.
F ( t )  e  i E t /   e  i t
Time dependent part of wavefunction for time independent Hamiltonian.
Time dependent phase factor used in wave packet problem.
Copyright – Michael D. Fayer, 2009
Total wavefunction
 E ( x , y , z , t )   E ( x , y , z )e  i E t / 
E – energy (observable) that
l b l state.
labels
Normalization
 E  E    *E  E d    E* e  i E t /   E e  i E t /  d    E*  E d
Total wavefunction is normalized if time independent part is normalized.
Expectation value of time independent operator S.
S   S     *E S  E d    E* e i E t /  S  E e  i E t /  d
S does not depend on t, e  i E t /  can be brought to other side of S.
 S     E* S  E d
Expectation value is time independent and depends only on the
ti
time
independent
i d
d t partt off the
th wavefunction.
f
ti
Copyright – Michael D. Fayer, 2009
Equation of motion of the Expectation Value in the Schrödinger Representation
Expectation value of operator representing observable for state S
A  S A S
example - momentum P  S P S
In Schrödinger
g representation,
p
, operators
p
don’t change
g in time.
Time dependence contained in wavefunction.
Want Q.M. equivalent of
ti
time
d
derivative
i ti off a classical
l i l dynamical
d
i l variable.
i bl
P
P
classical momentum goes over to momentum operator
P

P
 ? want Q.M. operator equivalent of time derivative
t
Definition:
The time derivative of the operator A , ii. e.,
e A
is defined to mean an operator whose expectation in
any state S is the time derivative of
p
of the operator
p
the expectation
A .
Copyright – Michael D. Fayer, 2009
Want to find
d A
dt
Use


S A S
t
H S  i

S
t
time dependent Schrödinger equation.
 



S A S 
S A S  S A
S
t
t
t




product rule
time independent – derivative is zero

S.
t
(operate to left)
S H  i 
Use the complex
p conjugate
j g
of the Schrödinger
g equation
q
Then

  i
 S
 S H
t


 

i
S 
H S

t
, and from the Schrödinger equation
From Schrödinger eq.

i
S A S   S H A S  S AH S 

t
From complex conjugate of Schrödinger eq.
Copyright – Michael D. Fayer, 2009
Therefore
d A
dt
This is the commutator of H A.

i
S H A  AH S

i
A  [ H , A]

The operator representing the time derivative of an observable is
i/ times the commutator of H with the observable.
Copyright – Michael D. Fayer, 2009
Solving the time independent Schrödinger equation
The free particle
momentum problem
P P p P
 p ( x) 
1
2
eik x
p  k
k  p/
Free particle Hamiltonian – no potential – V = 0.
2
P
H
2
2m
Commutator of H with P
3
3
P
P
[P, H ] 

0
2m 2m
P and H commute
P S  PPP S
3
Simultaneous Eigenfunctions.
Copyright – Michael D. Fayer, 2009
Free particle energy eigenvalue problem
H P E P
2  2
H 
2m  x 2
U momentum
Use
t
eigenkets.
i
k t
2  2  1 i k x 
H P 
e 
2 
2m  x  2


2
2 2m
2m
( ik )2 e i k x
2k 2  1 i k x 

e 

2m  2

p2

P
2m
2
p
Therefore, E 
2m
energy eigenvalues
i
l
Energy same as classical result.
Copyright – Michael D. Fayer, 2009
Particle in a One Dimensional Box
V= 
V= 
V=0
Infinitely high, thick,
impenetrable walls
-b
0
b
x
Particle inside box. Can’t get out because of impenetrable walls.
Classically
E is continuous. E can be zero. One D racquet ball court.
Q.M.
x p  /2
H  E 
E can’t be zero.
Energy eigenvalue problem
Schrödinger Equation
 2 d 2 ( x )

+V ( x )  ( x )  E  ( x )
2
2m d x
V ( x)  0
x b
V ( x)  
x b
Copyright – Michael D. Fayer, 2009
For
x b
 2 d 2 ( x )

 E  ( x)
2
2m d x
Want to solve differential Equation
Equation, but
solution must by physically acceptable.
Born Condition on Wavefunction to make physically meaningful.
1. The wave function must be finite everywhere.
2 The wave function must be single valued.
2.
valued
3. The wave function must be continuous.
4 First derivative of wave function must be continuous
4.
continuous.
Copyright – Michael D. Fayer, 2009
d 2 ( x )
2mE
 ( x)


2
2
dx

Functions with this property
Second derivative of a function equals a
negative constant times the same function.
sin and cos.
d 2 sin(ax )
2


a
sin(ax )
2
dx
d 2 cos(ax )
 a 2 cos(ax )
dx
These are solutions provided
2m E
a  2

2
Copyright – Michael D. Fayer, 2009
Solutions with any value of a don’t obey Born conditions.
Well is infinitely deep.
Particle has zero probability of being
found outside the box.
0

 = 0 for x  b
-b
x
0
b
Function as drawn discontinuous at
x  b
To be an acceptable wavefunction
  sin and cos  0 at x  b
Copyright – Michael D. Fayer, 2009
 will vanish at x  b if
n
a
 an n is an integer
2b
cos an x
n  1, 3, 5...
sin an x
n  2, 4, 6...
0
–bb
b
Integral number of half wavelengths
in box. Zero at walls.
Have two conditions for a2.
2 2
n

2mE
2
an 
 2
2
4b

Solve for E.
n 2 2  2 n 2 h2 Energy eigenvalues – energy levels, not continuous.
En 

2
8mb
8m L2
L = 2b – length
g of box.
Copyright – Michael D. Fayer, 2009
n2 2  2 n2 h2

En 
2
8mb
8m L2
Energy levels are quantized.
Lowest energy not zero.
1
2
n x
1
 n ( x )    cos
2b
b
1
2
n x
1
 n ( x )    sin
2b
b
x b
n  1, 3, 5 
x b
n  2, 4, 6 
L = 2b – length of box.
wavefunctions including
normalization constants
First
st few
ew wave
wavefunctions.
u ct o s.
Quantization forced by Born conditions
(boundary conditions)
n =3
0

n =2
0
Fourth Born condition not met –
first derivative not continuous.
Physically unrealistic problem because
the potential is discontinuous.
n =1
0
-b
0
x
b
Like classical string – has “fundamental”
and harmonics.
Copyright – Michael D. Fayer, 2009
Particle in a Box
Simple model of molecular energy levels.
Anthracene

L6 A
L
E2
S1
E
E1
 electrons – consider “free”
in box of length L.
Ignore all coulomb interactions.
S0
m  me  9  1031 kg
  6  1010 m
L6A
Calculate wavelength of absorption of light.
i in
i box energy level formula
f
Form particle
3h2
E  E2  E1 
8 2
8mL
 E  h
h  6.6  1034 Js
  E / h  7.64  1014 Hz
  c /   393 nm blue-violet
E  5.04  1019 J
Experiment  400 nm
Copyright – Michael D. Fayer, 2009
Anthracene particularly good agreement.
Other molecules, naphthalene, benzene, agreement much worse.
Important point
Confine a particle with “size” of electron to box
size
i off a molecule
l l
Get energy level separation, light absorption, in visible and UV.
Molecular structure, realistic potential
give accurate calculation, but
It is the mass and size alone that set scale.
Big molecules
Small molecules
absorb in red.
absorb in UV.
Copyright – Michael D. Fayer, 2009
Particle in a Finite Box – Tunneling and Ionization
Box with finite walls.
Ti
Time
independent
i d
d tS
Schrödinger
h ödi
E
Eq.
V(x)=V
V(x)=V
V(x)=0
-b
0
x
 2 d 2 ( x )

+V ( x )  ( x )  E  ( x )
2
2m d x
2m
V ( x)  0
x b
V ( x)  V
x b
b
Inside Box V = 0
 2 d 2 ( x )

 E  ( x)
2
2m d x
d  ( x)
2mE
E


 ( x)
2
2
dx

2
Second derivative of function equals
q
negative constant times same function.
Solutions – sin and cos.
Copyright – Michael D. Fayer, 2009
Solutions inside box
 ( x )  q1 sin
i
2mE
x
2

or
 ( x )  q2 cos
22mE
mE
x
2

Outside Box
d 2 ( x )
2m ( E  V )


 ( x)
2
2

dx
T o cases: Bo
Two
Bound
nd states,
states E < V
Unbound states, E > V
Bound States
d 2 ( x ) 2m (V  E )

 ( x)
2
2
dx

Second derivative of function equals
positive constant times same function.
Not oscillatory.
Copyright – Michael D. Fayer, 2009
Try solutions
d 2 e  ax
2 a x
a
e

2
dx
exp
p   ax 
Second derivative of function equals
positive constant times same function
function.
Then, solutions outside the box
1/ 2
 ( x)  e
 2 m (V  E ) 

 x
2



x b
Solutions must obey Born Conditions
 ( x ) can’t blow upp as x  
Therefore,,
1/ 2
 ( x )  r1 e
 2 m (V  E ) 

 x
2


xb
1/ 2
 ( x )  r2 e
 2 m (V  E ) 

 x
2


Outside box
Inside box
exp. decays
exp
oscillatory
x  b
Copyright – Michael D. Fayer, 2009
Outside box
Inside box

0
-b
0
x
exp. decays
oscillatory
The wavefunction and its first derivative
continuous
ti
att walls
ll – Born
B
Conditions.
C diti
b
Expanded view
For a finite distance into the material
finite p
probability
y of finding
gp
particle.

x
b
Classically forbidden region
V > E.
 0
probability
0
Copyright – Michael D. Fayer, 2009
Tunneling - Qualitative Discussion
Classically forbidden region.
Wavefunction not zero at far
side of wall.
0

Probability of finding particle small
but finite outside box.
X
A particle placed inside of box with not enough energy to go over the wall
can Tunnel Through the Wall.
Formula derived in book
mass = me
E = 1000 cm-1
V = 2000 cm-1
e
2 d [2 m (V  E ) /  2 ]1/2 Ratio probs - outside
Wall thickness (d) 1 Å
probability
0.68
vs. inside edges of wall.
10 Å
0.02
100 Å
3  10-17
Copyright – Michael D. Fayer, 2009
Electron in a rectangular potential exact result.
Infinite wall at x = 0. Finite wall between 4 Å and 6 Å.
barrier height 25 eV
electron energy 17 eV
No longer truly discreet single states.
Energy continuous with peaks.
2Å
energy no longer delta function
density o
of states
ampllitude
1.0
0.5
00
0.0
-0.5
-1.0
0
5
10
15
20
energy
amplitu
ude
1.0
Peaks in density of states near
particle in the box energies.
Higher energy – wider.
0.8
0.6
0.4
0.2
0.0
-0.2
2
3
4
5
6
7
8
9
10
11
Prof. Pat Vaccaro – analytical solution
Daniel Rosenfeld – program, numerical calculations
x (Å)
Copyright – Michael D. Fayer, 2009
Chemical Reaction
not enough energy to go over barrier
Temperature dependence of some
chemical reactions shows to much
product at low T. V > kT .
V
reactants
products
q
D
Decay
off wavefunction
f
ti iin classically
l i ll forbidden
f bidd region
i for
f parabolic
b li potential.
t ti l
  d (2mV /  2 )1/ 2
e-
tunneling
parameter
distance mass
light particles tunnel
barrier height
Copyright – Michael D. Fayer, 2009
Methyl Rotation
H
C H
H
Methyl groups rotate even
at very low T.
Radioacti e Deca
Radioactive
Decay
some probability outside nucleus
repulsive Coulomb interaction
nuclear attraction
strong interaction
Copyright – Michael D. Fayer, 2009
Unbound States and Ionization
E large enough - ionization
If E > V - unbound states
V( ) = V
V(x)
bound states
E<V
-b
b
b
Inside the box (between -b and b)
V=0
2mE
 ( x )  q1 sin
x
2

2mE
 ( x )  q2 cos
x
2

S l ti
Solutions
oscillatory
ill t
d 2 ( x )
2m ( E  V )


 ( x)
2
2

dx
Outside the box (x > |b|)
E>V
2m ( E  V )
 ( x )  s1 sin
x
2

 ( x )  s2 cos
2m ( E  V )
x
2

S l ti
Solutions
oscillatory
ill t
Copyright – Michael D. Fayer, 2009
unbound state

-b
b
To solve (numerically)
Wavefunction and first derivative
equal
q
at walls,, for example
p at x = b
In limit
2m ( E  V )
2mE
s
sin
b
q1 sin
b 1
2
2


 q1  s1
E V
(E V )  E
Wavefunction has equal
amplitude everywhere.
Copyright – Michael D. Fayer, 2009
For E >> V
 ( x )  sin
2mE
x
2

for all x
As if there is no wall.
Continuous range of energies - free particle
Particle has been ionized.
2mE

2

2mp 2

2
2m 
2m
 ( x )  sin k x
2k 2
k
2

Free p
particle wavefunction
Copyright – Michael D. Fayer, 2009
In real world potential barriers are finite.
Tunnelingg
Ionization
Copyright – Michael D. Fayer, 2009