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
p  i

x
one dimension
recall
three dimensions
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, 2007
Getting the Time Independent Schrödinger Equation
( x , y, z , t )
i
wavefunction

( 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, 2007
i
dF ( t )
dt  H ( x , y, z )  ( x , y, z )
F (t )
 ( x, y, z )
depends only
on t
depends only
on x, y, z
Can only be true for any x, y, z, t if both sides equal a constant.
Changing t on the left doesn’t change the value on the right.
Changing x, y, z on right doesn’t change value on left.
Equal constant
i
dF
dt  E  H 
F

Copyright – Michael D. Fayer, 2007
i
dF
dt  E  H 
F

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 on  get  back times a number.
’s are energy eigenkets; eigenfunctions; wavefunctions.
E
Energy Eigenvalues
Observable values of energy
Copyright – Michael D. Fayer, 2007
Time Dependent Equation (H time independent)
i
i
dF ( t )
dt  E
F (t )
dF ( t )
 E F (t )
dt
dF ( t )
i
  E dt .
F (t )
ln F  
iEt
Integrate both sides
C
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, 2007
Total wavefunction
 E ( x, y, z , t )   E ( x, y, z )e  i E t /
E – energy (observable) that
labels state.
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
time independent part of the wavefunction.
Copyright – Michael D. Fayer, 2007
Equation of motion of the Expectation Value in the Schrödinger Representation
Expectation value of operator representing observable for state S
A  S AS
example - momentum P  S P S
In Schrödinger representation, operators don’t change in time.
Time dependence contained in wavefunction.
Want Q.M. equivalent of
time derivative of a classical dynamical variable.
P
P
classical momentum goes over to momentum operator
P
P
 ? want Q.M. equivalent of time derivative
t
Definition:
The time derivative of the operator A , i. e., A
is defined to mean an operator whose expectation in
any state S is the time derivative of
the expectation of the operator A .
Copyright – Michael D. Fayer, 2007
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)
Use the complex conjugate of the Schrödinger equation S H   i
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, 2007
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, 2007
Solving the time independent Schrödinger equation
The free particle
momentum problem
P P p P
 p ( x) 
1
2
ei k x
p k
k  p/
Free particle Hamiltonian – no potential – V = 0.
2
P
H
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, 2007
Free particle energy eigenvalue problem
H P E P
2
H 
2m  x 2
2
Use momentum eigenkets.
 2  1 ik x 
H P 
e 
2 
2m  x  2

2
2

2 2m
( ik )2 e i k x
k2  1 ik x 

e 

2m  2

2
p2

P
2m
2
p
Therefore, E 
2m
energy eigenvalues
Energy same as classical result.
Copyright – Michael D. Fayer, 2007
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
d 2 ( x )

+V ( x )  ( x )  E  ( x )
2
2m d x
2
V ( x)  0
x b
V ( x)  
x b
Copyright – Michael D. Fayer, 2007
x b
For
d 2 ( x )

 E  ( x)
2
2m d x
2
Want to solve differential 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.
3. The wave function must be continuous.
4. First derivative of wave function must be
continuous.
Copyright – Michael D. Fayer, 2007
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
a 
2
2m E
2
Copyright – Michael D. Fayer, 2007
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, 2007
 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
–b
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.
n2 2 2 n2 h2 Energy eigenvalues – energy levels, not continuous.
En 

2
8mb
8mL2
L = 2b – length of box.
Copyright – Michael D. Fayer, 2007
n2 2 2 n2 h2
En 

2
8mb
8mL2
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 few wavefunctions.
Quantization forced by Born conditions
(boundary conditions)
n =3
0

n =2
0
Forth 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, 2007
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6 A
Calculate wavelength of absorption of light.
Form particle in box energy level formula
3h2
E  E2  E1 
8mL2
 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, 2007
Anthracene particularly good agreement.
Other molecules, naphthalene, benzene, agreement much worse.
Important point
Confine a particle with “size” of electron to box
size of a molecule
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, 2007
Particle in a Finite Box – Tunneling and Ionization
Box with finite walls.
Time independent Schrödinger Eq.
V(x)=V
V(x)=V
V(x)=0
-b
0
x
d 2 ( x )

+V ( x )  ( x )  E  ( x )
2
2m d x
2
V ( x)  0
x b
V ( x)  V
x b
b
Inside Box V = 0
d 2 ( x )

 E  ( x)
2
2m d x
2
d  ( x)
2mE
  2  ( x)
2
dx
2
Second derivative of function equals
negative constant times same function.
Solutions – sin and cos.
Copyright – Michael D. Fayer, 2007
Solutions inside box
 ( x )  q1 sin
2mE
2
x
or
 ( x )  q2 cos
2mE
2
x
Outside Box
d 2 ( x )
2m ( E  V )


 ( x)
2
2
dx
Two cases: Bound 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, 2007
Try solutions
d 2e  ax
2 a x

a
e
2
dx
exp  ax 
Second derivative of function equals
positive constant times same 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 up 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
oscillatory
x  b
Copyright – Michael D. Fayer, 2007
Outside box
Inside box

0
-b
0
x
b
exp. decays
oscillatory
The wavefunction and its first derivative
continuous at walls – Born Conditions.
Expanded view
For a finite distance into the material
finite probability of finding particle.

x
b
Classically forbidden region
V > E.
 0
probability
0
Copyright – Michael D. Fayer, 2007
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 )/
Wall thickness (d) 1 Å
probability
0.68
2 1/2
]
10 Å
0.02
Ratio probs - outside
vs. inside edges of wall.
100 Å
3  10-17
Copyright – Michael D. Fayer, 2007
Chemical Reaction
not enough energy to go over barrier
Temperature dependence of some
chemical reactions shows to much
product at low T. E > kT .
E
reactants
products
q
Decay of wavefunction in classically forbidden region for parabolic potential.
  d (2mV /
e-
tunneling
parameter
distance mass
2 1/ 2
)
light particles tunnel
barrier height
Copyright – Michael D. Fayer, 2007
Methyl Rotation
H
C H
H
Methyl groups rotate even
at very low T.
Radioactive Decay
some probability outside nucleus
repulsive Coulomb interaction
nuclear attraction
strong interaction
Copyright – Michael D. Fayer, 2007
Unbound States and Ionization
E large enough - ionization
If E > V - unbound states
V(x) = V
bound states
E<V
-b
b
Inside the box (between -b and b)
V=0
2mE
 ( x )  q1 sin
x
2
 ( x )  q2 cos
d 2 ( x )
2m ( E  V )


 ( x)
2
2
dx
2mE
2
x
Solutions oscillatory
Outside the box (x > |b|)
E>V
2m( E  V )
 ( x )  s1 sin
x
2
 ( x )  s2 cos
2m ( E  V )
2
x
Solutions oscillatory
Copyright – Michael D. Fayer, 2007
unbound state

-b
b
To solve (numerically)
Wavefunction and first derivative
equal at walls, for example at x = b
q1 sin
2mE
2
b  s1 sin
2m ( E  V )
2
b
In limit
E
V
(E V )  E
 q1  s1
Wavefunction has equal
amplitude everywhere.
Copyright – Michael D. Fayer, 2007
For E >> V
 ( x )  sin
2mE
2
x
for all x
As if there is no wall.
Continuous range of energies - free particle
Particle has been ionized.
2mE
2
2mp2


2
2m
 ( x )  sin k x
2
k2
2
k
Free particle wavefunction
Copyright – Michael D. Fayer, 2007
In real world potential barriers are finite.
Tunneling
Ionization
Copyright – Michael D. Fayer, 2007