Transcript Chapter08

Time Dependent Two State Problem
Coupled Pendulums
A
B
Weak spring
Two normal modes.
A
B
A
B
No friction. No air resistance. Perfect Spring
Copyright – Michael D. Fayer, 2007
A
B
Start A Swinging
A
B
Some time later B swings with full amplitude.
A stationary
Copyright – Michael D. Fayer, 2007
M+n
L
Electron Transfer
Electron moves between
metal centers.
M+m
v1 = 2
v2 = 3
v1 = 1
*
v2 = 2
v2 = 1
v1 = 0
Vibrational Transfer
Vibration moves between
two modes of a molecule.
v2 = 0
h
1
*
2
Electronic Excitation Transfer
Electronic excited state moves
between two molecules.
Copyright – Michael D. Fayer, 2007
Consider two molecules and their lowest two energy levels
A
B
EA
EB
Take molecules to be identical, so later will set
E A  EB  E0
States of System
A  Molecule A excited
B unexcited
B  Molecule B excited
A unexcited
A and B Normalized & Orthogonal
Copyright – Michael D. Fayer, 2007
Initially, take there to be NO interaction between them
No “spring”
H A  EA A
H B  EB B
H is time independent.
Therefore
A  e  iE At / 
B  e  iEB t / 
Time independent kets 
Spatial wavefunction
Time dependent
Part of wavefunction
Copyright – Michael D. Fayer, 2007
If molecules reasonably close together
Intermolecular Interactions
Coupling of states A & B
(Like spring in pendulum problem)
Then energy of molecule A is influenced by B.
Energy of A determined by both A & B
A Will no longer be eigenket of H
Then:
H A  He  iE At /   e  iE At / H 
But  &  are coupled
H   EA    
Coupling strength
Energy of interaction
Copyright – Michael D. Fayer, 2007
Coupling strength
Thus:
H A   EA    

e  iE At /
H B   EB    

e  iEB t /
Time dependent phase factors
Very Important
For molecules that are identical
E A  EB  E0
Pick energy scale so:
E0  0
Therefore
H   
H   
Copyright – Michael D. Fayer, 2007
Have two kets A & B describing states of the system.
Most general state is a superposition
t  C1 A  C2 B
Normalized
May be time dependent
Kets A & B have time dependent parts e  iEt /
For case of identical molecules being considered:
E A  EB  E0  0
Then:
A 
& B  
Any time dependence must be in C1 & C2.
Copyright – Michael D. Fayer, 2007
Substitute
t  C1   C 2 
into time dependent Schrödinger Equation:

i
t  H t  H C1   C 2  
t

 

i  C1   C2    C1   C 2 



C1 
 C1
t

C2 
Left multiply by 

i C1   C 2
Take derivative.
 &  t independent
 C2
t
normalized
 &  orthogonal
Left multiply by 

i C2   C1

Then:
i C1   C 2

i C2   C1
Eq. of motion of coefficients
 &  time independent. All time dependence in C1 & C2
Copyright – Michael D. Fayer, 2007
Solving Equations of Motion:
have:

i C1   C 2
Take
d
dt


i C1   C 2

C1  
but:

C2

i

2
C2  
then:
i
C1  
2
C1
C1
Second derivative of function equals
negative constant times function –
solutions, sin and cos.
C1  Q sin( t / )  R cos( t / )
Copyright – Michael D. Fayer, 2007
C1  Q sin( t / )  R cos( t / )
And:
C2 
i


C1
C2  i Q cos  t/  R sin  t/

Copyright – Michael D. Fayer, 2007
t normalized

t t  1  C1*   C 2* 
 C
1
  C2 

 C1*C1  C 2*C 2
C1*C1  C 2*C 2  1
Sum of probabilities equals 1.
This yields
R2  Q 2  1
To go further, need initial condition
Take for t = 0
C1  1 

C2  0 
Means: Molecule A excited at t = 0,
B not excited.
Copyright – Michael D. Fayer, 2007
C2  i Q cos  t/  R sin  t/
C1  Q sin( t / )  R cos( t / )

For t = 0
C1  1 

C2  0 

Means: Molecule A excited at t = 0,
B not excited.
R=1 & Q=0
For these initial conditions:
C1  cos( t / )
probability amplitudes
C 2   i sin( t / )
Copyright – Michael D. Fayer, 2007
Projection Operator:
t  C1 A  C 2 B  cos( t / ) A  i sin( t / ) B
Time dependent coefficients
Consider A A
Projection Operator
A A t  C1 A
Gives piece of t that is A
In general:
S   Ci i
i
k k S  Ck k
Coefficient – Amplitude (for normalized kets)
Copyright – Michael D. Fayer, 2007
Consider:
Closed Brackets  Number
S k k S  C k*C k  C k
2
Absolute value squared of amplitude of particular ket k in superposition S .
Ck
2
Probability of finding system in state k given
that it is in superposition of states S
Copyright – Michael D. Fayer, 2007
Projection Ops.  Probability of finding system in A or B
given it is in t
t  C1 A  C 2 B  cos( t / ) A  i sin( t / ) B
PA  t A A t  C1*C1  cos 2  t /
PB  t B B t  C 2*C 2  sin 2  t /
Total probability is always 1 since cos2 + sin2 = 1
  energy
 energy-sec

   rad
sec
Copyright – Michael D. Fayer, 2007
PA  t A A t  C1*C1  cos 2  t /
PB  t B B t  C 2*C 2  sin 2  t /
At t = 0
When
PA = 1
PB = 0
(A excited)
(B not excited)
 t/   /2
t  h / 4
PA = 0
PB = 1
(A not excited)
(B excited)
Excitation has transferred from A to B in time t  h / 4
At t  2h / 4
(A excited again)
(B not excited)
In between times  Probability intermediate
Copyright – Michael D. Fayer, 2007
Stationary States
Consider two superpositions of
1
 
A  B 

2
1
 
A  B 

2
H  
1
H A H B

2
1

 B  A

2
H   
A & B


H   
H   
Eigenstate,  Eigenvalue
Similarly
H    
Eigenstate,  Eigenvalue
Observables of Energy Operator
Copyright – Michael D. Fayer, 2007


2
Recall E0 = 0
If E0 not 0, splitting still symmetric
about E0 with splitting 2.
E0


2
Dimer splitting  2
E0
E=0
1
A  B

2
1
 
A  B

2
 


Delocalized States
pentacene in
p-terphenyl crystal
1.3 K
Probability of finding either molecule excited is equal
 A A 
1
2
Copyright – Michael D. Fayer, 2007
Use projection operators to find probability of being in eigenstate,
given that the system is in t

t   t   C1* A  C 2* B





1

A

B


  C .C .
2

1
1

C1*  C 2*
( C1  C 2 )
2
2
1
 [C1*C1  C 2*C 2  C1*C 2  C 2*C1 ]
2
complex conjugate
of previous expression
1
 [cos 2 ( t / )  sin 2 ( t / )  i cos( t / )sin( t / )  i sin( t / )cos( t / )]
2
1
t  t 
2
Make energy measurement 
Also
equal probability of finding  or -
1
t  t 
t is not an eigenstate
2
Copyright – Michael D. Fayer, 2007
Expectation Value
Half of measurements yield +; half 
One measurement on many systems  Expectation Value – should be 0.


t H t  C1* A  C 2* B H  C1 A  C 2 B

 C1*C1  H   C2*C1  H   C1*C 2  H   C 2*C 2  H 
Using
H   
H   
  C 2*C1   C1*C 2
   i sin( t
t H t 0
)cos( t
)  i sin( t
)cos( t
)
Expectation Value - Time independent
If E0  0, get E0
Copyright – Michael D. Fayer, 2007
Non-Degenerate Case
A
DE
EA
B
EB
DE 2  4 2
DE 2  2 2
2 2
PA  C C1 

cos
t
2
2
2
2
DE  4
DE  4
*
1

DE 2  4 2
2 2
PB  C C 2 
 1  cos
2
2
DE  4 
*
2

t


As DE increases  Oscillations Faster
Less Probability Transferred
Thermal Fluctuations change DE & 
Copyright – Michael D. Fayer, 2007
Crystals

Dimer splitting


Three levels
…
…
n levels
Using Bloch Theorem of Solid State Physics 
can solve problem of n molecules or atoms
where n is very large, e.g., 1020, a crystal lattice.
Copyright – Michael D. Fayer, 2007
Excitation of a One Dimensional Lattice

…
lattice spacing
…
j-1
j
j+1
j
ground state of the jth molecule in lattice (normalized, orthogonal)
 je
excited state of this molecule
Ground state of crystal with n molecules

g
  0 1  2   n1
Take ground state to be zero of energy.
Excited state of lattice, jth molecule excited, all other molecules in ground states
 j   0 1  2   je   n1
e
j
e
energy of single molecule
in excited state, Ee
set of n-fold degenerate eigenstates
in the absence of intermolecular interactions
because any of the n molecules can be excited.
Copyright – Michael D. Fayer, 2007
Bloch Theorem of Solid State Physics – Periodic Lattice Eigenstates
Lattice spacing - 
Translating a lattice by any number of lattice spacings, ,
lattice looks identical.
Because lattice is identical, following translation
Potential is unchanged by translation
Hamiltonian unchanged by translation
Eigenvectors unchanged by translation
Bloch Theorem – from group theory and symmetry properties of lattices
 p ( x   )  e 2 ip / L  p ( x )
 e ik  p ( x )
The exponential is the translation
operator. It moves function
one lattice spacing.
p is integer ranging from 0 to n-1.
L = n, size of lattice
k = 2p/L
Copyright – Michael D. Fayer, 2007
Any number of lattice translations produces an equivalent function,
result is a superposition of the kets with each of the n possible translations.
1 n1 ik j e
 (k ) 
e
j

n j 0
Bloch Theorem – eigenstates of lattice
Ket with excited state on jth molecule
Sum over all j possible positions (translations) of excited state.
Normalization so there is only a total
of one excited state on entire lattice.
In two state problem, there were two molecules and two eigenstates.
For a lattice, there are n molecules, and n eigenstates.
There are n different orthonormal
 (k )
arising from the n different values of the integer p,
which give n different values of k. k = 2p/L
Copyright – Michael D. Fayer, 2007
One dimensional lattice problem with nearest neighbor interactions only
H  H M  H j , j 1
Molecular Hamiltonian
in absence of intermolecular
interactions.
Intermolecular coupling between adjacent molecules.
Couples a molecule to molecules on either side.
Like coupling in two state (two molecule) problem.
H M  H M1  H M2    H M j    H Mn1
HM 
e
j
E 
e
H M  (k )  H M
Sum of single molecule Hamiltonians
The jth term gives Ee,
the other terms give zero because
the ground state energy is zero.
e
j
1 n1 ik j e
e
j

n j 0
1 n1 ik j
e

e
H


M
j
n j 0
 E  (k )
In the absence of intermolecular interactions,
the energy of an excitation in the lattice
is just the energy of the molecular excited state.
e
Copyright – Michael D. Fayer, 2007
Inclusion of intermolecular interactions breaks the excited state degeneracy.
H j , j 1  j    j 1    j 1
e
state with
jth
e
e
molecule excited
coupling strength
Operate H j , j 1 on Bloch states – eigenstates.
H j , j 1  ( k )  H j , j 1
1 n1 ik j e
e
j

n j 0
1 n1 ik j
e

e
H

j , j 1

j
n j 0
1 n1  ik j
e
e
ik j


e



e



j 1
j 1 

n j 0
Copyright – Michael D. Fayer, 2007
1 n1  ik j
e
e
ik j

H j , j 1  ( k ) 
e



e



j 1
j 1 

n j 0
Each of the terms in the square brackets can be multiplied by
e ik e  ik  1
1 n1  ik j ik  ik
e
e
ik j  ik ik

H j , j 1  ( k ) 
e
e
e



e
e
e



j 1
j 1 

n j 0
combining
combining
1 n1   ik ik ( j 1) e
e
ik
ik ( j 1)


e

e


e

e


j 1
j 1 

n j 0
j+1
j1
In spite of difference in indices, the sum over j is sum over all lattice
sites because of cyclic boundary condition.
Therefore, the exp. times ket, summed over all sites ( j )
 (k )
Copyright – Michael D. Fayer, 2007
Replacing exp. times ket, summed over all sites ( j ) with  ( k )
H j , j 1  ( k )  e  ik   ( k )  e ik   ( k )


  e ik  e  ik  ( k )
factor out  ( k )
 2 cos( k )  ( k )
Adding this result to
H M  (k )  E  (k )
e
Gives the energy for the full Hamiltonian.
E ( k )  E  2 cos( k )
e
The nearest neighbor interaction with strength  breaks the degeneracy.
Copyright – Michael D. Fayer, 2007
Result (one dimension, nearest neighbor interaction only, )
E ( k )  E  2 cos( k )
k  wave vector – labels levels
  lattice spacing
e
E ( k )  E  2 cos( k )
e
Exciton Band
Each state delocalized over entire crystal.
E(k) - Ee (units of )
2
1.5
1
0.5
0
-0.5
-1
-1.5
-2 


2
0

2


Brillouin zone
k
Quasi-continuous Range of energies from 2 to -2
Copyright – Michael D. Fayer, 2007
Exciton Transport
Exciton wave packet  more or less localized like free particle wave packet
Dispersion Relation:
 (k ) 
1
E (k ) 
1
 E e  2 cos( k ) 
Group Velocity:
Vg 
d ( k )
2

sin( k )
dk
Exciton packet moves with well defined velocity. Coherent Transport.
Thermal fluctuations (phonon scattering)
 localization, incoherent transport, hopping
Copyright – Michael D. Fayer, 2007