Transcript Slide 1

We claim – in our system all states are
localized. Why?
e
x
Few General Concepts
The physical scene we would be interested in
Creating electronic continuity
P
Wave-functions of first confined states
( probability to find electron at z = z0)
( Energy level of the state
)
E
E1
0
Z
Spatial proximity leads to wave-function overlap.
(a)
E12
E11
(b)
E12
(c)
E1
E1
E11
The distance determines the strength of
the overlap or DE=E12-E11.
E12
E1*
E1
Two states are equally shared by the sites
(Two identical pendulum in resonance)
E11
E12
E1
E1*
E11
Strong coupling overcomes minute
differences (low disorder)
E12
E1
E1*
E11
E1
E12
E11
E1 *
Two states are separate
(Very different pendulum do not
resonate - stronger disorder)
Lifshitz Localization
r1
+
E0
E0
E0
r2
+
E0
r2
E0
+
E0
E0
r3
+
E0
E1
E2
+
r1
E0
E3
E4
+
E0
No long range “resonance”
If there is a large disorder in the spatial coordinates
 no band is formed and the states are localized.
Conjugation length
Long
Short
Varying chain distance
Strong coupling
Weak coupling
Coupling also affected by relative alignment of the chains (dipole)
parallel
shift
tilt
Localization in “Soft” matter
What are conjugated polymers?
Polymers: carbon based long repeating molecules
-conjugation: double bond conjugation
poly[acetylene]
H
C
H
C
C
H
H
C
C
H
MEH-PPV
H
C
C
H
H
C
C
H
Molecular organic
Semiconductor
C
H
Conjugation
Molecular  levels
Z
p+
The phase of the
wave function
pAmplitude
Consider 2 atoms
p+
p-
p-
p+
p+
p+
=
=
p-
p-
+- +-
Anti-bonding
*
Less Stable state
+ -- +
Bonding

Stable state
4 atoms
LUMO
(Conduction)
HOMO
(Valence)
There is correlation between spatial
coordinates and the electronic
configuration!!
4 atoms
LUMO
(Conduction)
Energy
HOMO
(Valence)
Configuration coordinate
4 atoms
LUMO
(Conduction)
HOMO
(Valence)
4 atoms
LUMO
(Conduction)
HOMO
(Valence)
4 atoms
LUMO
(Conduction)
HOMO
(Valence)
Molecule’s
4 atoms
Length
LUMO
(Conduction)
Another coordinate system
c
c
(a) c
c
c
c
c
c
c
c

(c)
Sigma
Dimerised (1)
(d)
Dimerised (2)
Energy
(b)
(b)
Degenerate ground state
(c)
(d)
Bond Length
General or schematic configuration coordinate
Aromatic link
Quinoidal link
The potential at the bottom of the well is ~parabolic (spring like)
Spring Energy
Q0
E0spring
E=E0+B(Q-Q0)2
Simplistic approach
1.2 10
4
1 10
4
Elastic energy:
Eelast  BQ2
Eelast  BQ2
8000
E
6000
4000
2000
0
-2000
-40
-20
0
20
40
Q
Squeezed
c
c
c
c
Stretched
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
Equilibrium
c
c
c
c
c
c
c
c
c
Adding a particle will raise the
system’s energy by (m*g*h)
Q0
E0t= E0spring+E0elec
Here, the particle just entered the system
(molecule) and we see the state before the
environment responded to its presence
(prior to relaxation)
Q
The system relaxed to a new equilibrium state.
In the process there was an increase in elastic
energy of the environment and the electron’s
energy went down. On the overall energy was
released (typically) as heat.
On a 2D surface
The particle dug
himself a hole
(self localization)
A*
A
Q
Q0
E0spring
A*
6000
5000
4000
3000
If the potential
energy of the mass
would not depend on
its vertical position
E
A
2000
1000
0
-1000
-30
-20
-10
0
Q
10
20
30
40
A*
We’ll be interested in the phenomena
arising from the relation between the length
of the spring and the particle’s potential
energy.
Q
We’ll claim that due to this phenomenon
there the system (electron) will be stabilized
A*
6000
A’
5000
4000
If the potential
energy of the mass
would not depend on
its vertical position
E
3000
2000
1000
0
-1000
-30
-20
-10
0
Q
10
20
30
40
Stretch mode
En
En +dEn
L + dL
L
En 

2 2
2
n
2me L2
 2 2 2 
dEn    2
n  dL
3
 2me L

 F  dL
For small variations in the “size” of the molecule the electron phonon
contribution to the energy of the electron is linear with the
displacement of the molecular coordinates.
For -conjugated the atomic displacement is ~0.1A and F=2-3eV/A.
The general formalism:
Ee-ph=-AQ
E  E0  Eelast
6000
5000
E  E0  Eelast  Ee ph
4000
E
3000
Linear electron-phonon interaction:
Ee ph   AQ
2000
1000
E  E0  Eelast  Ee  ph
Eb
0
E  E0  BQ 2  AQ
-1000
-30
-20
-10
0
10
Q
20
Qmin
E
0
Q
2
Eb  BQmin
 AQmin
30
40
E0  E0 _ elast  E0 _ e ;  E0 _ e
2
En 
2
A
A
A
A





Qmin 
 B
  A


2 B 
4B 
 2B 
 2B 
The system was stabilized by DE through electron-phonon interaction 
Polaron binding energy
What is  ?
Molecule with e-ph relaxation
3000
3000
2500
2500
2000
2000
1500
1500
E
E
Molecule without e-ph relaxation
1000
1000
500
500
0
0
-20
-10
0
10
Q2
20
-20
30
-10
Q1
0
10
20
30
Q1
Qmin
What is the energy change, at Qmin, due to reorganization?
“stretch” the molecule to the configuration associated with the e-ph relaxation and see
how much is gained by the e-ph relaxation.
  BQ  Eb  2 Eb
2
min
Why all this is relevant to charge transport?
Molecule containing a charge
E
E-E0_Elect
Molecule without a charge
-20
-10
0
10
Q
20
30
-20
-10
0
10
Q
2
1
20
30
Qmin
Transfer will occur when by moving the electron from one molecule to
other there would be no change in total energy.
If the two molecules are identical and have the same E0 
The electron carries En+AQ1 and replace it with En+AQ2 
Transfer is most likely to occur when Q1=Q2=Q
Total excess energy to reach this state:
W  B  Q  Qmin   BQ 2
2
Transfer will occur when Q1=Q2=Q
Total excess energy to reach this state:
WMin 
W  B  Q  Qmin   BQ 2
2
 
1
2
B  Q  Qmin   BQ 2   0  Q  Qmin

Q 
2
To move an electron or activate the transport we need energy of:
Wa
WMin
2
1
1
A
1
2
 BQmin 
 Eb
2
2 4B 2
Electron transfer is thermally activated process
Typical number is:
Wa  200meV
e
Wa

kT / q
A

Q

 min 2 B 
E
Q
E
EC
Polaron Binding Energy
So far we looked into:
A  A*
Let’s look at the entire transport reaction:
A + D*  A* + D
Two separate molecules
E
E
Q1
Q2
One reaction or system
E
Q*
A system that is made of two identical molecules
1 10
4
8000
E
6000
4000
2000
0
-40
-20
0
20
40
Q*
As the molecules are identical it will be symmetric (the state where
charge is on molecule A is equivalent to the state where charge is
on molecule D)
E


Wa
If the reactants and the products have the same parabolic
approximation:
4Wa=2Eb
A system that is made of two identical molecules
6000
Products
Reactants
5000
E
4000
3000
D
A
D
A
2000
1000
Wa
0
-1000
Q*
As the molecules are identical it will be symmetric (charge on A is
equivalent to charge on D)
R   phonone
Average
attempt frequency
qWA

kT
P
Activation of the
molecular conformation
Probability of
electron to move
(tunnel) between
two “similar”
molecules
Requires the “presence” of phonons.
Or the occupation of the relevant phonons should be significant
What is a Phonon?
Considering the regular lattice of atoms in a uniform solid material, you would expect
there to be energy associated with the vibrations of these atoms. But they are tied
together with bonds, so they can't vibrate independently. The vibrations take the form of
collective modes which propagate through the material. Such propagating lattice
vibrations can be considered to be sound waves, and their propagation speed is the
speed of sound in the material.
The vibrational energies of molecules, e.g., a diatomic molecule, are quantized and
treated as quantum harmonic oscillators. Quantum harmonic oscillators have equally
spaced energy levels with separation DE = hu. So the oscillators can accept or lose
energy only in discrete units of energy hu.
The evidence on the behavior of vibrational energy in periodic solids is that the
collective vibrational modes can accept energy only in discrete amounts, and these
quanta of energy have been labeled "phonons". Like the photons of electromagnetic
energy, they obey Bose-Einstein statistics.
Considering a “regular” solid which is a periodic array of mass points, there
are “simple” constraints imposed by the structure on the vibrational modes.
Such finite size (L) lattice creates a square-well potential with discrete
modes.
Associating a phonon energy
vs is the speed of sound
in the solid
Energy
Energy
For a complex molecule with many degrees of freedom we use the
configuration co-ordinate notation:
2
2
1
1
0
0
Q
Configuration
Co-ordinate
Q
Configuration
Co-ordinate
E phonon  h phonon
For the molecule to reach larger Q – higher energy phonons states should be
populated
Bosons:
1
f ( E ) Bose  Einstein 
e
E
kT
1

1
e
h
kT
The relevance to our average attempt frequency:
1
f (h )effective 
1
h effective
e
Tphonon
1
kT
T
e
1
What will happen if T<Tphonon/2
6000
Products
Reactants
5000
In the context of:
E
4000
3000
A
B
A
2000
1000
Wa
0
-1000
Q
B
1
A system that is made of two identical molecules
6000
5000
E
4000
3000
A
B
A
B
2000
1000
Wa
0
-1000
Q
At low temperature the probability to acquire enough energy to
bring the two molecules to the top of the barrier is VERY low.
In this case the electron may be exchanged at “non-ideal”
configuration of the atoms or in other words there would be
tunneling in the atoms configuration (atoms tunnel!).
[D. Emin, "Phonon-Assisted Jump Rate in Noncrystalline Solids," Physical Review
Letters, vol. 32, pp. 303-307, 1974].
Would the electron transfer rate still follow exp(-qWa/kT)
High T regime:
1
kT  h phonon
3
~200k in polymers
Activation energy decreases with Temperature
[N. Tessler, Y. Preezant, N. Rappaport, and Y. Roichman, "Charge
Transport in Disordered Organic Materials and Its Relevance to ThinFilm Devices: A Tutorial Review," Advanced Materials, vol. 21, pp.
2741-2761, Jul 2009.]
Are we interested in identical
molecules?
(same A, B & E0)
e
Consider variations in E0
x
E
Effect of disorder or applied electric field on the two molecule system:


DG1
DG0
qR
VR  B  q  qR 
2
VP  B  q  qP   DG0
2
qc
qP
DG 0  1
qc 

2 B  q p  qR
 q p  qR
 
2

  B  qR  qP 
2
Energy activation for going to the lower site:
DG
1
DG1  B  qc  qR  
2
 DG
B
 2 B
0
2

 1  q p  qR
 qR  

 
q

q
2

 p R
q p  qR
 0

2
D
G

2
B
q

q

2
B
q

q
q
 p R 2
 p R R 
2 

8 B  q p  qR  
2
 DG 0  B  q  q 2  
p
R 
2 

8 B  q p  qR  
2
2
2
1
  DG 0 
0
 DG     
 1
4
4 

2
 DG 0

In the present case for going down in energy DG1  1 
4


For polaron transfer 2|Eb|) :



2
2
1
0
DG  2 Eb 
DG 
8Eb
1
In the present case for going down in energy
 DG 0
E
DG1  b 1 
2  2 Eb




2
Energy activation for going to the lower site:
2
Eb  DG 
1
0
DG  2Eb  
DG 
1 

8Eb
2  2 Eb 
0
2
1
DG  DG 



2
2
8 Eb
Eb
Ri  j  ui  j
0
0 2
  e  e   e  e 2 
 Eb 
 j i  j i  P
 exp  
exp


2
kT
2kT
8kTEb 




This term is usually negligible
e
j
 ei 
4 Eb
1
E
Effect of disorder or applied electric field on the two molecule system:


DG1
DG0=Ei-Ej
qi
Ri  j  ui  j
qc
qj
  e  e   e  e 2 
j
i
j
i
 Eb 

 P
 exp  
exp




2
kT
2kT
8kTEb 




Let’s consider a system characterized by:
Gaussian Distribution of States
E
1018cm-3
1017cm-3
e
x
Detailed Equilibrium
f  Ei  1  f  E j  uij  f  E j  1  f  Ei  u ji
f  Ei ,   1 1  exp   Ei    / kT 
 e j  e i  
uij

 exp  

u ji
kT 





exp   E j  Ei  / kT  E j  Ei
uij  u E j  Ei 
t 
1
else




Another form:

uij  u0 exp  R ij
P
-




exp   E j  Ei  / kT  E j  Ei

1
else


V. Ambegaokar, B. I. Halperin, and J. S. Langer, "Hopping Conductivity in Disordered Systems," Phys. Rev. B,
vol. 4, pp. 2612-&, 1971.
A. Miller and E. Abrahams, "Impurity Conduction at Low Concentrations," Phys. Rev., vol. 120, pp. 745-755,
1960.
Under which circumstances can we use:
d
J h   h nh E  Dh
nh
dx
1  and D are statistical quantities
A. Statistics has to be well defined
B. Variation in the structure/properties are slow
compared to the length scale we are interested in
Gaussian Distribution of States
E
1. Density and spatial regime
2. Carrier sampling DOS
1018cm-3
1017cm-3