phononless ac conductivity in coulomb glass

Phononless AC conductivity in Coulomb glass

Monte-Carlo simulations

Jacek Matulewski, Sergei Baranovski, Peter Thomas

Departament of PhysicsPhillips-Universitat Marburg, Germany

Faculty of Physics, Astronomy and InformaticsNicolaus Copernicus University in Toruń, Poland

Ráckeve, 30 VIII 2004

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Outline

1. Experimental results of AC conductivity measurements

2. Shklovskii and Efros’s model of zero-phonon AC hopping conductivity in the disordered system

3. “Coulomb term”

4. Simulation procedure

5. Results

a) Coulomb gap in states distribution

b) Pairs distribution

c) Conductivity

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Experimental resultsM. Lee and M.L. Stutzmann, Phys. Rev. Lett. 87, 056402 (2001)E. Helgren, N.P. Armitage and G. Gru:ner, Phys. Rev. Lett. 89, 246601 (2002)

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Experimental resultsM. Lee and M.L. Stutzmann, Phys. Rev. Lett. 87, 056402 (2001)E. Helgren, N.P. Armitage and G. Gru:ner, Phys. Rev. Lett. 89, 246601 (2002)

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Outline





5. Results


b) Pair distribution

c) Conductivity

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Shklovskii and Efros’s model of zero-phonon AC hopping conductivity of disordered system

i ij ij

ji

iii r

nnnEH

21

ji ij

j

ji ij

ji r

n

r

nEE 0

System of randomly distributed sites with Coulomb interaction:

• When T 0K, the Fermi level is present

• If frequency of external AC electric field is small, only pairs near the Fermi level contribute to conductivity (one site below and one over)

• Pair approximation

Site energy: (sites are identical)

Electron-electron interactionsare taken into account!

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Shklovskii and Efros’s modelPair of sites

)ˆˆˆˆ)((ˆˆ21

ˆˆˆ122112

12

21221112 aaaarI

rnn

nEnEH

Hamiltonian of a pair of sites:

2,1 1

1 j j

j

r

nE

Site energy is determined by Coulomb interaction with surrounding pairs

Overlap of site’s wavefunction

)exp()( 1212012 ararIrI

21,2112 ,,ˆ21

nnWnnH nn Notice that because of overlap I(r) “intuitive” states can be not good eigenstates

mmm WH 12ˆ

,

Anyway four states are possible a priori:

• there is no electron, so no interaction and energy is equal to 0

• there is one electron at the pair (two states)

• there are two electrons at the pair

0,0

1,1

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0

0

2

1

2

1

m

m

WEI

IWE


Only pairs with one electron are interesting in context of conductivity:

mmm WH 12ˆ

1,00,1

1,00,1)ˆˆˆˆ)((ˆˆ21

ˆˆ

21

21122112

212211

mW

aaaarIrnn

nEnE

1,00,1 21 m

The isolated sites base1

2

2

2

1 Normalisation

2122

1211

m

m

WIE

WIE

2121

EEE 2212 4IEE where

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Energy which pair much absorb or emit to move the electron between split-states(from to ):


2212 4IEEWWW

Source of energy: photons

20

2)(

i

iQ

And finally the conductivity: Shklovskii and Efros formula for conductivity in Coulomb glasses

r

r1

)( 4

02

lnI

ar

Numerical calculation (esp. for T > 0)

Energy which must be absorbed by pairs in unit volume due to change states

Q = QM transition prob.(Fermi Golden Rule)

prob. of finding“proper” pair· · prob. of finding photon

with energy equals to · )(4

2 2

re

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Outline





5. Results



c) Conductivity

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Additional Coulomb energy in transition

2212 4IEEWWW

Correction to sites energy difference

12 EEE

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A (all acceptors)

Di

Dj

ir jr

ij rrr +

rrE

i

beforei

11)( j

beforej r

E1)(

Site energies Energy of the system:

i

afteri r

E1

)( rr

Ej

afterj

11 )(

+

ij

beforei

afterj rr

EEE11)()(

In order to make the calculation possible we need to express the energy difference using sites energy values before the transition

rEE

rrEEE before

ibefore

jij

beforei

afterj

111 )()()()(

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• Physical cause of correction: changes in the Coulomb net configuration• Pair approximation: only one pair changes the state at the time• Unfortunately to obtain this term we need to forget about the overlap for a moment

In order to make the calculation possible we need to express the energy difference using sites energy values before the transition

rEE

rrEEE before

jbefore

jij

beforej

afterj

111 )()()()(

2

2

12 41

Ir

EEWWW

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Outline


2. Shklovskii and Efros’s model of zero-phonon AC hopping conductivity of disordered system



a) T = 0K (Metropolis algorithm)

b) T > 0K (Monte-Carlo simulation)

5. Results



c) Conductivity

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Simulation procedure (T = 0K)Metropolis algorithm: the same as used to solve the milkman problemGeneral: Searching for the configuration which minimise some parameterIn our case: searching for electron arrangement which minimise total energy

N=10K=0.5

Occupied donor

Empty donor

Occupied acceptor

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Simulation procedure (T = 0K)Metropolis algorithm for searching the pseudo-ground state of system

Step 01. Place N randomly distributed donors in the box2. Add K·N randomly distributed acceptors

Step 1 (-sub)3. Calculate site energies of donors4. Move electron from the highest occupied site to the lowest empty one

5. Repeat points 3 and 4 until there will be no occupied empty sites below any occupied (Fermi level appears)

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Step 2 (Coulomb term)6. Searching the pairs checking for occupied site j and empty i

If there is such a pair then move electron from j to i and call -sub (step 1) and go back to 6.

Effect: the pseudo-ground state (the state with the lowest energy in the pair approximation)

• Energy can be further lowered by moving two and more electrons at the same step (few percent)

01 ij

ijij rEEE

Simulation procedure (T = 0K)Metropolis algorithm for searching the pseudo-ground state of system

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Simulation procedure (T > 0K)Monte-Carlo simulations

Step 3 (Coulomb term)7. Searching the pairs checking for occupied site j and empty i

If there is such a pair Then move electron from j to i for sure Else move the electron from j to i with prob.

Call -sub (step 1).

Repeat step 3 thousands times

Repeat steps 0-3 several thousand times (parallel)

01 ij

ijij rEEE

kT

Eij

eTp

)(

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Outline





5. Results



c) Conductivity

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Coulomb gap in density of states for T = 0K

Coulomb gap created due to Coulomb interaction in the system

0

0.2

0.4

0.6

0.8

1

-4 -2 0 2 4

01 ij

ijij rEEE

Si:P

μEi

Nor

mal

ized

sin

gle-

part

icle

DO

S

Coulomb term, but not only ...

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Smearing of the Coulomb gap for T > 0K

0

1000

2000

3000

4000

5000

6000

7000

8000

9000

10000

-4 -2 0 2 4

T = 0 (0K)T = 0.3 (725K)T=1 (2415K)

μEi

Nor

mal

ized

sin

gle-

part

icle

DO

S

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Pair distribution (T = 0K)

ij

occupiedi

emptyj r

EE1

ijr

N=400, T=0K, NMonte-Carlo=1000, a=0.27

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Pair distribution (T > 0K)

ijr

ij

occupiedi

emptyj r

EE1

N=400, T=1/8 (300K), NMonte-Carlo=1000, a=0.27

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Pair distribution (T > 0K)

ijr

ij

occupiedi

emptyj r

EE1

N=400, T=1 (2415K), NMonte-Carlo=1000, a=0.27

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Pairs mean spatial distance (T = 0K)

2.4

2.6

2.8

3

3.2

3.4

3.6

3.8

4

0 0.05 0.1 0.15 0.2

pair

mea

n sp

atia

l dis

tanc

e

Mott’s formula

simulations

Distribution of pairs’ distances is very wide in contradiction to Mott’s assumption

N=1000, K=0.5, 2500 realisationsperiodic boundary conditions, AOER

02ln

Iar

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Pair energy distribution (T = 0K)

0

50000

100000

150000

200000

250000

0 1 2 3 4 5 6 7 8 9 10

We work here!!!

N=500, T=0, K=0.5, aver. over 100 real.N

umbe

r of

pai

rs

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Conductivity (T=0K)

1e-010

1e-009

1e-008

1e-007

1e-006

1e-005

0.001 0.01 0.1

Con

duct

ivit

y (a

rb. u

n.)

Helgren et al. (T=2.8K)n = 69%

simulations

N=500, T=0, K=0.5, aver. over 25k real.Δ(hw)=0.001 (blue), Δ(hw)=0.01 (green)

n = 69% of nC means a = 0.27 [l69%]

(in units of n-1/3)

fixed parameters for Si:P: a = 20Å, and nC = 3.52·1024 m-3 (lC = 65.7Å)

There is no crossover in numerical results!

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Conductivity (T=0K)

Con

duct

ivit

y (a

rb. u

n.)N=500, T=0, K=0.5, aver. over 25k real.Δ(hw)=0.001 (blue), Δ(hw)=0.01 (green)

1e-008

1e-007

1e-006

1e-005

0.0001

0.001 0.01 0.1

simulations

Helgren 69% Si:P

clearly visible crossover

a = 0.36

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Number of pairs for T > 0K

Num

ber

of p

airs

N=500, T=0, K=0.5, aver. over 100 real.

T = 0T = 0.125T = 0.3T = 0.5T = 1

0 0.5 1 1.5 2 2.5 3 0

50000

100000

150000

200000

250000

0 0.5 1 1.5 2 2.5 3

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Conductivity for T > 0K

Con

duct

ivit

y (a

rb. u

n.)N=500, T=0, K=0.5, aver. over 1000 real.Δ(hw)=0.01, T>0

0.00015

0.00025

0.00035

0.00045

0 0.05 0.1 0.15 0.2 0.25 0.3

T = 0.0T = 0.1T = 0.5T = 1.0

0

5e-005

0.0001

0.0002

0.0003

0.0004

0.0005

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1e-008

1e-007

1e-006

1e-005

0.0001

0.001

0.001 0.01 0.1 1

Conductivity for T > 0KC

ondu

ctiv

ity

(arb

. un.

)N=500, T=0, K=0.5, aver. over 1000 real.Δ(hw)=0.01, T>0

= 0.001 = 0.01 = 0.1 (0.5·1013 Hz)

temperature (dimensionless units)

Si:P 69%

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Shape of Coulomb gap for T = 0K (corresponding to conductivity in low frequency)

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

0 0.1 0.2 0.3 0.4 0.5

Nor

mal

ized

sin

gle

part

icle

-DO

S

this hump probably isonly a model artefact

hard gap

numerical simulations result

fitting of (Efros)xE

ae/0

fitting of (Baranovskii et al.) 47

0

0

ln

/

xE

xE

ae

iEx

Efros: many particle-hole excitations in whichsurrounding electrons were allowed to relax

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The End

phononless ac conductivity in coulomb glass

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