electronic structure of metal phthalocyanines on ag (100) · electronic structure of metal...

Electronic structure of metal phthalocyanineson Ag (100)

Cornelius Krull

Universitat Autònoma de Barcelona Institut Català de Nanotecnologia

Supervisors :Dr. Aitor MugarzaProf. Dr. Pietro Gambardella

Ph.D. Thesis

Facultat de física | 2012

12

±30

d

π d

TK = 29K 2eg

π

πd

π

EF

0.1− 1

20 10

E Φd

Ψ

Φ d

Ψ

�k �k′ �k′′ �k′

(�2

2mΔ+ V (r)

)Ψ = EΨ

Ψ

Ψ1 = eikz +A·e−ikz Ψ2 = B · eiκz + C · e−iκz

Ψ3 = D · eikz

k =

√2meE

�κ2 =

√2meΦ− E

�

me

T

T =|Ψ1|2|Ψ3|2

=A2

D2=

[(k2 + κ2

2kκ

)2

(κd)

]−1

Φ � E κd � 1

T ≈ 16k2κ2

(k2 + κ2)2·e−2κd

ITd

I ∝ T ∝ e−2κd

d ΦΦ =

5 2 Id

Mμ,ν

Ψμ Ψν

I =2πe

�

∑μ,ν

f(Eμ) [1− f(Eν + eV )] |Mμ,ν |2 δ(Eμ − Eν)

�r0R d

f(E) =1

1 + exp ((E EF )/kBT )

f(E) Mμ,ν

Eμ Eν

Ve ·V � Φ f(E)

f(E) = 1 E < EF EF

[1− f(Eν + eV )]

I =2π

�e2 V

∑μ,ν

|Mμ,ν |2δ(Eμ − EF ) δ(Eν − EF )

Mμ,ν

Mμ,ν = − �2

2m

ˆ (Ψ∗μ�∇Ψν −Ψν

�∇Ψ∗μ)d�S

Ψν = V−1/2

∑G

aG e

(−√κ2+|�k‖+�G|2z

)︸ ︷︷ ︸ · e(i[�k‖+�G]�x)︸ ︷︷ ︸

�G κ = �−1(2mΦ)

Φ �k‖Vsurface

Ψμ = V−1/2

κR eκR1

κ |�r − �r0|e−κ|�r−�r0|

Vtip R κΦ

Mμ,ν =�2

2m4πk−1V

−1/2kRekRΨν(�r0)

I = 32π3

�k4e2V Φ2R2e2kR

1

V

∑μν

|Ψν(�r0)|2 δ(Eμ − EF )δ(Eν − EF )

ρ (E) =1

V

∑μ

δ(Eμ − E)

ρ (E,�r0) =∑ν

|Ψν(�r0)|2 δ(Eν − E)

I ∝ V ρ (EF )ρ (EF , �r0)

�r0 EF

d E VT (d,E, eV )

I ∝EF+eVˆ

EF

ρ (EF − eV + ε)ρ (EF + ε)T (d, ε, eV )dε

T (d,E, eV )ρ T

ρ EF

EF + V

s d

Tρ

dI/dV

dI

dV∝ ρ (EF − eV ) · ρ (EF )

dI/dV

EF

I/VdI/dV d2I/dV 2 d2I/dV 2

dI/dV

EF dI/dV

dI/dV

dI/dV d2I/dV 2

Φt Φs

dI/dVI/V

(Vmod = vo (ωt) 1100

f = 1− 3

τ

I/V

Vout(t) =

ˆ t

t−Tc

(2πfref · s+ ϕ)Vin(s) · ds

τVmod τ

ττ

√2 · Vmod

dI

dV=

(· ·

)· 1

dI

dV=

[ ]

[ ][ ][ / ]/[ ] = [ / ] = [ ]

dI/dV

dI/dV

−0.6

dI/dV dI/dV

ττ

dI/dV

dI/dV

dI/dV dI/dV

p < 2 · 10−10

μm

μm

p < 2 · 10−10

+

p < 5 · 10−10

dI/dV

dI/dV

×

× × × × k

k

−ln(T )

∝ T 5

H = HBloch +HK

HK = −J �S · �s(r)HBloch

HK

�S �sr.

J−ln(T )

J

12

TK

T < TK

kBTK = De−1/2Jρo

D ρo

Ed

U

U

EF

HA =∑�kσ

Ekn�kσ︸ ︷︷ ︸+

1√N

∑�kσ

[V�kdc

†�kσ

cdσ + V ∗�kdc†dσc�kσ

]︸ ︷︷ ︸

+ Ed(nd↑ + nd↓)︸ ︷︷ ︸+ Und↑nd↓︸ ︷︷ ︸

Ek n�kσ = c†�kσc�kσc†�kσ c�kσ

Ed

nd↑ nd↓c†dσ cdσ

V�kσN

U > 0

U

Ed

EF Ed < EF

Ed + U > EF

Ed < Ed + U < EF

Ed

V�kσG0(E) = (E−HA)

−1

Und↑nd↓ ≈ Und↑ 〈nd↓〉 +Und↓ 〈nd↑〉

nd↑(↓)(E) =1

π

Γ(E − Ed − U

⟨nd↓(↑)

⟩)2+ Γ2

Γ = π∣∣V�kσ∣∣2 no(EF )

no(EF )Γ

〈nd↓〉 =ˆ EF

−∞nd↑(E)dE

〈nd↑〉 =ˆ EF

−∞nd↓(E)dE

Γ � U 〈nd↓〉 = 〈nd↑〉 = 12

Γ � U 〈nd↑〉 ≈ 1 〈nd↓〉 ≈ 0

πΓ = U

T > TK

Ed EF

Ed + U > EF

kBT � |Ed|Ed + U

�S �s

J �S�s

S =∑

k,σ,α=

Vk,d

εk − εαnαd,−σck.σdσ −H.C.

ε+ = εd + U

ε− = εd

n+d,−σ = nd,−σ

n−d,−σ = 1− nd,−σ

HA = H0 +H1

HSW = H0 +1

2[S,H1]

HSW H1

H0

J

J ≈ U

|Ed|(U − |Ed|) < 0

T = 0

12

εkεf

UVsf

(U = 0)εk εf Vsf

∣∣∣∣ εk − E Vsf

Vsf εf − E

∣∣∣∣ = 0

Ea Eb

|Vsf | � εk − εf = Δ

Eb = εf − |Vsf |2Δ

Ea = εk +|Vsf |2Δ

U = ∞

U = 0 εkεf

|Vsf | � εk − εf = Δ

ψ(S=0)1 = c†k↑c

†k↓ |0〉 ψ

(S=0)2

1√2

[c†k↑c

†f↓ − c†k↓c

†f↑]|0〉

Hm =

(2εk

√2Vsf√

2V ∗sf εk + εf

)

E = εk +1

2

[εk + εf ±

√(εk − εf )2 + 8 |Vsf |2

]

|Vsf | � εk−εf =Δ

E(S=0) = εk + εf − 2|Vsf |2Δ

E(S=0) = εk + εk +2|Vsf |2

Δ

Sz

c†k↑c†f↑ |0〉

1√2

[c†k↑c

†f↓ + c†k↓c

†f↑]|0〉 c†k↓c

†f↓ |0〉

E(S=0) = εk + εf

E(S=0) = εk + εf − 2|Vsf |2Δ

< E(S=1) = εk + εf

< E(S=0) = εk + εk +2|Vsf |2

Δ

2|Vsf |2Δ

kBTK =2|Vsf |2

Δ

U =∞

U =∞ εf

TK =1

2(ΓU)2exp

(πEd(Ed+U )

ΓU

)

Γ Ed U

T = 0kBTK

E = kBTK EF

ΓK

ΓK(T ) = 2√(πkBT )2 + 2(kBTK)2.

1/2TK

1/2

n n = 2S

J1,2 ΓTK1

TK2

TK1 < TK2

T � TK1< TK2

TK1 < T < TK2

1/2

GK

TK2

GK(T ) = G +G(0) ·[1 +

(T

TK

)ξ

· (21/α − 1)

]−α

ΓK GK

α ξ

ξ α

60

60

ΔE � TK

EF

EF

TK

EF

TK

EF

EF

TK

EF

dI

dV(ω) = a · (q + ε(ω))2

1 + ε(ω)2+ b+ c · ω ε =

ω − εkΓ

a, b, c ω q

6

5

4

3

2

1

0

Co

nd

uct

ance

[a.u

.]

-40 -20 0 20 40Energy

q=2

q=1

q=0.5

q=0

q=0.25

q=∞

q = 0(q = 1 q > 2 q = ∞

ΓΓ ≈ kBTK

dI/dV