Electronic structure of metal phthalocyanines on Ag (100) · Electronic structure of metal...
Transcript of Electronic structure of metal phthalocyanines on Ag (100) · Electronic structure of metal...
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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
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12
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±30
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d
π d
TK = 29K 2eg
π
πd
π
EF
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0.1− 1
20 10
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E Φd
Ψ
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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
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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)
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�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)︸ ︷︷ ︸
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�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
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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 )
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dI/dV
EF
I/VdI/dV d2I/dV 2 d2I/dV 2
dI/dV
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EF dI/dV
dI/dV
dI/dV d2I/dV 2
Φt Φs
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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
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dI
dV=
(· ·
)· 1
dI
dV=
[ ]
[ ][ ][ / ]/[ ] = [ / ] = [ ]
dI/dV
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dI/dV
−0.6
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dI/dV dI/dV
ττ
dI/dV
dI/dV
dI/dV dI/dV
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p < 2 · 10−10
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μm
μm
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p < 2 · 10−10
+
p < 5 · 10−10
dI/dV
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dI/dV
×
× × × × k
k
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−ln(T )
∝ T 5
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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
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Ed
U
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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
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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
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〈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
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Ed EF
Ed + U > EF
kBT � |Ed|Ed + U
�S �s
J �S�s
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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
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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 = ∞
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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
]
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|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
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U =∞
U =∞ εf
TK =1
2(ΓU)2exp
(πEd(Ed+U )
ΓU
)
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Γ Ed U
T = 0kBTK
E = kBTK EF
ΓK
ΓK(T ) = 2√(πkBT )2 + 2(kBTK)2.
1/2TK
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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
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ξ α
60
60
ΔE � TK
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EF
EF
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TK
EF
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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
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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