Nanoribbons, edges and quantum dots - University of...
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Nanoribbons, edges and
quantum dots
Dirk Wiedmann
13/06/13
Seminar: electronic and optical properties of graphene
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● Building quantum wells like
in semiconductors is
impossible in bulk graphene
because of KLEIN tunneling
Introduction
www.intenseco.com
● Solution: cut graphene into
ribbons / nanostructures
● Vacuum as infinite high
potential barrier
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● Summary of electronic properties of graphene
● Nanoribbons
– Calculation of electronic structure with Dirac
equation
– Calculation of electronic structure with Tight
Binding method
● Spin polarization at zigzag edges
● Conductance quantization
● Graphene quantum dots
Overview
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● TB Hamiltonian and Ansatz
where:
● leads to (if overlap integrals are neglected):
● with the geometrical factor
Summary: Graphene properties
P. Dietl, Diplomarbeit, KIT, 2009
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● Eigenvalues are
Summary: electronic structure
P. Dietl, Diplomarbeit, KIT 2009
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● Expansion of around K and K' points
● due to low energies the wave function is a LC of 4 terms
leading to a 4 dimensional space
● Dirac-like Hamiltonian
where
Summary: low energy limit
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● General solution for wave function consists of 4
terms
Summary: low energy limit
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● different shape of ribbons:
● armchair
● zigzag
● arbitrary
Graphene nanoribbons
M.I. Katnelson, Graphene Carbon in Two Dimensions, 2012
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● infinite dimension in y, but finite in x-direction
● Wavefunction must vanish at the atom sites next to
the edges
Armchair & ZigZag boundary conditions
P. Dietl, Diplomarbeit, KIT, 2009
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● If you have a potential well and a band structure, you
need to combine 2 states of the same energies with
different -vectors to create or
- functions, that fulfill the boundary conditions
● These 2 states can be in a single valley – like in the
case of GaAs with a direct band gap or in different
valleys like in Si or graphene valley mixing
Valley mixing
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● Valley mixing is different in
armchair and zigzag
● The structure is rotated by
30° and so the Brillouin
zone
● Armchair: to create a sin(x)
or cos(x), both valleys are
needed. They are coupled
● Zigzag: a cone lies
symmetric around .
2 states from same valley
no coupling between &
Valley mixing in GNRs
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● Calculation of dispersion relation by Dirac-
Hamiltonian in low energy limit
● … blackboard …
● if : no bandgap
Armchair ribbon
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● Dirac Hamiltonian changes because of the rotation of
the Brillouin zone, and .
● Graphical solution of
Zigzag ribbon
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● Solving the expression
● “choose” a value for , find discrete values for
● Plot the energy:
Zigzag ribbon
P. Dietl, Diplomarbeit, KIT, 2009
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● Treat ribbon as a 1D chain
● Unit cells are repeated
● Tight binding model for a chain:
● Using Bloch's Theorem
Zigzag – full electronic structure
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● Nearest neighbor coupling with
hopping parameter t
● Obtain energies by Eigenvalues of
Zigzag – full electronic structure
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Zigzag – full electronic structure
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Zigzag – full electronic structure
● For wider ribbons the dispersionless states
appear
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Armchair – full electronic structure
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Armchair – full electronic structure
A. Cresit et. al., Nano Res (2008)
● Low energy approximation just reasonable for wide
ribbons
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● TB model does not include spins / magnetism
adding a electron-electron interaction term to the
Hamiltonian
Hubbard-model:
● Mean field approximation:
Zigzag – edge states
O. Yazyev, Rep. Prog. Phys. 73, 056501 (2010)
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Zigzag – edge effects
O. Yazyev, Rep. Prog. Phys. 73, 056501 (2010)
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● Landauer formula
● Assume ballistic transport,
(holds for ribbons with smoothly varying )
● Conductance increases of additional as soon as
the energy reaches a new band
Conductance quantization
O. Roslyak et. al., Phys. Lett. A 374, 4061–4064 (2010)
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● Energy states can be calculated with TB method
(Hückel method)
Hamiltonian of the chain before - without the
neighbor unit cells
● Electrostatic effects
lead to the
Coulomb blockade
if tunnels into the dot
or vice versa is changed
by
Graphene quantum dots
B. Mandal et. al., J Nanopart Res 14, 1317 (2012)
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● There are armchair and zigzag ribbons showing a
different behavior due to different BCs and
corresponding valley mixing
● If the ribbons width is not too narrow, Dirac equation
gives a good band structure (for low energies)
● Zigzag edges show a spin polarization which could
be used for spin polarized transport
● The transmission probability for each channel is
Each band adds a once its energy is reached
● Quantum dots show coulomb blockade
single electron transistor possible
Conclusion
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● Petra Dietl, Numerical Studies of Electronic Transport
through Graphene Nanoribbons with Disorder, Diploma
Thesis, KIT, 2009
● Branislav K. Nikolić, Graphene Nanoribbons And Carbon
Nanotubes, University of Delaware, 2012
● Mikhail I. Katsnelson, Graphene Carbon in two Dimensions,
University Press Cambridge, 2012
● Oleg V. Yazyev, Emergence of magnetism in graphene
materials and nanostructures, Rep. Prog. Phys. 73,
056501 (2010)
● B. Mandal, Exploring the electronic structure of graphene
quantum dots, J Nanopart Res. 14, 1317 (2012)
References
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1 Band structure calculation of GNRs with Dirac equation
1.1 Armchair ribbon
We start with Dirac equation obtained by low energy approximation of the band structure of infiniteexpanded graphene.As qy is not restricted we can use plane waves as a solution. This part of the solution can be seperated
c(′)A,B = eiqyyϕ
(′)A,B For the solution in real space we therefore just need to replace qx = −i∂x. Applying
this to the Dirac equation one yieldsApplying the Dirac Equation on yields
(−i∂x + iqy)ϕA = ǫϕB
− (i∂x + iqy)ϕB = ǫϕA (1)
where ǫ = EhvF
Putting the second in the first equation (or vice versa) the following differential equation of secondorder is obtained
(
−∂2x + q2y
)
ϕ(′)A,B = ǫ2ϕ
(′)A,B (2)
We try to solve it with the Ansatz:
ϕB(x) = eiknx, ϕ′
B(x) = e−iknx (3)
And we find the energy to be ǫ = ±√
k2n + q2y as one would expect. But what are the allowed values
for kn ?We need to consider the boundary conditions, which are - according to the figure on slide 9 - thefollowing:
ΨK,K′(RA)|x=0 = 0 ΨK,K′(RB)|x=0 = 0 (4)
ΨK,K′(RA)|x=Wac+√3a = 0 ΨK,K′(RB)|x=Wac+
√3a = 0 (5)
where Ψ is the most general solution consisting of four terms, as mentioned before.By using the propertiers of the Wannier function that χ(RA−RA) = 1 and χ(RA−RB) = 0, we obtain
ϕA,B(x = 0) = ϕ′
A,B(x = 0) (6)
ϕA,B(x = Wac +√3a) = e−∆K(Wac+
√3a)ϕ
′
A,B(x = Wac +√3a) (7)
with ∆K the distance between K and K′
. By insert the Ansatz (Eq.(12)) we find the possible valuesof kn
kn =πn
Wac +√3a
+2π
3√3a
(8)
Now we can find an expression for the energy depending on qy and therefore the band structure of thearmchair ribbon:
E = ±
√
(
πn
Nac/2 + 1+
2π
3
)2
+(
ky√3a
)2√3
2t (9)
where Wac was replaced by Wac = (Nac − 1)√3a2 . As qy is 0 in the case of the armchair ribbon it can
be replaced by ky to make it clear that it starts from the middle of the Brillouin zone.
1
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1.2 Zigzag ribbon
As the Brillouin zone is rotated in this case we need to replace qx by −qy and qy by qx. The infinitedirection still remains in y-direction, where we again use plane waves.
−~ vf
0 qy + ∂x 0 0qy − ∂x 0 0 0
0 0 0 −qy + ∂x0 0 −qy − ∂x 0
cAcBc′Ac′B
= E
cAcBc′Ac′B
(10)
We get the equations
(−∂x + qy)ϕA = ǫϕB (11)
(∂x − qy)ϕB = ǫϕA (12)
we obtain the same differential as before in Eq. (2). As we have no valley mixing we need to make amore general Ansatz. For sublattice A and the K valley it is of the form
ϕA = αezx + βe−zx (13)
and z =√
q2y − ǫ2 can be real or imaginary.
For the B sublattice we find
ϕB(x) =α
ǫ(z − qy)e
zx − β
ǫ(z + qy)e
−zx (14)
With the boundary conditions
Ψ(RA)|x=0 = 0 Ψ(RB)|x=Wzz+2a = 0 (15)
we get
ϕA(x = 0) = ϕ′
A(x = 0) = 0 (16)
ϕB(x = Wzz + 2a) = ϕ′
B(x = Wzz + 2a) = 0 (17)
By inserting the Ansatz (24) & (25) into (27) & (28) we obtain a transcendental equation
qy − z
qy + z= e−2z(Wzz+2a) (18)
Graphical solution yield that there are nontrivial real solution just for qy > 1Wzz+2a . For these values a
almost flat band is obtained - the so called dispersionless states which corresponds to the edge states.If qy < 1
Wzz+2a a solution can be found for imaginary z = ikn. In this case equation (18) can betransformed into
qy =kn
tan(kn(Wzz + 2a)with ǫ = ±
√
q2y + k2n (19)
2