1

2

0 03 3 3( ) 1 4cos cos 4cos2 2 2x y yE k E J ak ak ak

Note the cones at K and K’ points

Band structure of Graphene

2

0' *

0

0

2 2 2 2 2 2

0

2

0 v ( )0 ( ),

v ( ) 0( ) 0

Expand band structure near the K'point, dropping E

3v 8402

This gives v ( ) v .

F x yk

F x y

F

F x y F x y

i q iqJ kH

i q iqJ k

J a Kms

E q q E q q

Expansion of band structure around K and K’ points

2 2 2 2 2 4recalls the relativistic for m=0 (massless Fermion)F x yE v q q E p c m c

But the 2 components are for the 2 sublattices

3

0*

0

2 2 2 2 2 2 2

Expand the band structure near the K point,with K=K'3 ( ) ( ); in theNow

0 v ( )0 ( ),

v ( ) 0( ) 0

again gives v ( )

same way2

v

,

F x y

x

kF x y

F

y

x y F x y

i q iqJ kH

i q iqJ k

E

ia

q q

q

E q q

q iq

*'

So the Hamiltonian is set in the Dirac-like form0 ( . ) 0

H= v .0 0 ( . )

(The matrices can take many forms, as long as they anticommute and have no Tr)

KF

K

H pH p

In the Dirac theory the 4 components are due to spin and charge degrees of freedom; here they are due to the two Fermi points and to the amplitude on

sites a and b. The analogy requires a massless Dirac particle.

4

Landau levels

Consider a Magnetic field B

00 0

2

fluxon. So, area such that B makes

magnetic leng

.

th

2

We define such that : .

he eB

LeB

L

v 2With v Fermi velocity, is a characteristic energy,

as is dimensionally obvious, which depends on B.

FF L

v 2 1 1v ,2 2

F FF

veBL L L

eB

Take B perpendicular to plane of Graphene.

Magnetic length

5

0

0

0

0

0

0

ˆ ˆv ( )ˆthat is : with ,ˆ ˆv ( )

ˆ ˆv [( ) ( )].

ˆ ˆv [( ) ( )]

ˆ ˆv [(Using the gauge ( ,0,

) ]ˆˆ ˆv [(

0

) ]

)

F x yk

F x y

F x y x y

F x y x y

F x yk

F x y

E i p ipH p p eA

i p ip E

E i p ip e A iAi p ip e A iA E

E i p ip eByH

i p i

A B

p eBy E

y

, independent of x.

1

2

( )[ , ] 0 is conserved : ( , ) .

( )xik x

k x x

f yH p p x y e

f y

0 1

0 2

v [ ] ( ).

v [ ] ( )

It is clear that the dynamics is along y.

F x yikxk

F x y

E i k ip eBy f yH e

i k ip eBy E f y

6

0 1

0 2

2

22

Rewrite the off-diagonal elements of v [ ] ( )

.v [ ] ( )

Inserting v and eB= , we find v [ ]2

[ ]2

F x yikxk

F x y

F F x y

x y

E i k ip eBy f yH e

i k ip eBy E f y

L k ip eByL

Ly k L i pL

20

01 [ ], where y .2 y x

y y iL p k LL

Recall the textbook elementary harmonic oscillator y: the annihilation operator is

00

0

1 [ ], characteristic length.2

Here the motion is along y, not x.

xx pxa i xx m

7

0 1 1†

0 2 2

( ) ( )( ) ( )

ikx ikxk

E i a f y f yH e Ee

i a E f y f y

0 1 10†

0 2 2

2 1† †

1 2

1

2

2

( ) ( )( ) ( )

1 and recalling that

1

1( )this is solved by , with

( )

Indeed

.

, (

n

E i a f y f yE E E

i a E f y f y

a n n ni af fi a f f a n n n

y nf yi y nf y

i af i a i y

n

1

†2

) 1 ,

1 .n

n

n a y n n y n f

i a y n i n y n f

8

Some Concepts from Topology

A convex set is a set of points containing all line segments between each pair of its points.

Euler’s Characteristic of a surface

V=number of vertices E=number of edges F=number of faces

V E F

9

From Wikipedia

10

Euler’s Theorem for convex polyhedra =2V E F

11

See e.g. http://www.solitaryroad.com/c775.html

Genus g of a surface is the largest number of non-intersecting closed curves that can be drawn on it withput separating it.

sphere g=0

torus g=1

double-hole doghnut g=2

Euler’s Theorem for general genus

=2(1-g)V E F

A graphene lattice with pbc and without holes has g=1. One can also insert pentagons and eptagonswithout changing g.

12

One can insert two heptagons and two pentagons without leaving the plane.

13

Each graphene vertex has 3 links. Let us consider only pentagonal or heptagonal deformations.

5 6 7

5 6 7

5 6 7 5 6 7

number of pentagons number of hexagons number of eptagons F=

5 6 7 5 6 7 E3 2

n n nn n n

n n n n n nV

5 6 7 5 6 75 6 7

5 6 7 5 6 7 5 6 7

5 7

=2(1-g)5 6 7 5 6 7

Therefore n =12(

=

1-g).

2(1-g)3 2

2(5 6 7 ) 3(5 6 7 ) 6( ) =12(1-g)

V E Fn n n n n n n n n

n n n n n n n n nn

Pentagons are balanced by equal number of eptagons.

5 70 and n 12 OK for g=0 (sphere).n

14

The insertion of a pentagon forces us to connect two sites that are of the same type, e.g. two white sites in the figure. Recall the structure of spinor:

,,

.',',

K aK bK aK b

from Jiannis Pachos cond-mat0812116

Non-Abelian Vector potential

15

In the magnetic case one introduces a vector potential A to allow the wave function to collect a phase factor. Here we want the wave function to collect a jump to the opposite components and this requires a non abelian vector potential such that

.C

Adr is off diagonal.

The zero modes of H are the eigenstates with zero eigenvalue in the limit of infinite systems. The Atiyah-Singer index theorem says that 2 times the number of zero modes is equal to the flux of the effective magnetic field. This gives insight on the low-energy sector in terms of the number of pentagons and heptagons for systems of any size.

One can make a unitary transformation such that the insertion of a pentagon or an eptagon introduces independent magnetic fields at K and K’.

16

G=1,N=0 G=0,N=2

G=2,N=0 G=0,N=4

Starting with a surface of genus g, =2(1-g) suppose we produce N=2 open faces by cutting along a line: then g'=g-1 and V,E,F

remain the same =2(1-(g'+ )). In general,2

the Euler

V E F

NV E F

5 7

theorem generalizes to =2(1-g))-N. Also, n =12(1-g) 6 . This is relevant for nanotubes.V E F n N

Open faces

17

AnyonsSee also Sumatri Rao, arXiv:hep-th/9209066,Jiannis Pachos,Introduction to Topological Quantum Computation

_In 3d, [S ,S ] , ( 1) ( 1) , 1

( 1) ( 1) 0 for m=S.

i j ijk ki S S S m S S m m S m m S

S S m m

_also, [S ,S ] , ( 1) ( 1) , 1

( 1) ( 1) 0 for m=-S.Hence, S and -S differ by an integer and S must be either integer or half integer.

i j ijk ki S S S m S S m m S m m S

S S m m

In 2d there is only the z axis, say, so no commutation relations, but the condition that the wave function be eigenfunction of Lz leads to integer angular momentum. However we shall see that this is violated of the particle has a flux ; then one finds

.4qS

18

Indeed consider a spinless particle with charge q orbiting around a thin solenoid at distance r. If the current in the solenoid vanishes ( i=0 ) then Lz=

integer. Now turn on the current i. The particle feels an electric field such that

2 2. .rotE nd r B nd rt t

( ) ( ). 22

vers z vers rE dl rE Et r t

Torque ( )2

L qr qE vers zt t

2zqL

Simple Model for an anyon

However this is too rough. The charge is actually being switched on at the same time

that the flux in the solenoid is being switched on, with q(t)=constant X (t).

.4zqL

q

r

19

Statistics

2 1

212 12

In 3d, one can deform to and to a trivial path continuously,

P 1 P 1.

212

2 2 i12 12

A double permutation P of identical particles restores the starting configuration.

P is equivalent to an adiabatic round trip. So, P brings e factor.

/ 212 In 2d this is not so, P is nontrivial.ie

20

Usual electromagnetism in 3+1 d

(4d) 4 2 20

1Action S ( )2

is the most general Lorentz and gauge invariant and yields Maxwell's equations.

d x E B AJ A

electromagnetism in 2+1 d

1 2 1 2 2 1A=(A ,A ), B= A A is a scalar

(3d) 4 2 20

0 1 2 2 1 1 2 0 0 2 2 0 1 1 0

1Action S ( )2

is the most general Lorentz and gauge invariant.

Chern-Simons term =m[A ( A - A )+A ( A - A )+A ( A - A )]

is Lorentz invariant ( i

d x E B AJ A m A A

m A A

s the antisymmetric tensor).

Chern-Simons anyons

21

(3d) 4 2 20

0 1 2 2 1 1 2 0 0 2 2 0 1 1 0

1Action S ( )2

is gauge invariant.Chern-Simons term:

=m[A ( A - A )+A ( A - A )+A ( A - A )]

If A A , with arbitrary function, has extra con

d x E B AJ A m A A

m A A

0 1 2 2 1 1 2 0 0 2 2 0 1 1 0

0 1 2 2 1 1 2 0 0 2 2 0 1 1 0

tributions

m[( )( A - A )+( )( A - A )+( )( A - A )]+m[A ( - )+A ( - )+A ( - )].The second line vanishes.

0 1 2 2 1 1 2 0 0 2 2 0 1 1 0

0 1 2 2 1 1 2 0 0 2 2 0 1 1 0

0 1 2 2 1 1 2 0 0 2 2 0 1 1 0

( )( A - A )+( )( A - A )+( )( A - A )=( ( A - A )) ( ( A - A )) ( ( A - A ))

[ (( A - A )) (( A - A )) (( A - A ))]but the last line vanishes. So the extra term is

a total derivativeand can be ignored.

22

11 2 1

22 1 2

3d "Maxwell" equations: BrotE+ 0,divt

(rotB)t

(rotB)t

E mB

E mE J

E mE J

2 2 2

The photon mass term makes the interaction short-ranged

div div

, but 0

and charged particles must carry a flux.

n nC C

E mB d x E m d xB d x Q

dlE m Q dlE

Qm

23

About Excitations of GrapheneShort account of current Theoretical work

We saw that in 1d the Peierls distortion

leads to a double minimum potential, that is to the existence of two vacua and to the possibility of charge fractionalization.

In 2d the analogous to the Peierls distortion ia s Kekule distortion

24

1/3 of the hexagons is undistorted

25

kiK. i(K' ).

Variation of hopping parameters in the Kekule' distortion(r)J(r,k)= e e is associated to a vorticity,3

(r)=slowly varying envelope function.

K r

1

2

3

26

Such a scalar field should not violate the symmetry between positive and negative energy states, that

really arise from expansions around K and K’, since K and K’ are treated in the same way.

**

So the Hamiltonian is set in the Dirac-like formv ( . )( . ) 0

H vv ( . )0 ( . )

FF

F

pppp

Hence it could produce a zero energy mode..

. This should correspond to a ½ charge excitation, in analogy to the charge fractionalization

mechanism in 1d.

27

See the book by Jiannis K. Pachos, "Introduction to topological Quantum Computation", Cambridge (2012)

Recalling the argument linking spin to flux

, the mechanism leading to 1/2 charges4

1should also lead to S= , that is, to anyon excitations.4

qS

These anyons are potentially useful: many-anyon states have quantum numbers that are expected to be robust against decoherence.