Fractional charge in 1d

(see e.g. R.Rajaraman, cond-mat/0103366)

Fractional Charge in Field Theory (in 1 and 3 d) was introduced by Jackiw-Rebbi, PRD (1976)

Consider case in 1+1 d , i.e. one space dimension + time

2

2 2 2

2

Let ( , ) Fermi-Dirac field and (x,t) = scalar field, coupled

through the mass, and the Lagrangian

L=L with:

1 1 L [( ) ( ) 1

2 2B

B F

g t x

x t

L

] (classical scal

L ( ( , )

)

)

ar

F i m x t

2 2

22 2 2 3

2 2 2

1 1L [( ) ( ) 1 ] nonlinear field equation ( - )= ( , ) ( , ) .

2 2B x t x t

g t x t x

1

2

22 2

3 2 23

3 32

[ ]( ) 1 1 ( ) 2Indeed, ( ) [ ] ,

2 2 [ ] [ ]2 2

[ ]( ) 2( ) [ ] ( ) ( ) =

2 [ ]2

xTanh

x d x d xx Tanh

x xdx dxCosh Cosh

xTanh

x sh sh ch sh shx Tanh x x

xch ch chCosh

2

2 23

2 2The nonlinear field equation ( - )= ( , ) ( , ) solved by

( , ) 1 . This is called vacuum sector, with 2 vacua : semiclassically,

( , ) 1 ( , ) 1 L ( ) L ( )F F

x t x tt x

x t

x t i m i mx t

Additional static (time-independent) solutions: soliton sector

23

2The nonlinear field equation - = ( , ) ( , ) solved by

( ) [ ].2

xx T

x t x t

anh

x

2

4 2 2 4

1.0

0.5

0.5

1.0

f x_ Tanhx

2

3

2 23

2 2

kink( ) [ ] is the static solution of ( - )= ( , ) ( , )

antikink2

xx Tanh x t x t

t x

kink

antikink

localized

state

-4 -2 2 4

-1.0

-0.5

0.5

1.0

-4 -2 2 4

-1.0

-0.5

0.5

1.0

4

5

There is an analogy between a field theory with two vacua and the much simpler problem of a particle in a double well potential, with molecules having a double configuration like ammonia and with polymers.

There is an analogy with domain walls in ferromagnets

Vacuum sector:classical solution for 1semi

6

2 1

In 1+1 d spinless case two components are enough;

we may take , = . Then,

(-i ) ( ) ( ) for positive energy spinor

(-i ) ( ) ( ) for negative energy spinor

x k k k

x k k k

m u x E u x

m u x E u x

2 2

( ( , )) ( , ) ( ) Dirac's Lagrangian

( , ) with , two component spinors, E =

,

0 0usually and

0 0

F

k

k k k

k

D

k

k

k

L i m x t x t i m

ux t u u k m

u

H p m

I

I

, since they anticommute.

Dirac’s theory in 3+1 dimensions

field in 1+1 dimensional theory

Vacuum sector

Electron-Positron field in Dirac’s theory

( . ( ) ) † ( . ( ) )

1,2

†

2 2 4

( , ) [ ( ) ( ) ( )v ( ) ]

annihilates electron with spin s, d creates positron with spin s,

(k)= ( ) .

i k r k t i k r k t

s s s s

k s

s s

r t b k u k e d k k e

b

c k m c

Field in 1+1 dimensional theory: there are stillparticles and antiparticles

†( , ) [ ]

Dirac's vacuum: v 0 d v 0

(no 2-component 'electrons', no 2-component 'positrons')

k kiE t iE t

k k k k

k

k k

x t b u e d u e

b

7

Charge conjugation in Dirac’s theory

1

2

3

4

If solves ( , ) ( , ) 0ieA mc

x t x tc

2 2

*

4

*

*3

2*

2

*

1

0 0 0 1

0 0 1 0C= K,

0 1 0 0

1 0 0 0

then = solves ( , ) ( , ) 0C C CieA mc

x t x tc

0

0

k

k

k

i

i

3 3

3 3 3 3

3

( , )

sends particles to antiparticles; in addition, [ , ] 0.

( ) (-i ) ( ) ( ) ( )

charge conjugation operator.

k

k

k k D

D k x k k k D k

ux t

u

u u H

H u x m u x E u x H u x

† †

_

† †

† † † †

Standard charge density in Dirac's theory:

1 ( , )= [ ( , ), ( , )] , with ( , ) [ ]

2

1ˆIndeed the charge is ( , ) ([ , ] [ , ])2

1{ (1 ) (1

2

k kiE t iE t

k k k k

k

k k k k

k

k k k k k k k

k

x t x t x t x t b u e d u e

Q dx x t b b d d

b b b b d d d

† †

)}

{ }.

number of particles - number of antiparticles

k

k k k k

k

d

b b d d

9

Charge conjugation in 1+1 dimensional theory

Charge operator in Dirac’s theory

4 2 2 4

1.0

0.5

0.5

1.0

Positive energy

continuum

Positive energy

continuum

negative

energy

continuum

localized

state

negative

energy

continuum

Soliton sector

10

Solving Dirac’s equation one finds a single, localized

0 mode (one-particle solution with E=0).

kink

antikink

-4 -2 2 4

-1.0

-0.5

0.5

1.0

- 4 - 2 2 4

- 1.0

- 0.5

0.5

1.0

Solving Dirac’s equation one finds a single, localized

0 mode (one-particle solution with E=0).

the two ground states must have charge ½ and - ½ .

Solution in Soliton Sector

12

There are two many-body Vacuum states with the zero mode filled or

unfilled.

These ground states:

differ by charge e

and are connected by the charge conjugation operator C since H

anticommutes with C=3

, therefore the charge conjugate of a g.s. must be a g.s. with opposite

charge.

0 HC 0CH HC V CH V

Fractional charge! Is this science fiction? Not quite.

2 1 3, = , anticommutes with C=DH p m

Fractional Charge in Polyacetilene, Su,Schrieffer and Heeger prl 1979

Unstable vacuum

Stable vacuum B

Stable vacuum A

13

An infinite Polyacetylene chain has alternating single and double bonds and hastwo ground states or ‘vacua’

-4 -2 2 4

-1.0

-0.5

0.5

1.0

-4 -2 2 4

-1.0

-0.5

0.5

1.0

Compare stable vacuum A and vacuum+2 solitons: perturbation is local

however solitons can be delocalized:

14

Starting from a vacuum

One can introduce an excitation consisting of two consecutive single bonds. Here is a couple of such excitations:

The excitations can propagate far apart by exchanging double bonds and single bonds

however solitons can be delocalized:

How many bonds change overall?

7

6

It is a matter of 1 electron (per spin) per bond, that is, ½ electron per soliton.15

The perturbed region has 1 less bond, that is one less charge; ½ propagates to leftand ½ to right.

16

Graphene

17

18

Corriere della sera 15 febbraio 2012

Lower resistivity than silver-Ideal for spintronics (no nuclear moment, little spin-orbit that

converts to charge current) , and breaking strength = 200 times greater than steel.

http://www.graphene.manchester.ac.uk/explore/what-can-graphene-do/

20

= basis

The lattice is bipartite

b a

Unit cell

a

(1,0)a

1 3( , )2 2

a

21 3 3 3area 6* * *

2 2 2a a a

a=1.42 Angstrom

3

2

3

22

1a

2a

1 2

1 2 1 2 1 2 1 2

and make an angle too; the triangle is equilateral and indeed,3

3 3 3 30 *3 | | | || | | || | sin( )

2 2 2 2 3

3 30

2 2

a a

i j k

a a k a a a a a a

Primitive vectors

1 2

Cartesian components are :

3 3 3 3( , ), ( , ),2 2 2 2

a a a a

1 2and so 3a a a

1

13

sin( )3 2

1cos( )

3 2

3

a

1a

3

23

Reciprocal lattice vectors

1 2

1

2

By definition, reciprocal lattice exp( . 1) . 0, 2 , 4 , , . 0, 2 , 4 ,....

and so the BZ must be hexagonal. Start by seeking vectors orthogonal to and vectors

orthogonal to

nG i G a G a G a

a

a

2 2 2 2 1

4 1 3Picking ( , ) . 2 , . 0.

3 2 2G G a G a

a

1 1 1 1 2

4 1 3 4 1 3 3 3Picking ( , ), we get . ( , ).( , ) 2 , . 0.

3 2 2 3 2 2 2 2G G a a G a

a a

1a

2a

1

1 2

2

Cartesian components are :

3 3 3 3( , ), ( , ),2 2 2 2

with 3 .

a a a a

a a a

24

3 1 4 2

5 2 5 1 5

There are 6 smallest equivalent reciprocal lattice vectors. We can take , ,

4moreover . 2 , . 2 with (1,0),

3

G G G G

G a G a Ga

6 5and finally .G G

To obtain the BZ draw the smallest G vectors and the orthogonal straight

lines through the centres of all the G vectors: the interior of the hexagon is

the BZ.

1a

2a

1

1 2

2

Cartesian components are :

3 3 3 3( , ), ( , ),2 2 2 2

with 3 .

a a a a

a a a

2

4 1 3( , )

3 2 2G

a

1

4 1 3 ( , )

3 2 2G

a

3G4G

BZ and important points.

1G2G K’

5G6G G

K

K’K’

K

K

M

1 2 3 4

5 6 5

From the reciprocal lattice vectors we obtain the BZ (Voronoi diagra

4 1 3 4 1 3 4 1 3 4 1 3( , ), ( , ), ( , ), ( , )

3 2 2 3 2 2 3 2 2 3 2 2

4(1,0),

3

m)

G G G Ga a a a

G G Ga

5

1 2(1,0)

2 3M G

a

26

2π 2πfrom M= (1,0), ΓM=

3a 3a

3since ' is equilateral ,

Distances and coordinates

'2

2 1 4' .

3 3 27

2

4' .

2

in the BZ

7

KK M K K

K Ka a

K K Ka

G G

G

1 2 3 4

5 6 5 5

4 1 3 4 1 3 4 1 3 4 1 3( , ), ( , ), ( , ), ( , )

3 2 2 3 2

Reciprocal lattice v

2 3 2 2 3 2 2

4 1 2(1,0), (1,0)

3

ector

3

s

2

G G G Ga a a a

G G G M Ga a

4

27a

K

K’K’

G

M

K

K

BZ and important points.

'K

4

27a

2

3a

5

1 2(1,0)

2 3M G

a

4' .

27K K

a

27

' 2 2K point on the right : ( , ) ( , )

2 3 27

4coordinates of K' point on top, ' (0, ) (0, )

2

Distances and coordinates in the

7

BZ

K KM

a a

K Ka

G

G

1 2 3 4

5 6 5 5

4 1 3 4 1 3 4 1 3 4 1 3( , ), ( , ), ( , ), ( , )

3 2 2 3 2

Reciprocal lattice v

2 3 2 2 3 2 2

4 1 2(1,0), (1,0)

3

ector

3

s

2

G G G Ga a a a

G G G M Ga a

K

K’K’

G

K

K

BZ and important points.

'K

4

27a

2

3a

M

4

27a

4' .

27K K

a

28

' 2 2K point on the right : ( , ) ( , )

2 3 27

4coordinates of K' point on top, ' (0, ) (0, )

2

Distances and coordinates in the

7

BZ

K KM

a a

K Ka

G

G

5

2 2 4K points at ( , ) by ( ,0) and are equivalent

3 33 3

K' points are alsoequivalent

connected Ga aa

1 2 3 4

5 6 5 5

4 1 3 4 1 3 4 1 3 4 1 3( , ), ( , ), ( , ), ( , )

3 2 2 3 2

Reciprocal lattice v

2 3 2 2 3 2 2

4 1 2(1,0), (1,0)

3

ector

3

s

2

G G G Ga a a a

G G G M Ga a

K

K’K’

G

K

K

BZ and important points.

'K

4

27a

2

3a

M

4

27a

5

4(1,0)

3G

a

4' '

27K K K K

a

G G

5

1 2(1,0)

2 3M G

a

29

Tight-binding model for the bands: denoting by a and b

the two kinds of sites the main hoppings are:

3

† † † †

0 0 , , 0

1

spin, 3js js js js j s js js j s

js js

H E a a b b J a b b a s J eV

1

2

3

(1,0)

1 3( , )

2 2

1 3( , )

2 2

a

a

a

1

2

3

b a

1 2angle between and

2 3 1 sin( )= , cos( )=-

3 2 2

Jean Baptiste

Joseph Fourier

. † † .( ) Inserting and ,ik j ik j

j k j k

k k

a a e a a e

30

3 3( )† † † ( )

, 1 1

1 2 3

3( ).

1

one finds that

1 3 1 3where (1,0) ( , ) ( , ).

2 2 2 2

Setting ( ) , using ( )

(secon

n n

n

ik j ihj iki k h j

j j k h k h

j khj n kh j n

ik i k h j

n j

a b a b e a b e e

a a a

k e e k h

† †

0 0

,

d orthogonality theorem with sum over classes),

we get the kinetic energy J J ( )j j k kj k

a b a b k

31

3

1

Let us evaluate ( ) . Setting 1,nik

n

k e a

1 3 1 3 3 3.( , ) .( , ) ( ) ( )

.(1,0) 2 2 2 2 2 2 2 2

/2

( )

32 cos .

2

x xy y

x

x x

k kik ik i k i k

ikik

ik ik

y

k e e e e e e

e e k

† †

† † † †

. ,

we get

we need besides, , but sinc

e

.

ik j

j kjs js js js

js

js js js js ks ks ks ksjs ks

k

a a b b

a a b

a a e

b a a b b

† †

0 0

,

J J ( )j j k kj k

a b a b k

3

† † † †

0 0 , ,

1

Since js js js js j s js js j s

js js

H E a a b b J a b b a

32

*

† † † †

0 0 .

But the spin index is unimportant.

ks ks ks ks ks ks ks ksjsks

H E a a b b J k a b k b a

s

0 02 † †

*

0 0

0 0

0

( ) ( )Dropping the spin index, ( ( ) b ( ))

( )( )

and for each k ands we have a separate problem.

( ) If is a normalized eigenvector

( )

( )of

(k

E J k a kH d k a k k

b kJ k E

a k

b k

E J kH

J

*

0

0 0† †

*

0 0

, with eigenvalue ,)

( ) ( )then ( ( ) b ( ))

( )( )

k

k

k E

E J k a ka k k

b kJ k E

Why 2 component? It is the amplitude of being in sublattice a or b.

33

2 2 2

0 0

0 0

The secular equation yields:

( ( )) | ( ) | 0

bands : ( ) | ( ) |

E E k J k

E k E J k

3/2

1

2

*

3Since ( ) 2 cos ,

2

3 3 3( ) ( ) 1 4cos cos 4cos

2 2 2

n x xik ik ik

y

n

x y y

k e e e k

k k k k k

0 0

*

0 0

( )

( )k

E J kH

J k E

2

0 0

upper band3 3 3( ) 1 4cos cos 4cos .

lower band2 2 2

Band structure:

x y yE k E J ak ak ak

34

2 2 2

0 0

0 0

The secular equation yields:

( ( )) | ( ) | 0

bands : ( ) | ( ) |

E E k J k

E k E J k

3/2

1

2

*

3Since ( ) 2 cos ,

2

3 3 3( ) ( ) 1 4cos cos 4cos

2 2 2

n x xik ik ik

y

n

x y y

k e e e k

k k k k k

( )

| ( ) |

i ke

k

0 0

0 0

11| ( ) | ,

2

1| ( ) | ,

12

upper component: atom of type a

lower component: atom of type

Eigenvectors

b

:

i

i

uE E J k

v e

u eE E J k

v

0 0

*

0 0

( )

( )k

E J kH

J k E

35

In[1]:= b x_, y_ : 1 4 Cos 3 x 2 Cos y 3 2 4 Cos y 3 2 ^2

band x_, y_ : b x, y bz x, y ;

0 0 0 0

,

(0) 9 3 extrema

At

E E J E J

G

(upper band)

36

• Band Structure of graphene

In[41]:= ContourPlot band x, y , x, 3, 3 , y, 3, 3 , ColorFunction Hue &

Out[41]=

37

2

0 0

3 3 3( ) 1 4cos cos 4cos

2 2 2x y yE k E J ak ak ak

Note the Dirac cones at K

and K’ points

38

0

/2

0( ) | ( )

Recall the band structure:

3with ( ) 2 c s| o

2x xiak iak

yE k E J k k e e ak

4Evaluateat K' point on top, ' (0,1) :

27K

3 4 2( ) 1 2cos( ) 1 2 ( ) 0

2 327k Cos

Absence of a gap

2 2K point on the right :( , )

3 27a a

/2 3 ( ) 2 cos

2x xiak iak

yk e e ak

0

0

At K and ' points, 0, ( ) , :

is the chemical potential , all low e-h excitations occur there.

The Fermi points are 2, K and K', the others are equivalent.

no gapKK E k E

E

/2 3So, ( ) 2 cos vanishes at K and K' points

2x xiak iak

yk e e ak

40

no gap

bands

bands

41

/2 3( ) 2 cos

2x xiak iak

yk e e ak

' ' ' '

3 3 3 3| , | ( ) ( )

2 2 2 2K Kk k x K x y K y

x y

ia a ia ak k k k

k k

Expansion of band structure around K and K’ points

0 0(

Band structure:

) | ( ) |E k E J k

' '

4Callk the ' point on top, (0,1); there, ( ) 0

27K K KK k k

3linear dependence ( ) ( ).

2

3In a similar way setting at K=-K' one finds ( ) ( )

Now set :

.2

K x y

K x y

iaq q iq

iaq k k q q iq

q k k

42

0

' *

0

0

22

*

0

0 v ( )0 ( ),

v ( ) 0( ) 0

3 v 840

2

0Eigenvalue

Expand bandstructure near the K'point,

s of : E - =0.0

Thi

dropp

s

ing E

F x y

k

F x y

F

i q iqJ kH

i q iqJ k

J a Km

s

2 2 2 2 2 2 2gives v ( ) v .F x y F x yE 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 ( Fmassless ermion)F x yE v q q E p c m c

3At K' ( ) ( )

2x y

iaq q iq

Recall: the 2 components are for the 2 sublattices

43

0

*

0

2 2 2 2 2 2 2

Expand the bandstructure near the K point,

3 ( ) ( ); in tNow

0 v ( )0 ( ),

v ( ) 0( ) 0

again gives v

he same way,

( ) v

2

F x y

k

F x y

F x y F y

x

x

y

i q iqJ kH

i q iqJ k

E q q E

iaq i

q

q q

q

44

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 in the

band structure and to the amplitude on sites a and b. The analogy

requires a massless Dirac particle.

'

It requires a rotation q , q to make the analogy evident.2

0 0v v v (

0 0

Relation to 4-component Dirac's equation

x y y x

x y x y

k F F F

x y x y

p p

iq q p ipH

iq q p ip

*

*

'

. )

0 0v v v ( . )

0 0

So the Hamiltonian is set in the Dirac-like form

0 ( . ) 0H= v .

0 0 ( . )

(The matrices can take many form, as long

x y y x

k F F F

x y y x

K

F

K

p

iq q ip pH p

iq q ip p

H p

H p

as they anticommute and have no Tr)

45

Landau levels in Graphene

Using the bandstructure near the K poin

0 v ( ) 0 v (p )

v ( ) 0 v (p ) 0

wit

t,

h ,

F x y F x y

k

F x y F x y

i q iq i ipH

i q iq i ip

p p eA

0

0

0

0

ˆ ˆv [( ) ( )]ˆ .

ˆ ˆv [( ) ( )]

ˆ ˆv [( ) ]ˆ , independent of x.

Using the gaug

ˆ ˆv

e ( ,0 )

( )

,0

[ ]

F x y x y

k

F x y x y

F x y

k

F x y

A

E i p ip e A iAH

i p ip e A iA E

E i p ip eByH

i p ip eBy E

B y

46

1

2

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

( )xik x

k x x x

f yH p p k x y e

f y

0 1

0 2

v [ ] ( ).

v [ ] ( )

It is clear that the dynamics is along y.

F x yikx

k

F x y

E i k ip eBy f yH e

i k ip eBy E f y

15 2

0

0

2

fluxon 4*10 in MKSA units (Tesla *m ).

So, for a given a B, one can define an area S such th

magnetic leng

at . That is,

. One defines such that : .

Then, using the Fermi vel

th

oci

BB

h

e

BS

hS L

eB eBL

vty v one can also define a frequency .F

F

BL

v 2We denote is a characteristic energy,which depends on B.F

BL

Consider a Magnetic field B perpendicular to plane of Graphene

Magnetic length

48

0 1

0 2

2

Rewrite the off-diagonal elements of band Hamiltonian near K

v [ ] ( ),

v [ ] ( )

inserting v and eB= . 2

We find in the off-diagonal

F x yikx

k

F x y

BF

B

E i k ip eBy f yH e

i k ip eBy E f y

L

L

22 0

2

0

element:

1 v [ ] [ ] [ ],

2 2

where y .

B BF x y x B y y

BB

x B

y yL iLk ip eBy y k L i p p

LL

k L

Recall the textbook elementary harmonic

oscillator y: the annihilation operator is

00

0

0 0

1ˆ [ ], characteristic length.

2

Here the motion is along y, not x, x , y shifts origin of oscillator.

x

B

x pxa i x

x m

L

49

0 1 1

†

0 2 2

The Schroedinger equation becomes

ˆ

ˆ

ikx ikx

k

E i a f fH e Ee

i a E f f

1 1

0†

2 2

2 1

†

1 2

ˆ0.

ˆ 0

ˆ .

ˆ

f fi aE E E

f fi a

i af f

i a f f

000

0

1ˆ [ ], characteristic length.

2

yx py ya i x

x m

2 1

†

1 2

†

2 † †

2 1

2 † 2

2 2

†

2

0

ˆ .

ˆ

ˆ times the first gives:

ˆ ˆ ˆ

ˆ ˆFrom the second .

ˆ ˆ the eigenvalues are obtained:

.

i af f

i a f f

i a

a af i a f

a af f

Setting a a n

n E n

51

Some Concepts from Topology

A convex set is a set of pointscontaining all line segments betweeneach pair of itspoints.

Euler’s Characteristic of a surface

V=number of vertices

E=number of edges

F=number of faces

V E F

52

(From Wikipedia)

53

Euler’s Theorem for convex polyhedra

=2V E F

54

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

Question: What is g for a graphene lattice with pbc ?

A graphene lattice with pbc and without holes has g=1. Actually

it is drawn over a Torus!

55

One can insert two heptagons and two pentagonswithout changing g or leaving the plane.

Euler’s Theorem for general genus

=2(1-g)V E F

Let us see why!

56

Each graphene vertex has 3 links. Let us consider onlypentagonal 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 E

3 2

n n n

n n n

n n n n n nV

5 6 7 5 6 75 6 7

5 6 7 5 6 7 5 6

5 7

7

=2(1-g)

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

3 2

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

Therefore n =12(1-g).

V E F

n n n n n nn n n

n n n n n

n

n n n n

5 7For g=1, any deformation with n is acceptable

Pentagons are balanced by equal number of epta s.

.

gon

n

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

57

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

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

Starting with a surface of genus g, =2(1-g) suppose we

produce N 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 th

V E F

NV E F

5 7

eorem generalizes to =2(1-g))-N.

Also, n =12(1-g) 6 . This is relevant for nanotubes.

(recall that Graphene with pbc is wound around a Torus).

V E F

n N

Open faces and nanotubes

2

The original ' equation

( )

( )

A A

B B

Dirac s

Ai eV c p e

t cmc

Ac p e i eV

c t

3

† † †

,

'

0 0

1

*

v ( . ) holds about K', with 2 components because of a and b amplitudes

Recall the relation to 4-component Dirac's equation:

js js js js j s j

j

k F

s

H p

H E a a b b J a b

†

, 0

'

v ( . ) holds about K, with 2 components because of a and b amplitudes

The Hamiltonian is set in the Dirac-like form by joining the two:

0 ( . ) 0H=

spin, 3

v0 0 ( .

k F

K

s js

K

s

j

F

j

s

H p

H p

H

b a s J eV

*.

)p

,

,Structure of the spinor: .

',

',

K a

K b

K a

K b

59

1

2

3

60

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.

from Jiannis Pachos cond-mat0812116

Non-Abelian Vector potential

61

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

.C

A dr 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. In this way one can

gain info on the low energy excitations of the deformed Graphenes.

What kind of excitations are they seeking? Anyons!

One can show that a unitary transformation exists such

that the insertion of a pentagon or an eptagon introduces

independent magnetic fields at K and K’.

62

Anyons

See 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 charge q and a flux ; then one finds

.4

qS

63

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

( ) ( ). 2

2

vers z vers rE dl rE E

t r t

Torque ( )2

L qr qE vers z

t t

2z

qL

Simple-minded 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).

2

0

1 1constant constant .

2 2 2 4

t

z

qL d

q

r

64

Statistics

2 1

2

12 12

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

P 1 P 1.

i /2

12

2 i

12

2

12

Suppose that an exchange P of two identical particles brings a factor e

factor to the wave function of the system.

So, P brings e factor.A double permutation

P of identical particles r

2

12

estores the starting configuration.

P is equivalent to an adiabatic round trip.

/ 2

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

65

Usual electromagnetism in 3+1 d22

(4d) 4

0

1Action S ( )

2 2

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

E Bd x AJ A

Let us reformulate electromagnetism in 2+1 d

1 2 1 2 2 1A=(A ,A ) is a vector, while B= A A is a scalar.

2 2(3d) 3

0

1Action S ( )

2 2

is the most general Lorentz and gauge invariant, m=photon mass (in reality this vanishes

but we assume that excitations at surfaces are dre

ssed and pho

E Bd x AJ A m A A

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

tons become massive). The 'new'

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

is Lorentz invariant ( is the antisymmetric tensor).

m A A

Chern-Simons anyons(See Pachos, Topological Quantum Computation Chapter 7)

66

(3d) 3 2 2

0

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

1The Action S ( )

2

is gauge invariant. Indeed, the Chern-Simons term:

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

if A A , with arbitrary func

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+m[A (

tion, has extra contributio

- )+A (

ns

m[( )( A - A )+(

- )+A ( - )].

The second line va

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

nishes.

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

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

We are left with:

( )( 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 ex

tra term is a total derivative

and since A =0 on the boundary it can be ignored.

See Kamath pra 40, 6791 (1989)

2 2

0Lagrangian densi ty 2

E Bm AA A AJ

1 2

0 1 2

mB J JA A A

0 1 2

0 1 1 1 2 1 1

1 11 2

2

( ) ( ) ( )A A A A

E BJ mE

t x

0 1 2

0 0 1 0 2 0 0

1 2

1 2

( ) ( ) ( )A A A A

E EmB

x x

0 1 2

0 2 1 2 2 2 2

22 1

2

( ) ( ) ( )A A A A

E BJ mE

t x

68

11 2 1

22 1 2

So one finds the 3d "Maxwell" equations (B is a pseudoscalar):

BrotE+ 0, div

t

(rotB) t

(rotB)t

E mB

EmE J

EmE J

2 2 2

The photon mass term makes the interaction short-ranged: at large distances

the flux of E vanishes, the flux of B in the flux tube remains.

div div

by the divergence theorem:

E mB d x E m d xB d x Q 2 div flux of E =0 because of short range

, and charged particles must carry a flux.

d x E

Qm Q

m

69

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 Peierlsdistortion ia s Kekule distortion

70

1/3 of the hexagons is undistorted

71

iK. i(K' ).

0

Variation of hopping parameters in the Kekule' distortion

(r)J(r,i)= e e is associated to a vorticity,

3

(r)=slo

th

wly varying envelope function, =positio

e hopping 3 modified by

i K r

i

J eV

ns of sites shown below.

1

2

3

72

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 form

v ( . )( . ) 0H v

v ( . )0 ( . )

F

F

F

pp

pp

But it could produce a zero energy mode, similar to jackiw-

Rebbi.

. This should correspond to a ½ charge excitation, in analogy to the charge fractionalization mechanism in 1d.

73

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.