Pumping 1. Example taken from P.W.Brouwer Phys. Rev.B 1998 Two parameter pumping in 1d wire back to...

Pumping

1

Example taken from P.W.Brouwer Phys. Rev.B 1998

Two parameter pumping in 1d wire

0x

back to phase 1

0x x L

Phase 2: very high tunneling barrier for x (0, x)

( )V x

length along wire

0x x L

Phase 3:raise barrier U for x ( x,L); carriers flow to right

( )V x

U

length along wire

0x x L

Phase 4:open barrier for x (0, x)

( )V x

U

length along wire

Phase 1: no potentiallength along

wire

Uniform conductor: no bias, no current

some charge shifted to left and right

Sequence:

0, 0 , 0 , 0,

___________________________________

U U U U

U

Circuit in parameter space

Berry phase associated to two-parameter pumping

2*

, ,,

Note the analogy with Landauer formula: I 2 .m n m nm n

eV t t

*

, ,1 2

, 1 2

Brouwer formula: X parameters, phase difference, =repetition frequency

sin I Im

2

i

m n m n

m n

t teX X

X X

There is a clear connection with the Berry phase (see e.g. Di Xiao,Ming-Che-Chang, Qian Niu cond-mat 12 Jul 2009).

Bouwer formulation for Two parameter pumping assuming linear response to parameters X1, X2

But this is not the only kind of pumping discovered so far.

A circuit is not enough: one needs singularities inside. The magnetic charge that produces the Berry magnetic field is made

of quantized Dirac monopoles arising from degeneracy. The pumping is quantized (charge per cycle= integer).

U

Circuit in parameter space

Mono-parametric quantum charge pumping ( Luis E.F. Foa Torres PRB 2005)

quantum charge pumping in an open ring with a dot embedded in one of its arms.

The cyclic driving of the dot levels by a single parameter leads to a pumped current when a

static magnetic flux is simultaneously applied to the ring.

The direction of the pumped current can be reversed by changing the applied magnetic field (imagine going to the other side of blackboard).

20 proportional to j

0

The response to the time-periodic gate

voltage is nonlinear.

time-periodic gate voltage

The pumping is not adiabatic.No pumping at

zero frequency.

The pumping is not quantized.

66

See also: Cini-Perfetto-Stefanucci,PHYSICAL REVIEW B 81, 165202 (2010)

77

Another view of same quantum effect described above

Bias U current in wires vortex magnetic moment of ring

Interaction with magnetic field proportional to U^3

This is Magnetic pumping

It must be possible to make all in reverse!

Interaction with B current vortex->magnetic moment of ring current in wires Bias

8

Model: laterally connected ring, same phase drop on all red bonds

( )t

0

( )2

†0

, 1

( ) ( )

( ) ( ) , ( ) ,

ring L R Tunnel

tN iN

ring mn m n mn hm n

H t H t H H H

hcH t h t c c h t t e

e

Different distribustions of the phase drop among the bonds are equivalent in the static case, but not here. This choice is simplest.

9

Half flux in and then out.

Charging of ring with no net pumping

We may avoid leaving the ring excited by letting it swallow integer fluxons

this is emf first clockwise then counterclockwise the ring remains

excitedthe ring remains charged

charge is sent to left wire similar charge is sent to right wire

time in units of hoppingt

10

Pumping by an hexagonal ring – insertion of 6 fluxons (Bchirality)

emf always same way ring returns to ground state

pumping is achieved

time in units of hoppingt

11

If the switching time grows the charge decreases. It is not adiabatic and not quantized!

Rebound due to finite leads

Pumping by an hexagonal ring – insertion of 6 fluxons (Bchirality)

effect of 6 fluxons in 100 time unitseffect of 6 fluxons in 200 time units

time in units of hoppingt

effect of 6 fluxons in 300 time units

12

What happened? We got 1-parameter pumping (only flux varies)

Charge not quantized- no adiabatic result

Linearity assumption fails and one may have nonadiabatic 1 parameter pumping

We got a strikingly simple and general case where linearity assumption that holds in the classical case fails due to quantum effects. In the present time-dependent problem the roles of cause and effect are interchanged.

Memory storage

Insertion of 3.5 flux quanta into a ring with 17 sides connected to a junction (left wire atoms have energy level 2 in units of the hopping integral th, right wire atoms have energy level 0). The figure shows the phase pulse and the geometry. Time is in units of the inverse of the hopping integral.

Right: expectation value of the ring Hamiltonian. The ring remains excited long after the pulse. It remembers.

Charge on the ring . The ring remains charged after the pulse. It remembers.

Fine! But memory devices must be erasable.

How can we erase the memory?

Same calculation as before performed in the 17-sided ring, but now with the A–B bond cut between times t = 30and t = 70.

The ring energy and occupation tend to return to thevalues they had at the beginning, and the memory of the flux isthereby erased.

16

Graphene

Unit cell

a

(1,0)a

1 3( , )2 2

a

21 3 3 36* * *

2 2 2area a a a

a=1.42 Angstrom

Lower resistivity than silver-Ideal for spintronics (no nuclear moment, little spin-orbit) and breaking strength = 200 times

greater than steel.

17

18Corriere della sera 15 febbraio 2012

19= basis

the lattice is bipartite

b a

20

1a

2a

1 2

1 2 1 2 1 2 1 2

angle too; 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

21

Reciprocal lattice vectors

1 2 1 2

3 3 3 3Here, ( , ), ( , ), with 3 .

2 2 2 2a a a a a a a

1 2By definition, reciprocal lattice . 0, 2 , 4 , , . 0, 2 , 4 ,....

and so the BZ must be hexagonal.

G G a G a

2 2 2 2 1

4 1 3( , ) . 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

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

BZ.

To obtain the BZ draw the smallest G vectors and the straight lines through the centres of all the G vectors: the interior of the hexagon is the

BZ.

1G2G

3G4G

5G6G

K’

K

K’K’

K

K

BZ and important points.

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 24

(1,0),3

m)

G G G Ga a a a

G G Ga

5

1 2(1,0)

2 3M G

a

23

2 2(1,0) ;

3 3

3' is equilateral '

22 1 4

'3 3 2

Distances and coordinates

724

' .2

in

7

the BZ

M Ma a

KK M K K

K Ka a

K K Ka

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 24 1 2

(1,0), (1,0)3

ector

3

s

2

G G G Ga a a a

G G G M Ga a

M

4

27a

K

K’K’

M

K

K

BZ and important points.

'K

4

27a

2

3a

24

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

2 3 274

coordinates of K' point on top, ' (0, ) (0, )2

Distances and coordinates in the

7

BZ

K KM

a a

K Ka

5

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

3 33 3K' points are also equivalent

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 24 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’

M

K

K

BZ and important points.

'K

4

27a

2

3a

M

4

27a

25

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 sjs 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 jj k j k

k k

a a e a a e

26

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 kikik

ik iky

k e e e e e e

e e k

1 2 3

1 3 1 3(1,0) ( , ) ( , )

2 2 2 2a a a

† † † †Besides, .js js js js ks ks ks ksjs ks

a a b b a a b b

3 3( )† † † ( )

, 1 1

3( ).

1

† †

,

one finds .

With ( ) , using ( ),

( )

n n

n

ik j ihj iki k h jj j k h k h

j khj n kh j n

ik i k h j

n j

j j k kj k

a b a b e a b e e

k e e k h

a b a b k

27

*† † † †

0 0 .

But 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 and s 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.

28

2 2 20 0

0 0

( ( )) | ( ) | 0

: ( ) | ( ) |

E E k J k

bands E k E J k

2

/2 *3 3 3 3( ) 2 cos ( ) ( ) 1 4cos cos 4cos

2 2 2 2x xik ik

y x y yk e e k k k k k k

( )

| ( ) |i k

ek

0 0

0 0

11| ( ) | ,

2

1| ( ) | ,

12

Eigenvectors:

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

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

29

In[1]:= bx_, y_ : 1 4 Cos3 x 2 Cosy 3 2 4 Cosy 3 2 ^2band x_, y_ : bx, y bzx, y ;

0 0 0 0

,

(0) 9 3 extrema

At

E E J E J

(upper band)

30

• Band Structure of graphene

31

2

0 0

3 3 3( ) 1 4cos cos 4cos

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

Note the cones at K and K’ points

32

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

Out[41]=

0

0

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

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

The points are 2, K and K', the others are equivale

o g

nt.

n apKK E k E

E

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

2x xiak iak

yk e e ak

0 0(

:

) | ( ) |E k

Ban

E J

d structure

k

/2 3( ) 2 cos

2x xiak iak

yk e e ak

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

27K

3 4( ) 1 2cos( )

2 272

1 2 ( ) 03

k

Cos

33

no gap

bands

bands

34

/2 3( ) 2 cos

2x xiak iak

yk e e ak

' ' ' '

3linear dependence ( ) ( ).

23

In a similar way setting at K=-K' one

3 3 3 3| , | ( ) ( )

2 2 2 2

Set :

finds ( ) ( ).2

K Kk k x K x

x y

K

y Kx

x y

yy

K

iaq q iq

iaq k k

ia a i

q q

a ak k k k

k

i

k

q k

q

k

Expansion of band structure around K and K’ points

0 0(

:

) | ( ) |E k

Ban

E J

d structure

k

' '

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

27K K KK k k

35

0' *

0

0

2 2 2 2 2 2

0

2

0 v ( )0 ( ),

v ( ) 0( ) 0

Expand bandstructure near the K'point, dropping E

3v 840

2

This gives v ( ) v .

F x y

kF x y

F

F x y F x y

i q iqJ kH

i q iqJ k

J a Km

s

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

3At K' ( ) ( )

2 x y

iaq q iq

But the 2 components are for the 2 sublattices But the 2 components are for the 2 sublattices

36

0*

0

2 2 2 2 2 2 2

Expand the bandstructure 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

*'

Relation to Dirac's equation

By a rotation q , q 2

0 0v v v ( . )

0 0

0 0v v v ( . )

0 0

So the Hamilt

x y y x

x y x y

k F F Fx y x y

x y y x

k F F Fx y y x

p p

iq q p ipH p

iq q p ip

iq q ip pH p

iq q ip p

*'

onian is set in the Dirac-like form

0 ( . ) 0H= v .

0 0 ( . )

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

KF

K

H p

H p