Hall effect

1

H y

z

x

J

VH

The Hall effect is the production of a potential

difference across a conductor by a magnetic field,

when an electric field produces a current along it. It

was in 1879 the earliest discovered nonlinear

electromagnetic phenomenon.

2 From Wikipedia

Hall effect in Drude Theory

In the steadystate : (rateof loss=rateof gain from field)

v0 v B , v drift velocity, scattering time

scattering field

dp dp

dt dt

mma e E

Taking (0,0, ), one applies along x, but charge piles up and the total field is

( , ,0). othing happens along , so forget about third component.

transient: using v B ((v B) ,(v B) ) (

x y

x y

B B E

E E E N z

v , v )y xB B

3

v vv

vv B

vv +v .

B and are given applied fields; in steady state, , so and v can be found.

v

v

v 0

x xx y x

y

y x y x

y

y

x yy x

m me E B eE B

me E

me E B eE B

m

E E

H y

z

x

J

VH

Thus, we obtain electric field versus v,but since one measures the steady current

v (n electrons per unit area) we rewrite in terms of J setting v= :

v

v

s s

x x

y y

JJ e n

ne

mB

E e

E mB

e

1

1 1, Drude mobility.

01en

x x

y ys s

m BBJ J ee

J Jmen mB B

e

(B)We may rewrite , where:

(B) 0

1 1Resistivity , ;

x xx xy x

y yx yy y

xx yx xy

s s

E J

E J

B

en en

4

v

.... in steady state and v can be found.

v .

xx

y x

y x

meE

E

eE B

5

is due to dissipation as usual.

due to Lorentz force.

xxx

x

y

yx

x s

E

J

E Bis

J en

It is proportional to magnetic field since the phenomenon is

due to the Lorentz force and the conductor is linear; it is inversely proportional to ns

since the Lorentz force goes with the velocity, not with J, and J=neV.

One calls R = the Hall coefficient .y

H

x

E

J B

One measures the voltage drop V along y and the current along x. V=RHI

, , the Hall field.0

1or , where is Drude mobility ,since ,

0 1( ) . This will be useful below, wh

1 0

x xx xy x

x xx x y yx x

y yx yy

xx

s

xx xy

yx yy s

E JE J E J is

E

eF

m en

BH

en

en

dealing with the quantum Hall effect.

6

4

3

1 2

4

(red arrow

, (blue a

s

r

)

r ows)H y

x xV

V V V E W W wid

V

t

V E L

h

H y

z

x

J

VH

Vx

3

21

Quantum theory of the 2d electron gas in (x,y) plane in magnetic field H parallel to z axis: Landau levels and DOS

second Landau Ga

( ,0,0) (does not treat x,y on equal footing)

(0, ,0) (also does not treat x,y on equal footing)

1( , ,0) ( treats x,y on equal fo

u

oting

Landau G

2

aug

ge

)

eA H y

A H x

A H y x

7

H

y

z

x

Main alternative Choices of the gauge

2

2

We choose the ( ,0,0) in a 2d box of sides L,

and introduce pbc along x

1ˆ

Land

2

2ˆ[ , ] 0 plane wavealong , , *

au Gauge

x y

x x x x

x

A H y

eHH p y p

m c

p H x p k k nL

We start with the Theory for spinless electrons

2

2

2 22 2 222 2 2

0

1ˆThen oscillator Hamiltonian along y,2

w cyclotronhere f

x y

x xx c

c

eHH p y p

m c

c k c keH eH eHy p y m y m y y

c c eH mc eH

eH

mc

7 2

0 0 0

0

0

, where flux quantum 4.13

req

610 *2 2

Dimensions : [ ] [ ] momentum [ ]

uency

1 [ ] [ ]

.

./

x x x

xx

c k k khc hcy Gauss cm

eH e H H e

keHy k y L

c eHy c

0

4

0

Width of gaussian

(If 1Tesla 10 Gauss 257Angstrom)

c

cl

eHm eHm

mc

H l

0

1 1LL (Landau levels) Amplitudes ( , ) ( )

2x

x

ik x harmonic

n c nk n

x

E n x y e y yL

8

0 is thecenter of oscillator, proportional to k .xy

2 2

Starting from the definition ( ) ( )2k

kE E

m

2 2

2 2

2

2

2 2( ) ( )

( ) ( )2

| |2

2( )

22 (

1 dimension:

) ( )2

mE mEk k

LE dk E

d k

dk m

mEk

L L mdk E E

k h E

m

Free electron density of states at B=H=0 (continuous model)

3 322

2 3 2 2

3

23

2

2( )

4 2( ) (4 ) ( ) ( ) ( )

2 (2 )| |

2

3 dimensions :

2 ( )

mEk

L L m mEE k dk E dkk k E

k

m

mL E E

h

Next, we find the DOS without B and point out the modification due to B in 2d

2 2

( ) ( )2k

kE E

m

Free electron density of states 2d at B=0 (continuous model)

22

2

2 2

2 2

Here we are interested in the 2 case :

2( )

( ) (2 ) ( )2

| |

2 2( ) ( ) ( ), the case of interest.

2

d

mEk

LE kdk E

k

m

L mE L mdk k E E

h

without H

E

2( ) ( )

x yL L mE E

h

( 1)

Integrating ( ) over one LL, the number of states is

degeneracyc

c

n

Ln

E

dE E D B

with B

E

0

With B, the DOS is:

1( ) ( ) ( ) ( ( ) ),

2L c

n

E D B E E n

11

where is the degeneracy of each Landau level

the same for all (see below).

LD B

0

0

1We saw:

2

and is thecenter of oscillator along y.

2But for the n-th state of the s-th LL, k

2and cannot exceed L .

2The condition L L L dete

s c

x

x

x

y

x

y x y

x

E s

c ky

eH

nL

cy n

eH L

c chn n

eH L eH

rmines the maximum n,

hence the degeneracy of the LL, the same for all the LL and

depends on H and on S=L L .x y

y

0

yL

Y coordinate of

center of oscillator

12

0

the degeneracy

The same for all

,

where . .. LL

x y

L

x y

L

eHL Ln D

hc

eHL L eHSD

hc hc

el

0

Let us put all the N electrons in the LLL.

All electrons fill exactly the LLL if each gets a fluxon!el L elN D N

The number of states in each LL = the number of fluxons over tha sample.

Each fluxon brings a state for each LL.

t

0 t 0

11 2 4

Given , what is the H such that all the N electrons are in the LLL ?

=N Threshold field H where .

At threshold, fill exactly the LLL

threshold fiel

.

Typical: 10 10 G u

a s

del el

elel x y

elt

N

NS L L

S

Ncm H

S

s. But we forgot about spin.

electron

fluxon

electron

fluxon

electron

fluxon

electron

fluxon

electron

fluxon

electron

fluxon

electron

fluxon

electron

fluxon

electron

fluxon

electron

fluxon electron

fluxon

13

0N.

2

At

with sp

H=

in, we redef

ine the thre

the LLL (cor

shold field such that th

responding to n=0) is fu

e spin up (down) ele

ecreasing

ctrons fill LLL:

H below the degeneracy lowers and on

l

l .

elt

t

tHD

HS

H

0

per spin.

the second LL n=1 is full.2

The number n=

Thedegeneracy of Landau Levels is

partially filled

estarts filling

bet

LL with 1.

A

ween and .1

1The levels g

t

ro

w2

L

t

t t

n c

D

HH

H HLL is H

E n

n

H

linearly with H since .c

eH

mc

t

0

Recall: H such that all the N electthreshold rons fill exactly the LLL: field

N

el

elt

spinl s

S

e s

H

Including spin

14

Chemical

potential

LL

Ground state energy of the 2d EG 1

For All electrons in LLL with n=0 .2 2

t c el el B el

eH H E N HN HN

mc

0

1 3For 0 and n=1 occupied other LL empty ( ) 2 ( ),

2 2 2

( number of states per spin) (factor 2 for spin).

tc L

L

HH n E D H

D

For the LL with is partially occupied.1

t tH HH n

0

The filled LL contribution to E amounts to

2 ( ) electrons for each filled LL

1Thus, the filled LL contribution to E= 2 ( ) ( )

2

L

L c

n

D H

D H n

15

Ht H

Including the energy contribution from a : since

1 there are 2 ( ) electro

partially filled

ns each with ene

LL

rgy ( + ) ,2

el L cN

n

D H

1

0

1 1total energy 2 (n+ ) [N -2 ]( + ) ,

2 2L c el L c

n

E D D

1

0

1 1Let us evaluate 2 (n+ ) [N -2 ]( + )

2 2L c el L c

n

E D D

21

0

1 ( 1)Since (n+ ) ,

2 2 2 2

1and the partly filled LL gives [N -2 ]( + ),

2

n

el LD

2

2

Inserting 2 and ,2

( ) (2 1) ( ) .

elc B L

t

B el t

t t

N HH D

H

H HE H N H

H H

2 21 12 [N -2 ]( + ) =N ( + ) 2 ( ).

2 2 2 2L el L el L

c c

E EN D D

0 el el

00

N NRecall:

2 2

2

L

t t

Lelt

DH S H

D HNH

S

0

The same for all Lw Lhere ...x y

L

eHL L eHSD

hc hc

17

2

2

2

2

2

(2 1) ( )

(2 1) ( )right of interval:

The minima haveequal values: indeed,

1

(2 1) (left

1

(

of interval:

)

)

1

(

1

B el t

t t

B e

t

t

tt

l

tH HH

E H

H

H HN H

H H

H NE H

H

H

H

2

)1

( 1)( ) B el tH NE H

18

2

2( )

minima

(2 1) )

( )

(B el t

t t

B el t

E H

E

H HN H

H H

N HH

2 2

2

Position of Maxima :

2 1 0

2 1 1 1 1midway

2 12

t

H dx x x

H dx

x

22

22 2

2 2

2 1 2 1(2 1) ( ) (2 1) ( ) 2 1

4 ( 1)2

Values of max

2

ima

( ) B el tB el t B el t

t t

N HH HN NE HH H

H H

Note!

All electrons contribute , not only the open LL

19

0

0

1Magnetic moment per unit surface

g.s.energy S=surface area

EM

S H

E

2

2(2 1) (( ) )B el t

t t

E HH H

N HH H

de Haas–van Alphen effect

20

Magnetic susceptivity per unit surfaceM

S MH

2

2(2 1) (( ) )B el t

t t

E HH H

N HH H

21 Лев Васи́льевич Шу́бников

Shubnikov-de Haas effect.

22

Fig 1 shows the Fermi energy EF located in

between[1] two Landau levels. Electrons

become mobile as their energy levels cross

the Fermi energy EF. With the Fermi

energy EF in between two Landau levels,

scatter of electrons will occur only at the

edges of a sample where the levels are

bent. The corresponding electron states

are commonly referred to as edge

channels.

Shubnikov-de Haas effect.

23

The Shubnikov-de Haas (SdH) effect

This completes our short discussion of the classical Hall effect and LL in 2d

electron gas and the phenomena known before 1980.But Nature had a big

surprise, once the necessary technology was made available..

The magnetoresistance

oscillates because te

increase of H brings LL at

the Fermi surace and the

electrons therein increase

the conductivity

24

http://www2.lbl.gov/Science-Articles/Archive/colossal-magnetoresistance.html Paul Preuss (2001)

This suggests that, even though the material is in many ways two-dimensional,

with conduction occurring in the thin manganese-oxide layers, some

conducting particles are confined to a single dimension -- a kind of

organization which has been called the "stripe phase."

pervskite=ossido doppio

The colossal magnetoresistance

( )since v

k

k

25

The giant magnetoresistance

Fert and Grünberg studied electrical resistance of structures

incorporating ferromagnetic and non-ferromagnetic materials. In

particular, Fert worked on multilayer films, and Grünberg in 1986

discovered the antiferromagnetic exchange interaction in Fe/Cr

films.[13]

K. v. Klitzing

Used the depletion layer of a

GaAs MOSFET as a 2d

electron gas

27

Quantized Hall effect of the two-dimensional electron gas in

GaAs- AlxGa1-xAs heterojunctions at low temperatures to 50 mK. In the small-

current and low-temperature limit s harp steps connect the quantized Hall

resistance plateaus.

The diagonal resistivity ρxx decreases with decreasing T at the Shubnikov—de

Haas peaks, as well as at the dips, and is vanishingly small at magnetic

fields above 40 kG

Hall resistance ρxy is proportional to magnetic field in Drude theory, however…..

28

3

1 2

4

(red arrow

, (blue a

s

r

)

r ows)H y

x xV

V V V E W W wid

V

t

V E L

h

H y

z

x

J

VH

Vx

3

21

The diagonal resistivity ρxx decreases with decreasing T at the Shubnikov—de

Haas peaks, as well as at the dips, and is vanishingly small at magnetic

fields above 40 kG

Oscillations of longitudina resistivity =Shubnikov-deHaas, minima close to 0. Plateaux in Hall resistivity =h/(ne2) with integer n correspond to the minima of resistivity

29

(From Datta page 25)

discovery: 1980

Nobel prize: 1985

K. v. Klitzing

30

Parallel

resistance

Hall resistance

To understand what happens, geometry is crucial. A rectangular conductor

very thin in z direction uniform in x direction is confined in y direction with B

in z direction.

Assuming for the sake of argument that the Hamiltonian H is separable, the

transverse (yz) dimensions yield many solutions that are called subbands:

let us assume that only one z subband is occupied and the confinement

along y is described by U(y).

22( )[ ( )] ( , ) ( , )

2 2

yxpp eBy

U y x y E x ym m

( , ) ( )ikxx y e y

22( )[ ( )] ( , ) ( , )

2 2

yxpk eBy

U y x y E x ym m

Origin of the Oscillations of longitudinal resistivity

x

y z

1- a paradoxical situation

22( )[ ( )] ( , ) ( , )

2 2

yxpk eBy

U y x y E x ym m

2

2 2

Expanding,

1[ ( ) ( ) ( )] ( , ) ( , ),

2 2

leads to the y subbands. The harmonic potential is modified by U(y).

y xk k

p keBm y y U y x y E x y y

m m eB

For =integer number of filled LL, there are only filled or empty LL The Fermi level is in a gap, no scattering should occur. It should be a band insulator. the experiment shows that xx resistivity =0 For =0.5,1.5,2.5,.. Half filled LL it should be a metal.

But experiment shows maximum xx resistivity

33

H

Filling LLL

Filling n=1

Filling n=2

Filling 3

xy

2

h

e

22

h

e

23

h

e

24

h

e

It should be a band insulator. the experiment shows that xx resistivity =0

For

=0.5,1.5,2.5,..

Half filled LL it

should be a

metal.

But experiment shows maximum xx

resistivity

34

No backscattering takes place. This situation when EF is between two LL, otherwise the LL at Fermi level carries current within the sample with scattering and maximum resistance (not an explanation but a description)

Simple picture involving edge states- the current is outside

Origin of zero resistance

Stairs are constant with better than ppb accuracy !

2The Quantum of conductance

2

h

e

in QHE mm-sized electron mean free paths because current carrying states in opposite directions are localized on opposite sides no backscattering

35

36

One gets the correct step hight from Drude theory, assuming integer number of occupied LL

and using the fact that the longitudinal resistance vanishes

Recall Hall resistance i

at

n

the ste

Drude

ps.

theory for Drude mobility , ( resistance)

0 1( ) , surface density

1 0| |

0 1 ( ( ) in cgs units) . Next, evaluate B at steps.

1 0

xx xy

s

yx yy s

xy

s

ezero

m

BH n

e n

B BH

enc en c

0

t t 0

t 0

The condition for exactly filled LLL (for each spin direction): D where D ,

threshold field for having all electrons in LLL: H 2

H . Since all LL have the same d2

L L

elel

el

N N

NN S

N

S

0 0 2

egeneracy,with exacty

1 1 1 1LL ar

H=

e filled. Then = = = .2 2 2 2

t elx

s

t

y

s

H N hc h

en c S en

H

c ec e ec e

Step height

2Why this works at steps? Why with =integer?

2xy

h

e

37

H

Filling LLL

Filling n=1

Filling n=2

Filling 3

xy

2

h

e

22

h

e

23

h

e

24

h

e

Points where results agree with Drude

38

Good, but why the stairs?

We did several crude approximations. Why is the result so precise)?

Why the plateaux? Why so exact?

R. Laughlin

39

Preliminary 1: gauge transformations and fluxons

Gauge transformation:

1 ( , )A'=A+ (x,t) V'=V- '(x,t)= (x,t)exp[ ].

ie x t

c t c

Gauge transformation:

A'= (x,t) (x,t) x

Adr

0

, this is no longer a gauge

2 since '(x,t)= (x,t)exp[ ]= (x,t)exp[ ].

If Adx

ie i

c

0However if this is again a gauge since

'(x,t)= (x,t)exp[2 ].

n

in

Why the plateaux? Why so exact?

40

2

Hall resistivity quantized with , =sharp integerxy

h

e

2

Hall conductivity quantized with , =sharp integerxy

e

h

Preliminary 2: Consider a Metal ribbon with long side along x, magnetic field along z.

z B

x

22

Assume noninteracting electrons. The electronic obeys:

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

2 2

Setting: ( , ) ( )

one obtains

yx

ikx

pp eByU y x y E x y

m m

x y e y

22

2

2 2

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

2 2

1 [ ( ) ( ) ( )] ( , ) ( , ),

2 2

yx

y xk k

pk eByU y x y E x y

m m

p keBm y y U y x y E x y y

m m eB

Laughlin does not even insert confining potential U(y) which plays no role in his argument.

y

Now add an electric field along y.

z B

x

E

42

22

0

2 22 2

0

0

2 2

0

( )

2 2

try solution ( )

where is a harmonic oscillator wave function;expanding,

1( ) ,

2 2 2

where . Next find y such that

1(

2

yx

ikx

kn n

n

y xc x c

c

c

pp eByH

m m

e y y

p kH m y k eE y

m m

eB

mc

m y

eE y

2 2 2 2

0 0 0

1 1) ( ) .

2 2

In this way we obtain just a shifted oscillator + a constant.

x c c ck eE y m y y m y

43

2 2 2 2 2

0 0 0

0 00 2

2 2

0 0

2 2 2 2

0

expand the r.h.s.

1 1( ) .

2 2

1This is solved by: [ ].

1 just shifts oscillator : ( ) .

2 2

This is OK for the

1 1

2 2

ribb

c x c c c

x

c c c

c

c c

c

n

mm y k eE y m y m yy

k eE Eky c

y m

m m m B

E n y

y

m

on in crossed fields.

Consider closing it as a ring pierced by a flux, with opposite sides connected to charge reservoirs of infinite capacity , each at the same potential as the side to which it is connected. The field B along z now becomes radial, and remains perpendicular to the ribbon. Replace E by a time-dependent flux inside, which produces an e.m.f. A current I flows around, the Hall potential VH exists between reservoirs.

Connect the opposite sides of the ring to charge reservoirs of infinite capacity , each at the same potential as the side to which it is connected.

Quantum Hall effect

( )t Reservoir

Reservoir

B

2Recall where = magnetic moment, S= r area ,

n normal, i=current,and E=- . / if B is normal and makes flux .

iSnc

B J c

45

46

The time-dependent flux t inside produces a e.m.f that excites a current I around (along x)

, total energyE

J c E

but the radial magnetic field then produces a Hall electric field VH, along the ribbon, that will transfer charge q from one reservoir to the other, contributing qVH to the energy of the whole system.

0

Current due to flux piercing the ring: J

for a cycle, using a macroscopic ring, assuming

(total energy) , since charge is transfered and system is restored

( )

H

H

E Ec c

E qV

qVEJ c c

hc

e

Hall conductivity .

H

J eq

V h

If the flux is one fluxon, the ring is completely restored in previous state .

47

In order to have the ribbon in same state as before the fluxon is applied, q=ne with n integer. Hence,

2

Hall conductivity !!is exactly quantizedH

J en

V h

Laughlin argues that the same holds true even in the presence of interactions.

Laughlin’s argument has been criticized on the grounds that different cycles of the pump may transport different amounts of charge, since q is not a conserved quantity. It is the mean transferred charge that must be quantized.

However some authors of the above criticisms have produced a remarkable

alternative explanation.

48

According to the Authors, Laughlin’s argument is short in one important step, namely, the inclusion of topological quantum numbers.

( , )

flux in the ring,

flux in the ammeter A that measures current along y

The Parameter space: ( , ).

Current flowing through A: J , c=constant

H H

c H

A

49

to the reservoirs is proportional to V

is proportional to which is proportional to J .

V J up to a constant: I show it is quantized.

Hall

x x

Hall Hall x x Hall

J

Et

Jt

Let function of whole system

if everything is done adiabatically.

wave

H i it t

50

H i it t

The reading of the ammeter does not change energy

0

Inserting J , setting c=1,

0 2 Re

2 Re 2 Im .

H H H H

c H

J H

J H i H

51

inserting and restoring c

2 2Im .

H it

J ic

it

it c t

2

So, , where 2 Im

curvature of the bundle of ground st

!

ates.

Recall:

Hall conductance

Hall

Hall

J K Kt

Jt

c K Berry

1integer where 2 Im

2

S= parameter space, is another Chern number! What is it?

SKdA K

52

Gauss and Charles Bonnet formula

12(1 )

2

curvature, genus

SKdA g

K g

Chern formula

1integer (Chern number)

2

2 Im

SKdA

K

The curvature of a sphere is positive,

The curvature of a saddle is negative, in a torus

it depends on the point. Chern produced a

quantum generalization of the Gauss-Bonnet

formula.

53

Arxiv: 1210.3200

http://www.riken.go.jp/lab-www/theory/colloquium/furusaki.pdf

Bi2Se3

is a 3d topological insulator

Topological Insulators

• (band) insulator with a nonzero gap to excited states

• topological number stable against any (weak) perturbation

• gapless edge mode which conducts current. At each energy there is a pair of

surface states connected by time reversal symmetry that have opposite spin and

momenta, scattering is strongly suppressed.

• Low-energy effective theory of

topological insulators

= topological field theory (Chern-Simons)

The QHE is the prototype topological insulator

http://wwwphy.princeton.edu/~yazdaniweb/Research_TopoInsul.php

55