The Quantum Hall Effects:Discovery, basic theory and open problems

K. Das GuptaIIT Bombay

Nanoscale Transport 2016, HRI (Feb 24 & 25, 2016)

Topics

The classical Hall voltageCurrent flow pattern in a Hall bar (How to solve)Discovery of the Quantum HallThe role of mobilityThe 2DEG in a MOSFETSetting up the Quantum Mechanical Hamiltonian (effective masses etc)Oscillation of the Fermi Level, Landau levelsGroup velocity of the eigenstatesChannels from a contact to anotherEdge states (why is this quantisation perfect)The logitudinal resistance oscillation

The Fractional Hall State (Discovery)Rotationally invariant solutionsMany body states in the first Landau LevelThe fake plasma analogy and the Laughlin statesCorrelations predicted by the Laughlin wavefunctionsWhy does one need to go beyond the Laughlin states

The Classical Hall effect is not entirely trivial

x

y

V=0

V=V0

W

L

∇ . j = 0j = σ EE = −∇V∇

2 V = 0

Current flowing through a rectangular block should satisfy:

What are the boundary conditions ?

V ( x ,0) = 0V ( x , L) = V 0

j x (W /2, y) = 0j x(−W /2, y) = 0

solution :

V ( x , y)= V 0

yL

The solution is simple becuase the relation between j and E is simple.What happens when we introduce B?The boundary conditions DO NOT change in any way.

( j xj y) = σ0

1+μ2 B2 (1 −μBμB 1 )(E x

E y)

equivalently

(E x

E y) = (ρ0 B/nq

−B/nq ρ0 )( j xj y)Notice ρxx = 0 ⇒ σxx = 0

The angle between E and j

tan δ =B/ nqρ0

= μ B

How does the current flow pattern look?Remember σ is now a 2×2 matrix

The Classical Hall effect is not entirely trivial (Boltzmann transport)

Rendell & Girvin, PRB 23, 6610(1981)

Can be solved exactly with a conformal map connecting the two shapes.

E y (x , y )+i E x( x , y) = exp [ f (Z )]

f (Z ) = ∑n odd

4δn π

sinh (nπ Z /L)cosh (nπW /L)

x

y

Z=x+iy

How do these solutions look?

typical value ~ 10-2 -10-3

But high mobilitywill lead to large angle

The Classical Hall effect is not entirely trivial : [left] d=0 and [right]d=0.95p/2

Current density vector j

Vectors denote the current flow. For hall angle close to 90, current will flow along equipotentials....each long edge becomes an equipotential!

The Classical Hall effect is not entirely trivial [top] d=0 and [bottom]d=0.95p/2

d=0

d=0.95p/2

The Classical Hall effect : Do we actually see d=0.95p/2 type states ?

Mobility and not conductivityfixes the Hall angle

For metals (''pure'' Cu, Ag at low temp) : n ~ 1029 m-3 r=10-9 Wm : so m ~ 0.1 m2/Vs

Si MOSFET : n~1015 m-2 m ~ 1-10 m2/Vs

It is only in semiconductors mB >> 1 is possible

In metals Hall voltage is useful for measuring carrier density etc...

''Anomalous Hall“ in ferromagnets

But large hall angle is not possible

Pfieffer & West, Physica E, 20, p57-64 (2003)

Thus two voltage probes placed at two points on the (long) side may measure no voltage drop.

The current density at the corner is very high.

All the current emanates from the edge of the ohmic contacts (the points of mathematicalsingularity of the series solution) and carries with it the potential of the contact.

We did not require much "quantum" physics to establish the important fact that almost dissi-pationless channels can arise in a strong magnetic field, irrespective of the amount of disorderinitially present. The zero field resistance dropped out of our consideration. The two-probe resistance measured between the ohmic-contacts and the Hall voltage measured between the opposite sides of the Hall bar will be the same. The Hall voltage measurement does not require that the Hall voltage probes be exactly opposite to each other.

Gives an idea why ohmic contact often fails in high magnetic fields.

The Classical Hall effect : What did we find for the very high field case ?

What did the experiments show?

The Hall voltage becomes flat at specific values:Samples from different labs/factories differing in microscopic details give the flat plateau at the same resistance value.

PRL 45, 494 (1980)

These data were from Si MOSFETS, today GaAs-AlGaAs heterostructure is more commonly

The exactness of the quantisations

Material independent constants are quite rare in Condensed matter

The resistance of the first quantum Hall plateau : 25.812807....kohms(Any 2D electron gas satisfying some basic criteria)

The Josephson frequency 483.5979 Mhz/microvolt(any superconducting weak link, irrespective of the material used)

Each of them correct to at least 8 significant figures...

Electrons in a band and the effective Schrödinger equation

A slowly varying potential is applied in addition to the lattice potential...

H = [ p2

2m0

+V l]+V slow

Ψ = ∫FBZ

d 3 k

(2π)3c (k ) u (k )e ik .r

H Ψ = ∫FBZ

d 3 k

(2π)3c (k ) H 0 [u (k )e

ik .r ]+V slowΨ=E Ψ

∫FBZ

d 3 k

(2π)3c(k ) E (k )[u(k )ei k . r ]+V slowΨ=EΨ

Approximation can be done after the KE operator has acted, not before..c (k )mostly peaked around k=0 and u(k )does not change wildy with k

u0 ∫FBZ

d 3 k

(2π)3E (k )c (k )e ik . r

+ V slowu0 ∫FBZ

d 3 k

(2π)3c (k ) e i k . r

=Eu0 ∫FBZ

d 3 k

(2π)3c (k )e ik .r

The only r dependence is in ei k . r : so E (k )e ik .r=E (−i∇)e ik .r

BlochHo

Free electron massNOT effective massVl is the fast varyinglattice potential

Call this F(r)

E (k ) is the band dispersion

simplest parabolic band : E (k )=ℏ2 k 2

2meff

: So

[ p2

2m eff

+V slow]F (r )=EF (r)

If there is a magnetic field, then

[ 12m eff

( p−q A)2+qV slow]ψ = Eψ

A remarkable simplification...

All effects of the complicated latttice potential somehow gets parametrized into a single effective mass.

The equation retains its structure!

Similar treatment can be done for graphene, which satisfies a different equation.

This is the Schrodinger equation we usually deal with.

Electrons in a band and the effective Schrödinger equation

Sheet of electrons in a magnetic field : Landau levels

TakeA = B(0, x ,0)

Then H =12m

[ px2+( py+eBx )2 ]

ψ(x , y) = eiky f k ( x)

[ px2

2m+

12m(ℏ k+eBx )2] f k (x ) = Ef k (x )

harmonic oscillator ω =

eBm

x0 =ℏ keB

[ px2

2m+

mω2

2(x+ x0)

2] f k (x) = Ef k (x)

En=(ν+12 )ℏω

A = B(0,x,0)

Different choice of A will help if rotational symmetry is useful.

A=B(-y/2, x/2,0)x0 =

ℏ keB

= kl2

l 2 = ℏeB

l ≈257 A

√B

Classical arguments gives sameCyclotron freq (independent of h)

Cannot give cyclotron radius, which depends on h

Degeneracy and other basic properties of the Landau levels

The eigenvalues are independent of k, leading to massive degeneracy.

The center of the ''oscillator'' must lie within the LxW sample

N =L2π∫0

W /l 2

dk

=LW2π

eBℏ

= Area×eBh

The states are actually strongly overlapping!

spatial width of nth harmonic oscillator state ≈ √n l

spatial separation between two states δ x0 =2πL

l 2

= 2π l( lL )

L=1 mm sampleB=1 Tesla

Calculate the separation & width

Periodic boundary Condition, think of the sheet as rolled upin a certain direction

L≫l Many body effects??

≃1014m−2

a very large degeneracy

Where is the Fermi level?

Consider N spinless particles per unit area (B=0)

D(E ) =m

2πℏ2

N =m

2πℏ2 E f (0)

Index of the highest occupied level when B≠0?

νmax = [ NeB/h ]

E f (B) = (νmax+12 )ℏω

E f (0)

E f (B)= (12+[1x ])x x=

ℏ eB/mE f (0)

Pudalov et al, JETP 62, 1079 (1985)

A voltmeter will measure difference in electrochemical potential

~20kB / Tesla in GaAs

Oscillation of the Fermi level with changing magnetic field

What happened to cyclotron motion? Where are these states physically located?

A state like e ik .r is everywhere

But not1

√Le iky H n (x+x0)e

−(x+ x0)

2

2 l2

x0=ℏkeB

: so value of k decides that

In presence of a potential U slowly varying on a scale of lE (x0) = (ν+1/2)ℏω+U (x0)

E (k ) = (ν+1/ 2)ℏω+U (kl 2)

Classical picture : charged particles have ''orbits'' in a magnetic field.But now.....calculate the current

⟨ J y ⟩ = −em⟨ψk | p+e A |ψk ⟩

= −eωc

√πm∫ dxe

−( x+kl 2

)2

l2

( x+kl2)

Will give zero, but the change of the sign on two sides of the gaussian is the ''remnant'' of classical circulation.

This is same as the Band bending approximationIn a solid, some fundamental Questions can be raised,But the approximation works!!

Increasing x (or k)

∂E∂ k>0

∂E∂ k=0

∂E∂ k<0

States near the edge and states in the bulk have different group velocities

The semiclassical complete closed and incomplete skipping orbits

The quantum mechanical view: v g =1ℏ∂ E∂ k y

y

A state cannot carry current, if......vg=0 Fermi level doesn't have a finite D(E)

Current starting from one lead must either be backscattered or end up in the other lead.

Those which start from a lead carry with them the electrochemical potential of that lead...till they equilibrate with some other lead.

Ohmic contact=Black body!

Conduction from one lead to another

Conduction from one lead to another

I L→R = e∫−∞

∞

dE D(E ) f (E ,μL)v g(E )T (E)

I R→ L = e∫−∞

∞

dE D(E ) f (E ,μR)vg (E )T (E )

I = e∫−∞

∞

dE D (E) [ f (E ,μ L)− f (E ,μR)]v gT (E )

I ≈e∫μL

μ R

dE D(E)vg (E )T (E)

D(E ) =1π

1dE /dk

v g =1ℏ

dEdk

I ≈e∫μ L

μR

dE D(E )vg (E )T (E )

=2e2

h(V R−V L)T (E )

In 1D the two factors cancel

What is the value of T(E)?

At the end: the current does not depend on how many carriers are there in the channel!

What should be the value of T(E) ?

lead inserts some excess carriers δn = ND(E)eδVI = (evg)ND (E)e δV

= Ne2

hδV

RH =δVI=

h

Ne2

D(E ) =1π

1dE /dk

v g =1ℏ

dEdk

T (E )=1T (E )≠1

Key features of the Quantum Hall state

Data: KDG, (Semiconductor Phy group, Cavendish Lab)

The oscillation of the longitudinal resistance

Basic question: How much is the density of states near the Fermi level?

If Fermi level does not lie in a continuum then a charge cannot accept a small amount of energy and accelerate.

Of course it can jump across a gap if energy is sufficient (1 kelvin = 2.08 x 1010 Hz)

The Fractional Hall State (Discovery)Rotationally invariant solutionsMany body states in the first Landau LevelThe fake plasma analogy and the Laughlin statesCorrelations predicted by the Laughlin wavefunctionsWhy does one need to go beyond the Laughlin statesThe even denominator states and the open questions

The Fractional Quantum Hall effect (1982)

Discovery of the Fractional Quantum Hall effect (1982)

Tsui, Stormer, Gossard PRL 48,1559 (1982)

Looking for Wigner crystal phases

Expectation: at high fields electrons maylocalise in a lattice

Prediction from 1930s by Wigner

The FQHE state occurs in sample with higher mobility: here

n~1011 cm-2

m~105 cm²/Vs

Willett et al PRL 59,1776 (1987)

FQHE: Initially the fractions all seen to have odd denominators...but

Why was this striking?

The proposed wavefunction for the FQHE states (Laughlin) can work for odd denominators only.

Also fractions like ½ , ¼ are not seen in the first Landau Level

FQHE: Closer look at the 5/2 even denominator state

J C Dean: PhD Thesis (McGill Univ, 2008) Notice the extremely low electron temperature necessary to see the state clearly

FQHE: The skyline of states known today.....odd, even

Pan et al PRL 88,176802 (2002)

n~1011 cm-2

m~107 cm²/Vs

Notice that higher mobility leads to discovery of more and more fractional states

So, where do we start …..

The states in the first Landau level are strongly overlapping: High magnetic field suppresses screening

''Quenches'' kinetic energy (meaning E ~ k, k2 no longer holds)

All these taken together means......

The repulsive Coulomb interaction will play an important role

We need to write a manybody wavefunction for electrons in LL=1

We didn't need to do this for IQHE (non-interacting picture was sufficient)

Also we are dealing with a no-small-parameter, non-perturbative problem!

Complex number trick (useful for 2D electrostatics) turns out to be very useful...

With A=B(0, x ,0) we had

ϕn( x , y) =1

√Le ikyH n( x+x0)e

−(x+ x0)

2

2 l2

With A=B(− y /2, x /2,0) we will get

ϕ0,m(x , y) =1

√2π l 2 2mm!zm e

−| z |2

4

The higher levels are generally derived by recurrence

ϕn ,m = (2 ∂∂ z

−z *

2l2)n

ϕ0,m

..... not needed if we are in LL1

Rotationally symmetric states (angular momentum eigenfn) for Landau levels …..

here z=x+iyl

instead of eiky

these have e imθ

The fixed angular momentum States are localised in both x & y Unlike the cartesian solution

So we have many states with in LL1 with different angular momenta. How do we put many particles in them?

Any polynomial f (z )× e−

| z |2

4

Will also be a valid wavefn in LL1But not with a fixed L z

Writing the wavefunction for many electrons together …..

Question:We have n states (orbitals) to put n electrons. How to do it ensuring the indishtiguishability of the particles?

Quantum statisticsThere should be no way to distinguish particle 1 from particle 2,3,4 etc.

Exchanging any two particles...

FERMIONS:…..results in the same wavefunction with a negative sign.

BOSONS:…...results in the same wavefunction

Many body state with all particles in LL1 …..

Let zidenote the co-ordinate of the ith particle

Ψ[ z ] = ∣(z1)0

( z2)0 ...... (zN )

0

(z1)1

( z2)1 ...... (zN )

1

⋮ ⋮ ....... ⋮

(z1)N−1 ( z2)

N−1 ...... (zN )N−1∣ × e

−14 ∑i

N

| zi |2

Ψ[ z ] = −∏i< j

N

(zi−z j)e−

14 ∑i

N

|zi |2

Determinant form automatically satisfies the antisymmetry.

There are no ''wasted zeros''

The Vandermonde form of the matrix...easy to see what the roots are

But there are no free parameters left for us to tweak!

g ( z) =N (N−1)

n2Z∫ d 2 z3…∫ d 2 zN |Ψ(0, z , z2 ,… , zN )|

2

Z = ⟨Ψ |Ψ⟩

g ( z) = 1−e−

12

|z |2

can be exactly calculated!

Question: Is there any kind of structure in the wavefunction we have written?

Inverted gaussian.No peaks or structure

Uniform structureless background positive charge

The lattice is abstracted away

Free electrons with Coulomb interaction move around

Background and average electron charge density cancel exactly

Useful theoretical construct

Try to get a feel of the terms in the wavefunction: analogy of a fake 2D jellium !

Jellium

Partition function of a fake 2D jellium !

Ψ[z ] = −∏i< j

N

(zi−z j)e−

14∑i

N

|zi |2

Z≡∫ d 2 z1…∫ d 2 zN |Ψ[ z ] |2

Imagine |Ψ[ z ]|2 as e−βU where

β=2m

and

U≡m2∑i< j

(−ln | zi−z j |) +m4∑k

| zk |2

charge neutrality will require :nm+ρB=0

since l 2=

ℏeB

m can be imagined as the filling fraction

!! hypothetical 2D electrostatics (in cgs units)!!

∮E . dl = 2πQCoulomb interaction is then given by

E(r )=Q rr

V (r )= Q(−lnrr0)

since r 0 is arbitrary, free to set r 0=lconsider each particle to have charge :m unitsinteraction energy can be summed pairwise:

m2∑i< j

ln (| zi−z j |)

For a point charge ∇2V = −2πδ(r )

so∇ 2 |z |2

4=

1

l 2

set the uniform background to be ρB=−1

2π l 2

...second term follows

Ψ1 /3[ z ] = −∏i< j

N

( z i−z j)3 e

−14∑i

N

| zi|2

The Laughlin wavefunction............it is an educated guess!

Antisymmtery is preserved. What two point correlations does it predict?Is there any structure now?

S. Girvin, Les Houches lecture notes 1998arXiv: 9907002

Solid lines : Monte-Carlo dataDots : Laughlin wavefnFor m > 7 this form isn't the best

The Laughlin wavefunction : How do we show there is an excitation gap?

Excitation above a many body state will be some collective excitation

Like Spin wave in Ferromagnet is diferent from flipping a single spin.

Δ(k ) = ⟨Ψ (k ) |H−E0 |Ψ(k )⟩ .1

S (k )

S (k )=N δk , 0+1+n∫d 3 r e ik . r [ g (r )−1 ]

Relating the static structure factor to excitation spectrum.

Feynman & BijlAbrikosov & Gorkov

But in this case we need to ensure that the excited state is still part of LL1.

So, need to ''project'' the density fluctuation on the LL1 and then calculate the quantitites

Turns out both quantites vary as ~ k4

Hence the gap, since the ratio stays finite

ℏ2 k 2

2Mif the sum wasnot restrictedto LL1

Is the FQHE after all IQHE (of some composite object) in disguise?

J C Dean: PhD Thesis (McGill Univ, 2008)

This question leads to the COMPOSITE FERMION picture (Jainendra Jain)PRL 63, 199 (1989).PRB 41, 7653 (1990).

Predicts FQHE at fractions:

ν =p

2mp±1m , p : small integers

But still it will have odd denominator fractions only

References:

S M Girvin, Les Houches Summer school Lecture notes (1998)

J H Davies, Physics of Low dimensional Semiconductors

J K Jain, Composite Fermions (Cambridge University Press)

K. von Klitzing, M. Pepper, G. Dorda, PRL 45, 494 (1980)

D. Tsui, H. Stormer, Gossard PRL 48,1559 (1982)