hall effect - istituto nazionale di fisica...
TRANSCRIPT
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