Download Chern-Simons theory and the fractional quantum Hall effect

Chern-Simons theory and the fractional quantum Hall effect
G. Luchini
Instituto de Fı́sica de São Carlos; IFSC/USP; Universidade de São Paulo
Caixa Postal 369, CEP 13560-970, São Carlos-SP, Brazil
This is a small seminar presented in the many-body physics course at IFSC-USP in 2011.
The fractional quantum Hall effect consists in the observation that the transversal conductivity 2
defined as in the classical effect - is quantized as σH ∝ eh , where the proportionality constant is a
rational number. The goal of these notes is to introduce the subject and discuss how one uses the
Chern-Simons theory as an effective theory for this scenario. Unfortunately I could not go deeper
and many nice things are missing. This is a reflection on the fact that the subject spreads out in
many directions. The notes are supposed to be a script for the presentation, where I shall give more
details.
PACS numbers:
A BRIEF REVIEW ON THE BACKGROUND
PHYSICS
The effect discovered by Edwin Hall in 1879, known
as the classical Hall effect (CHE), was at first of purely
academical interest and a complement to J.C. Maxwell’s
Treatise on Electricity and Magnetism, published six
years before. Beyond the expectations of Maxwell’s theory, it was observed that an electronic gradient appears
in the transversal direction of a current in a metallic slab
subjected to a perpendicular magnetic field. Then, a
resistance in that direction is measured and it is proportional to the applied field.
If the electrons motion can be confined into an effective
2 dimensional plane, and under a large magnetic field and
low enough temperatures one observes the quantum Hall
effect (QHE), characterized by the quantization of the
transversal resistance, which develops plateaux of constant values for certain ranges of the magnetic field. Remarkably the resistance is given uniquely in terms of universal constants, and does not depend upon the details
of the sample:
RH =
h 1
,
e2 ν
being
1
a proportionality constant.
ν
In 1980 the integer quantum Hall effect (IQHE) was observed by von Klitzing and others[1]: the factor ν of the
above formula corresponds to integer numbers. Surprisingly in 1983, Tsui and collaborators[2] observed the so
called fractional quantum Hall effect (FQHE), for which
ν = pq , p, q integers.
In these notes we discuss how one can use the ChernSimons gauge theory to understand or to deal with the
FQHE.
Dynamics of the 2DEG
There are three basic ingredients one needs in order
to observe the QHE: i) low temperature (<
∼ 4K); ii)
strong magnetic field (1 ∼ 30T ); iii) an effective 2 dimensional world for the electrons to live in. In particular this last requirement is achieved once it is possible to freeze the motion of the electrons in one direction in our 3 dimensional world. Quantum mechanics
allows that as follows. Take the sample to be a 3 dimensional quantum well, with height H in the z direction, which is the one we want to eliminate. With the
ansatz ψ(x, y, z) = X(x)Y (y)Z(z) the Schrodinger equation gives three independent
equations, and
2 2 differential
−h̄ d
in particular we have 2m dz2 + Vz (z) Z(z) = Ez Z(z),
being Vz (z) = z inside the sample and ∞ otherwise. The
n2
energy is then quantized as En ∝ H
2 and we see that
for very small heights the gap between energy levels is
very large, so if the system is at low temperatures, it has
no chance to produce excitations in the z direction and
becomes effectively 2 dimensional.
Another important observation is that the QHE appears for an electronic density nel ∼ eB
h . For metals the
typical electronic density is of the order 1018 /m2 , which
would imply a magnetic field of the order of 103 T , which
cannot be produced in laboratory. Semiconductors on
the other hand have electronic density about 1015 /m2 ,
which reduces the needed 3 orders of the magnetic field.
Thus, for the CHE one uses a conductor, while for the
QHE, a semiconductor. With these brief observations we
proceed to discuss the classical dynamics of a 2 dimensional free electron gas subjected to an electromagnetic
field.
The equation of motion of a electron under these circumstances reads
dp
1
m
= −e E + v × B − v
(1)
dt
c
τ
where τ is the mean diffusion time due to impurities of
the sample. The magnetic field is perpendicular to the 2
dimensional plane B = B k̂, and we consider the steady
2
scenario ṗ = 0, so that we get the set of equations
e
− eEx − vy B −
c
e
−eEy + vx B −
c
m
vx = 0
τ
m
vy = 0.
τ
motion are p˙x =
Now we multiply them by en
m , and after that introduce
2
eB
the definitions ωc = mc
and σ0 = nem τ . Then, with the
current vector j = (−nevx , −nevy ) one gets the
P conductivity tensor defined through the relation jµ = ν σµν Eν
as
σ0
1 −τ ωc
.
σ=
1
1 + τ 2 ωc2 τ ωc
The resistivity tensor is obtained by taking the inverse
(with the definition ρ0 ≡ σ0−1 ):
B
ρ0 nec
.
ρ=
B
ρ0
− nec
For convenience one may define the longitudinal conductivity σL = 1+τσ02 ω2 and the transverse or Hall conductivc
σ0 τ ωc
ity σH = 1+τ
2 ω 2 . A first remark is that under a SO(2)
c
transformation σ → RσR−1 the conductivity tensor remains unchanged. This reflects the fact that both x and
y directions are equivalent, and the conductor is homogeneous. Indeed, when no Hall effect is present (i.e.,
B = 0) then σ = σ0 1l; in the case of a homogeneous
conductor, with non-zero magnetic field, σxx = σyy and
σxy = −σyx . Now, for strong magnetic field and low
temperatures (kB T ωc h̄), ωc τ 1, and the longitudinal conductivity vanishes while the Hall conductivity
becomes σH → nec
B . Consequently, the longitudinal resistivity goes to zero and the transverse resistivity becomes
B
RH = nec
, directly proportional to the magnetic field[5].
A quick argumentation for the Hall effect is the following: a force appears in order to balance the Lorentz force
so that the charge carriers keep moving in the longitudiVH e
nal direction. Thus evB
c = L , where L is the distance
in the transversal direction of the sample between the
points one measures the voltage VH . Solving it for the
velocity of the charge carriers, plugging the result into
the current I = nLev and after that writing VH = RH I
we get exactly the previous result for the transverse resistance RH .
We are considering a 2 dimensional system of charged
and massive particles subjected to a constant magnetic
field in the perpendicular direction. Our next goal is
to find the classical trajectories of these particles. The
lagrangian describing such a system reads
L=
m 2 e
v − v·A
2
c
where B = ∇ × A locally, at least. The equations of
and p˙y =
e
p x = π x − Ax
c
(2)
(3)
∂L
∂x
&
∂L
∂y ,
where
e
py = πy − Ay
c
are the canonical momenta while πx = mẋ and πy = mẏ
are the kinetic ones.
With the definition of the cyclotron frequency ωc =
we get the coupled equations
ẍ + ωc ẏ = 0
eB
mc
ÿ − ωc ẋ = 0
which can be solved as follows: define z = x + iy
and calculate its second derivative, which gives a first
order differential equation in ż ≡ vz (t), whose solution is vz (t) = vz (0)eiωc t . Writing the integration
constant as vz (0) = v0 eiφt one then obtains vx =
v0 cos (ωc t + φ) and vy = v0 sin (ωc t + φ). Integration is
direct and the result is a circular trajectory centred at
R = (x0 , y0 ), called the guiding center: r = (x(t), y(t)) =
x0 + ωv0c sin (ωc t + φ), y0 − ωv0c cos (ωc t + φ) . An important quantity is the vector η = r − R. Comparing its
components with the components of the velocity vector
above one gets
πx = −mωc ηy
πy = mωc ηx .
We shall come to this result later. The hamiltonian
makes the bridge between the classical and quantum theory once the canonical quantization approach is considered, where roughly speaking the observables are promoted to operators acting on quantum states in a Hilbert
space, and the Poisson structure is mapped into a commutation relations between these operators: {∗, ∗} →
−ih̄[∗, ∗]. We have then Ĥ = ih̄∂t , p̂ = −ih̄∇ and x̂
and  act multiplicatively on the quantum states. The
hamiltonian reads
H=
1
1
e
πx2 + πy2 =
kp + Ak2
2m
2m
c
and the gauge freedom is apparent. So, in order to find
the wave function one needs to pick a particular gauge.
Two choices are relevant for our discussion: tha Landau
gauge AL = B(−y, 0, 0) and the symmetric gauge AS =
B
2 (−y, x, 0). They are related by AL = AS + ∇Λ with
Λ = − B2 xy.
In
the Landau gauge one gets H
=
(px − ec yB)2 + p2y . We choose periodic boundary
conditions along the y direction, i.e. ψ(0, y) = ψ(Lx , y),
(where Lx (analogously, Ly ) stands for the sample size
in x direction) which implies the quantization of the
corresponding wave vector component: km = L2πx m.
Noticing that [p̂x , Ĥ] = 0 we take the ansatz for the
wave function: ψ(x, y) = eikx φ(y). Then what happens
is that we “ factorize” our problem which becomes a
1
2m
3
Schrodinger equation problem in one dimension:
−
h̄2 2
m
2
∂y φ(y) + ωc2 (lB
k − y)φ(y) = Eφ(y).
2m
2
physics: we introduce the step operators[6]
lB
a = √ (πx − iπy )
2h̄
lB
a† = √ (πx + iπy ) (4)
2h̄
This is nothing but the quantum harmonic oscillator cench̄
ch̄
2
2
. The quantity lB
= eB
is called
tred in klB
= k eB
the magnetic length. The eigenstates above are given in
terms of Hermite polynomials and the energy spectrum
is
where the normalization constant was fixed so that the
bosonic commutation relation holds: [a, a† ] = 1. Inverting the above definitions and plugging the kinetic momenta in terms of the step operators into the
hamiltonian
one gets, as expected: H = h̄ωc a† a + 12 .
1
En = h̄ωc (n + ).
2
We use n to label the quantum state | n i, and the
action
of the step operators
√ on it is the usual a | n i =
√
n | n − 1 i, a† | n i = n + 1 | n + 1 i and the ground
state is defined by a | 0 i = 0. Thus, excitations to
n
(a† )
different LL are achieved by taking | n i = √n! | 0 i.
Each of these energy levels (each n) is called a Landau
Level (LL). The energy cost to go from one LL to another
is h̄ωc . Clearly we get a degeneracy in each LL, which
is associated to the value of km . We now proceed to calculate the maximum degeneracy. The restriction on the
size of the sample 0 < y < Ly together with the maxi2
mum value for y, i.e. the center of oscillation y = lB
k
Ly
implies 0 < km < l2 . Using the quantization condition
B
one gets kmax =
2πmmax
Lx
mmax =
=
Ly
2 .
lB
Then
Area of the sample
2
2πlB
measures the degeneracy of each LL. Notice that it is
directly proportional to the external applied magnetic
field. In fact, lets call is NB , the number of states in the
2 −1
)
as the density of
LL, then we identify nB = (2πlB
states in the LL and we can define the filling factor ν as
the ratio between the electronic density and the density
ch
of states in each LL:ν = nnBe = N
Φ Φ0 , where Φ0 ≡ e is
the unit flux and Φ ≡ B × Area is the magnetic flux. In
words:
ν=
number of eletrons
.
number of magnetic flux penetrating the sample
As promised before let us go back to the definition of
the kinetic momenta in terms of the η vector. The
2
2
2
hamiltonian becomes H = m
2 ωc ηx + ηy , and we notice that these quantities do not commute between them2
. Then we get also that the guidselves: [ηx , ηy ] = −ilB
ing center coordinates are also non-commutative, giving
2
[x0 , y0 ] = ilB
. This lead us to associate a kind of Heisenberg uncertainty principle to the two spatial directions,
2
where lB
plays a similar role of the h̄ in the usual relation
[x, p] = ih̄. The most precise measure one gets of these
2
coordinates is then given by ∆x0 ∆y0 = 2πlB
, which is
Area of the sample
a minimal area. Thus, the ratio
gives
∆x0 ∆y0
(again) the number of quantum states in each LL. Real
space resembles the phase space.
Let us consider a different approach to solve the hamiltonian which is more in the spirit of the many-body
Our description of the quantum state is not complete
yet. It is necessary to lift the degeneracy in each LL,
described before. One can also notice that something is
missing from the fact that the original hamiltonian - the
original description of the system- was given in terms of
two conjugate pairs, x, px and y, py , and so far we have
only one conjugate pair: a and a† .
The first step to solve this problem is to consider the
angular momentum Lz = xpy − ypx . Lets take the foleB
lowing substitutions: px = πx + eB
2c y, py = πy − 2c x,
c
c
x = x0 + eB πy and y = y0 − eB πx . Then we get
l2
B
Lz = − 2lh̄2 (x20 + y02 ) + 2h̄
(πx2 + πy2 ), which clearly comB
mutes with H. Now the second step is to notice that the
last part of Lz can be written in terms of a and a† , while
for the first part we define similarly the step operators
associated to the guiding center coordinates:
b= √
1
(x0 + iy0 )
2lB
b† = √
1
(x0 − iy0 ),
2lB
then
L̂z = h̄ a† a − b† b .
The quantum number associated to the number operator
b† b is m and the full quantum state is | n, m i =| n i⊗ |
m i, so that L̂z | n, m i = h̄(n−m) | n, m i. An arbitrary
state can be achieved from the vacuum | 0, 0 i by taking
(a† )n (b† )m
√
| n, m i = √
| 0, 0 i.
n!
m!
Now we are interested in construct the wave functions the
coordinate representation: ψn,m (x, y) = h x, y | n, mi. In
order to do so, lets extend the 2 dimensional plane to
a complex plane defining the two independent quantities
z = x + iy and z̄ = x − iy. In these new variables the
∂
step operators are given by (with ∂ = ∂z
and ∂¯ = ∂∂z̄ )
√
√
z
z̄
¯
a = −i 2
+ lB ∂
b = −i 2
+ lB ∂
4lB
4lB
4
∂ ∗
and the hermitian conjugate (remember that ( ∂z
) =
∂
− ∂ z̄ ).
A state in the lowest Landau level (LLL) is defined by
the condition a | n = 0, m i = 0, which, in the coordinate
representation becomes
z
+ lB ∂¯ ψn=0,m (z, z̄) = 0.
4lB
−
|z|2
4l2
B
,
The solution is very simple: ψn=0,m (z, z̄) = f (z)e
with f (z) any analytic function. Now a state with m = 0
at any LL is defined by b | n, m = 0 i, which reads
z̄
+ lB ∂ ψn,m=0 (z, z̄) = 0.
4lB
−
|z|2
2
The solution is given by ψn,m=0 (z, z̄) = f˜(z̄)e 4lB with
f˜(z̄) anti-analytic.
Naturally the wave-function for | n = 0, m = 0 i is just
the Gaussian term with a constant in the front (which is
then both analytic and anti-analytic), fixed by normaliza-
tion: ψn=0,m=0 (z, z̄) = √ 1
2
2πlB
−
e
|z|2
4l2
B
. The relevant state
for us is that for the LLL and arbitrary m:
√
m |z|2
− 2
im 2m
z
√
ψn=0,m = p 2
e 4lB .
2lB
2πlB m!
THE FQHE
The FQHE was first observed for a filling fraction of
ν = 13 , and then RH = eh2 ν = 3h
e2 . From the point of
view of kinetic energy the state with ν = 1/3 is highly
degenerate and there is no clear gap in the system: Pauli
principle does not forbid the electron to jump to another
LL (which costs h̄ωc ), but there is no need for it since
there is enough room in the lowest Landau level (LLL),
which is 1/3 filled. So, kinetic energy is not relevant to
this situation, and may drops out. What one needs here
is the Coulomb interaction between the electrons. This
makes the situation seriously different from that of the
IQHE: the strongly-correlated system cannot be treated
within the perturbation theory for Fermi liquids and one
needs to guess the ground state then. It is important to
2
remark that h̄ωc > VCoulomb ∼ leB .
We now discuss how to (more or less) get the Laughlin’s wave function[4], which was in fact a very well succeeded guess.
Lets start with the one-particle wave-function for the
LLL, which reads (absorbing the magnetic length in the
definition of z)
ψ ∼ z m e−
|z|2
4
,
i.e., and analytic function times a Gaussian part. The
quantum number m goes from 0 to NB − 1. Now, if
two particles are considered, with positions z1 and z2 ,
we write
ψ (2) (z, Z) ∼ Z M z m e−
(|z1 |2 +|z2 |2 )
4
where Z = (z1 + z2 )/2 and z = z1 − z2 . The quantum
number m stands for the relative angular momentum of
the two particles while M is related to the total angular
momentum of the system.
The fermionic character of the particles implies that m
is odd. It has to be an integer due to the analyticity. So,
m = 2s + 1, with s and integer. The Laughlin function
is a generalization of this:
P
Y
−
|zj |2 /4
j
ψm ({zj , z̄j }) ∼
(zk − zl )m e
.
(5)
k<l
This guess for the ground state must be now tested.
In the variational approach one bases the quality of
) |H| φ(τ ) i
≥
the trial function on the inequality h ψ(τ
h ψ(τ ) |ψ(τ )i
(Ground state energy). The goal is to minimize the l.h.s
of this inequality with respect to the parameters. In the
case here, Laughlin parameter is the angular momentum
quantum number m.
In the ansatz (5) one has that the maximum value for
Q
m(N −1)
the product k<l (zk − zl )m is zk
(as can easily
be tested), and we know that the highest power in the
complex position of the particle is fixed by NB , which
yields mN = NB , in the thermodynamic limit N, NB →
∞. Using the filling factor definition ν = N/NB we find
that the variational parameter m is entirely fixed by the
filling factor
ν=
1
1
=
,
m
2s + 1
and the Laughlin wave-function is a genuine candidate
for ground states with ν = 1, 13 , 15 , . . ..
The ν = 1 case
Notice that also the completely filled LL is described
by the Laughlin wave-function:
P
−
|zj |2 /4
j
ψ({zj }) = fN ({zj })e
,
and being non-degenerate can be written in terms of the
Slater determinant:
0
z1 . . . z1N −1 .. = Y(z − z ).
fN ({zj }) = ... ...
i
j
. z 0 . . . z N −1 i<j
1
N
5
Elementary excitations with fractional charge
In order to obtain elementary excitations one needs
1
. There are two
to vary the filing factor from ν = m
ways of doing so: changing the electronic density or
adding/removing magnetic flux. The later is related to
the number of zeros in the Laughlin wave-function, thus
one considers the ansatz
ψqh (z0 , {zj , z̄j }) =
N
Y
(zj − z0 )ψm ({zj , z̄j })
j
and effectively each quantum number is increased by
unit: mi → mi + 1, as can be verified just expanding
the Laughlin function. Each electron jumps from a state
with a given angular momentum to another (from a circular orbit to another with bigger radius if you want),
leaving an empty state with m = 0 behind, which is
the quasi-hole. In order to see that we can consider a
very naive but interesting argument. The filling factor
ν = 1/3 means that one electron shares 3 orbits. Then,
when we add a quantum of flux in the center each quantum number m, of each orbit, is shifted by one unit. One
considers that the last orbit is lost, once associated with
it is the maximum value of m, thus a charge of e/3 goes
out and what remains if a quasi-hole of charge −e/3 in
the center.
Notice that when a flux is introduced into the system
a radial electric field appears. Faraday’s induction gives
I
1 d
1 d
Φ ⇒ E2πR =
Φ.
E · dr =
c dt
c dt
Then a radial current density appears jr = σxy Eθ , giving
Z
Z
1
d
I = dθrjr = dθRσxy Eθ = σxy Φ,
c
dt
and the charge that goes out of the system is calculated
integrating I in time:
Z
σxy
h
Qout = dt I =
Φ0 = σxy .
c
e
So, we see that the fractional conductivity is directly associated to the fractional charge. Using the experimental
2
result σxy = ν eh one gets Qout = eν = 3e .
We now proceed to calculate the charge of the quasihole excitation in a more sophisticated way. The idea
is walk around with the quasi-hole, calculate its Berry
phase and then make it equal to the Aharonov-Bohm
phase.
The Berry phase is given by
Z tf
γ=i
dt h ψqh (t) | ψ̇qh (t)i,
ti
where we put a dynamics for the hole in the above ansatz
as z0 (t).
The time derivative of the
wave function is
P quasi-hole
d
ln(zi −z0 (t)) (in order to
easy to calculate and gives i dt
see that one can check the case of (z1 − z0 (t))(z2 − z0 (t)),
and try to write this same function again in the result).
Now, instead of this sum, we take the integral
Z
X d
d
ln(zi − z0 (t)) = d2 rρ(r) ln(z − z0 (t))
dt
dt
i
where the density (lots of delta functions) picks the points
zi . So we get
Z
Z
ż
γ = i d2 r dt
h ψqh | ρ(r) | ψqh i
z0 − z
and the integral in time can be substituted to the corresponding path:
I
dz0
= 2πi
z0 − z
if z is inside and it vanishes otherwise. Finally, after
plugging this result in what is left is the average density
times the area and we have
γ = −2πn × Area.
∗ H
The Aharonov-Bohm phase is given by eh̄ dl · A =
2πe∗ Φ
cΦ0 . When we make it equal to the Berry’s phase we
get the desired result e∗ = −νe.
ANYONS AND CHERN-SIMONS THEORY
In order to start the following discussion let us show
how a singular gauge transformation can map a fermionic
(or anyonic) wave function into a bosonic one.
Consider a system of two particles, and suppose the
effect of interchanging them is that the wave function
gains a factor eiθ : ψ(eiθ z) = eiθ ψ(z), z being the relative coordinate. A bosonic wave-function should satisfy
φ(eiπ z) = φ(z) and one can map both wave-functions by
the local gauge transformation φ(z) = eiη(z) ψ(z) with
η(z) = − πθ arg(z).
Now that we know how to map a fermionic wavefunction into a bosonic one we map the whole problem
of the 2DEG under strong magnetic field at low temperatures into an analogous problem involving bosons.
The hamiltonian reads
Ĥ = Ĥ0 + Ĥint
where
Z
Ĥ0 =
2
d2 r ψ † (r)
(−ih̄∇ + eA(r))
ψ(r)
2m
6
and Ĥint stands for the Coulomb interaction. We take
the gauge transformation
R 2 0
0
ψ(r) = e−iq d r θ(r−r ) φ(r)
where θ(r) is the angle between the vector r and the x
axis. This gauge transformation of the wave function
must be accompanied by a gauge transformation on the
vector potential. What we find out is that in order to
guarantee the gauge invariance of the theory one needs
to introduce the new field
Z
h̄
a(r) = − q∇ d2 r0 θ(r − r0 )ρ(r0 ),
e
called the Chern-Simons (CS) gauge field. One can show
that ∇2 θ(r) = 0 and therefore the CS field satisfies the
Coulomb gauge. Thus, the kinetic part of the hamiltonian becomes
Z
2
(−ih̄∇ + eA(r) + ea(r))
Ĥ0 = d2 r φ† (r)
φ(r).
2m
Associated with this gauge field one has the magnetic
field
h
b = ∇ × a = − qρ(r)k̂
e
(6)
.
for each field. In the case of φ:
(∂t +eA0 +ea0 )φ = −
1
(−ih̄∇−eA−ea)2 φ+(φ̄φ−ρ0 )φ,
2m
and one can verify that equations for a0 and a are exactly
(6) and (7). Remarkably the above equation shows that
the system we are dealing with consists of bosonic fields
interacting with the gauge field which is the sum of the
ordinary electromagnetic connection with CS field, and
there is also a self-interaction between the bosons. This
equation is known as the non-linear Schrodinger equation.
Uniform field solutions
√
Take the field to be uniform, given by φ = ρ0 . A
look at the equations of motion shows that one needs
A + a = 0 and a0 = 0, and equation (6) gives B = q he ρ0 .
From here one can extract the filling factor, which is
the ratio between the density of bosons ρ0 and the flux
1
.
density Be/h, getting ν = 1q = 2s+1
A comment on the conductivity
... unfiished... to appear soon!!
As showed in [3], the two problems, i.e. the original
one and that new one, with the new hamiltonian that
includes the CS gauge field and bosonic wave-functions
are equivalent only for q = 2s + 1.
Now we consider the time derivative of (6) and use
the continuity equation ∂t ρ + ∇ · j = 0 to get a total
divergence equation which implies (up to a constant)
µν ȧν =
h µ
qj .
e
(7)
We then realize that equations (6) and (7) completely determine the dynamics of CS gauge field, which is therefore
described by the lagrangian
L=
e µνρ
aµ ∂ν aρ − aµ j µ .
2hq
Our next step is to consider the theory given by
LCSLG =
2
h̄
e
∇−
(A + a) φ +
2mi
2m
2
1
e µνρ
+ δρ(r, t)V (r − r0 )δρ(r0 , t) +
aµ ∂ν aρ
2
2hq
The statistical properties of φ(r).
Using the known properties of the fermionic field
ψ(r) we can show that the
R Chern-Simons field φ(r) =
eiqη(r) ψ(r), where η(r) = d2 r0 θ(r − r0 )ρ(r0 ) has the following property:
φ(r1 )φ† (r2 ) + eiqπ φ† (r2 )φ(r1 ) = δ(r1 − r2 ).
This expression is the generalization of
ψ(r1 )ψ(r2 ) = eiαπ ψ(r2 )ψ(r1 ),
where q plays the role of the statistical angle α. For
bosons and fermions the statistical angle is 2 and
1 giving the phases 1 and −1, and therefore the
commutation/anti-commutation relations. The statistical angle being q = 2s + 1 implies a transformation of
fermions into bosons.
φ̄ (∂t + e(A0 + a0 )) φ + φ̄
where δρ(r) = |φ|2 − ρ0 .
It is quite direct to compute the equations of motion
[1] K. v. Klitzing, G. Droda, M. Pepper, Phys. Rev. Lett. 45,
494 (1980);
[2] D.C. Tsui, H. Stormer, A.C. Gossard, Phys. Rev. Lett.
48, 559 (1982);
7
[3] Zhang, S.C., Inter. Jour. Mod. Phys. B v.6, No.1, (1992)
25-58;
[4] Laughlin, R.B Phys. Rev. Lett. 50, 1395 (1983);
[5] Here is a good point to remark on the fact that in 2 dimensions the resistance and resistivity become the same
thing. In d dimensions they are related by R = ρL2−d , L
being the side of the hypercube.
[6] Notice that now we are interested in the symmetric gauge,
instead of the Landau gauge used previously.