Exchange symmetry

Jan Myrheim

Department of physics, NTNU

April 12, 2016

Contents

– Some general remarks about identical particles

– Two and three anyons in a harmonic oscillator potential

– Cluster and virial expansions for the anyon gas

– An approximate relation to exclusion statistics

Heisenberg and Dirac (1926)

Definition: N particles, numbered 1 to N, are identical if all observables aresymmetric under permutations of indices.

A symmetric operator X representing an observable preserves the symmetryof the wave function ψ,

ψ → Xψ ,

S → SS = S ,

A → SA = A .

We can have two different complete theories of quantum mechanics foridentical particles.

(I) Using only symmetric wave functions.

(II) Using only antisymmetric wave functions.

Heisenberg and Dirac argue that these two possibilities exist, but make noattempt to prove that other possibilities do not exist.

Statistics

In case (I), symmetric wave functions, counting of states leads toBose–Einstein statistics.

In case (II), antisymmetric wave functions, the Pauli exclusion principleholds, and counting of states leads to Fermi–Dirac statistics.

If we admit all wave functions, without imposing symmetry orantisymmetry, we get Maxwell–Boltzmann statistics.

In the thermodynamic limit we let N →∞ and the volume V → ∞with constant particle density n = N/V .

This limit exists for ideal gases of bosons and fermions, since their entropyis extensive (proportional to N at constant density).

With Maxwell–Boltzmann statistics, the number of states grows too fastwith N for the thermodynamic limit to exist.

Statistics

In both classical and quantum mechanics we need particles to be identicalbecause we want a thermodynamic limit.

In classical statistical mechanics the solution is to divide the phase spacevolume by N!

In quantum statistical mechanics the solution is to symmetrize orantisymmetrize the wave functions.

Except that we often do not.We antisymmetrize the wave function of the two electrons in a helium atom,but we do not antisymmetrize with the other 1026 electrons around.

Two electrons at different positions are identical, but distinguishable.We need not antisymmetrize if we, for example, scatter electrons in such away that we can follow one electron from the initial to the final state.

Scattering process 1 + 2→ 3 + 4

p

−p

p′

−p′

θ

detector

Incoming momenta p1 = −p2 = ±p.Outgoing momenta p3 = −p4 = ±p′.Scattering angle α = θ or π − θ, definition: p1 · p3 = |p||p′| cosα.Scattering amplitude f = f (α) because of rotational invariance.

f (α) = 〈out|F|in〉 .

Scattering p + n→ p + n

As an example for illustration, consider scattering of protons on neutrons.Disregard spin (assume spin independent interactions, a bad approximation).Treating them as different particles, we define the in and out states as

|in〉 = |pn〉|p,−p〉 , |out〉 = |pn〉|p′,−p′〉 .

We label, e.g., the incoming proton as particle 1 and call its momentum p.To measure the cross section

dσdΩ

= |f (θ)|2

we count the outgoing protons, but not the neutrons.If our detector counts both protons and neutrons, without distinction,we would have to redefine

dσdΩ

=12(|f (θ)|2 + |f (π − θ)|2

),

dividing by two to compensate for the double counting.

IsospinIn 1932, immediately after the discovery of the neutron, Heisenbergsuggested treating neutrons and protons as identical fermions.This is a natural idea because the nuclear forces do not distinguish them.

Then we have to antisymmetrize the in and out states.The isospin formalism is a technical aid.

We define the proton and neutron to be members of an isospin doublet.The isospin component I3 is +1/2 for protons and −1/2 for neutrons.

A state of two nucleons can have total isospin I = 0 or I = 1.There are three symmetric states |I, I3〉 with I = 1:

|1, 1〉 = |pp〉 , |1, 0〉 =1√2

(|pn〉+ |np〉

), |1,−1〉 = |nn〉 ,

and one antisymmetric state with I = 0:

|0, 0〉 =1√2

(|pn〉 − |np〉

).

Isospin

Define also symmetric and antisymmetric momentum states

|p+〉 =1√2

(|p,−p〉+ | − p, p〉

),

|p−〉 =1√2

(|p,−p〉 − | − p, p〉

),

and similarly for p′.

Then we have two possible proton neutron in states, with I = 1 or I = 0,

|in〉 = |1, 0〉|p−〉 or |0, 0〉|p+〉 ,

and two possible out states,

|out〉 = |1, 0〉|p′−〉 or |0, 0〉|p′+〉 .

Scattering

Conservation of isospin, or equivalently, conservation of the symmetry of themomentum state, implies that we have two different scattering amplitudes.One with isospin I = 1,

f−(θ) = 〈p′− |F|p−〉

=12(〈p′,−p′|F|p,−p〉 − 〈p′,−p′|F| − p, p〉

−〈−p′, p′|F|p,−p〉+ 〈−p′, p′|F| − p, p〉)

= f (θ)− f (π − θ) .

And one with isospin I = 0,

f+(θ) = 〈p′+ |F|p+〉

=12(〈p′,−p′|F|p,−p〉+ 〈p′,−p′|F| − p, p〉

+〈−p′, p′|F|p,−p〉+ 〈−p′, p′|F| − p, p〉)

= f (θ) + f (π − θ) .

Scattering

We compute the cross sections as

dσ±dΩ

=12|f±(θ)|2 =

12(|f (θ)|2 + |f (π − θ)|2

)± Re

(f (θ)f (π − θ)∗

),

dividing by two to compensate for double counting.

Conclusions:

Proton neutron scattering can take place via one of two channels: eithersymmetric in momentum (I = 0) or antisymmetric in momentum (I = 1).

The cross section we compute by treating them as different particles is theaverage of the symmetric and the antisymmetric channel,

dσdΩ

=12(|f (θ)|2 + |f (π − θ)|2

)=

12

(dσ+dΩ

+dσ−dΩ

).

Scattering

Even though a proton is different from a neutron, we may choose to preparethe initial state to be symmetric or antisymmetric in momentum.Then we observe bosonic or fermionic scattering, as we choose.

On the other hand, if we scatter identical particles we have no choice.They prepare themselves in symmetric or antisymmetric initial states.

If there is a mystery, it is this: how do electrons know that they are fermions?And α particles that they are bosons?

Photons know that they are bosons because they are quanta of harmonicoscillators. That is how we derive QED from Maxwell’s equations.

Or if we regard the quantum theory as more fundamental, we want photonsto be bosons because we want Maxwell’s equations as a classical limit.

Is isospin only a book keeping trick?

In the case of protons and neutrons, we may regard isospin as a tool forkeeping track of the symmetry or antisymmetry of spatial wave functions.

But isospin is a useful quantum number for all strongly interacting particles.For example, isospin conservation imposes strong constraints on thescattering of π mesons on nucleons.

In the quark model, 2I3 is the number of u quarks minus the number ofd quarks.

Isospin conservation can be understood as an approximate SU(2) symmetry,which is broken because the u and d quarks have different electric chargeand different mass.

We treat all six quarks u, d, c, s, t, b as identical fermions distinguished bydifferent values of quantum numbers, called flavours, similar to isospin.

However, the SU(6) symmetry extending the SU(2) isospin symmetry isbadly broken by the electroweak interaction and by the quark mass matrix.The quark masses are free parameters in the so called standard model.

Colour

The quark model nearly died in its infancy because it seemed to requirequarks to be bosons, instead of fermions as required by the spin–statisticstheorem.

The most dramatic case was the baryon decuplet, for example the ∆++

resonance made up of three u quarks. In the ground state of the uuu systemwe expect the quarks to have zero orbital angular momentum.

Therefore the spatial part of the wave function should be symmetric.The ∆++ has spin 3/2, requiring also a symmetric spin wave function.Conclusion: the u quarks seemed to be spin 1/2 bosons.

One idea at the time was that quarks were neither bosons nor fermions, butsatisfied parastatistics: the N-particle wave functions would belong to moregeneral representations of the permutation group SN .

Greenberg solved the problem in 1964 simply by postulating a newthree-valued quantum number, later called colour, now the basis of QCD.The quarks are fermions with an antisymmetric colour wave function, thenthe rest of the wave function must be symmetric.

Which particles are identical?

Consider the Hamiltonian of the hydrogen atom, again disregarding spin,

H =p 2

1

2m1+

p 22

2m2− k|r1 − r2|

, k =q1q2

4πε0.

It is symmetric under the interchange 1↔ 2, a fact which becomes evenmore obvious when we split off the centre of mass motion and write

H =P2

2M+

p2

2m− k|r|

,

with M = m1 + m2, m = m1m2/M,

P = p1 + p2 , r = r1 − r2 , p =m2p1 − m1p2

M.

Which particles are identical?

The Hamiltonian is symmetric, what about other observables?

The electric dipole moment is symmetric, it is

d = e(rp − re) = q1r1 + q2 r2 ,

independent of how we label the electron and the proton as particle 1 and 2.

The electron position is symmetric, it is

re = δ1e r1 + δ2e r2 .

All observables are symmetric under the interchange 1↔ 2.They have to be, because the labels are arbitrary, different labellings justcreate different mathematical descriptions of the same physical reality.

So in this sense the electron and proton are identical particles.Why don’t we symmetrize or antisymmetrize the wave function?

The configuration space

particle identity

particle position particle position

electron

proton

1

2

2

1

There is an electron at position re and a proton at position rp.This statement describes uniquely a configuration of the system.

If we label the electron as particle 1 and the proton as particle 2, we have acoordinate system covering the configuration space completely anduniquely, as the left part of the figure indicates.

The right part of the figure shows the alternative labelling, electron = 2 andproton = 1. It gives nothing new.

The configuration space

If we stick to one labelling, for example electron = 1 and proton = 2,then the question of symmetrizing or antisymmetrizing the wave functiondoes not arise.

It may seem that the essential point is that there is some discrete quantumnumber distinguishing protons from electrons.

Because of the discreteness, we get no nontrivial topology when we identifytwo parts of the configuration space differing only in the labelling ofparticles.

Then we use only one of the two separate copies of the configuration space,and forget about symmetrization or antisymmetrization.

The harmonic oscillator

One particle in two dimensions described by a complex coordinate.

H =p2

2m+

12

mω2r2 .

z =x + iyλ

, λ =

√~

mω,

∂ =∂

∂z=λ

2

(∂

∂x− i

∂

∂y

), ∂∗ =

∂

∂z∗=λ

2

(∂

∂x+ i

∂

∂y

).

a = ∂ +z∗

2, a† = −∂∗ +

z2,

b = ∂∗ +z2, b† = −∂ +

z∗

2.

[a, a†] = [b, b†] = 1 , [a, b] = [a, b†] = [a†, b] = [a†, b†] = 0 .

The harmonic oscillator

a†, a, b†, b create and annihilate quanta of energy and angular momentum,

H = ~ω(−2∂∂∗ +

|z|2

2

)= ~ω(a†a + b†b + 1) ,

L = −i~(

x∂

∂y− y

∂

∂x

)= ~(z ∂ − z∗∂∗) = ~(a†a− b†b) .

Ground state: ψ0 = e−|z|2

2 , aψ0 = bψ0 = 0 .

Excited states: ψjk = (a†)j (b†)k ψ0 .

Hψjk = (j + k + 1)~ω ψjk , Lψjk = `~ψjk , ` = j− k .

Maximally rotating states: (a†)j ψ0 = zj ψ0 , (b†)k ψ0 = (z∗)k ψ0 .

a†b† creates radial excitations without changing the angular momentum.

If j ≥ k then ψjk = (a†b†)k (a†)|`| ψ0 = (a†b†)k z|`| ψ0 .

If j ≤ k then ψjk = (a†b†)j (b†)|`| ψ0 = (a†b†)j (z∗)|`| ψ0 .

The harmonic oscillator

It is convenient to split off ψ0 = e−|z|2

2 from every wave function, writing

ψ = ψψ0 .

This amounts to a nonunitary transformation, the scalar product is

〈φ|ψ〉 =

∫d2z φ∗ψ =

∫d2z e−|z|

2 (φ)∗ψ .

The derivative operators are transformed as follows,

∂ → e|z|2

2 ∂ e−|z|2

2 = ∂ − z∗

2, ∂∗ → ∂∗ − z

2.

The annihilation and creation operators acting on ψ are

a = ∂ , a† = −∂∗ + z , b = ∂∗ , b† = −∂ + z∗ .

The harmonic oscillator

The transformed Hamiltonian and angular momentum operators are

H = ~ω(a†a + b†b + 1) = ~ω (−2∂∂∗ + z∂ + z∗∂∗ + 1) ,

L = ~(a†a− b†b) = ~(z ∂ − z∗∂∗) = L .

To simplify the notation, from now on we write a, b instead of a, b.Thus we redefine the annihilation and creation operators as

a = ∂ , a† = −∂∗ + z , b = ∂∗ , b† = −∂ + z∗ .

The harmonic oscillator: two anyons

H =1

2m

(p 2

1 + p 22

)+

12

mω2 (r 21 + r 2

2

).

Introduce the centre of mass and relative coordinates

Z =1√2

(z1 + z2) , z =1√2

(z1 − z2) ,

and the corresponding operators

a = ∂Z , a† = −∂ ∗Z + Z , b = ∂ ∗Z , b† = −∂Z + Z∗ .

c = ∂z , c† = −∂ ∗z + z , d = ∂ ∗z , d† = −∂z + z∗ .

The centre of mass moves like a two-dimensional harmonic oscillatordescribed by the a and b operators.

We subtract its contributions to the energy and angular momentum, andconsider only the relative motion, described by the Hamiltonian

Hrel = ~ω(c†c + d†d + 1) = ~ω (−2∂z∂∗

z + z∂z + z∗∂ ∗z + 1) .

Two anyons, relative motion

An anticlockwise interchange of the particles transforms z→ eiπz .

The effect on the wave function ψ should be that ψ → eiθψ .The angle θ defines the anyon statistics.

The angular momentum `~ of the relative motion can vary continuouslywhen θ varies continuously. The wave functions

ψ` = ψ`ψ0 = z`ψ0 for ` ≥ 0 ,φ` = φ`ψ0 = (z∗)−`ψ0 for ` ≤ 0

are the eigenstates of lowest energy for a given value of `.They are finite at z = 0 (zero when ` 6= 0) and have statistics angle θ = `π .Hence

` = even integer for bosons ,` = odd integer for fermions .

Two anyons, relative motion

All other energy eigenstates are radial excitations of ψ` and φ` created byrepeated applications of the operator c†d† .

They have angular momentum `~ and energy (1 + |`|+ 2k)~ωwith k = 0, 1, 2, . . ..

States with ` ≥ 0:

ψ`k = (c†d†)k ψ` = (d†)k (c†)k ψ`

= (−∂z + z∗)k (−∂ ∗z + z)k z` = (−∂z + z∗)k z`+k .

States with ` ≤ 0:

φ`k = (c†d†)k φ` = (c†)k (d†)k φ`

= (−∂ ∗z + z)k (−∂z + z∗)k (z∗)−` = (−∂ ∗z + z)k (z∗)−`+k .

They are all good wave functions since they are finite at z = 0 .

Note that ψ0k = φ0k .

Two anyons, relative motion

The operators c, c†, d, d† are odd under the interchange 1↔ 2, hence theychange the anyon statistics angle from θ to θ ± π.

In particular, they change bosons into fermions and vice versa.

They give for example that

c ψ` = ∂z z` = −` z`−1 = −` ψ`−1 ,

c†ψ` = (−∂ ∗z + z) z` = z`+1 = ψ`+1 ,

d ψ` = −∂ ∗z z` = 0 ,

d†ψ` = (−∂z + z∗) z` = −` z`−1 + z∗z` = ψ`−1,1 .

The wave function z`−1 is unphysical when 0 < ` < 1 , it diverges as z→ 0.

The product c†d† is even and preserves the statistics.

Two anyons, relative motion

Energies as functions of the angular momentum.The dashed lines represent radially excited states. = boson states , = fermion states .

The harmonic oscillator: three bosons or fermions

H =1

2m

(p 2

1 + p 22 + p 2

3

)+

12

mω2 (r 21 + r 2

2 + r 23

).

Define

η = e2πi

3 , η3 = 1 , η2 + η + 1 = 0 , η2 = η∗ = η−1 ,

and introduce the centre of mass and relative coordinates

Z =1√3

(z1 + z2 + z3) ,

u =1√3

(z1 + ηz2 + η2z3) ,

v =1√3

(z1 + η2z2 + ηz3) .

This is a discrete Fourier transform.

Three bosons or fermions

We consider only the relative motion, and split off the bosonic ground state

ψ0 = e−|u|2+|v|2

2 .

Annihilation and creation operators corresponding to u and v are

a = ∂u , a† = −∂ ∗u + u , b = ∂ ∗u , b† = −∂u + u∗ .

c = ∂v , c† = −∂ ∗v + v , d = ∂ ∗v , d† = −∂v + v∗ .

The Hamiltonian of the relative motion is

Hrel = ~ω(a†a + b†b + c†c + d†d + 2)

= ~ω (−2(∂u∂∗

u + ∂v∂∗

v ) + u∂u + v∂v + u∗∂ ∗u + v∗∂ ∗v + 2) .

The relative angular momentum is

Lrel = ~ω(a†a− b†b + c†c− d†d)

= ~ω (u∂u + v∂v − u∗∂ ∗u − v∗∂ ∗v ) .

Three bosons or fermionsRecall the definitions

u =1√3

(z1 + ηz2 + η2z3) , v =1√3

(z1 + η2z2 + ηz3) .

The cyclic permutation z1 → z2 → z3 → z1 gives

u→ η2u = η−1u , v→ ηv ,and hence

a† → η2 a† , b† → η b† , c† → η c† , d† → η2 d† .

A state of energy (p + q + r + s + 2)~ωand angular momentum (p− q + r − s)~ might be

ψpqrs = (a†)p (b†)q (c†)r (d†)s 1 .

The effect of the cyclic permutation is that ψpqrs → η−p+q+r−s ψpqrs .

Bosonic and fermionic three-particle wave functions must both besymmetric under cyclic permutations, hence they must have

p + s ≡ q + r (mod 3) .

Three bosons or fermions

In order to separate the bosonic and fermionic states we consider theinterchange z2 ↔ z3 , which is the same as u↔ v , or (p, q)↔ (r, s) .

In conclusion, when p, q, r, s are nonnegative integers withp + s ≡ q + r (mod 3) we have a bosonic state ψpqrs + ψrspq.

When (p, q) 6= (r, s) we have also a fermionic state ψpqrs − ψrspq.

The energy and angular momentum of these states are(p + q + r + s + 2)~ω and (p− q + r − s)~ .

The bosonic ground state is ψ0000 = 1 .

The fermionic ground state is also nondegenerate, it is

ψ1100 − ψ0011 = |u|2 − |v|2

= − i2√

3(z1z∗2 + z2z∗3 + z3z∗1 − z2z∗1 − z3z∗2 − z1z∗3) .

Three anyons

Define a statistics parameter ν = θ/π, allowed to vary continuously.

The eigenstates and eigenvalues of energy and angular momentum must varycontinuously with ν.

The angular momentum `~ must vary as ` = `0 + 3ν where `0 is an integer.

We see this by rotating an angle 2π, this gives a phase 2`π, or equivalently6θ = 6νπ since it is a double interchange of three particle pairs.

It is therefore natural to think in terms of trajectories, like Regge trajectories,where the energy varies with a continuously variable angular momentum.

From a boson (or fermion) point to the next fermion (or boson) point thechanges in ν and ` are ∆ν = 1 and ∆` = 3.The change in energy is ∆E = ±1 or ∆E = ±3.Hence the average slope over this interval is ∆E/∆ν = ±1 or ±3.

Three anyons

When the average slope is ±3, the energy varies linearly with ν and `, andthe wave functions are exactly known.

When the average slope is ±1, the energy varies nonlinearly, and there is nocase where the wave functions are exactly known.

This implies in particular that the anyonic states close to the fermionicground state are not exactly known.

The operator a†b† + c†d† produces radial excitations.It works on all states, both “linear” and “nonlinear”, building “towers” withenergies increasing in steps of 2~ω.

We may therefore simplify the classification of trajectories by consideringonly those that are bottoms of towers.

Three anyons

A trajectory can be followed through the degeneracies at boson and fermionpoints because the wave function must be a continuous function of θ.

Every trajectory has slope−3 for large negative ` and +3 for large positive `.

With ` increasing from −∞, at some boson point the slope changes to −1.

Next, at some boson or fermion point the slope changes to +1.

Finally, at some boson point the slope changes to +3.

There is one trajectory, through the boson ground state, that changes slopedirectly from −3 to +3.

Other trajectories skip one of the sections with slope ±1.

The following plots show the bottom of tower trajectories, worked out byStefan Mashkevich.

Three anyons: supersymmetryDiptiman Sen discovered the supersymmetry operator

Q = a†d − c†b = u∂ ∗v − v∂ ∗u ,

which preserves the energy but changes the angular momentum by two units.

It is symmetric under the cyclic permutation z1 → z2 → z3 → z1 ,which transforms u→ η2u, v→ ηv and ∂ ∗u → η2∂ ∗u , ∂ ∗v → η∂ ∗v ,but is antisymmetric under the transposition z2 ↔ z3, equivalent to u↔ v.

This operator annihilates some of the exactly known “linear” energyeigenstates, but Sen argued that for all other anyonic states it produces“supersymmetric” anyonic states with the same energy but statistics angleθ ± π instead of θ.

For example, the fermionic ground state |u|2 − |v|2 has supersymmetricbosonic partner states uv and (uv)∗.

Sen also pointed out that this (partial) supersymmetry of the three-anyonspectrum implies that the third virial coefficent for anyons is symmetricabout the semion point θ = π/2, midway between bosons and fermions.

Trajectories, with the bosonic ground state

Trajectories, with the fermionic ground state

Trajectories

Trajectories

Trajectories

Trajectories

The harmonic oscillator: partition functions

Notation: β = 1/(kT) , ξ = β~ω .

One particle in two dimensions: E = (1 + j + k)~ω , j, k = 0, 1, 2, . . ..

Z1(β) =∑

e−βE =1

4 sinh2( ξ2

) .Two bosons: E = ECM + Erel ,Erel = (1 + |`|+ 2k)~ω , ` = . . . ,−2, 0, 2, . . . , k = 0, 1, 2, . . . .

Z2B(β) = ZCM(β) Zrel(β) =1

4 sinh2( ξ2

) cosh ξ2 sinh2ξ

.

Two fermions: E = ECM + Erel ,Erel = (1 + |`|+ 2k)~ω , ` = . . . ,−3,−1, 1, 3, . . . , k = 0, 1, 2, . . . .

Z2F(β) = ZCM(β) Zrel(β) =1

4 sinh2( ξ2

) 12 sinh2ξ

.

Partition functions

Two distinguishable particles:

Z2D(β) = (Z1(β))2 = Z2B(β) + Z2F(β) .

Two anyons, θ = νπ: E = ECM + Erel ,Erel = (1 + |`|+ 2k)~ω , ` = . . . , ν − 2, ν, ν + 2, . . . , k = 0, 1, 2, . . . .

Z2(θ, β) = ZCM(β) Zrel(β) =1

4 sinh2( ξ2

) cosh((1− α)ξ)

2 sinh2ξ.

The two-anyon partition function is a function of the periodic sawtoothfunction α(θ), α = 0 for bosons and α = 1 for fermions.

Note that Z2 is an analytic function of θ at the fermion points.

Partition functions

We write the two-anyon partition function as

Z2(θ, β) =12(F11(θ, β) (Z1(β))2 + F2(θ, β) Z1(2β)

)where F11 is symmetric and F2 antisymmetric between bosons and fermions,

F11(θ, β) =cosh

((α− 1

2

)ξ)

cosh(ξ2

) , F2(θ, β) = −sinh((α− 1

2 )ξ)

sinh(ξ2

) .

In the path integral formalism the two particle paths wind around each other,and the functions F11 and F2 are probability generating functions for thedistributions of winding numbers.

Path integrals

Consider two distinguishable particles in two dimensions.

The position eigenstates |r1, r2〉 form a basis for the Hilbert space.They satisfy the orthogonality relation

〈s1, s2|r1, r2〉 = δ(2)(s1 − r1) δ(2)(s2 − r2)

and the completeness relation

I =

∫(R2)2

d2r1 d2r2 |r1, r2〉〈r1, r2| .

The boson and fermion position eigenstates

|r1, r2〉B,F =1√2

(|r1, r2〉 ± |r2, r1〉)

together form an alternative basis for the Hilbert space.

Path integrals

Consider two bosons.

The position eigenstates |r1, r2〉B satisfy the orthogonality relation

B〈s1, s2|r1, r2〉B = δ(2)(s1 − r1) δ(2)(s2 − r2) + δ(2)(s1 − r2) δ(2)(s2 − r1)

and the completeness relation, where IB is the projection on boson states,

IB =

∫(R2)2/S2

d2r1 d2r2 |r1, r2〉BB〈r1, r2|

=12

∫(R2)2

d2r1 d2r2 |r1, r2〉BB〈r1, r2|

=12

∫(R2)2

d2r1 d2r2(|r1, r2〉〈r1, r2|+ |r1, r2〉〈r2, r1|

).

(R2)2/S2 is the true configuration space, with (r1, r2) and (r2, r1) identified.

The fermionic projection IF is the same, but with − instead of +.Hence IB + IF = I.

Path integrals

The bosonic or fermionic partition function is

Z2B,F(β) = Tr(IB,F e−βH)

=12

∫(R2)2

d2r1 d2r2(〈r1, r2|e−βH|r1, r2〉 ± 〈r2, r1|e−βH|r1, r2〉

).

The matrix elements are two-particle propagators in imaginary time τ = ~β.They are products of one-particle propagators if the particles do not interact,

Z2B,F(β) =12

∫(R2)2

d2r1 d2r2(G(r1, r1; τ) G(r2, r2; τ)

±G(r2, r1; τ) G(r1, r2; τ))

=12

((∫R2

d2r G(r, r; τ)

)2

±∫R2

d2r G(r, r; 2τ)

)

=12((Z1(β))2 ± Z1(2β)

).

Path integrals

The one-particle propagator for the harmonic oscillator is

G(s, r; τ)mω

2π~ sinh(ωτ)exp(−mω

4~[

tanh(ωτ2 ) |s + r|2 + coth(ωτ2 ) |s− r|2]).

The limit ω → 0 is the free particle propagator

G(s, r; τ) =m

2π~τexp(− m

2~τ|s− r|2

).

We use these expressions to Monte Carlo generate paths, with a harmonicoscillator external potential or with no external potential.

For a closed path over an imaginary time interval of length L~β we generatethe first point r with a probability distribution G(r, r; L~β).

Given two points r1 at τ = τ1 and r2 at τ = τ2 we generate an intermediatepoint r at τ with a probability distribution G(r2, r; τ2 − τ) G(r, r1; τ − τ1).

N anyonsThe generalization from two to N anyons is the formula

ZN(θ, β) =∑P

FP(θ, β)∏

L

1νL!

(Z1(Lβ)

L

)νL

.

Here P = (ν1, ν2, ν3, . . .) is a partition of N, with νL = 0, 1, 2, . . . and∞∑

L=0

LνL = ν1 + 2ν2 + 3ν3 + · · · = N .

Example:

Z3(θ, β) =16

F111(θ, β) (Z1(β))3 +13

F3(θ, β) Z1(3β)

+12

F21(θ, β) Z1(2β)Z1(β) ,

from the partitions of 3:

(3, 0, 0, 0, . . .) = 1 + 1 + 1 ,(0, 0, 1, 0, . . .) = 3 ,(1, 1, 0, 0, . . .) = 2 + 1 .

Three anyons

The three terms of Z3 correspond to the permutations in the figure.

The first two are even permutations, the third is odd.

6τ

0

~β

-rj

j = 1 2 3

r r rr r r

-rj

1 2 3

HHHH

HH

r r rr r r

-rj

1 2 3

@@@r r rr r r

The winding number Qjk of a pair jk is their winding angle divided by π.

In case 1, all three winding numbers Q12, Q13, Q23 are even integers.In case 2, no single Qjk is an integer, but the sum Q = Q12 + Q13 + Q23 is aneven integer.In case 3, Q12 is an odd integer and Q(12)3 = Q13 + Q23 is an even integer.

Q = Q12 + Q13 + Q23 is an even/odd integer for an even/odd permutation.

Three anyons

The figure illustrates a three-particle path (r1(τ), r2(τ), r3(τ)) inducing acyclic permutation, and the same path represented as a closed one-particlepath over three times the imaginary time interval.

6τ

0

~β

2~β

3~β

-rj

HHH

HHHr r r

r r r-

r1H

HHHHH

r r rr r rr r rr r r

Three anyons

For a given partition P , and a given β, we can Monte Carlo generate paths tofind the probability distribution PP(Q, β) of the winding numbers Q.

To each path is associated the phase factor eiQθ.Thus FP is the probability generating function

FP(θ, β) =∑

Q

PP(Q, β) eiQθ .

Time reversal invariance ot the one-particle propagator implies thatP(−Q) = P(Q), and hence F is real.

Three anyons

Recall the exact results for two anyons,

F11(θ) =cosh

((α− 1

2

)ξ)

cosh(ξ2

) , F2(θ) = −sinh((α− 1

2 )ξ)

sinh(ξ2

) .

They define the probability distributions P11 for Q even and P2 for Q odd,

P11(Q) = −2ξ tanh

(ξ2

)ξ2 + (πQ)2 , P2(Q) = −

2ξ coth(ξ2

)ξ2 + (πQ)2 .

These distributions have very long tails, their standard deviations are infinite.The tails are there because F11(θ) and F2(θ) have discontinuous derivativesat the boson and fermion points.

In the Monte Carlo simulations we got the large winding numbers fromwindings at extremely small distances, down to 10−300, which is the limit ofthe computer hardware.

This shows that the theory does not apply directly to anyons in the realworld, which are collective excitations and definitely not point particles.

Three anyonsOne exact result for three anyons is also known,

F21(θ) = −sinh((α− 1

2 )3ξ)

sinh( 3ξ

2

) ω→0−−−→ 1− 2α .

Diptiman Sen was able to compute it exactly because it has contributionsonly from the “linear” states, with energies varying linearly with θ.

For F3 only the free-particle limit is exactly known,

F3(θ)ω→0−−−→ −1

8+

92

(α− 1

2

)2

= (1− 3α)

(1− 3

2α

).

Alain Dasnières de Veigy proved to all orders in perturbation theory (!) thegeneral formula for an L-cycle:

FL(θ)ω→0−−−→

L−1∏k=1

(1− Lα

k

).

This result is all the more remarkable because there are contributions from“nonlinear” states.

Free anyons

We consider now anyons in zero external potential.In order to have finite partition functions we impose periodic boundaryconditions in a square box of area A.

Define the thermal de Broglie wave length Λ = ~√

2πβ/m .The one-particle partition function is

Z1(β) =

[ ∞∑n=−∞

exp(−πn2Λ2

A

)]2

=AΛ2

[1 + 2

∞∑n=1

exp(−πn2A

Λ2

)]2

.

The last expression is a Poisson resummation, by Fourier expansion of

f (x) =

∞∑n=−∞

exp(−π(n + x)2Λ2

A

),

a periodic function of x. Up to exponentially small correction terms we have

Z1(β) =AΛ2 =

Am2π~2β

.

Statistical mechanics

Define z = eβµ where µ is the chemical potential.Then the grand canonical partition function is

Ξ(β, z) = 1 +

∞∑N=1

zNZN(β) .

Let CN be the set of all partitions of N, C = ∪∞N=0CN , C′ = ∪∞N=1CN .

We write one partition of N as P = (ν1, ν2, ν3, . . .) with∑

L LνL = N, then∑P∈C

=

∞∑ν1=0

∞∑ν2=0

∞∑ν3=0

· · ·

In this notation we have, with ν =∑

L νL,

ln Ξ =

∞∑ν=1

(−1)ν−1

ν

( ∞∑L=1

zLZL

)ν=∑P∈C′

(−1)ν−1(ν − 1)!∞∏

L=1

(zLZL)νL

νL!.

Statistical mechanicsFrom the grand canonical partition function we get the pressure P,

AβP = ln Ξ .

We get directly the cluster expansion

βP =

∞∑n=1

bnzn

with the cluster coefficients

bN =1A∑P∈CN

(−1)ν−1(ν − 1)!

∞∏L=1

Z νLL

νL!.

Introducing the number density

ρ = z∂(βP)

∂z=

∞∑n=1

nbnzn

we get the virial expansion

βP = ρ+

∞∑n=2

Anρn .

Statistical mechanics

The first cluster coefficients are given as

Ab1 = Z1 =AΛ2 ,

Ab2 = Z2 −Z 2

1

2,

Ab3 = Z3 − Z2Z1 +Z 3

1

3,

Ab4 = Z4 − Z3Z1 −Z 2

2

2+ Z2Z 2

1 −Z 4

1

4.

The first virial coefficients are given by the cluster coefficients as

A2 = −Λ4b2 ,

A3 = −2Λ6b3 + 4Λ8b 22 ,

A4 = −3Λ8b4 + 18Λ10b2b3 − 20Λ12b 32 .

Statistical mechanics

Remember now our formula for the N-anyon partition function in terms ofthe one-particle partition function:

ZN(θ, β) =∑P

FP(θ, β)∏

L

1νL!

(Z1(Lβ)

L

)νL

.

It gives a formula for the cluster coefficients:

bN =Z1(β)

A∑P∈CN

GP∞∏

L=1

1νL!

(Z1(Lβ)

LZ1(β)

) νL

in terms of a new set of coefficients

GP = (FP + · · · )(Z1(β))ν−1 .

Here · · · represents terms that are products of “F” coefficients.With the two-dimensional relation Z1(Lβ) = Z1(β)/L we get that

bN =1

Λ2

∑P∈CN

GP∞∏

L=1

1L2νL νL!

.

Statistical mechanics

The G coefficients for L-cycles are

GL = FLA→∞−−−−→

L−1∏k=1

(1− Lα

k

).

Other coefficients are

G11 = (F11 − 1) Z1A→∞−−−−→ α(α− 1) ,

G111 = (F111 − 3F11 + 2) Z 21 ,

G21 = (F21 − F2) Z1A→∞−−−−→ 2(1− 2α)α(α− 1) ,

G1111 = (F1111 − 4F111 − 3F 211 + 12F11 − 6) Z 3

1 ,

G211 = (F211 − 2F21 − F2F11 + 2F2) Z 21 ,

G22 = (F22 − F 22 ) Z1 ,

G31 = (F31 − F3) Z1 .

Statistical mechanics

All the G coefficients tend to finite limits when A →∞.To see why, take as an example

G211 = (F211 − 2F21 − F2F11 + 2F2) Z 21 ∝

∑q

p(q) eiqθ .

One Monte Carlo generated path, interchanging particles 1 and 2,contributes one odd integer winding number Q12 and three even integerwinding numbers

Q34 , Q(12)3 = Q13 + Q23 , Q(12)4 = Q14 + Q24 .

It contributes to F211 by the total winding number

Q = Q12 + Q34 + Q(12)3 + Q(12)4 .

It contributes to F21 by the two winding numbers

Qa = Q12 + Q(12)3 , Qb = Q12 + Q(12)4 .

It contributes to F2F11 by the winding number Qc = Q12 + Q34 .

And it contributes to F2 by the winding number Q12 .

Statistical mechanics

The formula

G211 = (F211 − 2F21 − F2F11 + 2F2) Z 21 ∝

∑q

p(q) eiqθ .

tells us to that for each Monte Carlo generated path we should update theFourier components p(q) as follows:

– add 1 to p(Q) ;

– subtract 1 from each of p(Qa) , p(Qb) , and p(Qc) ;

– add 2 to p(Q12) .

Recall that

Qa = Q12 + Q(12)3 , Qb = Q12 + Q(12)4 , Qc = Q12 + Q34 .

The net result of this updating is null if more than one of the three evenwinding numbers Q34,Q(12)3,Q(12)4 is zero.

For example, if Q(12)3 = Q34 = 0 , then Qa = Qc = Q12 and Qb = Q .

Statistical mechanics

In general, an N-particle path contributes to a coefficient GP only if all Nparticles wind in one cluster.

In this sense, GP is the connected part of FP .

As usual, the grand partition function Ξ is a sum of all diagrams, whereasln Ξ is a sum of connected diagrams.

Since the partition P consists of ν = ν1 + ν2 + · · · parts scattered withuniform probability over the area A, the probability for these parts stayingtogether goes to zero as A1−ν when A →∞.

This implies that GP goes to a finite limit.

Statistical mechanics

Explicit expressions for the first cluster coefficients are

Λ2b2 =G11

2+

G2

4,

Λ2b3 =G111

6+

G21

4+

G3

9,

Λ2b4 =G1111

24+

G211

8+

G22

32+

G31

9+

G4

16.

Our Monte Carlo method for computation actually proves the nontrivialresult that the cluster and virial coefficients are finite in the thermodynamiclimit.

Exact results

The second cluster and virial coefficients are exactly known, from the exactsolution of the two-anyon problem:

A2 = −Λ4b2 = Λ2(−1

4+

1− (1− α)2

2

).

Alain Dasnières de Veigy and Stéphane Ouvry computed the pressure P as afunction of z to second order in θ from the boson and fermion points.

Their result proves that all cluster and virial coefficients do depend on θ.

The θ dependence of the cluster coefficients is analytic (at least to secondorder) at the fermion point but nonanalytic at the boson point (because thefirst order correction is proportional to |θ|).

It is possible, however, that the θ dependence of the virial coefficients An

with n > 2 is analytic both at the fermion point and at the boson point (sincethere is no θ dependence to first order).

Approximate results

The most precise calculation of the third virial coefficient gave the result

A3 = Λ4(

136

+sin2θ

12π2 + a sin4θ

)with

a = −(1.652± 0.012)× 10−5 = − 1(621± 5)π4 .

The coefficient of the sin2θ term is the exact perturbative result.It is an important consistency check that it is reproduced numerically.

The numerical result is from a calculation of (many!) three-anyon harmonicoscillator energy levels carried out by Stefan Mashkevich.

All the θ dependence of A3 is due to the bottom of tower energy levels thatvary nonlinearly with θ.

The boson–fermion supersymmetry implies that A3 is analytic in θeverywhere, hence a Fourier expansion like this is to be expected.

Approximate results

A Monte Carlo calculation of the fourth virial coefficient gave the result

A4 = Λ6(

sin2θ

16π2

(1√3

ln(√

3 + 2)

+ cos θ)

+ (c4 + d4 cos θ) sin4θ

)with

c4 = −0.0053± 0.0003 , d4 = −0.0048± 0.0009 .

Again the exact perturbative result is incorporated.

Approximate results

The Monte Carlo calculation with four anyons suggested a polynomialapproximation for the G coefficients:

GP ' GP = Nν−2G ν−111

∏L

(LFL)νL

with

G11 = α(α− 1) , FL =

L−1∏k=1

(1− Lα

k

).

This is definitely only an approximation, but one can give heuristicarguments in support.

The four-particle approximate polynomials are

G1111 = 16G 311 , G211 = 8F2G 2

11 , G22 = 4F 22 G11 , G31 = 3F3G11 .

Monte Carlo, partition 4 = 1 + 1 + 1 + 1

G1111 − 16(α(α− 1))3 as a function of α. The curve marked “fit” is

− 3π2 α(α− 1) sin2(απ) .

Partition 4 = 2 + 1 + 1

G211 − 8(1− 2α)(α(α− 1))2 as a function of α. The curve marked “fit” is

− 23π2 (1− 2α) sin2(απ) .

Partition 4 = 2 + 2

G22 − 4(1− 2α)2α(α− 1) as a function of α. The curve marked “fit” is

2√3π2

ln(√

3 + 2) sin2(απ) cos2(απ) .

Partition 4 = 3 + 1

G31 − 3(1− 3α)(1− (3/2)α)α(α− 1) as a function of α. The curvemarked “fit” is

√3

4π2 ln(√

3 + 2) sin2(απ) cos2(απ) .

The cluster coefficient b4

The fourth cluster coefficient minus the polynomial approximation,Λ2(b4 − b4), as a function of α.The second order perturbation theory at α = 0 and α = 1 gives theparabolas shown.

The virial coefficient A4

The fourth virial coefficient, A4/Λ6, as a function of α.

The second order perturbation theory at α = 0 and α = 1 gives theparabolas shown.Two fits by Fourier series are shown, one with coefficients c4 = d4 = 0 andone “best fit” with c4 = −0.0053 , d4 = −0.0048 .

Approximate resultsThrough some remarkable polynomial identities, proved by Kåre Olaussen,these polynomial approximations for GP give the following polynomialapproximations for the cluster coefficients,

Λ2bN =1

N2

N−1∏k=1

(1− Ng

k

)with g = 1− (1− α)2 . These cluster coefficients give virial coefficientsthat are independent of the statistics parameter θ, or g, except for

A2 = Λ2(−1

4+

g2

),

which is the exact A2 for anyons.

It so happens that what we have now derived as an approximation for anyonsis the exact result for two-dimensional exclusion statistics with statisticsparameter g.

The quadratic dependence of g on θ is different from the linear dependencefor anyons in a strong magnetic field.

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