Exchange symmetry. Jan Myrheim. April 12, Department of physics, NTNU

Transcription

1 Exchange symmetry Jan Myrheim Department of physics, NTNU April 12, 2016

2 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

3 Heisenberg and Dirac (1926) Definition: N particles, numbered 1 to N, are identical if all observables are symmetric under permutations of indices. A symmetric operator X representing an observable preserves the symmetry of the wave function ψ, ψ Xψ, S SS = S, A SA = A. We can have two different complete theories of quantum mechanics for identical particles. (I) Using only symmetric wave functions. (II) Using only antisymmetric wave functions. Heisenberg and Dirac argue that these two possibilities exist, but make no attempt to prove that other possibilities do not exist.

4 Statistics In case (I), symmetric wave functions, counting of states leads to Bose Einstein statistics. In case (II), antisymmetric wave functions, the Pauli exclusion principle holds, and counting of states leads to Fermi Dirac statistics. If we admit all wave functions, without imposing symmetry or antisymmetry, 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 entropy is extensive (proportional to N at constant density). With Maxwell Boltzmann statistics, the number of states grows too fast with N for the thermodynamic limit to exist.

5 Statistics In both classical and quantum mechanics we need particles to be identical because we want a thermodynamic limit. In classical statistical mechanics the solution is to divide the phase space volume by N! In quantum statistical mechanics the solution is to symmetrize or antisymmetrize 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 electrons around. Two electrons at different positions are identical, but distinguishable. We need not antisymmetrize if we, for example, scatter electrons in such a way that we can follow one electron from the initial to the final state.

6 Scattering process detector p p θ p p Incoming momenta p 1 = p 2 = ±p. Outgoing momenta p 3 = p 4 = ±p. Scattering angle α = θ or π θ, definition: p 1 p 3 = p p cos α. Scattering amplitude f = f (α) because of rotational invariance. f (α) = out F in.

7 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 dσ = f (θ) 2 dω we count the outgoing protons, but not the neutrons. If our detector counts both protons and neutrons, without distinction, we would have to redefine dσ dω = 1 ( f (θ) 2 + f (π θ) 2), 2 dividing by two to compensate for the double counting.

8 Isospin In 1932, immediately after the discovery of the neutron, Heisenberg suggested 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 I 3 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, I 3 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 ).

9 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 +.

10 Scattering Conservation of isospin, or equivalently, conservation of the symmetry of the momentum state, implies that we have two different scattering amplitudes. One with isospin I = 1, f (θ) = p F p And one with isospin I = 0, = 1 2( 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). f + (θ) = p + F p+ = 1 2( 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).

11 Scattering We compute the cross sections as dσ ± dω = 1 2 f ±(θ) 2 = 1 ( f (θ) 2 + f (π θ) 2) ± Re ( f (θ)f (π θ) ), 2 dividing by two to compensate for double counting. Conclusions: Proton neutron scattering can take place via one of two channels: either symmetric in momentum (I = 0) or antisymmetric in momentum (I = 1). The cross section we compute by treating them as different particles is the average of the symmetric and the antisymmetric channel, dσ dω = 1 ( f (θ) 2 + f (π θ) 2) = 1 ( dσ+ 2 2 dω + dσ ). dω

12 Scattering Even though a proton is different from a neutron, we may choose to prepare the 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 harmonic oscillators. That is how we derive QED from Maxwell s equations. Or if we regard the quantum theory as more fundamental, we want photons to be bosons because we want Maxwell s equations as a classical limit.

13 Is isospin only a book keeping trick? In the case of protons and neutrons, we may regard isospin as a tool for keeping 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 the scattering of π mesons on nucleons. In the quark model, 2I 3 is the number of u quarks minus the number of d quarks. Isospin conservation can be understood as an approximate SU(2) symmetry, which is broken because the u and d quarks have different electric charge and different mass. We treat all six quarks u, d, c, s, t, b as identical fermions distinguished by different values of quantum numbers, called flavours, similar to isospin. However, the SU(6) symmetry extending the SU(2) isospin symmetry is badly broken by the electroweak interaction and by the quark mass matrix. The quark masses are free parameters in the so called standard model.

14 Colour The quark model nearly died in its infancy because it seemed to require quarks to be bosons, instead of fermions as required by the spin statistics theorem. 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 system we 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, but satisfied parastatistics: the N-particle wave functions would belong to more general representations of the permutation group S N. Greenberg solved the problem in 1964 simply by postulating a new three-valued quantum number, later called colour, now the basis of QCD. The quarks are fermions with an antisymmetric colour wave function, then the rest of the wave function must be symmetric.

15 Which particles are identical? Consider the Hamiltonian of the hydrogen atom, again disregarding spin, H = p p 2 2 k 2m 1 2m 2 r 1 r 2, k = q 1q 2. 4πɛ 0 It is symmetric under the interchange 1 2, a fact which becomes even more obvious when we split off the centre of mass motion and write H = P2 2M + p2 2m k r, with M = m 1 + m 2, m = m 1 m 2 /M, P = p 1 + p 2, r = r 1 r 2, p = m 2p 1 m 1 p 2 M.

16 Which particles are identical? The Hamiltonian is symmetric, what about other observables? The electric dipole moment is symmetric, it is d = e(r p r e ) = q 1 r 1 + q 2 r 2, independent of how we label the electron and the proton as particle 1 and 2. The electron position is symmetric, it is r e = δ 1e r 1 + δ 2e r 2. All observables are symmetric under the interchange 1 2. They have to be, because the labels are arbitrary, different labellings just create 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?

17 particle identity The configuration space proton 2 1 electron 1 2 particle position particle position There is an electron at position r e and a proton at position r p. 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 a coordinate system covering the configuration space completely and uniquely, as the left part of the figure indicates. The right part of the figure shows the alternative labelling, electron = 2 and proton = 1. It gives nothing new.

18 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 function does not arise. It may seem that the essential point is that there is some discrete quantum number distinguishing protons from electrons. Because of the discreteness, we get no nontrivial topology when we identify two parts of the configuration space differing only in the labelling of particles. Then we use only one of the two separate copies of the configuration space, and forget about symmetrization or antisymmetrization.

19 The harmonic oscillator One particle in two dimensions described by a complex coordinate. = z = λ 2 H = p2 2m mω2 r 2. z = x + iy, λ = λ ( x i y mω, ), = z = λ 2 a = + z 2, a = + z 2, b = + z 2, b = + z 2. ( x + i ). y [a, a ] = [b, b ] = 1, [a, b] = [a, b ] = [a, b] = [a, b ] = 0.

20 The harmonic oscillator a, a, b, b create and annihilate quanta of energy and angular momentum, ) H = ω ( 2 + z 2 = ω(a a + b b + 1), 2 ( L = i x y y ) = (z z ) = (a a b b). x Ground state: ψ 0 = e z 2 2, aψ 0 = bψ 0 = 0. Excited states: ψ jk = (a ) j (b ) k ψ 0. Hψ jk = (j + k + 1) ω ψ jk, Lψ jk = l ψ jk, l = j k. Maximally rotating states: (a ) j ψ 0 = z j ψ 0, (b ) k ψ 0 = (z ) k ψ 0. a b creates radial excitations without changing the angular momentum. If j k then ψ jk = (a b ) k (a ) l ψ 0 = (a b ) k z l ψ 0. If j k then ψ jk = (a b ) j (b ) l ψ 0 = (a b ) j (z ) l ψ 0.

21 The harmonic oscillator It is convenient to split off ψ 0 = e z 2 2 from every wave function, writing ψ = ψψ 0. This amounts to a nonunitary transformation, the scalar product is φ ψ = d 2 z φ ψ = d 2 z e ( φ) z 2 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 ã =, ã = + z, b =, b = + z.

22 The harmonic oscillator The transformed Hamiltonian and angular momentum operators are H = ω(ã ã + b b + 1) = ω ( 2 + z + z + 1), L = (ã ã b b) = (z z ) = L. To simplify the notation, from now on we write a, b instead of ã, b. Thus we redefine the annihilation and creation operators as a =, a = + z, b =, b = + z.

23 The harmonic oscillator: two anyons H = 1 ( p 2 2m 1 + p2 2 ) 1 ( + 2 mω2 r1 2 + r2 2 ). Introduce the centre of mass and relative coordinates Z = 1 2 (z 1 + z 2 ), z = 1 2 (z 1 z 2 ), 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 oscillator described by the a and b operators. We subtract its contributions to the energy and angular momentum, and consider only the relative motion, described by the Hamiltonian H rel = ω(c c + d d + 1) = ω ( 2 z z + z z + z z + 1).

24 Two anyons, relative motion An anticlockwise interchange of the particles transforms z e iπ z. The effect on the wave function ψ should be that ψ e iθ ψ. The angle θ defines the anyon statistics. The angular momentum l of the relative motion can vary continuously when θ varies continuously. The wave functions ψ l = ψ l ψ 0 = z l ψ 0 for l 0, φ l = φ l ψ 0 = (z ) l ψ 0 for l 0 are the eigenstates of lowest energy for a given value of l. They are finite at z = 0 (zero when l 0) and have statistics angle θ = lπ. Hence l = even integer for bosons, l = odd integer for fermions.

25 Two anyons, relative motion All other energy eigenstates are radial excitations of ψ l and φ l created by repeated applications of the operator c d. They have angular momentum l and energy (1 + l + 2k) ω with k = 0, 1, 2,.... States with l 0: ψ lk States with l 0: = (c d ) k ψl = (d ) k (c ) k ψl = ( z + z ) k ( z + z) k z l = ( z + z ) k z l+k. φ lk = (c d ) k φl = (c ) k (d ) k φl = ( z + z) k ( z + z ) k (z ) l = ( z + z) k (z ) l+k. They are all good wave functions since they are finite at z = 0. Note that ψ 0k = φ 0k.

26 Two anyons, relative motion The operators c, c, d, d are odd under the interchange 1 2, hence they change 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 l = z z l = l z l 1 = l ψ l 1, c ψl d ψ l = ( z + z) z l = z l+1 = ψ l+1, = z z l = 0, d ψl = ( z + z ) z l = l z l 1 + z z l = ψ l 1,1. The wave function z l 1 is unphysical when 0 < l < 1, it diverges as z 0. The product c d is even and preserves the statistics.

27 Two anyons, relative motion Energies as functions of the angular momentum. The dashed lines represent radially excited states. = boson states, = fermion states.

28 The harmonic oscillator: three bosons or fermions Define H = 1 ( p 2 2m 1 + p2 2 + p 3 2 ) 1 ( + 2 mω2 r1 2 + r2 2 + r3 2 ). η = e 2πi 3, η 3 = 1, η 2 + η + 1 = 0, η 2 = η = η 1, and introduce the centre of mass and relative coordinates This is a discrete Fourier transform. Z = 1 3 (z 1 + z 2 + z 3 ), u = 1 3 (z 1 + ηz 2 + η 2 z 3 ), v = 1 3 (z 1 + η 2 z 2 + ηz 3 ).

29 Three bosons or fermions We consider only the relative motion, and split off the bosonic ground state ψ 0 = e u 2 + v 2 2. Annihilation and creation operators corresponding to u and v are a = u, c = v, a = u + u, b = u, b = u + u. c = v + v, d = v, d = v + v. The Hamiltonian of the relative motion is H rel = 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 L rel = ω(a a b b + c c d d) = ω (u u + v v u u v v ).

30 Recall the definitions Three bosons or fermions u = 1 3 (z 1 + ηz 2 + η 2 z 3 ), v = 1 3 (z 1 + η 2 z 2 + ηz 3 ). The cyclic permutation z 1 z 2 z 3 z 1 and hence gives u η 2 u = η 1 u, v ηv, a η 2 a, b η b, c η c, d η 2 d. A state of energy (p + q + r + s + 2) ω and angular momentum (p q + r s) might be ψ pqrs = (a ) p (b ) q (c ) r (d ) s 1. The effect of the cyclic permutation is that ψpqrs η p+q+r s ψ pqrs. Bosonic and fermionic three-particle wave functions must both be symmetric under cyclic permutations, hence they must have p + s q + r (mod 3).

31 Three bosons or fermions In order to separate the bosonic and fermionic states we consider the interchange z 2 z 3, which is the same as u v, or (p, q) (r, s). In conclusion, when p, q, r, s are nonnegative integers with p + s q + r (mod 3) we have a bosonic state ψ pqrs + ψ rspq. When (p, q) (r, s) we have also a fermionic state ψ pqrs ψ rspq. 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 ψ 0000 = 1. The fermionic ground state is also nondegenerate, it is ψ 1100 ψ 0011 = u 2 v 2 = i 2 3 (z 1z 2 + z 2 z 3 + z 3 z 1 z 2 z 1 z 3 z 2 z 1 z 3).

32 Three anyons Define a statistics parameter ν = θ/π, allowed to vary continuously. The eigenstates and eigenvalues of energy and angular momentum must vary continuously with ν. The angular momentum l must vary as l = l 0 + 3ν where l 0 is an integer. We see this by rotating an angle 2π, this gives a phase 2lπ, or equivalently 6θ = 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 the changes in ν and l are ν = 1 and l = 3. The change in energy is E = ±1 or E = ±3. Hence the average slope over this interval is E/ ν = ±1 or ±3.

33 Three anyons When the average slope is ±3, the energy varies linearly with ν and l, and the wave functions are exactly known. When the average slope is ±1, the energy varies nonlinearly, and there is no case where the wave functions are exactly known. This implies in particular that the anyonic states close to the fermionic ground 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 with energies increasing in steps of 2 ω. We may therefore simplify the classification of trajectories by considering only those that are bottoms of towers.

34 Three anyons A trajectory can be followed through the degeneracies at boson and fermion points because the wave function must be a continuous function of θ. Every trajectory has slope 3 for large negative l and +3 for large positive l. With l 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 slope directly 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 by Stefan Mashkevich.

35 Three anyons: supersymmetry Diptiman 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 z 1 z 2 z 3 z 1, which transforms u η 2 u, v ηv and u η 2 u, v η v, but is antisymmetric under the transposition z 2 z 3, equivalent to u v. This operator annihilates some of the exactly known linear energy eigenstates, but Sen argued that for all other anyonic states it produces supersymmetric anyonic states with the same energy but statistics angle θ ± π instead of θ. For example, the fermionic ground state u 2 v 2 has supersymmetric bosonic partner states uv and (uv). Sen also pointed out that this (partial) supersymmetry of the three-anyon spectrum implies that the third virial coefficent for anyons is symmetric about the semion point θ = π/2, midway between bosons and fermions.

36 Trajectories, with the bosonic ground state

37 Trajectories, with the fermionic ground state

38 Trajectories

39 Trajectories

40 Trajectories

41 Trajectories

42 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,.... Z 1 (β) = e βe 1 = 4 sinh 2( ). ξ 2 Two bosons: E = E CM + E rel, E rel = (1 + l + 2k) ω, l =..., 2, 0, 2,..., k = 0, 1, 2,.... Z 2B (β) = Z CM (β) Z rel (β) = 1 4 sinh 2( ξ 2 ) cosh ξ 2 sinh 2 ξ. Two fermions: E = E CM + E rel, E rel = (1 + l + 2k) ω, l =..., 3, 1, 1, 3,..., k = 0, 1, 2, Z 2F (β) = Z CM (β) Z rel (β) = 4 sinh 2( ) ξ sinh 2 ξ.

43 Two distinguishable particles: Partition functions Z 2D (β) = (Z 1 (β)) 2 = Z 2B (β) + Z 2F (β). Two anyons, θ = νπ: E = E CM + E rel, E rel = (1 + l + 2k) ω, l =..., ν 2, ν, ν + 2,..., k = 0, 1, 2, cosh((1 α)ξ) Z 2 (θ, β) = Z CM (β) Z rel (β) = 4 sinh 2( ) ξ 2 sinh 2. ξ 2 The two-anyon partition function is a function of the periodic sawtooth function α(θ), α = 0 for bosons and α = 1 for fermions. Note that Z 2 is an analytic function of θ at the fermion points.

44 Partition functions We write the two-anyon partition function as Z 2 (θ, β) = 1 2 ( F11 (θ, β) (Z 1 (β)) 2 + F 2 (θ, β) Z 1 (2β) ) where F 11 is symmetric and F 2 antisymmetric between bosons and fermions, F 11 (θ, β) = cosh (( ) ) α 1 2 ξ cosh ( ), F ξ 2 (θ, β) = sinh((α 1 2 )ξ) 2 sinh ( ). ξ 2 In the path integral formalism the two particle paths wind around each other, and the functions F 11 and F 2 are probability generating functions for the distributions of winding numbers.

45 Path integrals Consider two distinguishable particles in two dimensions. The position eigenstates r 1, r 2 form a basis for the Hilbert space. They satisfy the orthogonality relation s 1, s 2 r 1, r 2 = δ (2) (s 1 r 1 ) δ (2) (s 2 r 2 ) and the completeness relation I = d 2 r 1 d 2 r 2 r 1, r 2 r 1, r 2. (R 2 ) 2 The boson and fermion position eigenstates r 1, r 2 B,F = 1 2 ( r 1, r 2 ± r 2, r 1 ) together form an alternative basis for the Hilbert space.

46 Path integrals Consider two bosons. The position eigenstates r 1, r 2 B satisfy the orthogonality relation B s 1, s 2 r 1, r 2 B = δ (2) (s 1 r 1 ) δ (2) (s 2 r 2 ) + δ (2) (s 1 r 2 ) δ (2) (s 2 r 1 ) and the completeness relation, where I B is the projection on boson states, I B = d 2 r 1 d 2 r 2 r 1, r 2 BB r 1, r 2 (R 2 ) 2 /S 2 = 1 d 2 r 1 d 2 r 2 r 1, r 2 BB r 1, r 2 2 (R 2 ) 2 = 1 d 2 r 1 d 2 ( r 2 r1, r 2 r 1, r 2 + r 1, r 2 r 2, r 1 ). 2 (R 2 ) 2 (R 2 ) 2 /S 2 is the true configuration space, with (r 1, r 2 ) and (r 2, r 1 ) identified. The fermionic projection I F is the same, but with instead of +. Hence I B + I F = I.

47 Path integrals The bosonic or fermionic partition function is Z 2B,F (β) = Tr(I B,F e βh ) = 1 d 2 r 1 d 2 ( r 2 r1, r 2 e βh r 1, r 2 ± r 2, r 1 e βh r 1, r 2 ). 2 (R 2 ) 2 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, Z 2B,F (β) = 1 d 2 r 1 d 2 ( r 2 G(r1, r 1 ; τ) G(r 2, r 2 ; τ) 2 (R 2 ) 2 ±G(r 2, r 1 ; τ) G(r 1, r 2 ; τ) ) ( ( = 1 ) 2 ) d 2 r G(r, r; τ) ± d 2 r G(r, r; 2τ) 2 R 2 R 2 = 1 (Z1 (β)) 2( 2 ± Z 1 (2β) ).

48 Path integrals The one-particle propagator for the harmonic oscillator is mω ( G(s, r; τ) 2π sinh(ωτ) exp mω [ tanh( ωτ 4 2 ) s + r 2 + coth( ωτ 2 ) s r 2]). The limit ω 0 is the free particle propagator G(s, r; τ) = m ( 2π τ exp m 2 τ s r 2). We use these expressions to Monte Carlo generate paths, with a harmonic oscillator external potential or with no external potential. For a closed path over an imaginary time interval of length L β we generate the first point r with a probability distribution G(r, r; L β). Given two points r 1 at τ = τ 1 and r 2 at τ = τ 2 we generate an intermediate point r at τ with a probability distribution G(r 2, r; τ 2 τ) G(r, r 1 ; τ τ 1 ).

49 N anyons The generalization from two to N anyons is the formula Z N (θ, β) = F P (θ, β) ( ) νl 1 Z1 (Lβ). ν P L L! L Here P = (ν 1, ν 2, ν 3,...) is a partition of N, with ν L = 0, 1, 2,... and Lν L = ν 1 + 2ν 2 + 3ν 3 + = N. Example: L=0 Z 3 (θ, β) = 1 6 F 111(θ, β) (Z 1 (β)) F 3(θ, β) Z 1 (3β) from the partitions of 3: F 21(θ, β) Z 1 (2β)Z 1 (β), (3, 0, 0, 0,...) = , (0, 0, 1, 0,...) = 3, (1, 1, 0, 0,...) =

50 Three anyons The three terms of Z 3 correspond to the permutations in the figure. The first two are even permutations, the third is odd. β 0 τ r j j = r j r j The winding number Q jk of a pair jk is their winding angle divided by π. In case 1, all three winding numbers Q 12, Q 13, Q 23 are even integers. In case 2, no single Q jk is an integer, but the sum Q = Q 12 + Q 13 + Q 23 is an even integer. In case 3, Q 12 is an odd integer and Q (12)3 = Q 13 + Q 23 is an even integer. Q = Q 12 + Q 13 + Q 23 is an even/odd integer for an even/odd permutation.

51 Three anyons The figure illustrates a three-particle path (r 1 (τ), r 2 (τ), r 3 (τ)) inducing a cyclic permutation, and the same path represented as a closed one-particle path over three times the imaginary time interval. 3 β 2 β β 0 τ r j r 1

52 Three anyons For a given partition P, and a given β, we can Monte Carlo generate paths to find the probability distribution P P (Q, β) of the winding numbers Q. To each path is associated the phase factor e iqθ. Thus F P is the probability generating function F P (θ, β) = Q P P (Q, β) e iqθ. Time reversal invariance ot the one-particle propagator implies that P( Q) = P(Q), and hence F is real.

53 Three anyons Recall the exact results for two anyons, F 11 (θ) = cosh (( ) ) α 1 2 ξ cosh ( ), F ξ 2 (θ) = sinh((α 1 2 )ξ) 2 sinh ( ). ξ 2 They define the probability distributions P 11 for Q even and P 2 for Q odd, P 11 (Q) = 2ξ tanh ( ) ( ξ 2 ξ 2 + (πq) 2, P 2ξ coth ξ ) 2 2(Q) = ξ 2 + (πq) 2. These distributions have very long tails, their standard deviations are infinite. The tails are there because F 11 (θ) and F 2 (θ) have discontinuous derivatives at the boson and fermion points. In the Monte Carlo simulations we got the large winding numbers from windings at extremely small distances, down to , which is the limit of the computer hardware. This shows that the theory does not apply directly to anyons in the real world, which are collective excitations and definitely not point particles.

54 Three anyons One exact result for three anyons is also known, F 21 (θ) = sinh((α 1 2 )3ξ) sinh ( ω 0 ) 1 2α. 3ξ 2 Diptiman Sen was able to compute it exactly because it has contributions only from the linear states, with energies varying linearly with θ. For F 3 only the free-particle limit is exactly known, ( α 1 2 = (1 3α) 2) F 3 (θ) ω (1 32 α ). Alain Dasnières de Veigy proved to all orders in perturbation theory (!) the general formula for an L-cycle: F L (θ) ω 0 L 1 k=1 ( 1 Lα k This result is all the more remarkable because there are contributions from nonlinear states. ).

55 Free anyons We consider now anyons in zero external potential. In order to have finite partition functions we impose periodic boundary conditions in a square box of area A. Define the thermal de Broglie wave length Λ = 2πβ/m. The one-particle partition function is [ Z 1 (β) = exp ( πn2 Λ 2 ) ] 2 [ = AΛ ( ) ] 2 A exp πn2 A Λ 2. n= The last expression is a Poisson resummation, by Fourier expansion of f (x) = exp ( π(n + x)2 Λ 2 ), A n= a periodic function of x. Up to exponentially small correction terms we have Z 1 (β) = A Λ 2 = Am 2π 2 β. n=1

56 Statistical mechanics Define z = e βµ where µ is the chemical potential. Then the grand canonical partition function is Ξ(β, z) = 1 + z N Z N (β). N=1 Let C N be the set of all partitions of N, C = N=0 C N, C = N=1 C N. We write one partition of N as P = (ν 1, ν 2, ν 3,...) with L Lν L = N, then = P C ν 1=0 ν 2=0 ν 3=0 In this notation we have, with ν = L ν L, ln Ξ = ( ( 1) ν 1 ) ν z L Z L = ( 1) ν 1 (z L Z L ) νl (ν 1)!. ν ν L! P C ν=1 L=1 L=1

57 Statistical mechanics From the grand canonical partition function we get the pressure P, AβP = ln Ξ. We get directly the cluster expansion βp = b n z n n=1 with the cluster coefficients b N = 1 ( 1) ν 1 (ν 1)! A P C N L=1 Introducing the number density we get the virial expansion ρ = z (βp) z βp = ρ + = nb n z n n=1 A n ρ n. n=2 Z νl L ν L!.

58 Statistical mechanics The first cluster coefficients are given as Ab 1 = Z 1 = A Λ 2, Ab 2 = Z 2 Z 2 1 2, Ab 3 = Z 3 Z 2 Z 1 + Z 3 1 3, Ab 4 = Z 4 Z 3 Z 1 Z Z 2Z1 2 Z The first virial coefficients are given by the cluster coefficients as A 2 = Λ 4 b 2, A 3 = 2Λ 6 b 3 + 4Λ 8 b 2 2, A 4 = 3Λ 8 b Λ 10 b 2 b 3 20Λ 12 b 3 2.

59 Statistical mechanics Remember now our formula for the N-anyon partition function in terms of the one-particle partition function: Z N (θ, β) = F P (θ, β) ( ) νl 1 Z1 (Lβ). ν P L L! L It gives a formula for the cluster coefficients: b N = Z 1(β) ( 1 Z1 (Lβ) A ν L! LZ 1 (β) G P P C N L=1 in terms of a new set of coefficients G P = (F P + )(Z 1 (β)) ν 1. ) νl Here represents terms that are products of F coefficients. With the two-dimensional relation Z 1 (Lβ) = Z 1 (β)/l we get that b N = 1 1 Λ 2 L 2νL ν L!. G P P C N L=1

60 Statistical mechanics The G coefficients for L-cycles are Other coefficients are G L = F L A L 1 k=1 ( 1 Lα k ). G 11 = (F 11 1) Z 1 A α(α 1), G 111 = (F 111 3F ) Z1 2, G 21 = (F 21 F 2 ) Z 1 A 2(1 2α)α(α 1), G 1111 = (F F 111 3F F 11 6) Z1 3, G 211 = (F 211 2F 21 F 2 F F 2 ) Z1 2, G 22 = (F 22 F2 2 ) Z 1, G 31 = (F 31 F 3 ) Z 1.

61 Statistical mechanics All the G coefficients tend to finite limits when A. To see why, take as an example G 211 = (F 211 2F 21 F 2 F F 2 ) Z 2 1 q p(q) e iqθ. One Monte Carlo generated path, interchanging particles 1 and 2, contributes one odd integer winding number Q 12 and three even integer winding numbers Q 34, Q (12)3 = Q 13 + Q 23, Q (12)4 = Q 14 + Q 24. It contributes to F 211 by the total winding number Q = Q 12 + Q 34 + Q (12)3 + Q (12)4. It contributes to F 21 by the two winding numbers Q a = Q 12 + Q (12)3, Q b = Q 12 + Q (12)4. It contributes to F 2 F 11 by the winding number Q c = Q 12 + Q 34. And it contributes to F 2 by the winding number Q 12.

62 Statistical mechanics The formula G 211 = (F 211 2F 21 F 2 F F 2 ) Z 2 1 q p(q) e iqθ. tells us to that for each Monte Carlo generated path we should update the Fourier components p(q) as follows: add 1 to p(q) ; subtract 1 from each of p(q a ), p(q b ), and p(q c ) ; add 2 to p(q 12 ). Recall that Q a = Q 12 + Q (12)3, Q b = Q 12 + Q (12)4, Q c = Q 12 + Q 34. The net result of this updating is null if more than one of the three even winding numbers Q 34, Q (12)3, Q (12)4 is zero. For example, if Q (12)3 = Q 34 = 0, then Q a = Q c = Q 12 and Q b = Q.

63 Statistical mechanics In general, an N-particle path contributes to a coefficient G P only if all N particles wind in one cluster. In this sense, G P is the connected part of F P. As usual, the grand partition function Ξ is a sum of all diagrams, whereas ln Ξ is a sum of connected diagrams. Since the partition P consists of ν = ν 1 + ν 2 + parts scattered with uniform probability over the area A, the probability for these parts staying together goes to zero as A 1 ν when A. This implies that G P goes to a finite limit.

64 Statistical mechanics Explicit expressions for the first cluster coefficients are Λ 2 b 2 = G G 2 4, Λ 2 b 3 = G G G 3 9, Λ 2 b 4 = G G G G G Our Monte Carlo method for computation actually proves the nontrivial result that the cluster and virial coefficients are finite in the thermodynamic limit.

65 Exact results The second cluster and virial coefficients are exactly known, from the exact solution of the two-anyon problem: ( A 2 = Λ 4 b 2 = Λ 2 1 ) 1 (1 α) Alain Dasnières de Veigy and Stéphane Ouvry computed the pressure P as a function 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 second order) at the fermion point but nonanalytic at the boson point (because the first order correction is proportional to θ ). It is possible, however, that the θ dependence of the virial coefficients A n with n > 2 is analytic both at the fermion point and at the boson point (since there is no θ dependence to first order).

66 Approximate results The most precise calculation of the third virial coefficient gave the result ( ) 1 A 3 = Λ sin2 θ 12π 2 + a sin4 θ with a = (1.652 ± 0.012) = (621 ± 5)π 4. The coefficient of the sin 2 θ 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 harmonic oscillator energy levels carried out by Stefan Mashkevich. All the θ dependence of A 3 is due to the bottom of tower energy levels that vary nonlinearly with θ. The boson fermion supersymmetry implies that A 3 is analytic in θ everywhere, hence a Fourier expansion like this is to be expected.

67 Approximate results A Monte Carlo calculation of the fourth virial coefficient gave the result ( sin A 4 = Λ 6 2 ( θ π 2 ln ( ) ) ) + cos θ + (c 4 + d 4 cos θ) sin 4 θ with c 4 = ± , d 4 = ± Again the exact perturbative result is incorporated.

68 Approximate results The Monte Carlo calculation with four anyons suggested a polynomial approximation for the G coefficients: G P G P = N ν 2 G ν 1 11 (LF L ) νl with L 1 ( G 11 = α(α 1), F L = 1 Lα k L k=1 ). This is definitely only an approximation, but one can give heuristic arguments in support. The four-particle approximate polynomials are G 1111 = 16G 3 11, G 211 = 8F 2 G 2 11, G 22 = 4F 2 2 G 11, G 31 = 3F 3 G 11.

69 Monte Carlo, partition 4 = G (α(α 1)) 3 as a function of α. The curve marked fit is 3 π 2 α(α 1) sin2 (απ).

70 Partition 4 = G 211 8(1 2α)(α(α 1)) 2 as a function of α. The curve marked fit is 2 3π 2 (1 2α) sin2 (απ).

71 Partition 4 = G 22 4(1 2α) 2 α(α 1) as a function of α. The curve marked fit is 2 3 π 2 ln( 3 + 2) sin 2 (απ) cos 2 (απ).

72 Partition 4 = G 31 3(1 3α)(1 (3/2)α)α(α 1) as a function of α. The curve marked fit is 3 4π 2 ln( 3 + 2) sin 2 (απ) cos 2 (απ).

73 The cluster coefficient b 4 The fourth cluster coefficient minus the polynomial approximation, Λ 2 (b 4 b 4 ), as a function of α. The second order perturbation theory at α = 0 and α = 1 gives the parabolas shown.

74 The virial coefficient A 4 The fourth virial coefficient, A 4 /Λ 6, as a function of α. The second order perturbation theory at α = 0 and α = 1 gives the parabolas shown. Two fits by Fourier series are shown, one with coefficients c 4 = d 4 = 0 and one best fit with c 4 = , d 4 =

75 Approximate results Through some remarkable polynomial identities, proved by Kåre Olaussen, these polynomial approximations for G P give the following polynomial approximations for the cluster coefficients, Λ 2 b N = 1 N 1 N 2 k=1 ( 1 Ng k with g = 1 (1 α) 2. These cluster coefficients give virial coefficients that are independent of the statistics parameter θ, or g, except for ( A 2 = Λ g ), 2 which is the exact A 2 for anyons. It so happens that what we have now derived as an approximation for anyons is the exact result for two-dimensional exclusion statistics with statistics parameter g. The quadratic dependence of g on θ is different from the linear dependence for anyons in a strong magnetic field. )

76 Norwegian Easter holiday