Exchange symmetry Jan Myrheim Department of physics, NTNU April - PowerPoint PPT Presentation

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 are symmetric under permutations of indices. A symmetric operator X representing an observable preserves the symmetry of the wave function ψ , ψ → X ψ , 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.

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.

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 10 26 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.

Scattering process 1 + 2 → 3 + 4 detector p ′ θ p − p − p ′ Incoming momenta p 1 = − p 2 = ± p . p 3 = − p 4 = ± p ′ . Outgoing momenta p 1 · p 3 = | p || p ′ | cos α . Scattering angle α = θ or π − θ , definition: 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 | out � = | pn �| p ′ , − p ′ � . | in � = | pn �| p , − p � , We label, e.g., the incoming proton as particle 1 and call its momentum p . To measure the cross section d σ d Ω = | 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 � | f ( θ ) | 2 + | f ( π − θ ) | 2 � d σ d Ω = 1 , 2 dividing by two to compensate for the double counting.

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 , 1 � = | pp � , | 1 , 0 � = √ | pn � + | np � , | 1 , − 1 � = | nn � , 2 and one antisymmetric state with I = 0: � � 1 √ | 0 , 0 � = | pn � − | np � . 2

Isospin Define also symmetric and antisymmetric momentum states � � 1 √ | p + � = | p , − p � + | − p , p � , 2 � � 1 | p −� = √ | p , − p � − | − p , p � , 2 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 ′ −� | 0 , 0 �| p ′ + � . or

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 −� � = 1 � p ′ , − p ′ | F | p , − p � − � p ′ , − p ′ | F | − p , p � 2 � −�− 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 + � � = 1 � p ′ , − p ′ | F | p , − p � + � p ′ , − p ′ | F | − p , p � 2 � + �− p ′ , p ′ | F | p , − p � + �− p ′ , p ′ | F | − p , p � = f ( θ ) + f ( π − θ ) .

Scattering We compute the cross sections as � | f ( θ ) | 2 + | f ( π − θ ) | 2 � � f ( θ ) f ( π − θ ) ∗ � d σ ± d Ω = 1 2 | f ± ( θ ) | 2 = 1 ± Re , 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, � d σ + � � | f ( θ ) | 2 + | f ( π − θ ) | 2 � d σ d Ω = 1 = 1 d Ω + d σ − . d Ω 2 2

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.

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, 2 I 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.

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.

Which particles are identical? Consider the Hamiltonian of the hydrogen atom, again disregarding spin, H = p 2 + p 2 k k = q 1 q 2 1 2 − | r 1 − r 2 | , . 2 m 1 2 m 2 4 πǫ 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 = P 2 2 M + p 2 2 m − k | r | , with M = m 1 + m 2 , m = m 1 m 2 / M , p = m 2 p 1 − m 1 p 2 P = p 1 + p 2 , r = r 1 − r 2 , . M