Physics 562: Statistical MechanicsSpring 2003, James P. SethnaHomework 3, due Wednesday Feb. 26Latest revision: February 10, 2003, 11:53 amReadingPathria, chapters 6, 7 & 8.Feynman, “Lectures on Physics” volume III, chapter 4Yeomans, chapter 1, 2.1-2.4, 3.1-3.4, 6.1-6.2(3.1) Ensembles and Statistics: Three Particles, Two Levels.A system has two single-particle quantum states, with energies (measured in degreesKelvin) E0/kB= −10 and E2/kB= 10. Experiments are performed by adding threenon-interacting particles to these two states, either identical spin 1/2 fermions, identicalspinless bosons, distinguishable particles, or spinless identical particles obeying Maxwell-Boltzmann statistics. Please make a table for this problem, giving your answers for thefour cases (Fermi, Bose, Dist., and MB) for each of the three parts. Calculations may beneeded, but only the answers will be graded.-30 -20 -10 0 10 20 30E/kB (degrees K)-2-10123S/kBLog(1)Log(2)Log(1/2)Log(3)Log(4)Log(8)Log(1/6)ABCDEConstant E, NEntropies for Three Particles in Two StatesACBDEFigure (3.1.A)(a) The system is first held at constant energy. In figure (3.1.A), which curve representsthe entropy of the fermions as a function of the energy? Bosons? Distinguishableparticles? Maxwell-Boltzmann particles?1020406080T (degrees K)-30-20-1001020Average Energy E/kB (degrees K)ABCDEConstant T, NEnergies of Three Particles in Two StatesABCDEFigure (3.1.B)(b) The system is now held at constant temperature. In figure (3.1.B), which curverepresents the mean energy of the fermions as a function of temperature? Bosons?Distinguishable particles? Maxwell-Boltzmann particles?020406080Temperature T (degrees K)-80-60-40-20020406080Chemical Potential µ/kBABCDEFConstant T, µChemical Potentials for Three Particles in Two StatesABCDEFFigure (3.1.C)(c) The system is now held at constant temperature, with chemical potential set to holdthe average number of particles equal to three. In figure (3.1.C), which curve repre-sents the chemical potential of the fermions as a function of temperature? Bosons?Distinguishable? Maxwell-Boltzmann?2(3.2) Phonons a nd Photons are Bosons.Phonons and photons are the elementary, harmonic excitations of the elastic and electro-magnetic fields. We’ve seen in an earlier problem that phonons are decoupled harmonicoscillators, with a distribution of frequencies ω. A similar analysis shows that the Hamil-tonian of the electromagnetic field can be decomposed into harmonic normal modes calledphotons.This problem will explain why we think of phonons and photons as particles, instead ofexcitations of harmonic modes.(a) Calculate the partition function for a quantum harmonic oscillator of frequency ω.(b) Calculate the grand canonical partition function for bosons multiply filling a singlestate with energy ¯hω.Forµ = 0, show that the grand canonical partition functionfor bosons is the same as the canonical partition function for a quantum harmonicoscillator up to a shift in the arbitrary zero of the total energy of the system.The Boltzmann filling of a harmonic oscillator is therefore the same as the Bose-Einsteinfilling of bosons into a single quantum state, except for an extra shift in the energy of¯hω/2. This extra shift is called the zero point energy.(c) A system has N harmonic modes with frequencies ωn. Show that the partition functionis the same as that of a grand-canonical system with zero point energyNn=1¯hωn/2and N distinguishable, non-interacting boson states with energy ¯hωnand chemicalpotential µn=0.Often, the label for the N states or modes in part (c) is the wavevector k (e.g., for phononsin crystals and photons in vacuum or crystals). In this case, instead of thinking of thesystem as N bosons each with one state, we can think of it as one boson with N possiblestates, with a dispersion relation ω(k). Real, massive bose particles like He4in free spacehave a similar dispersion relation Ek=¯h2k2/2m: phonons and photons are thus justanother kind of Bose particles with dispersion relation Ek=¯hωk.(d) Do phonons or photons Bose condense at low temperatures? Can you see why not?(Hint: is their number conserved?)Be careful not to get confused: Bose atoms in a harmonic potential can Bose condense(problem 3.3), but a harmonic vibrational excitation of a collection of atoms cannot. Theformer has a fixed number of bosons with a variety of (equally spaced) available energystates: the latter is viewed as a variable number of bosons occupying a single energy state.3(3.3) Bose Condensation in a Parabolic Potential.“Observation of Bose-Einstein Condensation in a Dilute Atomic Vapor”, M.H. Anderson,J.R. Ensher, M.R. Matthews, C.E. Wieman, and E.A. Cornell, Science 269, 198 (1995).http://jilawww.colorado.edu/bec/Wieman and Cornell in 1995 were able to get a dilute gas of rubidium-87 atoms to Bosecondense. (The first author was the post-doc Anderson: Ensher and Matthews were thegraduate students.)(a) Is rubidium-87 (37 protons and electrons, 50 neutrons) a boson or a fermion?(b) At their quoted maximum number density of 2.5 × 1012/cm3, at what temperatureTpredictcdo you expect the onset of Bose condensation in free space? They claim thatthey found Bose condensation starting at a temperature of Tmeasuredc= 170nK. Isthat above or below your estimate? (Useful constants: h =6.6262 × 10−27erg sec,mn∼ mp=1.6726 × 10−24gm, kB=1.3807 × 10−16erg/K.)Bose-Einstein Condensation at 400, 200, and 50 nano-Kelvin, from Jila (above reference).The pictures are spatial distributions 60ms after the potential is removed; the field of viewof each image is 200µm ×270µm. The left picture is roughly spherically symmetric, and istaken before Bose condensation; the middle has an elliptical Bose condensate superimposedon the spherical thermal background; the right picture is nearly pure condensate. I believethis may not be the same experiment as described in their original paper.The trap had an effective potential energy that was harmonic in the three directions,but anisotropic with cylindrical symmetry. The frequency along the cylindrical axis wasf0=120Hz so ω0∼ 750Hz, and the two other frequencies were smaller by a factor of4√8: ω1∼ 265Hz. The Bose condensation was observed by abruptly removing the trappotential,∗and letting the gas atoms spread out: the spreading cloud was imaged 60mslater by shining a laser on them and using a CCD to image the shadow.For your convenience, the ground