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Chapter 8 Fermi Perfect Fermi GasIn this chapter, we study a gas of non-interacting, elementary Fermi par-ticles. Since the particles are non-interacting, the potential energy is zero, andthe energy of each Fermion is simply related to its momentum by = has one-half integral spin, which we denote by s. The state of theFermion depends on the orientation of this spin (with respect to an appliedmagnetic field) as well as on its location in phase space. For spin s, there are2s+1 spin states or orientations ofs.

= gB( F) is the density of states at the Fermi surface and δ ∼ kT is, from (8.14), the region of energy affected at temperature kT. Also, the excitation energy of each Fermion is roughly the classical thermal energy. Thus, the energy required to heat the Fermi gas to temperature kT (the thermal

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Transcription of C:/Documents and Settings/yilmaz/Desktop/latex-Physics ...

1 Chapter 8 Fermi Perfect Fermi GasIn this chapter, we study a gas of non-interacting, elementary Fermi par-ticles. Since the particles are non-interacting, the potential energy is zero, andthe energy of each Fermion is simply related to its momentum by = has one-half integral spin, which we denote by s. The state of theFermion depends on the orientation of this spin (with respect to an appliedmagnetic field) as well as on its location in phase space. For spin s, there are2s+1 spin states or orientations ofs.

2 In the absence of a magnetic field, these2s+ 1 states all have the same energy and serve simply to expand the numberof states available to the Fermion. Thus the density of states in volume elementd of phase space, including the spin states , isgsd h3wheregs= (2s+ 1).Thestatistical mechanics of the Fermi gas follows directly from the Grand CaronicalPartition function ( ) and the Fermi function n( ) =(e ( )+ 1) 1( )which gives the expected number of Fermions in energy state . The expectednumber of Fermions in energy range to +d is thendN= nsgsd h3ordN( ) = n( )gsg( )d where, as in ( ),g( ) = 2 V(2mh2)3/2 1 consider specifically a fixed numberNof Fermions in volumeVhavingnumber densityn=N/V.

3 ThenN= s ns=gs nd h3orN=gs2 V(2mh2)3/2 0d 1/2(e ( )+ 1).( )The integral here cannot be further reduced because we don t know the chemicalpotential (T). However, sinceNis fixed and the factors outside the integral12 Statistical constant, we do know that (T) must vary withTso as to keep the integralconstant. In fact, we will use the relation to determine (T). We shall see thatatT= 0 K is equal to the Fermi energy, F, and as T is increased decreasesuniformly until it joins on to its classical limit = kTlog (Z0/N) at very hightemperature.

4 Similarly, the internal energy isU= s nsgs=gs n( )g( )d orU=gs2 V(2mh2)3/2 0d 3/2(e ( )+ 1).( )and we see that the chemical potential plays a central role in determining theproperties of a Fermi now specialize to spin 1/2 Fermions for whichgs= 2. In the presenceof a magnetic field (say along the z axis) the spin now aligns either parallel oranti-parallel to the magnetic field,sz= 1/2, to make up the two possible spinstates. Our gas is then a model of conduction electrons in a metal or of the case of conduction electrons, the model neglects the interactionbetween the electrons and the interaction of the electrons with the ions inthe metal.

5 The electron-electron interaction is simply the Coulomb interac-tionv(r) =e2/r. This varies very slowly withrand has a very long range sothat each electron interacts with many other electrons. If the electron densityis high, a given electron interacts with so many electrons that the total poten-tial energy it experiences is almost flat. The potential energy is then almostconstant and can, to a good approximation, be subtracted off as a model also neglects the interaction between the conduction electronsand the ions in the metal.

6 This electron-ion interaction is a sensitive balancebetween the electron-nucleus attraction and the electron-ion core electron re-pulsion. In suitable cases, these two interactions largely cancel leaving theconduction electrons effectively free. Examples where this cancellation is nearlycomplete are the alkali metals and aluminum, and in these cases, a Fermi gas isa good model of the conduction the case of liquid3He, we simply ignore the3He-3He atom TemperatureAtT= 0 K, the Fermi gas will be in its ground state.

7 This is formed byfilling up the lowest energy states until all theNFermions are there are two spin states per space state, this requiresN/2 space statesin phase space. The configuration space part of phase space is just the volumeV. Thus, we must fill up a sphere in momentum space of volume 4 p3F/3 suchthat1h3(V4 p3F3)=N2( )whereh3is the volume of phase space taken up by one state. The momentumpFof the most energetic Fermion is called the Fermi momentum. We couldChapter this physically using our Particle Hilton discussed in Section TheFermion Hilton has two rooms per floor, one must be occupied by a spin upFermion (sz= 1/2) and the other by a spin down Fermion (sz= 1/2).

8 Toform the ground state in the Hilton, we fill up the lowestN/2 floors with oneFermion per room. This sea of occupied states is often called the Fermi result ( ) may be obtained more formally usingN= s nswhere ns= 1 up to the last occupied state and ns= 0 beyond, ,N= s1 = 2 d h3 1= 2V4 h3 pF0p2dp=2h3(V4 p3F3).The Fermi momentum is, therefore, simply related to the particle densitynbypF=~(3 2n)1 higher the density (the smaller the volumeV), the larger the sphere in mo-mentum space required to occupy all the Fermions.

9 The corresponding energyof the most energetic Fermion is F=p2F2m=(~22m)(3 2n)2/3,( )called the Fermi Energy. This could also have been obtained directly asN= s ns= 2 F0d g( ) 1= 2 (2 V)(2mh2)3/223 3/2F( )Comparison of eqs. ( ) and ( ) suggests a convenient expression forg( ) interms of Fas,g( ) =34N F( F)1/2( )It is also convenient to define a density of states including directly both spinstatesgB( ) 2g( ) =32N F( F)1/2( )To locate the chemical potential, we recall that atT= 0 K the expectedoccupation is n( ) = 1 for states of energy < Fand all higher states areempty.

10 Hence,4 Statistical Mechanics. n( ) =1(e ( )+ 1={1 < F0 > F( )Since atT= 0 K, the Fermi function can fulfill this condition only if = , atT= 0 K , is equal to the Fermi energy. This is consistent withthe thermodynamic relation =( U N)N= Nfor to add an additional Fermion(beyond the fixed number N), we must place this particle at energy Fsince alllower states are 0 K, the statistical properties of the Fermi gas are given simply byeq. ( ). This distribution is displayed in Fig. along with the distributionof Fermions over energy, n( )g( ).)}


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