2.(10分 (1)费米能级:c=B2n2x (4分 n维球体积:Vs~mmax (4分) EF N(N/)2/n (2分) (2)玻色凝聚温度有右式确定:N wsg(e)de T (3分) 2xm\32 态密度:9O=(h2 (1分) (1分) mk7)3/2 V=> kTc m(. (3分) mINi/V=m2 N2/V N2/V=N1/ (1分) Ta/T=32/3 (1分) 3.(16分)Cm=/8S M (4分) aS as 7/s OM aM/ aT (3分) 由 Maxwell关系:(S (4分) aM M=1→(m=C (2分)+(2分) M 整理:CH-CM (1分) 4.(18分)T=0K时,H=G+TS=G (3分) 因此:G=N=2a2>0=÷1>0 (2分) 对玻色子,<0=0,因此体系为费米子体系。 (5分) 对费米子,在T=0K时,每个单粒子态占据一个粒子。 (4分 因此,基态粒子数为u (4分)2. £10 © ×2¤ (1) ¤’U?µF = h 2n 2 max 8mL2 (4©) n ‘¥NȵVs ∼ n n max =⇒ N ∼ n n max (4©) F ∼ (N/V ) 2/n (2©) (2) ÀÚvà§Ýkmª(½µN = Z ∞ 0 ωsg() d e /kT − 1 (3©) ݵg() = 2 √ π  2πm h 2 3/2 V √  (1©) ωs = 2s + 1 (1©) N ∼ ωs  mkT h 2 3/2 V =⇒ kTc ∼ 1 m  N ωsV 2/3 (3©) m1N1/V = m2N2/V =⇒ N2/V = N1/V (1©) Tc1/Tc2 = 32/3 (1©) 3. £16 ©¤ CH = T  ∂S ∂T  H , CM = T  ∂S ∂T  M (4©) CH − CM = T ∂S ∂T  H −  ∂S ∂T  M  = T  ∂S ∂M  T  ∂M ∂T  H (3©) d Maxwell 'Xµ  ∂S ∂M  T = −  ∂H ∂T  M (4©) M = C T H =⇒  ∂H ∂T  M = M C ,  ∂M ∂T  H = − C T 2 H (2©) + (2©) nµCH − CM = MH T (1©) 4. £18 ©¤ T = 0 K ž§H = G + T S = G (3©) ÏdµG = Nµ = 2a 2 > 0 =⇒ µ > 0 (2©) éÀÚf§µ < 0 = 0§ÏdNX¤’fNX" (5©) 餒f§3 T = 0 K ž§z‡üâfÓâ‡âf" (4©) Ïd§Äâfê ω (4©)