i Γ

Transcription

1

2 i Γ h N van Leeuwen

3 ii N : N : N : N : Planck

4 iii Appendix: ζ

5 1 S W S = k B log W 1) ) 3),, 4) 5) ) 6) 1) 5)

6 1 1.1 x f(x) x x x = i x i f(x i ) (1.1) x x P (x) x x+dx P (x)dx x = xp (x)dx (1.) P (x) 1 P (x) = ( πσ exp (x x ) 0) σ (1.3) x = < (x x ) >= P (x) xp (x)dx = x 0 (1.4) (x x ) P (x)dx = σ (1.5)

7 1 3 P(x) x Problem 1.1 x x 0 σ x 0 + σ 68% x 0 σ x 0 + σ 95% y n 1 n σ iy i = x n 1/n 1. x = i x if(x i ) ϵ { } lim P x 1 + x + + x n x n n > ϵ = 0 (1.6) n S n = n i x i S n = n x var(s n ) def = (S n n x ) n = (x i x ) + (x i x )(x j x ) i i>j = nσ σ = i (x i x ) P (x i ) (1.7) σ (x i x ) P (x i ) t P (x) = t P { x x t} (1.8) x i x t x i x t σ P { x x t} (1.9) t

8 1 4 x S n, x n x, σ nσ t = nϵ nσ t P { S n n x t} (1.10) σ nϵ P { S n x ϵ} (1.11) n n n S n { lim P Sn n x n σ n } β < β = Φ(β) def / = dx e x (1.1) π S n n x nσ S n X P (S n = X) = dx 1 dx dx n P (x 1 )P (x ) P (x n )δ(s n X) (1.13) δ(s n X) = 1 π dke ik(s n X) (1.14) P (X) = 1 π = 1 π ( dke ikx dx 1 P (x 1 )e ikx 1 dke ikx Q n (k) = 1 π Q(k) Q(k) = dxp (x)(1 + ikx k x ) ( ) dx n P (x n )e ikx n ikx+n log Q dke ) = 1 + ik x k x +

9 1 5 ) n log Q = n log (1 + ik x k x + ) = n (ik x k x (ik) x + ) = n (ik x k x + P (X) = 1 ) dk exp (ik(n x X) nk x π ik(n x X) nk x x = σ P (X) = 1 π ( π nσ exp = n x ( k i n x X ) (X n x ) n x n X ) (X n x ) 1 = ( nσ πnσ exp ) (X n x ) nσ (1.15) (1.16) x i P ( ) X n = n πσ exp ( ) (X/n x ) σ /n (1.17) x n X/n x x σ/ n (1.14) πδ(x) = dke ikx = lim e ikx ηk dk η +0 ( ikx ηk = η k i x ) x η 4η π /4η lim η +0 η e x dke ikx (1.18)

10 1 6 x = 0 x > 0, x < 0 0 π π dx lim /4η η +0 η e x = lim 4πη = π η +0 η dke ikx π x = p q (p, q) (p, q) p + dp q + dq P (p, q)dpdq q i p i (p 1,, p N, q 1,, q N ) dγ N dγ = dp 1 dp dp N dq 1 dq dq N (1.19) p, q (Liouville) dpdq Liouville (p, q) (P, Q) dγ = dp 1 dp dp N dq 1 dq dq N = dp 1 dp dp N dq 1 dq dq N (1.0) J J = (P 1, P,, P N, Q 1, Q,, Q N ) (p 1, p,, p N, q 1, q,, q N ) = 1 (1.1)

11 1 7 J = (P 1, P,, P N, Q 1, Q,, Q N ) (P 1, P,, P N, q 1, q,, q N ) J = (Q 1, Q,, Q N ) (q 1, q,, q N ) / (p1, p,, p N, q 1, q,, q N ) (P 1, P,, P N, q 1, q,, q N ) / (p1, p,, p N ) P (P 1, P,, P N ) q F (P 1, P,, P N, q 1, q,, q N ) p i = F q i (1.) Q i = F P i (1.3) (i, k) p i / P k = F/ q i P k (i, k) Q i / q k = F/ q k P i J = :

12 8.1 Γ N Γ (p 1,, p N, q 1,, q N ) Γ i ϵ i E N ϵ i = p i m + mω q i (.1) E N = N ϵ i = E N (p 1, p,, p N, q 1, q,, q N ) (.) i=1 6N Γ N (Principle of equal weight) Γ E N (p 1, p,, p N, q 1, q,, q N ) (p 1, p,, p N, q 1, q,, q N ) 6N 1 ϵ

13 9 ϵ ϵ + ϵ Γ ϵ dγ P (Γ)dΓ P (Γ)dΓ = dγ ϵ<e N <ϵ+ ϵ dγ P (Γ) = 1 ϵ<e N <ϵ+dϵ dγ (.3) P (Γ) A(Γ) = A((p 1, p,, p N, q 1, q,, q N )) A = dγa(γ)p (Γ) (.4).3 h h dpdq p q = h /h f (f, 3 N 3N) i P i 3 1 P i = 1 N cell = P i = 1 N cell = h hn cell = h3 h 3 N cell = dµ ϵ ϵ+δϵ dµ (.5) dµ ϵ ϵ+δϵ dµ (.6) 1 h h

14 10.4 Γ W ϵ<e W = dγ N <ϵ+dϵ (.7) h f f (degrees of freedom) (Boltzmann) 1 N f = N N f = N 3 N f = 3N S def = k B log W (.8) (p 1, p, p 3, q 1, q, q 3 ) (p 1, p 3, p, q 1, q 3, q ), (p 3, p, p 1, q 3, q, q 1 ) W = ϵ<e N <ϵ+dϵ dγ h f N! N! ( ) 1 S T = E V (.9) (.10) N H = N i p i m (.11)

15 11 E N Ω(E) Ω(E) = 1 h 3N N! i p i /m E i N dp i i=1 p i m E N i=1 dq i = V N h 3N N! i p i me me 3N N dp i (.1) n R n C n R n S n (R) nc n R n 1 I n = I n = dx 1 x n exp( x 1 x n) = π n (.13) 0 drs n (r)e r = nc n drr n 1 e r (.14) r = t 0 dtt 1/ t (n 1)/ e t / = Γ(n/)/ I n = nc n Γ(n/) nc ( n n ) Γ = π n, C n = 0 π n i=1 Γ ( (.15) n + 1) Problem.1 n = 1,, 3 Ω = V N π 3N 3N me Γ ( 3N + 1) (.16) N!h 3N E E + E W W = Ω(E + E) Ω(E) = dω de E = Ω3N E E (.17)

16 1 S S = k B log W = k B [ log V N πme 3N h 3N N!Γ ( 3N + 1) + log 3N ] E E (.18) N! Γ (Stirling) Γ(M + 1) = 0 dze z z M = 0 z+m log z dze z = M (Taylor) z + M log z = f(z) = f(m) 1 (z M) M Γ(M + 1) = e M+M log M πm log M! = log Γ(M + 1) = log πm M + M log M (.19) M 10 3 log 10 3 M + M log M 10 3 log M! = M + M log M (.0) S = k B [N log V h 3 + 3N log(πme) + N N log N + 3N 3N log 3N 1 log(π 3N/) + log(3n ] E E ) N, N log N ( S = k B N log V h log(πme) + 1 log N log 3N )

17 13 S = k B N ( 5 + log V h 3 N + 3 ) 4πmE log 3N (.1) 1 T = S E = k BN 3 E, E = 3N k BT (.) T ds = de + P dv P T = S V = k BN 1 V, P V = Nk BT (.3).5. ±ϵ ϵ E M E = Mϵ N + = 1 (N + M) N = 1 (N M) N + +ϵ N ϵ E = Mϵ W M W M = N! ( 1 (N M))! ( 1 (N + M))! ( S = k B log W M = k B N log N 1 N M (N M) log 1 T = S E = 1 S ϵ M = k B ϵ log N M N + M 1 N + M (N + M) log (.4) ) (.5) (.6)

18 14 β def = 1 k B T (.7) e βϵ = N M N + M, M = Ntanhβϵ (.8) E E = N ϵtanhβϵ (.9) S/N M/N = tanhβϵ S N = k B (log N 1 (1 + thβϵ) log N (1 + thβϵ) 1 (1 thβϵ) log N ) (1 thβϵ) = k B (log N 1 (1 + thβϵ) log N 1 (1 thβϵ) log N 1 (1 + thβϵ) log 1 (1 + thβϵ) 1 (1 thβϵ) log 1 ) (1 thβϵ) ( = k B log 1 (1 + thβϵ) log(1 + thβϵ) 1 ) (1 thβϵ) log(1 thβϵ) ( = k B log 1 log(1 th βϵ) 1 ) (1 + thβϵ) thβϵ log (1 thβϵ) S N = k B(log(coshβϵ) βϵthβϵ) (.30) βϵ 1 k B log βϵ 1 0 Problem. (.30) βϵ 1.5.3

19 15 a L = na n W W = N C N/+n = N! ( N + n)! ( N n)! (.31) N 1, n N S = k B log W = k B [log + N log N N ( ) ( )] N N + + n + n = k B [ log + N log N N log ( N ( ) ( ) ( ) ( ) N N N N + n log + n n log n ) 4 n n log ( ) N log 4 n log(n/) 4n N ( ) N + n n log 4n N n N ( )] N + n N n ( ) S = k B log + N log n L = k B (log + N log ) k B N a N ( ) f T df = SdT fdl (.3) L = ( ) S L T = k BL Na (.33) f = c k BL Na T (.34)

20 16.6 E E Ω(E, V, N, ) = dγ N <E h f W = Ω E = Ω(E + E) Ω(E) = D(E) E E D(E) N Ω exp (Nϕ(E/N, V/N, )) (.35) ϕ(e/n, V/N, ) = ϕ(u, v, ), u, v, ϕ > 0, ϕ > 0, ϕ > 0 D N 1 N ϕ u enϕ 1 (.36) ( ) D ϕ E N + ϕ e Nϕ 1 (.37) Ω, D E D(E) E < Ω(E) < D(E)E (.38) log Ω log D(E)E = log ϕ E O(ln N) log(d(e) E) log D(E)E = log E O(ln N) E log W log Ω O(log N) N log N/N log 10 3 / W Ω.7 I II

21 17 I E I D I (E I ) II E II D II (E II ) E D(E) W W = D(E) E = D I (E I )de I D II (E II )de II E<E I +E II <E+ E E+ E EI de I D I (E I )D II (E E I ) de II = E E E I D(E) = de I D I (E I )D II (E E I ) de I D I (E I )D II (E E I ) (.39) E < E I + E II < E + E I E I E I E f(e I ) = D I(E I )D II (E E I ) E D(E) E S I E I + S II(E E I ) E I = 0 1 T I = S I E I = S II E II = 1 T II (.40)

22 18 3 T N N 3.1 A B B A A ν, E ν, P ν A B E tot E tot + E E tot E ν E tot E ν + E A ν B D B D B (E tot E ν ) E D tot E tot E ν E tot E ν + E D tot (E tot ) E P ν = D B(E tot E ν ) E D tot (E tot ) E (3.1) S = k B log D B E D B E = e S/k B

23 3 19 ( ) S(Etot E ν ) P ν exp k B k ( B S(Etot ) exp E ) ν k B T k B e βe ν k B P ν = A exp( βe ν ) (3.) ) E ν ±ϵ P (ϵ) = Ae βϵ P ( ϵ) = Ae βϵ P (ϵ) + P ( ϵ) = 1 A = 1 e βϵ + e βϵ E = ϵp (ϵ) + ( ϵ)p ( ϵ) = ϵ e βϵ e βϵ = ϵtanhβϵ e βϵ + e βϵ C C = d E dt C = ϵ 1 k B T cosh βϵ (3.3) 3. T

24 Ck B T ϵ βϵ P ν = 1/W S = k B log W = k B P ν log(w ) = k B P ν log(1/p ν ) ν S = k B P ν log P ν (3.4) 1 F ν ν F = E T S = ν E ν P ν + k B T ν P ν log P ν = ν (E ν P ν + k B T P ν log P ν ) F P ν P ν ν P ν = 1 F = F + λ( ν P ν 1) F F = 0, > 0 P ν Pν k B T/P ν > 0 F P ν = λ + E ν + k B T (1 + log P ν ) = 0 log P ν = λ + k BT + E ν, P ν = Ae βeν k B T (3.) 1

25 3 1 Problem 3.1 P ν k B ν P ν log P ν E E = E = ν E ν P ν S = S λ( ν E ν P ν E) Λ( ν P ν 1) 0 = S P ν = k B (1 + log P ν ) λe ν Λ k B log P ν = k B + λe ν + Λ P ν = Ae λeν/k B λ = 1/T λ Problem 3. Maxwell ) P (v) = A exp ( β mv (3.5) A v x, v x, v, v, 1/v 3.4 Z Z def = ν e βeν (3.6)

26 3 Z = 1 dγ N! h f e βe(γ) (3.7) 1 = ν P ν = ν Ae βeν = AZ Z Z E = E E = ν E ν P ν = 1 E ν e βeν Z ν = β ( ν e βeν ) / Z β Z = Z E = log Z (3.8) β E = dγ h f E(Γ)e βe(γ) dγ h f e βe(γ) Z β = Z = β log Z 3.4. F = E T S F = ν = ν = ν E ν P ν + k B T ν E ν P ν + k B T ν E ν P ν + k B T ν P ν log P ν ( e βe ν ) P ν log Z P ν ( βe ν log Z) k B T β = 1 F = k B T log Z (3.9)

27 3 3 Z = e βe ν = ded(e)e βe (3.10) ν Z = βe+log D(E) dee f(e) f(e) = βe + log W = βe + S k B E 0 f(e) = f(e 0 ) + f (E 0 ) (E E 0 ) + = βe 0 + S 0 + f (E 0 ) (E E 0 ) + k B f N, E N f N 1 ( ) Z = e βe 0+S 0 /k B f (E 0 ) de exp (E E 0 ) = e βe 0+S 0 /k B π f log Z = βe 0 + S 0 k B + log F = k B T log Z π f = β(e 0 T S 0 ) + O(log N) 3.5 C V C V = δq δt ( ) F S = T V = T V ( ) S T V = T ( k BT log Z) C V = k B T (T log Z) T = k B T ( log Z + T ) T T log Z = k B T T log Z + k BT T log Z = ( k B T ) T T log Z (3.11)

28 3 4 (3.8) T = β T C V = T C V = β = 1 k B T β ( ) β log Z ( ) E T V (3.1) (3.7) Z = 1 N! dγ h f e βe(γ) dγ = dp 1 dp dp N dq 1 dq dq N, E(Γ) = i Z = 1 N!h 3N p i m dp 1 dp dp N dq 1 dq dq N e β 1 m (p 1 +p + ) Z = V N (πmk B T ) 3N/ (3.13) N!h 3N λ T def = h πmk B T (3.14) ( V ) N (3.15) Z = 1 N! λ 3 T F = k B T log Z [ = k B T N log V N log N + N log h 3N + 3N ] log(πmk BT )

29 3 5 = k B T N [ log V N log πmk ] BT h E = β log Z = β log(aβ 3N/ ) = 3N k BT (3.16) S N = 1 F (3.17) N[ T = k B log V N log πmk BT + T 3 ] (3.18) h T [ = k B log V N log πmk ] BT (3.19) h E/ T 3Nk B / Problem 3.3 5Nk B / H = p m + 1 I ( p θ + 1 sin θ p ϕ ) 3.6. H = p m + mω x (3.0) ( ϵ n = n + 1 ) hω (3.1) n N E = N i=1 ( n i + 1 ) hω

30 3 6 { exp β n 1 =0 ( n Z = = = n 1 =0 n =0 { exp ( n 1 =0 1 sinh β hω n N =0 [ β ) N e β N i=1 (n i+1/) hω ( n ) ] } N hω ) } hω = e β hω/ 1 1 e = 1 β hω e β hω/ e = 1 β hω/ sinh β hω Z F = k B T log Z ( F = Nk B T log sinh β hω ) E = log Z/ β E = N log sinhβ hω β = N hω cothβ hω k B T hω k B T hω (3.) (3.3) coth β hω = eβ hω + 1 e β hω 1 { kb T hω (k BT hω) 1 + e β hω, (k B T hω) E { Nk B T, N(e β hω + 1/) hω, (k B T hω) (k B T hω) Z classical Z = 1 { ( )} p N dx dp exp β h N m + mω x = ( kb T hω ) N (3.4) Z k B T hω sinh(β hω/) β hω/ Problem 3.4 Problem 3.5

31 R/ 5R/ E = 1 f c i ξi ξ i A f Z = A dξ i e β f i=1 c iξi / (πk B T ) = A f i c i i=1 E = log Z β ( E = β log A i=1 (πk B T ) f i c i 1 c iξi = A f i=1 dξ ie β f i=1 c iξi / 1c iξi A f i=1 dξ ie β f = 1 i=1 c iξi / k BT ) = f β = k BT f (3.5) k B T/ 3.8 van Leeuwen q A ϕ(r) H = 1 m (p qa) + Qϕ(r) (3.6) Z = A dpdre βh p = p qa A = 0 van Leeuwen

32 3 8 L L = mṙ qϕ(r) + qṙ A(r) (3.7) d dt ( L ẋ d L dt ẋ L x = 0 (3.8) ( ϕ = q x + q ẋ A x x + ẏ A y L x ) = ddt (mẋ + qa x) = m d dt x + q m d dt x + q ϕ x + q ), x + ż A z x ( t A x + ẋ x A x + ẏ y A y + ż z A z [ ( t A Ax x + ẏ y A ) ( y Ax ż x z A )] z = 0 x ) E = gradϕ(r) A, B = rota (3.9) t m d r = q [E + v B] (3.30) dt p def = L ṙ p = mṙ + qa (3.31) H = p ṙ L (3.3) H = (mṙ + qa) ṙ L ( mṙ = mṙ + qṙ A = 1 m (p qa) + qϕ(r) ) qϕ(r) + qṙ A(r)

33 /N! V A N A T V B N B T A m A B m B (3.13) Z A = V N A A (πm Ak B T ) 3N A/ N A!h 3N A Z B = V N B B (πm Bk B T ) 3N B/ N B!h 3N B Z A+B = (V A + V B ) N A+N B (πm A k B T ) 3N A/ (πm B k B T ) 3N B/ N A!N B!h 3(N A+N B ) (3.33) F = k B T (log Z A+B log Z A log Z B ) ( = k B T N A log V A + V B + N B log V A + V B V A V B ) S = F T = k B ( N A log V A + V B V A + N B log V ) A + V B V B (3.34) Z A+B = (V A + V B ) N A+N B (πmk B T ) 3(N A+N B )/ (N A + N B )!h 3(N A+N B ) (3.35) [ ] F = k B T log (V A + V B ) (N A+N B ) log V N A NB A VB log (N A + N B )! N A! N B! = k B T [(N A + N B ) log(v A + V B ) N A log V A N B log V B (N A + N B ) log(n A + N B ) + N A log N A + N B log N B ] [ = k B T N A log V A + V B N A + N A log V A + V B N B V A N A + N B V B N A + N B F = 0, S = 0 ]

34 V 0 x i N V (r 1,, r N ) = i<j V ( r i r j ) (3.36) dr 1 dr N = V N def Q N = dr 1 dr N exp( β V ( r i r j )) i<j = dr 1 dr N (1 + f ij ) i<j f ij = (e βv ( r i r j ) 1) (3.37) f ij Q N = dr 1 dr N (1 + f ij + f ij f kl + ) i<j ijkl dr 1 dr N (1 + f ij ) i<j V N + V N(N 1) N dr 1 dr f 1 r 1 r = r r 1 [ Q N = V N 1 + N 1 ] drf r v v = V/N ( N 1 Q N V N exp v drf r ) (3.38) Z N Z N = V ( N N 1 exp N!λ 3N T v F = F 0 k B T N 1 v drf r ) (3.39) drf r (3.40)

35 3 31 F 0 (3.16) P = F/ V P = Nk BT k BT drf V v r = Nk ( BT 1 1 ) V v 4π drr f r 4π drr f r = 4π 3 (a)3 = 3π 3 a3 P = Nk BT V ( v ) 16πa 3 3 ) P (V N 16πa3 = Nk B T (3.41) 3 N 16πa3 3 4πa3 N 4πa h +0 e βh ρ def = x e βh x (3.4) x >= U ν > Trρ = x x e βh x = ν ν e βh ν = ν e βe ν = Z (3.43) ) ( h H i x, x ϕ ν (x ) = E ν ϕ ν (x ) (3.44) TrU AU = TrAUU = TrA

36 3 3 x e βh x = ν x ν e βeν ν x = ν e βeν ϕ ν (x )ϕ ν(x ) e βh ϕ ν (x ) = e βeν ϕ ν (x ) (3.45) x e βh x = e βh ν ϕ ν (x )ϕ ν(x ) ϕ ν (x )ϕ ν(x ) = δ(x x ) (3.46) ν x e βh x = e βhδ(x x ) (3.47) δ(x) = 1 π dkeikx N ( [ N ( ) x e βh x 1 = exp β m i x i + V (x 1,, x N )]) δ(x 1 x 1) δ(x N x N) = 1 (π) 3N 1 i f( 1 i i dk 1 x (eikx Φ(x)) = e ikx dk N e β[ ] e i N i ( k + 1 i k i (r i r i ) ) Φ(x) (3.48) x ( x )(eikx Φ(x)) = e ikx f k + 1 ) Φ(x) (3.49) i x ( h ) x e βh x 1 = dk (π) 3N 1 dk N ( [ e i N N ( ) i k i (r i r i ) 1 exp β hk i + V (x 1,, x N )]) m i x i hk = p h / x Z = Trρ = dx 1 dx N x e βh x i Z = 1 (π h) 3N dx 1 dx N (3.7) dp 1 dp N exp [ β ( N i p i m + V (x i,, x N ) )] (3.50)

37 S = k B log W k B log Ω W E Ω(E) W = D E = Ω(E + E) Ω(E) E D. S/ E = 1/T 3. T E E/ V = p µ E/ N = µ 1. ν P ν P ν e βϵν. Z Z = ν e βϵν P ν = e βϵν /Z 3. S = k B ν P ν log P ν 4. F = E T S = ν ϵ νp ν + k B T ν P ν log P ν F = k B T log Z E = log Z/ β 5. F/ T = S, F/ V = P, F/ N = µ 1 ( ϵ n = hω n + 1 )

38 3 34 N E = hω N i=1 ( n i + 1 ) 3.1. N : E = hω N i=1 ( ni + 1 ) W W W = M = E N hω hω n 1 =0 n N =0 δ n1 +n + +n N,M (3.51) δ m,m = 1 π dθe i(m m)θ (3.5) π W = n 1 =0 n N =0 π = 1 π = 1 π π π π π 1 π dθe i(m (n 1+n + +n N ))θ π π ( ) N dθ e inθ e imθ ( dθ Taylor ( ) N 1 = 1 + Nx + 1 x S = k B log W n=0 1 1 e iθ N(N + 1) x + +! ) N e imθ W = 1 N(N + 1) (N + M 1) π π M! N(N + 1) (N + m 1) x m + (3.53) m! = (N + M 1)! M!(N 1)! = k B [(N + M 1) log(n + M 1) M log M (N 1) log(n 1)] [ = k B (N 1) log N + M 1 + M log N + M 1 ] N 1 M [ ( k B N log 1 + M ) + MN ( N log 1 + N )] M (3.54)

39 3 35 M = Ē hω N S = k B N [ log M/N = n ( 1 + E ) ( E + N hω N hω 1 ) ( log 1 + S = k B N S/ E = 1/T [ log(1 + n) + n log ( )] n = k B N [(n + 1) log(1 + n) n log n] 1 T = S E = n S E n = k B hω log n + 1 n 1 E N hω 1 )] n + 1 n = e β hω, n = 1 e β hω 1 E = N hω e β hω 1 + N hω ( 1 = N hω e β hω ) (3.55) (3.56) C = E T hω e β hω ( x ) = N hω k B T (e β hω 1) = k 1 BN sinh (x/) (3.57) x = β hω N : Z = = = n 1,n,,n N e βe n 1 =0 e β hω(n 1+1/) ( e β hω/ 1 e β hω e β hω(n +1/) n =0 ) N = ( e β hω/ e β hω 1 ) N

40 3 36 F F = k B T log Z = Nk B T ( log e β hω/ log(e β hω 1) ) ( = N hω ) + k BT log(e β hω 1) F = N ) ( hω + k BT log(1 e β hω ) (3.58) E = β log Z = N [ ] β hω log(e β hω 1) β ] [ hω hωeβ hω = N e β hω 1 [ ] 1 = N hω + 1 e β hω 1 S = F T = β T F β [ N = 1 k B T β log(1 e β hω ) + 1 ] hωe β hω β 1 e β hω ] = Nk B [ log(1 e β hω 1 ) + β hω e β hω 1 n = 1/(e β hω 1) S = k B N [(n + 1) log(1 + n) n log n] Problem N : H = p 1 m + p m + p N m + mω q1 + mω q + + mω q N (3.59)

41 3 37 E Ω(E) Ω = 1 dp h N 1 dp N dq 1 dq N (3.60) x i def = H(p,q)<E mω q def i, y i = 1 m p i (3.61) r N r N π N /N! Ω = = = ( ) N mω 1 1 dx m h N 1 dx N dy 1 dy N x 1 + +y 1 ( ) <E N πe 1 hω N! ( ) N E 1 hω N! S = k B log 1 ( ) N E k B N N! hω [ log E ] hωn + 1 (3.6) 1 T = S E = Nk B E, E = Nk BT (4 ) Problem N : Z Z = 1 h N exp [ β dp 1 dp N dq 1 dq N ( p 1 m + p m + p N m + mω q1 + mω q + + mω q N )] Z = 1 ( ) N πm π = h N β βmω ( kb T hω F = k B T log Z = k B T N log(k B T/ hω) ) N

42 3 38 (3.58) [ hω F = N + k BT log N (1 1 + β hω (β hω) [ hω + k BT log β hω + k B T log(1 β hω/) Nk B T log k BT hω )] + ]

43 T T µ ν N ν E ν B P ν D B(N tot N ν, E tot E ν ) E D tot (N tot, E tot ) E [ ] 1 exp (S(N tot N ν, E tot E ν ) S(N tot, E tot )) k B S(N tot N ν, E tot E ν ) = S(N tot E tot ) de = T ds pdv + µdn S N ν S E ν N tot E tot ds = de + pdv µdn T S(N tot N ν, E tot E ν ) = S + µ T N ν E ν T P ν exp( β(e ν µn ν )) (4.1) N = P ν N ν, E = P ν E ν, 1 = P ν (4.) ν ν ν

44 4 40 S = k B P ν log P ν λ E ( P ν E ν E ) λ N ( P ν N ν N ) λ 0 ( ν ν ν ν S/ P ν = 0 P ν 1) (4.3) k B (log P ν + 1) λ E E ν λ N N ν λ 0 = 0 [ P ν exp 1 ] (λ E E ν + λ N N ν ) k B λ E = 1 T, λ N = µ T P ν e β(e ν µn ν ) (4.4) (4.5) 4. P ν = 1 Ξ e β(eν µnν), P ν = 1 ν Ξ def = ν e β(eν µnν) = N e βµn Z N (4.6) Z N N Z/ β ν β log Ξ = (E ν µn ν )e β(eν µnν) Ξ = ν (E ν µn ν )P ν = E + µ N E = T S P V + G = T S P V + µn P V T S = log Ξ (4.7) β

45 4 41 ν µ log Ξ = β N νe β(e ν µn ν ) Ξ = ν βn ν P ν = β N S = k B ν P ν log(e β(e ν µn ν ) /Ξ) P V = T S E + µ N N = k B T log Ξ (4.8) µ = k B T ν P ν log e β(eν µnν) Ξ E + µ N = k B T ν P ν log Ξ P V = k B T log Ξ (4.9) H( p) = p x + p y + p z m z = V dp h 3 x dp y dp z exp( βh( p)) h. λ T = πmk B T z = V (πmk BT ) 3/ h 3 z = V λ 3 T

46 N Ξ = Z N = zn N! e βµn Z N = N=0 N=0 βµn zn e N! x = e βµ z ( )] Answer: Ξ = exp [e βµ 4. N = k B T µ ( Answer: N = e βµ V λ 3 T V λ 3 T log Ξ N ), (5.69) 5. pv = k B T log Ξ = N k B T

47 N N N 1 H ν >= ϵ ν ν > (5.1) ϵ ϵ < ϵ ν < ϵ + ϵ ρ(ϵ) ϵ ρ(ϵ) ϵ ν < ϵ N(ϵ) 1 Θ(x) 1, x > 0 Θ(x) = 1/, x = 0 0, x < 0 (5.) N(ϵ) = ν Θ(ϵ ϵ ν ) (5.3) 1 N Ω

48 5 44 ρ(ϵ) = dn(ϵ) dϵ δ(x) = dθ(x) dx = ν (5.4) δ(ϵ ϵ ν ) (5.5) N(ϵ) ρ(ϵ) E p L 1 N(E) = 1 h p m < E (5.6) me < p < me (5.7) me me ρ(e) = dpdq = ml π h mel π h (5.8) 1 E (5.9) h m x ψ ν = ϵ ν ψ ν (5.10) ψ ν (x + L) = ψ ν (x) ψ ν = 1 L e ikx, k = πn L (5.11) n ( ) ϵ n = h π n (5.1) m L ϵ n < E n n < me L π h (5.13) n n me L π h (5.8)

49 E p p x + p y m < E (5.14) p 0 = me (5.15) S N(E) = 1 dpdq = Sp 0 h 4π h = SmE π h (5.16) p x+p y<p 0 1 ρ(e) = Sm π h (5.17) E Si-MOS GaAs- AlGaAs p p x + p y + p z m < E (5.18) p 0 = me (5.19) V N(E) = 1 dpdq = V p3 0 h 3 6π h 3 = V me 6π h 1 p x+p y+p z<p 0 3 (5.0) ρ(e) = V m 3 E (5.1) 4π h Problem E 1/, E 0, E 1/ d E (d )/ Problem Z = ν e βϵ ν = 4 (3.13) 0 deρ(e)e βe (5.)

50 He (Pauli) 5.3 N H N = N h i, i=1 h i = h m r i + V (r i ) (5.3) h i i ν > ϵ ν E N ν > n ν E = ϵ ν n ν, N = n ν (5.4) ν ν {n ν } ( ) P ({n ν }) P ({n ν }) = 1 Ξ e β(e({nν}) µn) = 1 Ξ e β ν (ϵν µ)nν (5.5) {n ν } e β ν(ϵν µ)nν = ν ( n ν e β ν (ϵν µ)nν ) (5.6) n ν n ν = 0, 1 n ν = 0, 1,, fm

51 n ν n ν = {nν } P ({n ν })n ν = = {n ν } n νe β ν (ϵ ν µ)n ν {n ν } e β ν (ϵ ν µ)n ν n ν=0,1 n νe β ν (ϵ ν µ)n ν n ν e β ν (ϵ ν µ)n ν = 0 e0 + 1 e β(ϵν µ) 1 + e β(ϵ ν µ) f F (ϵ ν ) def = n ν = 1 e β(ϵ ν µ) + 1 (5.7) f F (ϵ ν ) ϵ ν T = 0 f F (ϵ) = Θ(µ ϵ) ϵ F k B T F = ϵ F p F /m = ϵ F k F = p F / h N 1/ ±1/ N = V h 3 dp x dp y dp z = V 4πp 3 F p <p F h 3 3 = V 3π k3 F (5.8)

52 5 48 n = N/V k F = (3π n) 1/3 (5.9) v F = p F m = h(3π n) 1/3 m p F = h(3π n) 1/3 (5.30) ϵ F = h m (3π n) /3 (5.31) Problem 5.3 n = m 3 1. Fermi Fermi ( m 1, m). Fermi (7.1 ev) 3. Fermi (8. ) Problem 5.4 N k F (5.1) N = ϵf 0 (5.3) ρ(ϵ)dϵ (5.33) 5.5 ( III) ev 1eV = K k B (5.34) E N = dϵρ(ϵ)f F (ϵ) (5.35) E = dϵρ(ϵ)f F (ϵ)ϵ (5.36)

53 [ ( )] 5.1: 1/ cosh β(ϵ µ) ev 5eV, 500K ϵ F µ = ϵ F + AT + BT + 1/(e C/T + 1) (Sommerfeld) dϵg(ϵ)f F (ϵ; µ, T ) = µ dϵg(ϵ) + π 6 (k BT ) g (µ) + O(T 4 ) (5.37) ( dϵg(ϵ) f ) F = G(µ) + π ϵ 6 (k BT ) G (µ) + O(T 4 ) (5.38) G(x) = x dx g(x )G()f F () = G( )f F ( ) = 0 g g(x) = G(x) = 0, x < 0 f F (x) x g f F ϵ f F µ f F ϵ = βeβ(ϵ µ) (e β(ϵ µ) + 1) = β 1 [ ( cosh β(ϵ µ) )] (5.39) ϵ µ 0 0

54 5 50 G(ϵ) ϵ µ G µ ( dϵg(ϵ) f ) ( F = dϵ f ) { } F G(µ) + G (µ)(ϵ µ) + G (µ) (ϵ µ) + ϵ ϵ (5.40) ( dϵ f ) F (ϵ µ) n = (k B T ) n e x dx ϵ (e x + 1) xn = (k B T ) n I n (5.41) I 0 = e x dx (e x + 1) = [ ] 1 = 1 (5.4) e x + 1 I 1, I 3 0 I = π 3 ( dϵg(ϵ) f ) F = G(µ) + G (µ) π ϵ 3 (k BT ) + O(T 4 ) ( ) I J(k) = e ikx dx (e x + 1)(e x + 1) = (ik) m I m (5.43) m! I m = 1 d m J(k) i m dk m (5.44) k=0 I m J(k) J(k) = m=0 πk sinh πk 1 π k 6 I = π /3 ζ(x) = n=1 (5.45) 1 n x (5.46) x =, 4, J(k) Appendix

55 µ(t ) µ(t ) 1 N = dϵρ(ϵ) e β(ϵ µ) + 1 [ µ ] N dϵρ(ϵ) + π 6 (k BT ) ρ (µ) 0 [ ϵf 0 dϵρ(ϵ) + (µ ϵ F )ρ(ϵ F ) + π 6 (k BT ) ρ (ϵ F ) N = ϵ F 0 dϵρ(ϵ) ], (µ ϵ F )ρ(ϵ F ) = π 6 ρ (ϵ F )(k B T ) (5.47) µ = ϵ F π 6 D D (k BT ) (5.48) D D = d log ρ(ϵ F) dϵ F = 1 ϵ F (5.49) µ = ϵ F [ ( 1 π kb T 1 ϵ F ) ] (5.50) Problem 5.5 (5.49) (5.35) E = dϵρ(ϵ) e β(ϵ µ) + 1 ϵ [ µ ] dϵρ(ϵ)ϵ + π 0 6 (k BT ) (ϵρ(ϵ)) µ [ ϵf ] dϵρ(ϵ)ϵ + ϵ F ρ(ϵ F )(µ ϵ F ) + π 6 (k BT ) (ρ(ϵ F ) + ϵ F ρ (ϵ F )) 0

56 5 5 ϵf dϵρ(ϵ)ϵ = E(T = 0) (5.51) 0 (5.47) π ϵ F ρ(ϵ F )(µ ϵ F ) = ϵ F 6 ρ (ϵ F )(k B T ) E(T ) = E(T = 0) + π 3 ρ(ϵ F)(k B T ) (5.5) C = ( ) E T V = π 3 k Bρ(ϵ F )T = γt (5.53) γ = π 3 k B ρ(ϵ F) E(T = 0) ρ(ϵ) = A ϵ N = ϵf 0 dϵρ(ϵ) = A 3 ϵ F 3/ N = 4 3 ϵ Fρ(ϵ F ) (5.54) ϵf ϵf E(T = 0) = dϵρ(ϵ)ϵ = A dϵϵ 3/ = A ϵ F 5/ E(T = 0) = 3 5 Nϵ F (5.55) 3ϵ F /5 T ϵ F ± k B T k B T ρ(ϵ F ) k B T E = E(T ) E(0) B(k B T ) ρ(ϵ F ) B T 3Nk B / (5.54) C fermi = π k B T (5.56) C classical 3 ϵ F k B /ϵ F k B T/ϵ F 10

57 F = G pv G = Nµ, pv = k B T log Ξ F = G pv = Nµ k B T log Ξ = Nµ k B T log ν (1 + e β(ϵ ν µ) ) = Nµ k B T ν log(1 + e β(ϵν µ) ) = Nµ k B T 3 = = = β dϵρ(ϵ) log(1 + e β(ϵ µ) ) dϵρ(ϵ) log(1 + e β(ϵ µ) ) dϵn (ϵ) log(1 + e β(ϵ µ) ) ( dϵn(ϵ) d dϵ dϵf F (ϵ)n(ϵ) ) log(1 + e β(ϵ µ) ) F = Nµ dϵf F (ϵ)n(ϵ) (5.57) µ ϵf 0 dϵf F (ϵ)n(ϵ) dϵn(ϵ) + π 3 (k BT ) N (µ) dϵn(ϵ) + µ ϵ F dϵn(ϵ) + π 3 (k BT ) N (µ) N(ϵ) = Bϵ 3/ N(ϵ F ) = N 1 ϵf dϵn(ϵ) = B 0 5 ϵ F 5/ = 5 Nϵ F 3 [N log(1 + e β(ϵ µ) )] 0 log 0

58 5 54 (µ ϵ F )N(ϵ F ) = N(µ ϵ F ) F = 3 5 Nϵ F π 3 (k BT ) ρ(ϵ F ) (5.58) S = F T = π 3 k Bρ(ϵ F )T (5.59) log T 3 C = T ds/dt = S pv = 5 Nϵ F + µ (5.50) µ ϵ F ϵ F dϵn(ϵ) + π 3 (k BT ) ρ(ϵ F ) dϵn(ϵ) N(µ ϵ F ) = N π (k B T ) 1 ϵ F (5.54) π 3 (k BT ) ρ(ϵ F ) = π 4ϵ F (k B T ) N [ ( ) ] pv = Nϵ F 5 + π kb T 6 ϵ F (5.60) pv = Nk B T Nϵ F B z E ± = p m µ HB (5.61)

59 5 55 µ H µ µ H ±µ H B N = = = = ϵf 0 µ(b) ρ(ϵ)dϵ µ H B µ(b)+µh B 0 µ(b) 0 ρ(ϵ + µ H B)dϵ + ρ(ϵ)dϵ + µ(b) µ H B µ(b) µh B 0 ρ(ϵ)dϵ + O(B ) ρ(ϵ µ H B)dϵ ρ(ϵ)dϵ M = µ Hρ(ϵ F )B (5.54) χ µ(b) = ϵ F + O(B ) (5.6) µ(b) µ(b) M = µ H ρ(ϵ + µ H B)dϵ µ H ρ(ϵ µ H B)dϵ µ H B µ H (µ H B)ρ(ϵ F ) M classical µ H B 3 µ H B M = Nµ H (5.63) ϵ F χ def = M B (5.64) 3 µ H χ = Nµ H (5.65) ϵ F e βµ HB e βµ HB M classical = N(µ H P + µ H P ) = Nµ H e βµ HB + e N µ H B βµ HB k B T (5.66) M fermi = 3 k B T (5.67) M classical ϵ F

60 βµ 1 µ f F e β(ϵ µ) (5.68) 1 ρ(ϵ) = AV ϵ N = AV n = N/V 0 dϵ ϵe β(ϵ µ) = AV e βµ (k B T ) 3/ dx xe x = πav e βµ (k B T ) 3/ 0 n = πae βµ (k B T ) 3/ (5.69) log n = log( πa) + βµ + 3 log(k BT ) µ = k B T [ n log 3 ] πa log(k BT ) (5.70) Problem 5.6 (5.70) (8.17) n ν 1 1 n ν n 1ν 1 n ν = T (5.71) A N = AV ϵf 0 dϵ ϵ = 4 3 AV ϵ F 3/ A [ 4 µ = k B T log 3 π 3 log k ] BT = k B T log ϵ F [ 4 3 π ( TF T ) 3/ ] (5.7) µ 3 T log T (5.73)

61 n ν = {nν } P ({n ν })n ν = = {n ν } n νe β ν (ϵ ν µ)n ν {n ν } e β ν (ϵ ν µ)n ν n ν =0 n νe β(ϵν µ)nν n ν=0 e β(ϵν µ)nν = 0 e0 + 1 e β(ϵν µ) + e β(ϵν µ) e β(ϵ ν µ) + e β(ϵ ν µ) + n ν = 1 β = 1 β ϵ ν log n ν =0 e β(ϵ ν µ)n ν 1 log ϵ ν 1 e β(ϵν µ) f B (ϵ ν ) = n(ϵ ν ) = 1 e β(ϵ ν µ) 1 (5.74) 1. f B < 0 µ minϵ ν µ 0. N N = ν f B(ϵ ν ) µ A = m 3 4π h D = A ϵ 3 A N = AV 0 ϵ dϵ e β(ϵν µ) 1 (5.75) n = N V x = A(k B T ) 3/ dx e x βµ 1 = A(k B T ) 3/ I(βµ) 0 I(y) = 0 x dx e x y 1 (5.76)

62 5 58 ξ def = e βµ I(βµ) = = = = = xe x ξ dx 0 1 e x ξ dx xξ n e nx n=1 0 ξ n dx xe nx n=1 0 π ξ n n=1 π n=1 1 n 3/ ξ n 1 n 3/ def = ϕ(z, ξ) (Appel) ξ = 1 n=1 ξn 1 n z (Riemann) ζ(z) def µ 0 I(βµ) I(0) = = π ζ(3/) = n=1 1 n z Appendix π.61 (5.77) π n = A(k B T ) 3/ I(βµ) A(k B T ) 3/ ζ(3/) (5.78) T T c k B T c = ( ) /3 h n (5.79) πm ζ(3/) T < T c ϵ = 0 ϵ = 0 N 0 ϵ N 0 = N AV dϵ 0 e β(ϵ ν µ) 1 π = AV ζ(3/)((k BT c ) 3/ (k B T ) 3/ ) T = T c N 0 0 [ ( ) ] 3/ T N 0 = N 1 (5.80) T c

63 5 59 Problem He T c T F K K 4 He T c 8.5 (5.70) [ µ = k B T log n 3 ] πa log(k BT ) (5.81) 1 π n = A(k B T c ) 3/ ζ(3/) (5.8) µ = k B T log [ ζ(3/) ( ) ] 3/ Tc T (5.83) µ k B T log T (5.84) f( ω = πf) ϵ = hω (5.85) E({n ν }) = ν ϵ ν n ν (5.86) Z = {n ν } e β ν ϵνnν = = ( ) ( ) e βϵ 1n 1 e βϵ n n 1 =0 n =0 1 1 e 1 βϵ 1 1 e βϵ

64 5 60 n ν n ν = {n ν } n νe β ν ϵ ν n ν = 1 log Z (5.87) {n ν } e β ν ϵ ν n ν β ϵ ν n ν = 1 e βϵν 1 (5.88) (Planck) k N(k) = 4πk3 3 ϵ = hck V (π) 3 (5.89) N(ϵ) = V ( ϵ ) 3 (5.90) 3π hc ρ(ϵ) = d dϵ N(ϵ) = V π ϵ ( hc) 3 (5.91) E = ν n ν ϵ ν = = = 0 dϵρ(ϵ)f B (ϵ)ϵ V dϵϵ 3 1 π ( hc) 3 0 e βϵ 1 V π ( hc) (k BT ) 4 dxx e x dxx 3 1 e x 1 0 dxx 3 1 e x 1 = (Appendix ) 0 dxx 3 e x (1 + e x + e x + ) = 6ζ(4) = π4 15 E = E V = V π 15( hc) 3 (k BT ) 4 (5.9) π 15( hc) 3 (k BT ) 4 (5.93)

65 5 61 u u = E π/ V c cos θ sin θdθ 0 π sin θdθ = c 4 0 E V (5.94) u = π k 4 B 60 h 3 c T 4 (5.95) (Stefan Boltzmann) π k 4 B 60 h 3 c = J/m sec K 4 (5.96) T 4 ϵ = hck ϵ = hvk v 10 3 m/s E = V π 10( hv) 3 (k BT ) 4 (5.97) 10 C V = E T T 3 (5.98) (Debye) γt K T 3 100K 3R 3 3R (Delong-Putit) Planck Planck ϵ 1, ϵ (> ϵ 1 ) hω = ϵ ϵ 1 n ϵ i (i = 1, ) N i dn dt = AnN 1 + A(n + 1)N (5.99)

66 5 6 0 N /N 1 = exp( hω/k B T ) 1 n = exp( hω/k B T ) 1 (5.99) N > N 1 n n + 1 n n exp(a(n N 1 )t)

67 63 6 (1970 ) µ S m = µs H = i B m i = Bµ i S z i (6.1) Z = e βµbsz i = {Si z} N S e βµbsz i=1 Si z= S S e βµbsz i = e βµbs 1 (eβµb ) S+1 1 e βµbs Si z= S = e βµbs e (S+1)βµBS 1 e βµbs sinh βµb(s + 1/) = sinh βµb/ i (6.) ( ) N sinh βµb(s + 1/) Z = (6.3) sinh βµb/ F = k B T N log sinh βµb(s + 1/) sinh βµb/ (6.4)

68 6 64 Problem 6.1 ( z ) M {S zi } i µsz i e βh M = i m i = µ i S z i = {S zi } i e βh (6.5) M = k B T N [ βµ = [ S + 1 NSµ S M = k B T log Z B ( S + 1 ) { ( coth βµb S + 1 )} ] coth S + 1 S x 1 S coth x S βµ coth ( )] βµb (6.6) x = βsµb (Brillouin) B S (x) def = S + 1 S + 1 coth S S x 1 S coth x (6.7) S M = NSµB S (x) (6.8) Sµ S lim S B S (x) coth x 1 x (6.9) (Langevin) Problem 6. S = 1/ B s (x) = coth x coth x = 1 + tanh x tanh x 1 tanh x = tanh x (6.10) M = Nµ tanh βµb coth z 1 z + z 3 (6.11) (6.1)

69 B s (x) = S + 1 ( S 1 S S + 1 x + 1 ) S S x 1 ( S S x + 1 ) x 3 S = (S + 1) 1 x = S + 1 1S 3S βsµb χ M = NSµ S µb k B T = 1 k B T χ def = ( ) M B T χ = Nµ S(S + 1) 1 3k B T S(S + 1) Nµ B (6.13) 3 (6.14) (6.15) (Curie) lim z coth z = 1 ( S + 1 M = NSµ 1 ) = NSµ (6.16) S S 6. M 1.. m i = µs i 1K 1000K 1 coth z = 1/ tanh x 1+z / z+z 3 /6 = 1 z (1 + z /3)

70 6 66 (Heisenberg) Ψ = χ(σ 1, σ )ϕ(r 1, r ) (6.17) χ ϕ(r 1, r ) H = J ij S i S j (6.18) ij (Ising) H = J ij S z i S z j (6.19) 6.3 H = J ij S z i S z j (6.0) ij (nearest neighbor) i h i = JSi z Sj z (6.1) j:n.n.

71 6 67 j:n.n. of i Sz j i Sj z Sz j def = m h MFA i = zjmsi z (6.) H = J ij [(S z i m)(s z j m) m + m(s z i + S z j )] = J ij (S z i m)(s z j m) + JzNm Jm z i S i H MFA = Jzm i S i + JzNm (6.3) Jzm = Bµ 9.1 s = 1/ B s=1/ (x) = tanh x (6.4) M = B s=1/ (SβµB) Nµ = Nµ tanh βzjm M M = Nµ Si z = Nµm (6.5) m m = 1 tanh βzjm (6.6) βzj < 1 m = 0 βzj > 1 m = 0, m = ±m 0 zj k B T c = 1 (6.7) T c T > T c 0 T < T c m = m 0

72 6 68 m 0 T c m 0 = 1 tanh βzjm 0 = 1 ( βzjm ) 3 (βzjm 0) 1 = 1 ( βzj 1 1 ) 3 (βzjm 0) = T c (1 13 ) T (βzjm 0) (6.8) 1 T T c = 1 3 (βzjm 0) = 1 3 m 0 = 3 T T c ( Tc T ) m 0 Tc T T c (6.9) m T < T c m = 0, ±m 0 m m = 0 F = k B T N[log( cosh βzjm) βzjm ] (6.30) F para = k B T N log (6.31) [ F ferro k B T N log + log (1 + (βzjm 0) + (βzjm ) ] 0) 4 βzjm 0 4 log(1 + ax + bx ) = ax + ) (b a x + O(x 3 ) (6.3)

73 6 69 = log (1 + (βzjm 0) + (βzjm ) 0) 4 βzjm 0 4 ( 1 ) ( (βzjm0 ) 1 + βzj 4 1 ) (βzjm 0 ) 4 8 ) (1 TTc (βzjm0 ) 1 1 (βzjm 0) 4 (6.9) F = k B T N βzjm 0 = 3 T c T T c def = 3 t (6.33) [ ( 3 log + t 3 )] 4 t = k B T N [log + 34 t ] (6.34) m = 0 m = m 0 3Nk B T t /4 T < T c T = T c C = T S T = T F T { kb T N log, T > T c F = k B T N [log + 3(Tc T ) 4T c ], T T c (6.35) { 0, T > Tc C = 3Nk B (T +(T T c ))T (6.36), T T Tc c C = 3Nk B (6.37) z, J n n

74 : : 3 : T c x x T T c T c y (6.38) y T c (universality) m 0 ( Tc T m 0 T c ) β, β = 1 (6.39) 1/ ( ) α Tc T C, α = 0 (6.40) T c 0 χ = dm/db H MFA = Jzm i S i + JzNm h i S i = (Jzm + h) i S i + JzNm (6.41) h 3 h, m( h) m = 1 tanh β(zjm + h) (6.4) m 1 β(zjm + h) β, 1/k B T 3 h = gµ B B g g- µ B

75 6 71 ( m 1 T ) c = hβ T m = 1 h k B (T T c ) χ γ χ m h 1 T T c (6.43) χ (T T c ) γ, γ = 1 (6.44) m (6.4) T = T c β c zj/ = 1 m m + β ch 4 ( m + β ) 3 ch (6.45) 3 3β c h = ( m + β ) 3 ch (6.46) m h h 3 h m h δ (6.46) m h 1/δ, δ > 1 (6.47) m h 1/3, δ = 3 (6.48)

76 7 7 N N N 7.1 C V = ( ) E T V = T ( log Z ) β Z = ν exp( βe ν) (4 ) C V = = = = 1 log Z k B T β 1 k B T β 1 Z k B T 1 k B T ( T = 1 k B T β ν exp( βe ν)e ν Z ν exp( βe ν)e ν ( ν exp( βe ν)e ν ) [ ν exp( βe ν)e ν Z Z ) ( ν exp( βe ν)e ν Z ) ] (7.1) ν P ν P ν = exp( βe ν )/Z C V = 1 k B T [ ν P ν E ν ( ν P ν E ν ) ] = 1 k B T ( E E ) (7.) E E =< (E E ) >= E C V = 1 k B T E (7.3) Problem

77 df = df (B = 0) M db (7.4) 1 M = ( ) F B T (9.1 ) = k B T B log Z = k BT Z Z (7.5) χ = M B = k BT ZZ Z Z (7.6) H 0 H(B) = H 0 µb i S i = H 0 MB (7.7) ( z z ) Z = {S i } e β(h 0 µb i S i) (7.8) Z Z = (βµ i S i)e β(h 0 µb i S i) Z {S i } Z Z = (βµ i S i) e β(h 0 µb i S i) Z {S i } = β M (7.9) = β M (7.10) χ = β( M M ) = 1 k B T M (7.11) 1 M M = µ i S i M

78 N 3 3N 3N 6N N N A (= mol 1 ) T P ρ V 6N 8. (thermodynamics): 1) T P ρ V S E F ) 1 de = Q W (8.1) 3)

79 8 75 ds > Q T (8.) Q T ds (8.3) 4) 8.3 E F =E T S G =F + P V H =E + P V S V T P µ N ρ E, F, G, H de = T ds P dv (8.4) df = SdT P dv (8.5) dg = SdT + V dp (8.6)

80 8 76 E, F, G, H E S V F V T G T P H S P 8. dh = ds + V dp (8.7) de + P dv T ds (8.8) 1. Q = 0 0 ds ds = 0. df + SdT = d(e T S) + SdT de Q = P dv dt = 0 df P dv 3. F 4. de = d(g+st P V ) = Q P dv T ds P dv dg+sdt V dp 0 G S F G dn µdn dg = V dp SdT + µ 1 dn 1 + µ dn (8.9) dn 1 + dn = 0 G, P, T µ 1 = µ (8.10)

81 ν 1 A 1 + ν A + + ν m A m ν 1B 1 + ν B + + ν nb n (8.11) H + O H O (8.1) ν 1 =, ν = 1, ν 1 = A 1 = H, A = O, B 1 = H O ν 1 A 1 + ν A + + ν m+1 A m+1 + ν m+ A m+ + = 0 (8.13) ν m+i = ν i, A m+i = B i dg = m+n i=1 N i ν i µ i dn i = 0 (8.14) dn i = ν i (8.15) dg = 0 = m+n i=1 m+n i=1 µ i ν i (8.16) µ i ν i = 0 (8.17)

82 de = ( E T P V = f(t ) (8.18) ( ) E V T = 0 (8.19) de = T ds P dv ds = de T + P dv (8.0) T ) dt V ds = ( ) E T V dt T + f(t ) T dv V (8.1) f(t ) T = const def = R (8.) R (= J/K mol) (Boltzmann) k B k B = R N A = J/K (8.3) R 8.5 G dg = SdT + V dp + µdn (8.9 ) λ G(T, P, λn) = λg(t, P, N) (8.4)

83 8 79 λ G(λN) λ = G(λN) λ=1 (λn) Problem 8.1 G = µn G = µn λg λ = G (λn) λ = µn G = µn (8.5) E(λS, λv, λn) = λe(s, V, N), F (T, λv, λn) = λf (T, V, N) dg = Ndµ + µdn = SdT + V dp + µdn Ndµ = SdT + V dp (8.6) Maxwell ( ) f df(x, y) = dx + x A, B ( A ) y x y = f(x, y) y x ( ) f dy = A(x, y)dx + B(x, y)dy (8.7) y x = f(x, y) x y = ( ) B x y de = T ds P dv ( ) ( ) T P = V S S V (8.8) (8.9) Maxwell C C V C V = Q T = T S T = T F T = ( ) E T V (8.30) C P ( ) ( ) E V C P = + P (8.31) T P T P C P = C V + R C V C P

84 E W S = k B log W E 1 T = S E Stirling N! = πnn N e N N- Ω N Ω N = π N Γ( N + 1) 1 W = dp (π h) f 1 dp dp f dq 1 dq dq f N! (f 3 f = 3N ) E E + de

85 9 81 E 1, E,..., E s,... E s P (E s ) P (E s ) exp ( βe s ), (β = 1 k B T ) E s Maxwell f(v x, v y, vz) exp ( β m v ) S = k B P s ln P s Z Z s Z = s exp ( βe s ) 1 Z = dp (π h) f 1 dp dp f dq 1 dq dq f exp ( βe(p 1,, q 1 )) N! Z F F = k B T log Z E ( E 1, E,, E s, ) E = log Z β C V C V = T S T = T ( F ) V = k B T (T log Z) V T T T

86 9 8 E i, N i i µ P (i) exp ( β(e i µn i )) Ξ Ξ Ξ = i exp ( β(e i µn i )) N Z(N) Ξ = N Z(N) exp (βµn) P V = k B T log Ξ < N > < N >= k B T log Ξ µ E E = µ < N > log Ξ µ N(E) E ϵ ν N(E) = E de δ(e ϵ ν ) N(E) = dx (p x + )/m E dp x (π h) d (d ) D(E) D(E) = dn(e) de

87 9 83 ν < n ν >= 1 exp (β(ϵ ν µ)) + 1 = f F (ϵ ν ; β, µ) (β ) θ(µ ϵ) µ ϵ F ϵ F = µ(t = 0) ϵ F ( µ) ev 1eV 1 ( K) ν < n ν >= 1 exp (β(ϵ ν µ)) 1 = f B(ϵ ν ; β, µ) 0 µ = 0 1 < n ν >= exp (βϵ ν ) 1 = f P (ϵ ν ; β) µ 0 < n ν >= Ae βϵ ν < N > < E > < N >= ν < n ν >= ded(e) < n ν > < E >= ν ϵ ν < n ν >= ded(e)e < n ν > < N > µ < E > T C V f F (E) deg(e)f F (E; β, µ) = µ deg(e) + π 6 (k BT ) g (µ) + O((k B T/µ) 4 )

88 9 84 ( deg(e) f F (E) E ) = G(µ) + π 6 (k BT ) G (µ) + O((k B T/µ) 4 ) < N > g(e) = D(E) < E > g(e) = ED(E) ( ( µ = ϵ F 1 π kb T ) ) 1 ϵ F < E >= 3 5 Nϵ F + π 6 D(ϵ F )(k B T ) < N > µ G = µ < N > pv = k B T log Ξ F = G pv ( ) ( ) n n - M T c M (T c T ) β χ χ (T c T ) γ β, γ < E > < M > C = 1 k B T < E > χ = 1 k B T < M > C χ

89 85 10 Appendix: ζ ζ(s) = s s s + = n=1 1 n s (10.1) n z(s) = s 3 1 s 4 + = ( 1) n 1 s n=1 n s (10.) ζ z z(s) = = 1 (s 1)! 1 (s 1)! ζ(s) = z(s) (10.3) 1 1/s 1 dxx s 1 e x (1 e x + e x e 3x + ) dxx s 1 1 e x + 1 (10.4) z(s) = 0 ( ) x s 1 dx s! e x + 1 = 0 dx xs e x (10.5) s! (e x + 1) s = m(m 0 ) (ik) m (ik) m z(m) = m=0 = 1 m=0 0 m=0 = 1 Re (ikx) m (m)! (ikx) m (m)! e x (e x + 1) dx e x (e x + 1) dx e ikx e x dx (10.6) (e x + 1) J(k) = e ikx e x dx (10.7) (e x + 1)

90 10 Appendix: ζ 86 [, + ] [ + πi, + + πi] J(k)(1 e πk e ikx e x ) = πires (e x + 1) (10.8) x=iπ (x iπ) x x = iπ ike πk J(k) = πk sinh(πk) = 1 π 6 k + 7π4 360 k4 31π k6 + (10.9) (ik) m z(m) = 1 πk sinh(πk) = 1 m=1 ) (1 π 6 + 7π π (10.10) z() = π /1, z(4) = 7π 4 /70, z(6) = 31π6 (10.3) 3040 ζ() = z() = π 6, (10.11) ζ(4) = 1 z(4) 1 1/ = z(4) = π4 90, (10.1) ζ(6) = z(6) 1 3 = z(6) 1 1/5 31 = π6 945 (10.13)

91 H = p m + mω x 1. 1 z ( n + 1 ) hω n 0 n 6. 1 z Z E F C V Z Z = exp( βe ν ) ν (β, ν, E ν 1/k B T, ) E = log Z β 3 Na n =.7 10 cm 3

92

93 Z E Z Z = ν exp( βe ν ) E = log Z β (β, ν, E ν 1/k B T, ) z = 1 h H(x, p) = p m + mω x dx dp exp( βh(x, p)) E n E n = ( n + 1 ) hω n 0 n z = n= z 7. 1 e βen

94 H( p) = p x + p y + p z m z = V dp h 3 x dp y dp z exp( βh( p)) h. λ T = πmk B T z = V (πmk BT ) 3/ h 3 z = V λ 3 T 3. N Ξ = Z N = zn N! e βµn Z N = N=0 N=0 βµn zn e N! x = e βµ z 4. N = k B T log Ξ N µ 5. pv = k B T log Ξ = N k B T

95 11 91, 3 dxe αx = π α ν P ν Z P ν = 1 Z exp( βe ν) Z = ν exp( βe ν ) E F E = log Z β, F = k B T log Z (β, ν, E ν 1/k B T, ) 1. S = F T S = k B P ν log P ν. C = E T E E ν z = 1 h H(x, p) = p m + mω x dx dp exp( βh(x, p))

96 E n E n = ( n + 1 ) hω n 0 n z = n= z e βen 3 N! N! = 0 dxx N e x x N e x = e f(x) f(x) = df(x) dx x 0 = f(x) x 0 = 0 x 0 f(x) 1 (x x 0) N N N! N log N N πne log N N log N!

97 Fermi ev. 3. C V = 1 k B T E 1. E = E E. ±ϵ P (ϵ) = E, E 3. e βϵ e, P ( ϵ) = e βϵ + e βϵ βϵ e βϵ + e βϵ 3 1. Fermi Bose Boltzmann. Bose 4 Planck f(ϵ) = 1 e βϵ 1 1. V D(ϵ) = V ϵ π ( hc) 3. V 3.

98 11 94 SEA Z Z = ν exp( βe ν ) β, ν, E ν 1/k B T, E = log Z β n = m /10 1. Fermi Fermi. Fermi 3. Fermi 4. Sommerfeld 5. Pauli 3 1. Fermi Bose Boltzmann, Planck. Fermi Fermi k B T 3. Bose 4. Planck E/V E V = k (k BT ) x ( hc) y k x, y 5.

99 11 95 SEA Fermi 1eV 1. Fermi kg Pauli 1. Fermi Fermi k B T. Bose 3. Planck E/V E V = k (k BT ) x ( hc) y k x, y E N(E) p = (p x, p y, p z ) p x + p y + p z m < E p p 0 = V N(E) = 1 dpdq = p 3 h 3 0 = p x +p y +p z <p 0 E 1 D(E) D(E) =

100 11 96 SEA n = m /10 1. Fermi Fermi. Fermi 3. Fermi 4. Sommerfeld 5. Pauli 1. Fermi Bose Boltzmann, Planck. Fermi Fermi k B T 3. Bose 4. Planck E/V E V = k (k BT ) x ( hc) y k x, y E N(E) p = (p x, p y ) p x + p y m < E p p 0 = S N(E) = 1 dpdq = p h 0 = p x +p y <p 0 E 1 D(E) D(E) =

101 11 97 dxe αx = π α , 4 1 Z E Z Z = ν exp( βe ν ) E = log Z β (β, ν, E ν 1/k B T, ) z = 1 h H(x, p) = p m + mω x dx dp exp( βh(x, p)) E n E n = ( n + 1 ) hω n 0 n z = n=0 e βen

102 z ±ϵ 1. +ϵ P +, ϵ P N! N! = 0 dxx N e x x N e x = e f(x) f(x) = df(x) dx x 0 = f(x) x 0 = 0 x 0 f(x) 1 (x x 0) N N N! N log N N πne log N N log N!

103 11 99 SEA Na n =.7 10 cm 3 = m 3 = 10 1/10 1. Fermi Fermi. Fermi 3. Fermi 4. Fermi Fermi k B T 5. Sommerfeld 1. Bose Boltzmann, Planck. Bose 3. Planck E/V E V = k (k BT ) x ( hc) y k x, y E N(E) p = (p x, p y, p z ) p x + p y + p z m p p 0 = V N(E) = 1 dpdq = p 3 h 3 0 = p x+p y+p z<p 0 < E E 1 D(E) D(E) =

104 Z E F Z Z = ν exp( βe ν ) (β, ν, E ν 1/k B T, ) 1. ν P ν exp( βe ν )/Z. E = ν E νp ν E = log Z β 3. F = E T S S = k B ν P ν log P ν F k B, T, Z z = 1 h H(x, p) = p m + mω x dxe αx = π α dx dp exp( βh(x, p))

105 E n E n = ( n + 1 ) hω n 0 n z = n= z e βen 3 ±ϵ 1. +ϵ P +, ϵ P

106 11 10 SEA E N(E) p = (p x, p y ) p x + p y m p p 0 = S N(E) = 1 dpdq = p h 0 = p x+p y<p 0 < E E 1 D(E) D(E) = Si-MOS FET cm 10 1/10 1. Fermi Fermi. Fermi Fermi k B T 3. Sommerfeld 3 1. Bose Boltzmann, Planck. Bose 4 Planck E/V k N(k) = ( 4πk 3 V ϵ = hck N(ϵ) = V ϵ ) 3 3 (π) 3 3π hc 1. D(E). E = ν n νϵ ν = 1 dϵd(ϵ) ϵ 0 exp(βϵ) 1 π 4 E/V 15 dxx 3 1 = 0 e x 1

107 Z E F Z = ν exp( βe ν ) E F E = log Z β, F = k B T log Z (β, ν, E ν 1/k B T, ) π dxe αx = α 1 1 H(x, p) = p m + mω x E n E n = ( n + ) 1 hω n 0 n z = e βe n n= z ±ϵ 1. +ϵ P +, ϵ P. 3.

108 H( p) = p x + p y + p z m 1. 1 z = V h 3 dp x dp y dp z exp( βh( p)) z = V (πmk B T ) 3/ h 3. λ T = h πmk B T z = V λ 3 T 3. N Ξ = Z N = zn N! e βµn Z N = N=0 N=0 βµn zn e N! x = e βµ z 4. N = k B T log Ξ N µ 5. pv = k B T log Ξ = N k B T 4 N! N! = 0 dxx N e x x N e x = e f(x) f(x) = df(x) dx x 0 = f(x) x 0 = 0 x 0 f(x) 1 (x x 0) N N N! N log N N πne log N N log N!

109 SEA E N(E) p = (p x, p y ) p x + p y m < E p p 0 = S N(E) = 1 dpdq = p h 0 = p x+p y<p 0 E 1 D(E) D(E) = n m 1. Fermi Fermi. Fermi 3. Fermi 3 f F (ϵ; µ, T ) Sommerfeld dϵg(ϵ)f F (ϵ; µ, T ) = µ dϵg(ϵ) + π 6 (k BT ) g (µ) + O(T 4 ) 4 1. Bose Boltzmann, Planck. Bose

110 Z E F ν P ν P ν = exp( βeν) Z Z = ν exp( βe ν ) E F E = log Z β, F = k B T log Z (β, ν, E ν 1/k B T, ) π dxe αx = α 1 E = log Z β z = 1 h H(x, p) = p m + mω x dx dp exp( βh(x, p))

111 E n E n = ( n + 1 ) hω n 0 n z = n= z e βen 3 3 H( p) = p x + p y + p z m 1. 1 z = V h 3 dp x dp y dp z exp( βh( p)) z = V (πmk B T ) 3/ h 3. λ T = h πmk B T z = V λ 3 T 3. N Ξ = Z N = zn N! e βµn Z N = N=0 N=0 βµn zn e N! x = e βµ z 4. N = k B T log Ξ N µ 5. pv = k B T log Ξ = N k B T

112 SEA n m 1. Fermi Fermi. Fermi 3. Fermi 4. Sommerfeld 5. Pauli Bose Boltzmann, Planck E N(E) p = (p x, p y, p z ) p x + p y + p z m p p 0 = V N(E) = 1 dpdq = p 3 h 3 0 p x+p y+p z<p 0 < E E 1 ρ(e) ρ(e) = 4 1/ ρ(e) = A = n n = A 0 1 exp(β(ε µ)) + 1 dε µ β = 1/k B T, T 1. ε F n, A. exp( β(ε µ)) dx e x e x + 1 = log(e x + 1) µ T, ε F