19 1 14 007 3 1 1 Ising 4.1................................. 4................................... 5 3 9 3.1........................ 9 3................... 9 3.3........................ 11 4 14 4.1 Legendre.............................. 14 4. Landau.......................... 14 1
1 s 1 H(s 1 ) N s 1, s,, s N H({s 1,, s N }) = N H(s k ) k=1 Z N =Tr {s1,,s N }e βh({s 1,,s N }) =Tr s1 Tr s Tr sn e β P k H(s k) N = Tr sk e βh(s k) k=1 1 f e βf =Tr s e βh(s) N f N e βf N =Z N f N N = f = f N 1 N N N:( ) P.W.Anderson More is different (P.W.Anderson)
z B N 1 e βf = H 0 =µbs z, S z = ± 1 S z =± 1 e βµbsz f = 1 β log ( cosh βµb N N H N =µb Sk, z Sk z = ± 1 k=1 e βf N = S z 1 =± 1 f N N f = f S z N =± 1 k = cosh βµb e βµbsz k ) ( βµb ) N = cosh m χ m = µ k Sz k N = 1 N 1 Z = f B = µ = µ = 1 N Z ( βb) = 1 N sinh βµb cosh βµb βµb tanh χ = m B = µ B=0 βµ = µ 4 β 1 T 1 Z Tr (e βh k log Z ( βb) 1 cosh βµb B=0 1/T 3 S z j )
1 Ising.1 Ising H =H 0 + H I H 0 =µb k S z k H I = J k S z ks z k+1 N SN+1 z = Sz 1 Z N =Tr {Sk }e βh =Tr S1 =±1/ Tr SN =±1/ exp{ β k ( JS z ks z k+1 + µb (Sz k + S z k+1)} =Tr S1 =±1/ Tr SN =±1/ T S1 S T S S 3 T SN 1 S N T SN S 1 =Tr T N T ab (transfer matrix) (ab) T ab = exp{βjab βµb (a + b)}, a, b = ±1 ( ) T = e K h e K e K e K+h K = βj 4 h = βµb λ ± (λ λ + ) 1 1 U Z N =e βf N = λ N + + λ N = λ N + T =UDU, D = diag(λ +, λ ) ( ) { N λ } 1 + λ + Tr T N =Tr UD N U = Tr UUD N = Tr D N 4
f N f = lim N N = 1 β log λ + det(t λi) =(e K h λ)(e K+h λ) e K =λ λe K cosh h sinh K = 0 λ ± =e K( cosh h ± sinh h + e 4K) f = 1 { ( K + log cosh h + sinh h + e β 4K)}. U = H = 1 Z Tr He βh = β log Z = β (βf) λ ± =e K cosh h ± e K cosh h e K + e K =e K( cosh h ± cosh h 1 + e 4K) =e K( cosh h ± sinh h + e 4K) 5
3 f = 1 log cosh K β E e = lim N N = J 4 tanh K e T = J 4 = J 4 K T K tanh K 1 kt cosh K C J 0.10 0.08 0.06 0.04 0.0 0.5 1.0 1.5.0 T J 1: 3 f = 1 β (K + log(1 + e K ) = 1 β (K + log(e K (e K + e K )) = 1 log cosh K β E N log cosh K β = K log cosh K β K = J 4 tanh K 6
m 4 m = f B = βµ f h = µ sinh h sinh h + e 4K -m/(µ/) 1.0 0.8 h=1 0.6 0.4 h=0.001 h=0.01 h=0.1 0. 0.0 0.5 1.0 1.5.0 T/J : lim lim m = µ h 0 T 0 lim sinh h h 0 sinh h = µ sgn h lim lim m =0 T 0 h 0 4 m = f B = βµ f h = µ sinh h + sinh h cosh h sinh h+e 4K cosh h + sinh h + e 4K = µ sinh h sinh h + e 4K 7
5 χ = m B = µ 4 B=0 = βµ 1 k B T exp( k B T ) m h h=0 χ T 3: 5 m h m = µ (1 + e 4K sinh h) 1/ m h = µ ( 1/)(1 + e 4K sinh h) 3/ ( e 4K ) sinh 3 h cosh h = µ e 4K (1 + e 4K sinh h) 3/ sinh 3 h cosh h = µ e 4K (sinh h + e 4K ) 3/ (sinh h) 3/ sinh 3 h cosh h = µ e 4K (sinh h + e 4K ) 3/ cosh h = µ h=0 ek 8
3 3.1 Ising H =µb i S z i J ij S z i S z j z ( z). i S z j, j ij S = S z j S z j j J ij S z i S z j J i S z i Sj z j ij = Jz S i S z i H H MF H MF =µb Si z J Sz i i =µb eff i S z i S z i B eff =B Jz µ S = B Jz µ m m =µ S ( ) B eff 3. f MF = 1 β log ( cosh βµb eff ) 9
m = f = µ B eff tanh βµb eff = µ tanh βµ ( B Jz µ m) B = 0 m = µ tanh βjz µ m T < TC T > TC m 4: m 0 µ βjz µ βjz 4 1 T < T C k B T C = Jz 4 m (phase transition) (order parameter) m m (Spontaneous symmetry breaking) 10
3.3 (Universality) T C B = 0 T < T C, (T T C ) 6 m = µ βjz tanh µ m µ ( βjz µ m 1 ) (βjz 3 µ m) 3 = β m (βjz)3 m 3 β C 48 m C( β β C 1) 1/ C ( TC T T C ) γ, γ = 1 m (TC-T) 1/ T TC 5: 6 tanh x x 1 3 x3 11
T > T C ( ) B m 7 χ = dm ( ) γ T TC db, γ = 1 B=0 T C χ (TC-T) -1 0 1 TC T 6: 7 m = µ µ 4 βµ Jz tanh (B µ m) β(b Jz µ µ 4 β CB + β m β ( ) C β 1 m =CB β C ( ) 1 T TC m B T C m) = µ 4 βb + Jz 4 βm χ = dm ( ) γ T TC db, γ = 1 B=0 T C 1
β = β C = 4/(Jz) m, B m = µ tanh β ( Cµ Jz (B µ m) = µ tanh βc µ B β ) CJz µ m = µ ( tanh βc µ B m ) µ µ { βc µ B m µ 1 (β C µ 3 B m } ) 3 µ β Cµ 4 B + m 1 m 3 6 µ B m 3 = m δ, δ = 3 β, γ, δ (critical exponents) 13
4 4.1 Legendre f(b) B e Nβ f =Tr e βh m = f B Legendre d f = f B db F =f(b) mb F F(m, B) (m, B) df =d f mdb Bdm ( ) f = B m db Bdm = Bdm F B, F = F(m) m 4. Landau B = 0 B = F m df dm =0 F m F(m) m m F m F(m) =am + bm 4 14
F T < TC T > TC m 7: m F infty b > 0 a { finite T < T C m = 0 T > T C m = { a < 0 a > 0 T < T C T > T C b =const. a =a 0 (T T C ), a 0 > 0 df dm =am + 4bm3 = 0 T < T C m = a b =C T T C, C = a0 b 15