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1 000 (N =000) 50 ( N(N ) / = ) μm.5 g cm -3.5g cm 3 ( 0 6 µm) 3 / ( g mo ) ( mo ) =

2 (0 6 ) ~ ( 0 6 ) ~ 0 9

3 q R q, R q q E = 4πε 0 R R (6.) -6 (a) (b) (c) (a) (b) (c) (a) (b) (c) (a) (b) (b) (c)

4 a, b, c q R φ( r,ρ) = π exp ( r ρ ) (6.) r = r f ( r,η) = φ r ( n aa nbb ncc,ρ ) n a = n b = n c = = exp r n aa nbb ncc ρ (6.3) π n a = n b = n c = r f f r,ρ r,ρ r (6.4) = F hk exp π ha * + kb * + c * h= k= = F hk a, b, c a a * =, a b * = 0, a c * = 0 b a * = 0, b b * =, b c * = 0 c a * = 0, c b * = 0, c c * = a *, b *, c * F hk F hk = f ( r, ρ) exp π ( ha * + kb * + c * ) r d r (6.5) Ce (6.3) F hk = φ r ( n aa nbb ncc,ρ ) exp π ( ha * + kb * + c * ) r d r Ce n a = n b = n c = = φ( r,ρ) exp π ha * + kb * + c ( * ) ( r + n aa + nbb + ncc ) d r ha * + kb * + c ( * ) ( n aa + nbb + ncc ) = hn + kn + n a b c F hk = φ ( r, ρ) exp π ( ha * + kb * + c * ) r d r (6.6) (6.)

5 F hk = π exp ( r ρ )exp π( ha * + kb * + c * ) r d r = π ρ exp π ha * + kb * + c * 3 ρ f r,ρ = π = π ρ 3 n a = n b = n c = h= k= = exp r n a a nb b nc c ρ exp π ha * + kb * + c * ρ exp π ( h a * + kb * + c * ) r f ( r,ρ) = exp r ρ = π exp π g π ρ 3 ρ g g = ha * + kb * + c * (6.7) (6.8) exp( π g r ) (6.9) r = π exp r ρ d ρ (6.0) 0 r = π exp r ρ d ρ + π exp r ρ d ρ (6.) 0 (6.) π exp r ρ d ρ = 0 (6.9) = π exp π g ρ 3 ρ exp π g r 0 g = exp π g r g 0 0 π exp r ρ d ρ π ρ exp π g 3 ρ d ρ = exp( π g r ) g πg exp π g ρ = exp π g r g (6.) 0 πg exp π g (6.) d ρ

6 π exp r ρ d ρ = r erf r ρ = r erfc r (6.3) r = exp π g r g N πg exp π g + r erfc r (6.4) erfc( x) compementary error functon erfc x π e t x dt (6.5) x (6.4) (6.4) = 0 r = 0 (6.4) = 0 erfc r r erfc r r r = πr r exp( z )d z π = r exp( z ) πr d z π r 0 0 (6.6) = 0 q R V = q q 4πε 0 R R = q q 4πε 0 R R + r = exp π g r (6.6) g erfc r r r π ( r 0) 0 q R R (6.7) πg exp π g + r erfc r (6.8)

7 V = q 4πε 0 g πg exp π g q π + q R R erfc R R q exp π g + 0 R R q R R erfc R R ( ) (6.9) g = ha * + kb * + c * = n aa + nbb + ncc g, g πg exp π g = s g s g h= s g s g h k= s g + h q exp π g R R exp π g πg =± s g h k + q π + q R R erfc R R 0 = q π + q R R erfc R R + s s n a = s s n a n b = s + n a n c =± ( s n a n b ) q q cos πg R R R R erfc R R q R R erfc R R (6.0) (6.) g g = 0 q q = 0 g g = 0 g

8 a R 0 = 0 R = 0.5b + 0.5c R = 0.5a + 0.5c R 3 = 0.5a + 0.5b R 4 = 0.5a R 5 = 0.5b R 6 = 0.5c R 7 = 0.5a + 0.5b + 0.5c R 0 = 0 4πε 0 a V 0 = 7 q = R = =.906 n a + n b + n c = V = 7 = q + R n a n a = n b = + n a n a n b n c = + n a + n b ( n a,n b,n c ) ( 0,0,0) 7 =0 q R + n a a + nb b + nc c = n a + n b + n c = V = 7 = q + R n a n a = n b = + n a n a n b n c = + n a + n b ( n a,n b,n c ) ( 0,0,0) 7 =0 q R + n a a + nb b + nc c = n a + n b + n c = 3 V 3 = V = V = V = V =

9 V 8 = V 9 = V 0 = R 0 = 0 4πε 0 a V = 7 N N h N h k π g exp π ghk k= N+ h = N+ h + k hk h= N q + erfc R = R + n n n a n n a n b 7 q exp( π g hk R ) q π R + erfc R + na n b n c n b = n+ n a =0 na n b n c n a = n n c = n+ n a + n b ( n a,n b,n c ) ( 0,0,0) g hk = h a * + k b * + c * na n b n c = n a a + nb b + nc c } v Ce = ( a b ) c = π v /3 Ce max { n n n Ewad a b c } max{ h + k + }

10 max{ n a, n b, n c } = 4 max { n a, n b, n c } 4 ( 4 +) ( 4 +) ( 4 +) = 9 3 = 4389 max h, k, { } = 3 max{ n a, n b, n c } = 4 max{ h, k, } 3 ( 3 +) ( 3 +) ( 3 +) = 7 3 = 343 max{ n a, n b, n c } 4 ( 4 +) ( 4 +) ( 4 +) = 9 3 = 79

11 a R 0 = 0 R = 0.5a + 0.5b + 0.5c 4πε 0 a V 0 = =.547 n a + n b + n c = V = q + R n a n a = n b = + n a n a n b n c = + n a + n b ( n a,n b,n c ) ( 0,0,0) =0 q R + n a a + nb b + nc c = q + R = n a n a = n b = + n a n a n b n c = + n a + n b ( n a,n b,n c ) ( 0,0,0) n a a + nb b + nc c R + n aa + nbb + ncc n a + n b + n c = V = q + R n a n a = n b = + n a n a n b n c = + n a + n b ( n a,n b,n c ) 0,0,0 =0 q R + n a a + nb b + nc c

12 = n a + n b + n c = 3 V 3 = V 4 = V 5 = V 6 = V 7 V 8 = 0.460V 9 = V 0 = = max{ n a + n b + n c } max{ h + k + }

13 実は NaC 型の単位構造は八重極モーメントまでゼロであるという特殊な構造なので たまたま普通の計算方法 単純な格子和 でもうまく行ったのです CsC 型の構造は四 重極モーメントまではゼロですが 八重極モーメントがゼロでない値を取るので格子和の 収束の速さに大きな違いが現れます 普通の物質は NaC 型構造のように特殊な構造をと りませんから エバルト法は3次元の物質でクーロン相互作用を計算するために必須なも のだといえます 6 3 セル多重極子展開法 reengard-rokhn method セル多重極子展開法は基本セルの中の原子の数が多いときに 基本セル内での原子間ポ テンシャルを評価するために有効な方法だと言われています この方法では まず基本セルを下図に示すように小さいセルに分割します 下図で太い 線で区切られているのが周期的境界条件を満たす基本セルであり 細い線で区切られてい るのが多重極子展開のためにさらに細かく区切ったセルを意味します 原子 が斜線部のセルにいるとして 原子 に作用する力を 白い箱の隣接セルにある原 子からの直接相互作用と ドットをつけた部分にある原子からの多重極展開の項に分けて 考えます V ( r ) = q near ρ r r + V (r r ) poe far

14 r r V poe poe V r ( r ) = Z R + µ R p R + Q β R R β O βγ R R β R + γ + p+ R p+4 R p+6 β β γ r r β γ x, y, z Z, µ, Q β, O βγ M C f C f C f C f C f C f C f C f C f C f C f C f C f C f C f C f C f C f C f C f C f C f C f C f C f C f C f C n C n C n C f C f C f C f C f C n C 0 C n C f C f C f C f C f C n C n C n C f C f C f C f C f C f C f C f C f C f C f C f C f C f C f C f C f C f poe V r ( r ) = Z R + µ R p R + Q β R R β O βγ R R β R + γ + p+ R p+4 R p+6 β R = r r r r = x, y,z β = x, y,z γ = x, y,z Z = q µ = p β γ q r

15 Q β = p q p + O βγ = p p + r r β δ β r 6 ( p + 4)r r β r γ ( r δ βγ + r β δ γ + r γ δ β )r r d d / R d / R </ C 0 C n C 0 7N / M = V far +V r near V r r poe = V r r + q r r p far near C f C 0 N / M V far r C f C 0 C 0 V far r C 0

16 = V (0) + V () r poe V r r + V () β r r β + β V T ( r ) r r C 0 V (0) V () V () β C 0 N / M r Z r r = Z p r + pz p p+ r ( r r ) + pz ( r r ) r r p + p+4 r µ r ( r ) µ r r r = µ + p+ p+ p+ r r p + ( µ r ) r r p+4 r + + V (0) V (0) = Z r p µ r r p+ + V () r V () = pz r r + µ ( p+ p+ r p + )( µ r )r p+4 + r T V C0 04M + 8N M 4 = =4 6 = = 40 4 = 44 3 = 4 6 = 4 µ x µy µ z r

17 Q β = p q p + r r β δ β r Q β = Q β Q xx Q xy Q xz Q yy Q yz Q zz O βγ = p p + 6 ( p + 4)r r β r γ ( r δ βγ + r β δ γ + r γ δ β )r O βγ = O γβ = O βγ = O βγ = O γβ = O γβ O xxx O xxy O xxz O xyy O xyz O xzz O yyy O yyz O yzz O zzz V (0) = ( V ) r =0 V () = V r V β r =0 () = V r r β r =0 8N M N / 4 ~ 60N

18 L M = 8 L N M = N / 4L L = og 8 N 4 Z = q µ = pq r Q β = p q p + O βγ = p p + r r β δ β r q p + 4 / r r β r γ ( r δ βγ + r β δ γ + r γ δ β )r / 6 r = r x r y r z 8 Z ( ) ( ) = q k = Z k k= 8 k=

19 8 µ ( ) ( = q k r k + R k = µ ) ( k + Z ) k R k k= 8 k= Q ( ) β = 8 k= ( ) Q kβ ( µ ) ( k R kβ + Z ) k R k R kβ poe V r ( r ) = V (0) + V () () (3) r + V β r r β + V βγ r r β r γ + V (0) V () V β () V (3) βγ poe V r ( r ) = Z β R + µ R p R + Q β R R β O βγ R R β R + γ + p+ R p+4 R p+6 β β β γ γ R = r r r Z r r p µ r r r r p+ = Z r p + pz r p+ ( r r ) + pz µ r = µ + p+ p+ r r p + ( r r ) r r p + p+4 r ( µ r ) r r p+4 r + + r V (0) V () V () β V (3) βγ

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