Re: [請益] 不知道是數學還是物理問題了 = =

看板Physics作者Frobenius (▽．(▽×▽φ)=0)時間14年前 (2010/04/26 03:43)推噓2(2推 0噓 3→)

留言5則, 3人參與討論串2/3 (看更多)

※ 引述《h888512 (衝)》之銘言： : Helium atom的perturbation : (0) 1 : E = ∫∫dτdτ' Ψ*(r1,r2) ---Ψ(r1,r2) : r12 : Z^6 : = --- ∫e^(-2zr) 4πr^2 dr ∫e^(-2zr')r'^2 dr' ........ : π^2 Let ρ1 = (2Z/a) r1 , ρ2 = (2Z/a) r2 , dρ1 = (2Z/a) dr1 , ρ2 = (2Z/a) dr2 r1 = (a/2Z) ρ1 , r2 = (a/2Z) ρ2 |r1 - r2| = (a/2Z) ρ12 => 1 / |r1 - r2| = (2Z/a) / ρ12 2 3 2 dτ = 4πr1 dr1 = 4π (a/2Z) ρ1 dρ1 2 3 2 dτ'= 4πr2 dr2 = 4π (a/2Z) ρ2 dρ2 (0) 1 Z 3/2 -(Z/a)r1 1 Z 3/2 -(Z/a)r2 Ψ1s^2 (r1,r2) = ---- ( - ) e ---- ( - ) e √π a √π a 1 Z 3/2 -ρ1 /2 1 Z 3/2 -ρ2 /2 = ---- ( - ) e ---- ( - ) e √π a √π a (1) (0)* ^ (0) E = ∫∫dτdτ' Ψ (r1,r2) H' Ψ (r1,r2) 2 (1) (0) e' (0) E = ∫∫dτdτ' Ψ1s^2 (r1,r2) ---- Ψ1s^2 (r1,r2) r12 2 Z^6 e' -2(Z/a)r2 -2(Z/a)r1 1 = -------- ∫e dτ' ∫e --------- dτ π^2 a^6 |r1 - r2| 2 Z^6 e' -ρ2 3 2 -ρ1 2Z/a 3 2 = -------- ∫e 4π(a/2Z) ρ2 dρ2 ∫e ------ 4π(a/2Z) ρ1 dρ1 π^2 a^6 ρ12 2 e' Z -ρ2 2 -ρ1 1 2 = ---- - ∫e ρ2 dρ2 ∫e ----- ρ1 dρ1 2 a ρ12 ----------------------------------------------------------------------------- 電荷位置ρ1在導體球殼ρ2內之電位為定值 1 1 即 ρ1 < ρ2 => ----- = --- ( ρ1 = 0 ~ ρ2 ) ρ12 ρ2 電荷位置ρ1在導體球殼ρ2外之電位等同導體球殼上之電荷集中在球殼中心 1 1 即 ρ1 > ρ2 => ----- = --- ( ρ1 = ρ2 ~ ∞ ) ρ12 ρ1 ----------------------------------------------------------------------------- 2 e' Z ∞ -ρ2 2 ρ2 -ρ1 1 2 ∞ -ρ1 1 2 = ---- - ∫ e ρ2 [ ∫ e --- ρ1 dρ1 + ∫ e --- ρ1 dρ1 ] dρ2 2 a 0 0 ρ2 ρ2 ρ1 2 e' Z ∞ -ρ2 2 1 ρ2 -ρ1 2 ∞ -ρ1 = ---- - ∫ e ρ2 [ --- ∫ e ρ1 dρ1 + ∫ e ρ1 dρ1 ] dρ2 2 a 0 ρ2 0 ρ2 2 e' Z ∞ -ρ2 2 1 ρ2 -ρ1 2 ∞ -ρ1 = ---- - ∫ e ρ2 --- [ ∫ e ρ1 dρ1 + ρ2 ∫ e ρ1 dρ1 ] dρ2 2 a 0 ρ2 0 ρ2 ----------------------------------------------------------------------------- http://mathworld.wolfram.com/IncompleteGammaFunction.html (1)(2)(3)(5)(8)(9) k ρ2 -ρ1 2 -ρ2 2 ρ2 ∫ e ρ1 dρ1 = γ(3,ρ2) = 2! ( 1 - e Σ --- ) 0 k=0 k! -ρ2 2 -ρ2 2 = 2 [ 1 - e ( 1 + ρ2 + ρ2 /2 ) ] = 2 - e ( 2 + 2 ρ2 + ρ2 ) k ∞ -ρ1 -ρ2 1 ρ2 -ρ2 ∫ e ρ1 dρ1 = Γ(2,ρ2) = 1! e Σ --- = e ( 1 + ρ2 ) ρ2 k=0 k! ρ2 -ρ1 2 ∞ -ρ1 ∫ e ρ1 dρ1 + ρ2 ∫ e ρ1 dρ1 0 ρ2 -ρ2 2 -ρ2 = 2 - e ( 2 + 2 ρ2 + ρ2 ) + ρ2 e ( 1 + ρ2 ) -ρ2 2 -ρ2 2 = 2 - e ( 2 + 2 ρ2 + ρ2 ) + e ( ρ2 + ρ2 ) -ρ2 = 2 - e ( 2 + ρ2 ) ----------------------------------------------------------------------------- 2 e' Z ∞ -ρ2 2 1 -ρ2 = ---- - ∫ e ρ2 --- [ 2 - e ( 2 + ρ2 ) ] dρ2 2 a 0 ρ2 2 e' Z ∞ -ρ2 -ρ2 = ---- - ∫ e ρ2 [ 2 - e ( 2 + ρ2 ) ] dρ2 2 a 0 2 e' Z ∞ -ρ2 -2ρ2 -2ρ2 2 = ---- - ∫ ( 2 e ρ2 - 2 e ρ2 - 2 e ρ2 ) dρ2 2 a 0 http://mathworld.wolfram.com/LaplaceTransform.html 2 e' Z 1! 1! 2! = ---- - [ 2 ------- - 2 ------- - ------- ] 2 a 1^(1+1) 2^(1+1) 2^(2+1) 2 2 2 e' Z 1 1 e' Z 8 2 1 e' Z 5 = ---- - ( 2 - - - - ) = ---- - ( - - - - - ) = ---- - ( - ) 2 a 2 4 2 a 4 4 4 2 a 4 2 5 2 Z 5 e Z = - e' - = - ------ - 8 a 8 4πε0 a -- http://mathworld.wolfram.com/SphericalHarmonicAdditionTheorem.html -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 118.160.212.42

推

chungweitw

04/26 07:25, , 1^F

04/26 07:25, 1^F

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chungweitw

04/26 07:26, , 2^F

04/26 07:26, 2^F

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Frobenius

04/26 07:56, , 3^F

04/26 07:56, 3^F