2.1.波函数的统计解释 8/86 (x,y,z+L)来代替实际的边界条件。 ll dr Adt=a2l3=1 VEL 所以:A= VV VL3 归一化波函数为:k(,D)=e1k=0 边界条件则限制了波失k的取值 krL=2nr,kyL= 2nyI,k,L=2nzT 即 2 2丌 k k h2 E=ho(k)= k m (n2+n2+n2),自由粒子能谱 ● First●Prev●Next●Last● Go Back● Full Screen●cose●Quit• First • Prev • Next • Last • Go Back • Full Screen • Close • Quit §2.1. żêÚO)º 8/86 ψ(x, y, z + L)5O¢S>.^‡" Z ∞ |ψ| 2 dτ = Z V=L3 A 2 dτ = A 2L 3 = 1 ¤±µA = 1 √ V = 1 √ L3 8zżꏵψk(r, t) = 1 √ V e i(k·r−ωt) >.^‡K› ÅkŠµ kxL = 2nxπ, kyL = 2nyπ, kzL = 2nzπ =µ kx = nx 2π L , ky = ny 2π L , kz = nz 2π L E = ~ω(k) = p 2 2m = ~ 2 2m ￾ k 2 x + k 2 y + k 2 z
=  2π L 2 ~ 2 2m ￾ n 2 x + n 2 y + n 2 z
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