势阱内(0<x<a) d'y, 2mE dx2 h 2y=0 k2- 2me dy +hy=0 2 dx 2 方程的通解为 Y(x)=Asinh+ Bcos ac y(0)=0→>B=0 y(a)=0→> Asina=0势阱内 0 2 2 2 2 +  =   mE dx d 2 2 2  mE k = 0 2 2 2 +  =  k dx d 方程的通解为 (0) = 0 → B = 0 (a) = 0 → Asinka = 0 Ψ(x) = Asinkx + Bcoskx (0  x  a)