§27势垒贯穿 U。(0<x<a) U(x) 10(x<0,x>a) 先讨论E>U情况 2me 2+k=0,k1= (x<0,x>a) h 2m(E-U0) d2+k2y=0,k2= (0<x<a)
§2.7 势垒贯穿
§27势垒贯穿 f,=Ae +Ae i1r (xo) 92=B 22x+B e-k2(0<x<a) 93=Cel a A+A=BB k1a-kiA=k2B-k2B Beira+B Ceil k2Be一k2Be=Ck1e
§2.7 势垒贯穿
§27势垒贯穿 4k,k,e-ikla (k21+k2)2e2-(,-1、2A A 2i(ki-k2)sin k2a (k1-k2) (k1+k2)2e A
§2.7 势垒贯穿
§27势垒贯穿 d dx A Jn=故1(C2 R JA
§2.7 势垒贯穿
§27势垒贯穿 R-| 1A12(k3-k32)sin?k2a A|12(k2-k2)2sin2k2a+4k2 D=Jin|[2(i-k2)2sin k2a+4kik2 C kika D+R=1
§2.7 势垒贯穿