Homework 6, due December 6, 1996 Problem 1 Part a) At x ? a2 (x) = AeiKx + (Ar + Bt)e?iKx At x a2 (x) = (At + Br)eiKx + Be?iKx a From (8.68) ( a2 ) = eika (? a2 ) a= eika(Ae?aiK 2 + (Ar + Bt)eiK a2 ) which is equal to (At + Br)eiK 2 + Be?iK 2 which gives eika(A + (Ar + Bt)eiKa ) = B + (At + Br)eiKa or (eika(1 + reiKa ) ? teiKa )A + (eika teiKa ? 1 ? reiKa )B = 0 Similarly from the derivative condition: eika(AK ? (Ar + Bt)KeiKa ) = ?BK + (At + Br)KeiKa leads to (eika(1 ? reiKa ) ? teiKa )A + (?eika teiKa + 1 ? reiKa )B = 0 Set the determinant equal to zero: (eika(1 + reiKa ) ? teiKa )(?eika teiKa + 1 ? reiKa )? (eikateiKa ? 1 ? reiKa )(eika(1 ? reiKa ) ? teiKa ) = 0 Combine terms in the exponent of big K: e2iKa(t2 ? r2)2eika ? 2t(1 + ei2ka)eiKa + 2eika = 0 Multiply by e?ika e?iKa to get eiKa (t2 ? r2 ) ? t(e?ika + eika ) + e?iKa = 0 and combine exponents in a cosine: 2t cos(ka) = eiKa (t2 ? r2 ) + e?iKa For v=0 we have r=0, t=1, and indeed we get k=K. Part b)2 dw = d 21 2 ? 1 d2 22 dx dx dx use2 Schrodinger's equation to show h dw 2m dx = (v ? E )12 ? 1 (v ? E )2 = 0 Part c) w(l ; l ) = jtj22iK for x a2 and w(l ; l ) = 2iK (1 ? jrj2) for x ? a2 . These have to be the same, hence (8.72) is correct. Part d) w(l ; r ) = 2iKtr for x a2 and also w(l ; r ) = ?2iKt r for x ? a2 hence tr = ?t r and rt is purely imaginary. This gives (8.75) Part e) 2t cos(ka) = eiKa (t2 ? r2 ) + e?iKa becomes 2jtjei cos(ka) = eiKa (jtj2ei2 + jrj2ei2 ) + e?iKa or using (8.72) 2jtjei cos(ka) = eiKa ei2 + e?iKa multiply by e?i to get 2jtj cos(ka) = ei(Ka+) + e?i(Ka+) 1 and we have (8.76) Part f) Work in the gap containing Ka = n ? and write Ka = n ? + Ka Solve for cos(Ka + ) = jtj This gives cos(n + Ka) = jtj The sign is determined by the value of n and we get for small values: p 1 ? 21 (Ka)2 jtj or Ka 2(1 ? jtj) Now jrj2 = 1 ? jtj2 = (1 ? jtj)(1 + jtj) 2(1 ? jtj) and hence Ka jrj The energy gap is h2 2Kajrj which gives (8.78). Note the factor of two gap 2hm2 2K 2K = ma 2 because we have to go from minus to plus K . Part g) If jtj is very small we see that cos(Ka + ) = jtj tells us that Ka is close to half integral values of pi. Write Ka = n + 2 ? +Ka to get sin(n +Ka) = jtj Again the sign follows from the value of n. This gives Ka jtj and now the band width is very narrow and of order jtj. Part h) A delta function potential. We have ? 2hm2 ddx2 2 + g (x) = E This is a standard example. Use l to get x < 0 (x) = AeiKx + Are?iKx x > 0 (x) = AteiKx The wave function is continuous, hence 1+r = t. By integrating Schrodinger's equation we nd 2 d h ? 2m ( dx (0+ ) ? ddx (0? )) + g (0) = 0 or ( ddx (0+ ) ? ddx (0? )) = 2hmg2 (0) This gives iKt ? (iK ? iKr) = 2hmg2 t or t + r ? 1 = ?i h2mg 2K t Substitute r to get t(1 + i hmg 2K ) = 1 or t = 1+i1hmg 2K This gives for the phase 2 real(t) = ? h K cot( ) = imaginary (t) mg and for the norm: jtj2 = 1+( 1mg2 )2 = 1+tan1 2 () = cos2 ( ) h K 2