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