8.4. THEFINITESQUAREWELL:BOUNDSTATES 137
substituteinto(8.74)
Ae−ika+Beika =
[
1
2
(1+
k
q
)ei(k−^2 q)a+
1
2
(1−
k
q
)ei(k+2q)a
]
E
k(Ae−ika−Beika) = q
[
1
2
(1+
k
q
)ei(k−^2 q)a−
1
2
(1−
k
q
)ei(k+2q)a
]
E (8.76)
SolvingforEintermsofA,wefind
2 kAe−ika =
[
(k+q)
1
2
(1+
k
q
)ei(k−^2 q)a+(k−q)
1
2
(1−
k
q
)ei(k+2q)a
]
E
=
[
(k+
k^2
2 q
+
q
2
)ei(k−^2 q)a+(k−
k^2
2 q
−
q
2
)ei(k+2q)a
]
E (8.77)
sothat
E
A
=
2 e−ika
(1+ 2 kq+ 2 qk)ei(k−^2 q)a+(1− 2 kq− 2 qk)ei(k+2q)a
=
e−^2 ika
cos(2qa)+ 21 ( 2 kq+ 2 qk)(e−^2 iqa−e^2 iqa)
=
e−^2 ika
cos(2qa)− 2 i(kq+qk)sin(2qa)
(8.78)
Inasimilarway,onecansolveforBintermsofA
B
A
=
1
2 ie
− 2 ika(q
k−
k
q)sin(2qa)
cos(2qa)−^12 i(qk+kq)sin(2qa)
(8.79)
Fromtheseratios,weobtainthetransmissionandreflectioncoefficients
T =
∣∣
∣∣E
A
∣∣
∣∣
2
=
1
cos^2 (2qa)+^14 (kq+qk)^2 sin^2 (2qa)
R =
∣∣
∣∣B
A
∣∣
∣∣
2
=
1
4 (
q
k−
k
q)
(^2) sin^2 (2qa)
cos^2 (2qa)+^14 (kq+qk)^2 sin^2 (2qa)
(8.80)
anditiseasytocheckthat
R+T= 1 (8.81)
asrequiredbyconservationofprobability.
NownoticethatsomethinginterestinghappenswhentheenergyEoftheincoming
particleissuchthat
2 qa= 2
√
2 m(E+V 0 )
a
h ̄
=nπ (8.82)