Numbering the three regions from left to right,
u 1 (x) =eikx+Re−ikx
u 2 (x) =Aeik′x
+Be−ik′xu 3 (x) =TeikxAgain we have assumed no wave incident from the right (but we could add that solution if we
wanted).
We now match the wave function and its first derivative at the two boundaries yielding 4 equations.
That’s good since we have 4 constants to determine. Atx=awe have 2 equations which we can
use to eliminateAandB.
Teika=Aeik′a
+Be−ik′aikTeika=ik′Aeik′a
−ik′Be−ik′ak
k′Teika=Aeik′a
−Be−ik′aAeik′a
=1
2
Teika(
1 +
k
k′)
Be−ik′a
=1
2
Teika(
1 −
k
k′)
Atx=−awe have 2 equations which can now be written in terms ofRandTby using the above.
e−ika+Reika=Ae−ik′a
+Beik′aike−ika−ikReika=ik′Ae−ik′a
−ik′Beik′ae−ika+Reika=1
2
Teika[(
1 +
k
k′)
e−^2 ik′a
+(
1 −
k
k′)
e^2 ik′a]
ke−ika−kReika=1
2
Teikak′[(
1 +
k
k′)
e−^2 ik′a
−(
1 −
k
k′)
e^2 ik′a]
e−ika−Reika=1
2
Teika[(
k′
k+ 1
)
e−^2 ik′a
−(
k′
k− 1
)
e^2 ik′a]
We can add equations to eliminateR.
e−ika+Reika=1
2
Teika[(
1 +
k
k′)
e−^2 ik′a
+(
1 −
k
k′)
e^2 ik′a]
e−ika−Reika=1
2
Teika[(
k′
k+ 1
)
e−^2 ik′a
−(
k′
k− 1
)
e^2 ik′a]
2 e−ika=1
2
Teika[(
2 +
k′
k+
k
k′)
e−^2 ik′a
+(
2 −
k′
k−
k
k′)
e^2 ik′a]
4 e−^2 ika=T[
4 cos(2k′a)− 2 i(
k
k′+
k′
k)
sin(2k′a)]
T=
2 e−^2 ika
2 cos(2k′a)−i(k
k′+k′
k)
sin(2k′a)T=2 kk′e−^2 ika
2 kk′cos(2k′a)−i(k^2 +k′^2 ) sin(2k′a)