Applying the Boundary Conditions
In order to calculate the transmission probability Twe have to apply the appropriateboundary conditions to (^) I, (^) II, and (^) III. Fig. 5.14 shows the wave functions in regions
I, II, and III. As discussed earlier, both and its derivative xmust be continuous
everywhere. With reference to Fig. 5.14, these conditions mean that for a perfect fit at
each side of the barrier, the wave functions inside and outside must have the same
value and the same slope. Hence at the left-hand side of the barrier
(^) I (^) II (5.87)
(5.88)
and at the right-hand side
(^) II (^) III (5.89)
(5.90)
Now we substitute (^) I, (^) II, and (^) IIIfrom Eqs. (5.75), (5.81), and (5.85) into the
above equations. This yields in the same order
ABCD (5.91)
ik 1 Aik 1 Bk 2 Ck 2 D (5.92)
Cek^2 LDek^2 LFeik^1 L (5.93)
k 2 Cek^2 Lk 2 Dek^2 Lik 1 Feik^1 L (5.94)
Equations (5.91) to (5.94) may be solved for (AF) to give
e
(ik 1 k 2 )L
e
k^1 (ik^1 k^2 )L (5.95)
k 2
k 2
k 1
i
4
1
2
k 1
k 2
k 2
k 1
i
4
1
2
A
F
d (^) III
dx
d (^) II
dx
d (^) II
dx
d (^) I
dx
The Tunnel Effect 195
x = 0 x = L
I II III
x
Figure 5.14At each wall of the barrier, the wave functions inside and outside it must match up
perfectly, which means that they must have the same values and slopes there.
Boundary conditions
at x 0 x^0
Boundary conditions
at xL
}
xL
}
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