This is independent of coordinate, as required by the fact that it
is proportional to the conserved Wronskian. We identify
¯
hα
j + (x ≥ 0) = J[ y(x) ] =
|b + |
2
(5.63)
πm
as the tunneling current propagating from left to right for x ≥ 0.
This result will prove useful later.
Next we must match the solutions and their derivatives at x = 0,
which is the emission surface. For x ≤ 0 inside the bulk material
we form the solution and its first derivative as
+ikx
−ikx
u(x) = a + e
+ a − e
+ikx − a − e
−ikx ].
u
� (x) = ik [ a + e
(5.64)
For x ≥ 0 in the vacuum we form
u(x) = Y [ y(x) ]
u
� (x) = −α Y
� (y).
(5.65)
Matching the solutions and first derivatives at x = 0, equivalently
y = αβ, we have two simultaneous equations,
a + + a − = Y (αβ)
a + − a − =
iα Y
� (αβ).
(5.66)
k
Solving for a + and a − we find
a + =
1
2
Y (αβ) +
iα Y
� (αβ)
2k
iα
a − =
1
2
Y (αβ) −
Y
� (αβ).
(5.67)
2k
Substituting for Y and Y
� above, we find
a + = b +
1
2
[ Bi(αβ) + iAi(αβ) ] +
iα [ Bi
� (αβ) + iAi
� (αβ) ]
2k
iα
a − = b + 2
1 [ Bi(αβ) + iAi(αβ) ] −
[ Bi
� (αβ) + iAi
� (αβ) ] .
2k
(5.68)
315
5.4. Field emission