Page 58 - Science at the nanoscale

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June 9, 2009
Brief Review of Quantum Mechanics
48
T are respectively given by
2
|Q|
R =
2
|P|
2
|S|
(3.48)
T =
2
|P|
Similarly, for region II, the Schr¨odinger equation is given by
Eq. (3.37). The wavefunction is therefore
κx
−κx
+ Ve
ψ II (x) = Ue
(3.49)
To determine the coefficients in the wavefunctions, we make use
of boundary conditions again.
At x = 0,
dψ I
dψ II
ψ I (0) = ψ II (0) and
=
dx
dx
P + Q = U + V and ik(P − Q) = κ(U − V)
At x = W,
dψ III
dψ II
ψ II (W) = ψ III (W) and
=
dx
dx
−κW
ikW
ikW
κW
and ikSe
+ Ve
= Ue
Se
= κ(Ue
)
which can be simplified to the following equations:
2
2
(k + κ )(e
Q
2κW
− 1)
(3.50)
=
2
2κW
P
e
(k + iκ) − (k − iκ)
−ikW κW
S RPS: PSP0007 - Science-at-Nanoscale 2 κW − Ve −κW (3.47) ch03
e
4ikκe
= 2κW 2 2 (3.51)
P e (k + iκ) − (k − iκ)
Thus the reflection coefficient R, (Eq. 3.47) and the transmission
coefficient T, (Eq. 3.48) can be expressed as
−1
" #
4E(V o − E)
R = 1 + (3.52)
2 2
V sinh (κW)
o
−1
" #
2 2
V sinh (κW)
o
T = 1 + (3.53)
4E(V o − E)
x
x
e −e −x e +e −x
Note that sinh(x) = 2 and cosh(x) = 2 .