Diploma thesis

3 Complex square well potential
or
ψ s 1(−x) = ψ s 3(x) and ψ s 2(x) = ψ s 2(−x), (3.4)
ψ a 1(−x) = −ψ a 3(x) and ψ a 2(−x) = −ψ a 2(x). (3.5)
Therefore it is sufficient to solve (2.9) only in area 1 and 2, since then ψ 3 is determined via
symmetry according to (3.4) and (3.5). The Schrödinger equation (2.9) and the Hamilton operator
(2.2) with the considered potential yields that this solution ψ has to be continuous at x = ±L
and differentiable at x = ±w. Making the ansatz
ψ 1,2 = A 1,2 e −ik 1,2x + B s,a
1,2e ik 1,2x
(3.6)
and demanding continuity provides
ψ s (x) = 2iA s e iks 1 L ⎧⎪ ⎨
⎪ ⎩
− sin [k s 1(x + L)]
sin[k s 1 (w−L)]
cos(k s 2 w) cos(k s 2x)
sin [k s 1(x − L)]
sin [k1(x ⎧⎪ a + L)]
⎨
, ψ a (x) = −2iA a e ika 1 L sin[k a(w−L)]
1
sin(k aw) sin(k2x)
a
2
⎪ ⎩
sin [k a 1(x − L)]
(3.7)
with A = A 1 and the complex wavenumbers k 1 and k 2
k 2 1 = 2M 2 (E R + iE I ) , k 2 2 = 2M 2 [E R + i(E I + C)] , (3.8)
where the indices R and I denote the real and imaginary part of the quantities, respectively.
Normalizing these solutions, that is demanding
∫ ∞
−∞
∫ w
∫ L
|ψ(x)| dx = 2 |ψ 2 (x)| dx + 2 |ψ 1 (x)| dx = 1, (3.9)
0
w
yields for the absolute value of the constants A s and A a
|A s | =
{
2e −2ks I,1 L [ sin 2k
s
R,1 (w − L)
k s R,1
− sinh 2ks I,1(w − L)
+ cosh 2ks I,1(w − L) − cos 2k s R,1(w − L)
cosh 2k s I,2w + cos 2k s R,2w
k s I,1
( sinh 2k
s
I,2 w
k s I,2
+ sin )]} −
1
2ks 2
R,2w
kR,2
s
(3.10)
and
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