Diploma thesis

3.1 Static solutions of Schrödinger equation
|A a | =
{
2e −2ka I,1 L [ sin 2k
a
R,1 (w − L)
k a R,2
− sinh 2ka I,1(w − L)
+ cos 2ka R,1(w − L) − cosh 2k a I,1(w − L)
cos 2k a R,2w − cosh 2k a I,2w
k a I,2
( sinh 2k
a
I,2 w
k a I,2
− sin )]} −
1
2ka 2
R,2w
. (3.11)
kR,2
a
Thus we can fix them only up to a phase factor e iϕ , ϕ ∈ [0, 2π). Knowing the particular wave
function it is possible to calculate the density ρ = |ψ(x)| 2 and the current j(x) from (2.5) at some
fixed time. The symmetric and antisymmetric wave functions (3.7) are continuous at x = ±w and
±L. Since (2.9) yields that the second derivative of the wave function ψ ′′ only has a finite jump
discontinuity at x = ±w and an infinite one at ±L, its first derivative ψ ′ has to be continuous
at ±w, too, but not at ±L. Thus we obtain a relation between k 1 and k 2 and thus via (3.8) an
additional condition the energy E has to fulfill, which is our quantization condition
⎡ √ ⎤
√ 2M
Es cot ⎣(w − L)
2 Es ⎦ + √ ⎡
⎤
E s + iC tan ⎣w√
2M
2 (Es + iC) ⎦ = 0 (3.12)
in the symmetric case and in the antisymmetric case
⎡ √ ⎤
√ 2M
Ea cot ⎣(w − L)
2 Ea ⎦ − √ ⎡
⎤
E a + iC cot ⎣w√
2M
2 (Ea + iC) ⎦ = 0. (3.13)
The solutions of these two transcendental equation allow to determine all important physical
quantities E, ρ, j etc. of the problem.
For consistency one can evaluate the real limit by setting C = 0, which directly yields
⎛
⎞
⎛
cos ⎝√
2M
2 Es L⎠ = 0 and sin ⎝√
2M
2 Es L⎠ = 0. (3.14)
⎞
This can only be fulfilled by real arguments, since cosh(x) has no real root and cos(x) and sin(x)
no mutual one:
cos(x + iy) = cos(x) cosh(y) − sin(x) sinh(y) = 0 ⇒ x =
(
n + 1 )
π, y = 0 (3.15)
2
sin(x + iy) = sin(x) cosh(y) + cos(x) sinh(y) = 0 ⇒ x = nπ, y = 0. (3.16)
Therefore E has to be real and is given by
E s = 2 π 2 (n + 1 ) 2
2ML 2 2
and E a = 2 π 2
2ML 2 n2 . (3.17)
(
This is followed by k1 s = k2 s =: kn s = π L n +
1
2
)
and k
a
1 = k a 2 =: k a n = π L n and the identities 9