Diploma thesis

4.1 Static solutions of Schrödinger equation
since lim ω→∞ Γ (a, ω 2 ) = 0. Before inserting this into the wave function (4.68) in area 2 we have
to consider that this is only the absolute value of N s and also includes only the absolute value
of R s which are in general complex numbers. Thus we have to add an arbitrary phase factor
N s = |N s | e iϕ , R s = |R s | e iϑ . We already know ϕ = nπ from the real limit, when area 2 vanishes,
and, since R s only contains real functions for ω → ∞, we can state |R s | = R s . This seems to be
right since exactly ϑ = 0 leads us to the correct wave function
ψ2,∞ s (χ) = ψ∞ s (χ) = (−1) n (2n)!
√
(
1
√ 1
n! π2
2n
(2n)! e− 2 χ2 1F 1 −n; 1 )
2 ; χ2
√
1
= √ 1 π2
2n
(2n)! e− 2 χ2 H 2n (χ) . (4.70)
Thus, we have
√ ψ∞ s √ λ 1
(√ )
(x) =
1 π 2 2n (2n)! e− 2 λx2 H 2n λx = ψ
s
0 (x). (4.71)
So finally we got in some sense the same result as for the complex square well potential since
the wave function and the real part of the energy yield nothing else than the real limit and only
the imaginary part of the energy provides something new. Thus the imaginary potential does
not influence the wave function qualitatively but yields an imaginary part of the energy which is
linear in c. The interesting issue is that this seems to be independent of the particular shape of
the imaginary potential since in both cases, square well in the limit ω → π/2 as well as harmonic
potential for ω → ∞, we get the same result for the imaginary part.
4.1.2 Antisymmetric states
Let us now evaluate along similar lines the antisymmetric wave function (4.23), that means the
wave functions ψ a 1, ψ a 2, ψ a 3 have to fulfill
ψ a 1 (−χ) = −ψ a 3(χ) , ψ a 2 (−χ) = −ψ a 2 (χ) . (4.72)
Next we derive conditions for A a 2 and ε by demanding continuity and differentiability of ψ a at
χ = −ω. But first of all we change slightly the notation for ψ 1 , which seems to be reasonable in
the antisymmetric case. Instead of expressing A a 1 via A s 1 we now do it vice verca and get for the
constant of the antisymmetric part
A s 1 = A a 1 lim χ→∞
χ 1 F 1
(
a1 + 1 2 ; 3 2 ; χ2)
1F 1
(
a1 ; 1 2 ; χ2 ) . (4.73)
Performing an analogous discussion as we did for A s 1 provides correspondingly
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