Diploma thesis

4 Complex harmonic potential
more. Taking the limit ω → ∞ in (4.37) we have to use the limit of the confluent hypergeometric
functions (4.27) and neglect the first term since the exponential function in the second grows
much stronger than every polynomial in ω. Furthermore we calculate from (4.20) that θ → 1
and multiply the whole expression with the denominator of R s so that all remaining terms have
the denominator Γ(a 1 )Γ(a 2 ). This may be also reasonable since in the considered limit we have
ψ s 2 = ψ s on the whole axis so that it should be lim χ→∞ ψ s 2 (χ) = 0. Otherwise the calculation of
R s in (4.34) could include some division by zero. Since the prefactor of the right-hand side, which
is then the numerator of R s , vanishes in the highest order and all terms of the order 2(a 1 + a 2 ) − 1
in ω cancel each other, there remains only one term of order 2(a 1 + a 2 ) − 3 which has to be equal
to zero. Thus the quantization condition for ω → ∞ reads
lim
ω→∞
ω 2(a 1+a 2 )
Γ(a 1 )Γ(a 2 )
= 0. (4.65)
We want to extract a condition for a 2 from that, which is fulfilled (4.65) for all a 1 , since it is not
realized any more in this system for ω → ∞. Eq. (4.65) is always ensured if a 2 is equal to some
negative integer number a 2 = −n. This yields nothing else than the "real limit" for area 2 since
it is the same result for a 2 like for a 1 in the limit c → 0 or ω → 0 when area 2 disappeared.
Nevertheless it is followed by a different energy which is still complex and contains an imaginary
part which is linear in the depth c of the well:
lim a 2 = 1 ω→∞ 4 (1 − ε ∞ − ic) = −n ⇔ ε s ∞ = 4n + 1 − ic (4.66)
For the wave function this provides
⇔
ψ s 2,∞ (χ) = ψ s ∞ (χ) = N s e − 1 2 χ2 1F 1
(
−n; 1 2 ; ω2 )
(
E∞ s = Ω 2n + 1 )
− iC. (4.67)
2
= (−1) n N s n!
(2n)! e− 1 2 χ2 H 2n (χ) . (4.68)
Evaluating the normalization constant (4.45) in the limit ω → ∞ yields
|N s | = 1
|R s |
⎡
⎢
n∑
⎣
µ,ν=0
(−n) ν (−n) µ
ν!µ! ( ) (
1 1
2 2)
ν
µ
(
Γ µ + ν + 1 ) ⎤ −1/2
⎥ ⎦
2
(4.41)
= 1 [ ∫ ∞
∣ (
2 e − 1 ∣∣∣
|R s 2 χ2 1F 1 −n; 1 )∣ ]
∣∣∣ 2 −1/2
| 0
2 ; χ2 dχ
(4.48)
= (2n)!
n! |R s |
(4.56)
= (2n)!
n! |R s |
[∫ ∞
−∞
√
] −1/2
e − 1 2 χ2 |H 2n (χ)| 2 dχ
1
√ π2
2n
(2n)! , (4.69)
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