Diploma thesis

4 Complex harmonic potential
4.1.1 Symmetric states
With (4.22) the continuity condition ψ s 1(−w) = ψ s 2(−w) provides
[ (
A s 1e − 1 2 ω2 1F 1 a 1 ; 1 )
2 ; ω2 − 2ω Γ ( )
a 1 + 1 2
Γ (a 1 )
1F 1
(
a 1 + 1 2 ; 3 2 ; ω2 ) ] = A s 2e − 1 2 θω2 1F 1
(
a 2 ; 1 2 ; θω2 )
. (4.33)
With this we reexpress A s 2 in terms of A s 1, which then exhibits the normalization constant of the
whole symmetric wave function, so that we call it from now on N s := A s 1. Then the continuity
condition reads
A s 2 = N s e 1 2 (θ−1)ω2 1 F 1
(
a1 ; 1 2 ; ω2) − 2ω Γ(a 1+ 1 2)
Γ(a 1 ) 1F 1
(
a1 + 1 2 ; 3 2 ; ω2)
1F 1
(
a2 ; 1 2 ; θω2 ) =: N s R s , (4.34)
where R s = R s (ω, ε, c) represents an abbreviation. We can now write down the wave functions in
all 3 areas obeying (4.14), which yields the symmetric wave function ψ s :
[ (
e
⎧⎪ − 1 2 χ2 1F 1 a1 ; 1; 2 χ2) + 2χ Γ(a 1+ 1 (
2)
Γ(a 1 ) 1F 1 a1 + 1; 3; 2 2 χ2)] , χ < −ω
⎨ (
ψ s (χ) = N s R s e − 1 2 θχ2 1F 1 a2 ; 1; 2 θχ2)
, −ω ≤ χ ≤ ω
⎪ ⎩
. (4.35)
e − 1 2 χ2 [
1F 1
(
a1 ; 1 2 ; χ2) − 2χ Γ(a 1+ 1 2)
Γ(a 1 ) 1F 1
(
a1 + 1 2 ; 3 2 ; χ2)] , χ > ω
4.1.1.1 Quantization condition
Now we have to ensure the continuity of the first derivative of ψ s at χ = −ω
d
dχ ψs 1 (χ)
=
∣ χ=−ω
d
dχ ψs 2 (χ)
, (4.36)
∣ χ=−ω
which yields with (4.35)
(
ω 1 F 1 a 1 ; 1 )
(
2 ; ω2 − 4a 1 ω 1 F 1 a 1 + 1; 3 )
2 ; ω2 +
2 Γ ( )
a 1 + 1 [
2 (1
− ω
2 ) 1F 1
(a 1 + 1 Γ (a 1 )
2 ; 3 )
2 ; ω2 + 4 (
a 1 + 1 )
ω 2 1F 1
(a 1 + 3 3 2
2 ; 5 )]
2 ; ω2
(
= 1 F 1 a1 ; 1 2 ; ω2) − 2ω Γ(a1+ 1 2) (
Γ(a 1) 1F 1 a1 + 1 2 ; 3 2 ; ω2) [ (
(
1F 1 a2 ; 1 2 ; θω2) θω 1 F 1 a 2 ; 1 )
(
2 ; θω2 − 4a 2 θω 1 F 1 a 2 +1; 3 )]
2 ; θω2 . (4.37)
Here we used the derivative of the confluent hypergeometric functions with respect to the last
argument
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