54
D. J. Fernández
u j (x) = (a −
ω ) j −1 u 1 (x), ε j = ε 1 − (j − 1)ω, j = 1, · · · , k. As a consequence,
L + = P k−1 (
H ) l + , where l ± are third-order differential ladder operators fulfilling
[
H , l
±
] = ±ω l
± ,
l
+ l
−
=
H −
ω
2
H − ε 1 − ω
H − ε k
.
The roots of the polynomial l + l − suggest the following 3 extremal states
ψ E 1 ∝ B
+
k exp
−
ωx 2
2
,
E 1 =
ω
2
,
ψ E 2 ∝ B
+
k a
+
ω u 1 ,
E 2 = ε 1 + ω,
ψ E 3 ∝
W (u 1 , . . . , u k−1 )
W (u 1 , . . . , u k )
,
E 3 = ε k = ε 1 − (k − 1)ω.
In order to generate the PIV transcendents, we have to scale the Hamiltonian
H ,
as well as the involved factorization energies, and introduce the variable z =
√
ωx,
as follows:
H =
H
ω = −
1
2ω
d 2
dx 2 +
1
2 ωx 2 −
1
ω
d 2
dx 2 [ln W (u 1 , · · · , u k )]
= −
1
2
d 2
dz 2 +
z 2
2 −
d 2
dz 2 [ln W (u 1 , · · · , u k )] .
The corresponding PIV transcendents are simply calculated through
g(z) = −z −
d
dz ln
ψ E 3
z
√
ω
.
Some results are shown in Tables 1 and 2 for k = 1 and k = 2, respectively.
Table 1 PIV transcendents for k = 1, ε 1 =
5ω
2 , u 1 (x) = φ 2 (x)
ψ E 3
1
u 1
B + φ 0
B + a +
ω u 1
E 3
ω
5
2
1
2
7
2
g(z)
6z−4z 3
2z 2 −1
−
2z 2 +1
z−2z 3
4z
−4z 4 +4z 2 +3
8z 6 −4z 4 +6z 2 −3
a
−2
4
−5
b
−18
−2
−8
Table 2 PIV transcendents for k = 2, , 1 =
5ω
2 , u 1 (x) = φ 2 (x), u 2 = a −
ω u 1
ψ E 3
u 1
W (u 1 ,u 2 )
B + φ 0
B + a +
ω u 1
E 3
ω
3
2
1
2
7
2
g(z)
8z 5 +6z
1−4z 4
4z
2z 2 +1
−
4z 4 +3
4z 5 +8z 3 +3z
a
0
3
−6
b
−18
−8
−2