292 6. Applications of Determinants in Mathematical Physics
which is equivalent to (b). This completes the proof of Lemma 6.20.
Exercise.Prove that
(
ω∂
∂ρ−
p−q− 1ρ)
A
pq
n =−A
(n+1)
n+1,qAn∂A
pn
n∂z+A
1 q
n∂
∂z(
A
(n+1)
p+1, 1An)
+
∂A
p,q− 1
n∂z,
ω∂A
pq
n∂z=
A
(n+1)
n+1,qAn(
∂
∂ρ−
nρ)
A
pq
n−A
1 q
n(
∂
∂ρ−
1
ρ)
A
(n+1)
p+1, 1An−
(
∂
∂ρ−
q− 1ρ)
A
p,q− 1
n−
(
p+1ρ)
A
p+1,q
n(ω2
=−1).Note that some cofactors are scaled but others are unscaled. Hence, prove
that
(ω∂
∂ρ−
n− 2ρ)
En− 1An=
EnAn∂
∂z(
An− 1An)
−
An− 1An∂
∂z(
EnAn)
,
ω∂
∂z(
En− 1An)
=(−1)
n
EnAn(
∂
∂ρ−
nρ)
En− 1An+
An− 1An(
∂
∂ρ−
1
ρ)
EnAn.
6.10.3 The Intermediate Solutions
The solutions given in this section are not physically significant and are
called intermediate solutions. However, they are used as a starting point in
Section 6.10.5 to obtain physically significant solutions.
Theorem.Equations (6.10.1) and (6.10.2) are satisfied by the function
pairsPn(φn,ψn)andP
′
n(φ′
n,ψ′
n), wherea.φn=ρn− 2
An− 1An− 2=
ρn− 2A
11
n− 1,
b.ψn=
ωρn− 2
En− 1An− 2=
(−1)
n
ωρn− 2E
n− 1 , 1
n− 1=
(−1)
n− 1
ωρn− 2
A 1 nAn− 2,
c. φ′
n=
A
11ρ
n− 2,
d.ψ
′
n=(−1)
n
ωA
1 nρ
n− 2(ω2
=−1).The first two formulas are equivalent to the pairPn+1(φn+1,ψn+1), where
e. φn+1=ρn− 1A^11
,
f.ψn+1=(−1)
n+1
ωρn− 1E
n 1