Determinants and Their Applications in Mathematical Physics

290 6. Applications of Determinants in Mathematical Physics

Lemma 6.19.

a.

∂epq

∂ρ

+ω

∂apq

∂z

=

(

q−p

ρ

)

epq,

b.

∂apq

∂ρ

+ω

∂epq

∂z

=

(

p−q+1

ρ

)

apq (ω

2

=−1).

Proof. Ifp≥q−1, then, applying (6.10.3) withr→p−q,

(

∂

∂ρ

+

p−q

ρ

)

epq=

(

∂

∂ρ

+

p−q

ρ

)

(ω

p−q+1

up−q+1)

=−

∂

∂z

(ω

p−q+1

up−q)

=−ω

∂apq

∂z

.

Ifp<q−1, then, applying (6.10.4) withr→q−p,

(

∂

∂ρ

+

p−q

ρ

)

epq=

(

∂

∂ρ

−

q−p

ρ

)

(ω

q−p− 1

uq−p− 1 )

=

∂

∂z

(ω

q−p− 1

uq−p)

=−ω

∂apq

∂z

,

which proves (a). To prove (b) withp≥q−1, apply (6.10.4) withr→

p−q+ 1. Whenp<q−1, apply (6.10.3) withr→q−p−1.

Lemma 6.20.

a.E

2 ∂E

n 1

∂ρ

+ωA

2 ∂A

n 1

∂z

=

(n−1)E

2

E

n 1

ρ

,

b.A

2

∂A

n 1

∂ρ

+ωE

2

∂E

n 1

∂z

=

(n−2)A

2

A

n 1

ρ

(ω

2

=−1).

Proof.

A=|apq|n,

n

∑

p=1

apqA

pr

=δqr,

E=|epq|n,

n

∑

p=1

epqE

pr

=δqr.

Applying the double-sum identity (B) (Section 3.4) and (6.10.12),

∂E

n 1

∂ρ

=−

∑

p

∑

q

∂epq

∂ρ

E

p 1

E

nq

,

∂A

n 1

∂z

=−

∑

p

∑

q

∂apq

∂z

A

pn

A

1 q