Page 305 - Determinants and Their Applications in Mathematical Physics

290 6. Applications of Determinants in Mathematical Physics

Lemma 6.19.

q − p
a. ∂e pq + ω ∂a pq = e pq ,
∂ρ ∂z ρ

p − q +1 2
b. ∂a pq + ω ∂e pq = (ω = −1).
∂ρ ∂z ρ a pq
Proof. If p ≥ q − 1, then, applying (6.10.3) with r → p − q,
∂ p − q ∂ p − q p−q+1

+ e pq = + (ω u p−q+1 )
∂ρ ρ ∂ρ ρ
∂ p−q+1
= − (ω u p−q )
∂z
= −ω ∂a pq .
∂z
If p<q − 1, then, applying (6.10.4) with r → q − p,

∂ p − q ∂ q − p q−p−1
+ e pq = − (ω u q−p−1 )
∂ρ ρ ∂ρ ρ
∂ q−p−1
= (ω u q−p )
∂z
= −ω ∂a pq ,
∂z
which proves (a). To prove (b) with p ≥ q − 1, apply (6.10.4) with r →
p − q + 1. When p<q − 1, apply (6.10.3) with r → q − p − 1.
Lemma 6.20.
2
∂E n1 2 ∂A n1 (n − 1)E E n1
2
a. E + ωA = ,
∂ρ ∂z ρ
2
∂A n1 2 ∂E n1 (n − 2)A A n1 2
2
b. A + ωE = (ω = −1).
∂ρ ∂z ρ
Proof.
n

A = |a pq | n , a pq A pr = δ qr ,
p=1
n

E = |e pq | n , e pq E pr = δ qr .
p=1
Applying the double-sum identity (B) (Section 3.4) and (6.10.12),
∂E n1 p1
= − ∂e pq E E ,
nq
∂ρ ∂ρ
p q
∂A n1
pn
= − ∂a pq A A 1q
∂z ∂z
p q