462
E Quaternions
∞
−∞
dx j S
N
1,1 (x j , x j ) = N and
∞
−∞
dx j ¯
σ N 1 (x j , x j ) = N
1 0
0 1
.
(E.36)
Also,
∞
−∞
dx k ¯
σ N 1 (x i , x k )· ¯
σ N 1 (x k , x j ) = ¯
σ N 1 (x j , x j ) + ¯
A N (x i , x j ),
(E.37)
where
¯
A N (x i , x j ) =
0
S N
1,2 (x i , x j )
−S N
2,1 (x i , x j )
0
.
(E.38)
Below, we compute some integrals used to obtain Eqs. (E.36)–(E.38).
Useful Integrals
If we use the recursion relation for harmonic eigenstates φ n (x),
φ 2n+1 =
2n
2n + 1
φ 2n−1 −
2
2n + 1
φ
2n ,
(E.39)
where φ
n (x)≡
dφn(x)
dx , we can write
∞
−∞
dzφ 2p (z)
dφ 2q (z)
dz
=
∞
−∞
dzφ 2p (z)
√
q φ 2q−1 (z)
−
2q + 1
2
φ 2q+1 (z)
= 0.
(E.40)
The integrand is zero because φ 2p (z) is an even function of z and φ 2q−1 (z) and φ 2q+1 (z)
are odd functions of z, making the integrand an odd function of z. Next note that
∞
−∞
dzφ 2p (z)
z
0
dt φ 2q (t) = 0
(E.41)
because
z
0 dt φ 2q (t) is an odd function of z, thus making the integrand odd. Finally, note
that
∞
−∞
dz
dφ 2p (z)
dz
z
0
dt φ 2q (t) =
∞
−∞
dt φ 2p (t) φ 2q (z)
z=∞
z=−∞
−
∞
−∞
dz φ 2p (z)φ 2q (z) = −δ p,q .
(E.42)
If we now integrate the quaternion determinant of ¯
Q N over energies x j , we
obtain the relations
∞
−∞
dx n+1 . . .
∞
−∞
dx N Qdet[ ¯
Q N ] = (N − n)! Qdet[ ¯
Q n (N)],
(E.43)