160
6 Quantum Dynamics and Random Matrix Theory
which expresses the invariant measure in terms of the matrix elements of the
Hamiltonian.
We can also write the invariant measure of ¯
H R in terms of its eigenvalues
and eigenvectors, and in doing so, we will be able to show that the eigenvalues
and eigenvectors are statistically independent. This is called the polar form of the
invariant measure.
To obtain the polar form of the invariant measure, we first take the differential of
Eq. (6.5),
d ¯
H R = d ¯
O· ¯
E R · ¯
O
T
+ ¯
O·d ¯
E R · ¯
O
T
+ ¯
O· ¯
E R ·d ¯
O
T ,
(6.10)
and introduce the differential
δ ¯
O ≡ ¯
O
T
·d ¯
O = −d ¯
O
T
· ¯
O = −δ ¯
O
T ,
(6.11)
where we have used the fact that ¯
O T · ¯
O = ¯
1 and d ¯
O T · ¯
O + ¯
O T ·d ¯
O = 0. The matrix
δ ¯
O is antisymmetric.
If we now multiply Eq. (6.10) on the left by ¯
O T and on the right by ¯
O, we find
δ ¯
F ≡ ¯
O
T
·d ¯
H R · ¯
O = δ ¯
O· ¯
E R − ¯
E R ·δ ¯
O + d ¯
E R .
(6.12)
The (i, j )th element of the matrix δ ¯
F can be written
δ ¯
F ij = de i δ i,j + (e j − e i )δ ¯
O i,j .
(6.13)
Note that the diagonal elements of δ ¯
O don’t contribute. The invariant metric of ¯
H R
can now be written
(ds)
2
H R
= Tr(d ¯
H R ·d ¯
H
T
R ) = Tr(δ ¯
F ·δ ¯
F
T )
=
N
i=1
(de i )
2
+ 2
1≤i
(e j − e i )
2 (δ ¯
O i,j )
2 ,
(6.14)
so Det[ ¯
g] = 2 N(N−1)/2
1≤i
symmetric matrix, H R , can then be written
dd H R =
1≤i
|e j − e i |
de 1 de 2 × . . . ×de N δδ δO ,
(6.15)
where
δδ δO = 2
N(N−1)/4 δO 1,2 δO 1,3 × . . . ×δO N −1,N
(6.16)
6 Quantum Dynamics and Random Matrix Theory
which expresses the invariant measure in terms of the matrix elements of the
Hamiltonian.
We can also write the invariant measure of ¯
H R in terms of its eigenvalues
and eigenvectors, and in doing so, we will be able to show that the eigenvalues
and eigenvectors are statistically independent. This is called the polar form of the
invariant measure.
To obtain the polar form of the invariant measure, we first take the differential of
Eq. (6.5),
d ¯
H R = d ¯
O· ¯
E R · ¯
O
T
+ ¯
O·d ¯
E R · ¯
O
T
+ ¯
O· ¯
E R ·d ¯
O
T ,
(6.10)
and introduce the differential
δ ¯
O ≡ ¯
O
T
·d ¯
O = −d ¯
O
T
· ¯
O = −δ ¯
O
T ,
(6.11)
where we have used the fact that ¯
O T · ¯
O = ¯
1 and d ¯
O T · ¯
O + ¯
O T ·d ¯
O = 0. The matrix
δ ¯
O is antisymmetric.
If we now multiply Eq. (6.10) on the left by ¯
O T and on the right by ¯
O, we find
δ ¯
F ≡ ¯
O
T
·d ¯
H R · ¯
O = δ ¯
O· ¯
E R − ¯
E R ·δ ¯
O + d ¯
E R .
(6.12)
The (i, j )th element of the matrix δ ¯
F can be written
δ ¯
F ij = de i δ i,j + (e j − e i )δ ¯
O i,j .
(6.13)
Note that the diagonal elements of δ ¯
O don’t contribute. The invariant metric of ¯
H R
can now be written
(ds)
2
H R
= Tr(d ¯
H R ·d ¯
H
T
R ) = Tr(δ ¯
F ·δ ¯
F
T )
=
N
i=1
(de i )
2
+ 2
1≤i
2 (δ ¯
O i,j )
2 ,
(6.14)
so Det[ ¯
g] = 2 N(N−1)/2
1≤i
dd H R =
1≤i
de 1 de 2 × . . . ×de N δδ δO ,
(6.15)
where
δδ δO = 2
N(N−1)/4 δO 1,2 δO 1,3 × . . . ×δO N −1,N
(6.16)
