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The Diagonalizing transformations XI(n) and PI(n)
for Q(n) and P(n) respectively; diagonalizing the
Fourier transformation UPSILON(n).

   In  [Section IX], the matrix elements of the unitary diagonalizing
   transformation XI(n) for Q(n) where found.

        Q(n)  ->  XI!(n) Q(n) XI(n)
        |q(n, k)>  =  XI!(n) |n, k>

        |n, k>   =  XI(n) |q(n, k)>

   where |q(n, k)> are the eigenvectors of Q(n).

The matrix XI(n) is real, so the diagonalizing transformation is actually orthogonal. The diagonalizing transformation PI(n) for P(n), is then quite naturally related by the Fourier transform Fr(n): Define PI(n) as the transformation that maps the eigenbasis of N(n) to the eigenbasis of P(n) so,

        P(n)  ->  PI!(n) P(n) PI(n)                                    (10.1a)

   so P(n) is diagonalized and

        |p(n, k)>  =  PI!(n) |n, k>                                    (10.1b)

        |n, k>   =  PI(n) |p(n, k)>                                    (10.1c)

   where |p(n, k)> are the eigenvectors of P(n) and

        PI(n) |p(n, k)>  =   p(n, k) |p(n, k)>                         (10.1d)

        p(n, k)  =  q(n, k)                                            (10.1e)

   Then using the properties of XI(n) and Fr(n), it is not
   difficult to see that

        PI(n)  =  Fr(n) XI(n) Fr!(n)                                   (10.2a)

   is the diagonalizing transformation for P(n), and then

        PI!(n)  =  Fr(n) XI!(n) Fr!(n)                                 (10.2b)

   Although XI(n) is orthogonal, PI(n) is not, yet it is,
   of course, unitary.

The second Fourier transform UPSILON(n) arises as the diagonalizing transformation of C(n) the cyclic operator used in the proper polar decomposition of B(n) and B!(n). Diagonalizing UPSILON(n) is apparently not so simple an enterprise, even though the eigenvalues, sans multiplicity, are obvious. The following are pertinent results of [Mehta 1987]:


       f_jk  =  f_jk(n)  =

       SIGMA exp( - (pi/n)(mn + j)^2 ) H_k(sqrt(2 pi/n)(mn + j))       (10.3)

   where H_k(z) is the kth unnormalized Hermite polynomial as defined
   by the Rodrigues formula  (9.4).  Then,

        SIGMA <n, j| UPSILON(n) |n, m> f_mk(n)  =  i^k f_jk(n)         (10.4)

   So the second index of f_jk labels eigenvectors of
        <n, j| UPSILON(n) |n, m>
   that are attached to the eigenvalues i^k.  f_jk is a periodic
   function of both indicies with period n, so f_0k = f_nk,
   and f_j0 = f_jn.  More generally:

        f_jk(n)  =   (-1)^k f_(n-jk)(n)  =  f_(j+nk)(n)                (10.5)


        SIGMA  f_(jm)(n) f_(mk)(n)  =  0,  i^j not= i^k                (10.6)

   Eigenvectors with different eigenvalues are orthogonal.  The f_jk(n)
   provide an infinite number of eigenvectors; the problem is to choose
   a linearly independent set of them.  Mehta makes the following
   conjecture based on the numerical evidence of special cases.

   The f_jk(n) are linearly independent for:

        k = 0, 1, ..., n-1        if n is odd
        k = 0, 1, ..., n-2, n     if n is even

A bothersome and somewhat surprising (to me) event is that the eigenvectors are expressed as a doubly sided infinite summation. It would be considerably more convenient to have a finite linearly independent set expressed in closed form.

These three diagonalizing transformations are all outside the group that preserve the form of G(n). They will all however, preserve the matrix elements <psi|phi> and <psi|G(n)|phi>.

We wish to establish relative to the indexing implicitly chosen in [Section IX], which eigenvalue belongs to the index k of q(n, k). By inspection, as k increases, the real value of the eigenvalue decreases. This can be rigorized; it is messy. The operator Fr^2(n) maps Q(n) to -Q(n), and therefore reverses the signs of the spectral values of Q(n). Therefore, the matrix elements of Fr^2(n) in the basis

        |q(n, k)> = X^(t(n) |n, k>

   must be,

        <q(n, k)| Fr^2(n) |q(n, j)>  =  delta_k(n-j)                   (10.7)

   When diagonalizing Q(n) by XI(n) q(n, 0) is always the maximal
   eigenvalue; further, if n is odd, q(n, n-1/2) = 0.  If n is even,
   we've seen that there is no zero eigenvalue for Q(n).  For any n,
   the eigenvalues of Q(n) when diagonalized by XI(n), are uniformly
   decreasing down the diagonal.  The above is then also true for P(n)
   diagonalized by PI(n).

   Since the second Fourier transform UPSILON(n) diagonalizes the cyclic
   operator C(n) and,

        C(n)  =  exp( i omega_T t(n) )  =  exp( i2 pi/n F(n) )

   where F(n) is the "phase matrix" and t(n) is the time operator for
   a harmonic oscillator, then UPSILON(n) also diagonalizes t(n) and F(n).

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Created: August 1997
Last Updated: May 28, 2000