Operator Trace Formulae

FCCR Table of Contents


Trace Formulae

   This appendix contains useful formulae expressing the traces of various
   operators used in the paper, and various traces of their products.

   Q(n), P(n), G(n) and N(n) are defined in [Section II].
   The S_a(n) are defined in [Section XIV].

   The formulae mostly depend on the results proven in [Section VIII].

   Some of the formulae are calculated here, others should be obvious from
   inspection and general principles.
   
   
        Tr( G(n) )  =  0
        Tr( G^2(n) )  =  n(n-1)
        Tr( G^3(n) )  =  -n(n-1)(n-2)  <  0
   
        Tr( G^k(n) )  =  (n-1) + (-1)^k (n-1)^k
   
        Tr( Q(n) )  =  0
        Tr( P(n) )  =  0
        Tr( Q^2(n) )  =  n(n-1)/2
        Tr( P^2(n) )  =  n(n-1)/2
        Tr( N(n) )  =  n(n-1)/2
        Tr( N^2(n) )  =  n(n-1)(2n-1)/6
        Tr( N^2(n) G(n) )  =  n(n-1)(2n-1)/6 - n(n-1)^2
                          =  n(n-1)[ (2n-1)/6 - (n-1) ]
                          =  n(n-1)[ (2n-1)/6 - (6n-6)/6 ]
                          =  n(n-1)[ (-4n+5)/6 ]
                          =  -n(n-1)(4n-5)/6
   
        Q(n) G(n) Q(n)  =  Q^2(n) - n(n-1)/2 |n, n-2><n, n-2|
        P(n) G(n) P(n)  =  P^2(n) - n(n-1)/2 |n, n-2><n, n-2|
   
        Q(n) G(n) Q(n) + P(n) G(n) P(n)   =
         Q^2(n) + P^2(n) - n(n-1) |n, n-2><n, n-2|
         = N(n) + (1/2)G(n) - n(n-1) |n, n-2><n, n-2|
         = N(n) + (1/2)I(n) - n(n-1) |n, n-2><n, n-2| - (n/2) |n, n-1><n, n-1|
        
   
        Tr( Q(n) G(n) Q(n) )  =  0
        Tr( P(n) G(n) P(n) )  =  0
   
        Tr( Q(n) G^2(n) Q(n) )  =  n^2(n-1)/2
        Tr( P(n) G^2(n) P(n) )  =  n^2(n-1)/2
   
        Tr( Q(n) G^k(n) Q(n) )  =  (1/2)[(n-1)^k(-1)^k + (n-1)^2]
        Tr( P(n) G^k(n) P(n) )  =  (1/2)[(n-1)^k(-1)^k + (n-1)^2]
   
        Tr( N(n) )  =  n(n-1)/2
        Tr( N(n) G(n) N(n) )  =  -n(n-1)(4n-5)/6  <  0
   
        Tr( N(n) G^2(n) )  =  +n(n-1)(2n-3)/2  >  0
   
        Tr( (N(n) + (1/2)G(n)) G(n) (N(n) + (1/2)G(n)) )  =
             Tr( N(n) G(n) N(n) ) +
             Tr( N(n) G^2(n) ) +
             (1/4)Tr( G^3(n) )
   
             =  [n(n-1)/2][(n-2)/6]  =  n(n-1)(n-2)/12
   
        Tr( M(n) G(n) M(n) )  =
        Tr( (N(n) + G(n)) G(n) (N(n) + G(n)) )  =
             Tr( N(n) G(n) N(n) ) +
             2 Tr( N(n) G^2(n) ) +
             Tr( G^3(n) )
   
             =  [n(n-1)/2][(2n-1)/3]  =  n(n-1)(2n-1)/6
   
   
        Tr( M(n) G(n) N(n) )  =
        Tr( (N(n) + G(n)) G(n) N(n) )  =
  	Tr( N(n) G(n) N(n) ) + Tr( G^2(n) N(n) )  =
            -n(n-1)(4n-5)/6 +n(n-1)(2n-3)/2  =  n(n-1)(n-2)/3  >  0
   
        Tr( Q(n) P(n) )  =  Tr( P(n) Q(n) )  =  0
   
   
        Tr( Q(n) G(n) P(n) )  =  -in(n-1)/2
        Tr( P(n) G(n) Q(n) )  =  +in(n-1)/2
   	Tr( G(n) N(n) ) =  (1/2)(n-2)(n-1)) - (n-1)^2
   	                =  -(n(n-1)/2)
   
   
   	Tr( G(n)[ N(n) + (1/2)G(n) ] )  =  Tr( G(n)N(n) ) + (1/2)Tr( G^2(n) )
   	              =  Tr( G(n)N(n) ) + (1/2)( (n-1) + (n-1)^2 )
   	              =  (1/2)(n-2)(n-1)) - (n-1)^2
   	                                + (1/2)( (n-1) + (n-1)^2 )
   	              =  (1/2)(n-2)(n-1)) + (1/2)( (n-1) - (n-1)^2 )
   	              =  (1/2)(n-1)[ (n-2) + 1 - (n-1) ]
   	              =  (1/2)(n-1)[ (n-1) - (n-1) ]
                         =  0
   
   Then for the oscillator energy, Cf. [Theorem 8.10],
   
   	E(n)  =  N(n) + (1/2)G(n)  =  (1/2)(Q^2(n) + P^2(n) )
   
   	Tr( E(n) G(n) )  =  0
   
   	Then also,
   
   	Tr( G(n)P^2(n) )  =  - Tr( G(n)Q^2(n) )
   
   Direct computation also shows that
   
   	Tr( P(n) )  =  Tr( Q(n) )  =  Tr( G(n) )  =  0
   
   	Tr( P(n)G(n)P(n) )  =
   	     Tr( P^2(n)G(n) )  =
   	     Tr( G(n)P^2(n) )  =  0
   and
   	Tr( P(n) )  =  0  =  Tr( Q(n) )
   	Tr( G(n)P(n) )  =  0  =  Tr( G(n)Q(n) )
   
   	Tr( P^2(n) )  =  n(n-1)/2  =  Tr( Q^2(n) )
   	Tr( G^2(n) )  =  (n-1) + (n-1)^2
   	             =  (n-1)(1 + (n-1))
   	             =  n(n-1)
   
   	Tr( E(n) )  =  Tr( N(n) ) + (1/2)Tr( G(n) )
   	            =  Tr( N(n) )
   	            =  n(n-1)/2
   
   	Tr( E^2(n) )  =  Tr( N^2(n) ) + Tr( N(n)G(n) ) + (1/4) Tr( G^2(n) )
                      =  n(n-1)(2n-1)/6  -  n(n-1)/2  +  (1/4)n(n-1)
                      =  n(n-1)(4n-5)/12
   
   	Tr( E^2(n) G(n) )  =
               Tr( N^2(n) G(n) ) + Tr( N(n)G^2(n) ) + (1/4) Tr( G^3(n) )
               =  Tr( N^2(n) G(n) ) + Tr( N(n)G^2(n) ) + (1/4) Tr( G^3(n) )
               =  -n(n-1)(4n-5)/6 + n(n-1)(2n-3)/2 - (1/4)n(n-1)(n-2)
               =  n(n-1)[ -(4n-5)/6 + (2n-3)/2 - (n-2)/4 ]
               =  n(n-1)[ -2(4n-5) + 6(2n-3) - 3(n-2) ]/12
               =  n(n-1)[ -8n + 10 + 12n - 18 - 3n + 6 ]/12
               =  n(n-1)[ n + 16 - 18 ]/12
               =  n(n-1)(n-2)/12
   
   	Tr( S_3^2 )  = (n/4)(n-1)^2 - (n-1)Tr( N(n) ) + Tr( N^2(n) )
   	           = (n/4)(n-1)^2 - (n/2)(n-1)^2 + Tr( N^2(n) )
   	           = -(n/4)(n-1)^2 + Tr( N^2(n) )
   	           = -(n/4)(n-1)^2 + n(n-1)(2n-1)/6
   	           = (n/2)(n-1)( -(n-1)/2 + (2n-1)/3 )
   	           = (n/2)(n-1)( (4n - 2 - 3n + 3)/6 )
   	           = (n/2)(n-1)( (n + 1)/6 )
   	           = (n/12)(n^2-1)
   
   
   By unitary equivalence
   
   	Tr( S_1^2 )  = Tr( S_2^2 )  = Tr( S_3^2 )
   
   
   The trace of the Casimir operator must then be
   
   	3 Tr( S_1^2 ) = (n/4)(n^2-1)
   
   and indeed from equation (14.8)
   
   	Tr( (n^2-1)/4 I(n) )  =   (n/4)(n^2-1)
   
   Trace is also used to define inner products and norms on the algebra of
   operators.  Tr( A! G(n) B ) defines an inner product of A and
   B with respect to the ground form G(n).  See [Appendix A].  A Euclidean
   inner product is defined by Tr( A! B ).  This then induces a norm
   
   	Tr( A! A )
   
   which in turn induces a metric, d( A, B ) defined by
   
   	Tr( A! - B!)(A - B) )  =
   
   		Tr( A! A ) + Tr( B! B )
   		- Tr( A! B ) - Tr( B! A )
   
   	=    Tr( A! A ) + Tr( B! B )
   	   - Tr( A! B ) - Tr( (A! B)! )
   
   	=    Tr( A! A ) + Tr( B! B )
   	     - 2 Re( Tr( A! B ) )
   
   the last term being a measure of the angle between the operators. (?)
   
   If B = aA, then
   
   	d( A, B ) =  (1 + aa* Tr( A! A )
   	     - 2 Re( a Tr( A! A ) )
   
   	= ( 1 + aa* - 2 Re( a ) ) Tr( A! A )
   
   Cf. [Appendix A].
   ------------------------------------------------
   
   	Tr( N(n) Q(n) )  =  0
   		In fact Diags of NQ are zero
   	Tr( N(n) P(n) )  =  0
   Also
   	Tr( N(n) G(n) Q(n) )  =  0
   		In fact Diags of NGQ are zero

   ------------------------------------------------
   Trace and orthogonality w.r.t. standard Euclidean ground form
   and w.r.t G(n)
   ------------------------------------------------
   Re asymptotic conformal factor between S_a(n) and Q_a(n) expressed
   and calculated by Theorem 14.1 we have:
   
   	|| S_a ||^2  =  Tr( S_a^2 )  =  (n/12)(n-1)(n+1)
   	|| Q_a ||^2  =  Tr( Q^2(n) )  =  n(n-1)/2
   
   Therefore,
   	|| Q_a ||^2  =  [6/(n+1)] || S_a ||^2
   or
   
   	|| Q_a ||  =  [6/(n+1)]^1/2 || S_a ||
   
   
   In terms of the a *different* norm of an operator which in finite
   dimensional spaces is equal to the value of the maximal eigenvalue,
   our refined spacing rule is in fact, exact.
   
   	q_max  =  [6/(n+1)]^(1/2) s_max
   
   but s_max  = (n-1)/2,
   
   This however is not that "spectral norm" but a "trace norm". It is
   exactly what is done in the [Appendix A] only in different language.
   The trace norm is not connected to the spectral norm, and does not allow
   as many sequences of operators to converge to an operator on an infinite
   dimensional Hilbert space.  The points of convergence must be those of
   "trace class".  Cf. [Appendix A].

   The trace norm is the root of the sum of the squares of eigenvalues, a
   kind of n-dimensional Euclidean norm, not a norm in the spectral plane.

   While
   
   	q_max  =  [6/(n+1)]^(1/2) (n-1)/2
   
   How does this fit with
   
   	|| S_a ||  =  [(n/12)(n^2-1)]^(1/2)
   
   which grows as n^(3/2), while s_max is linear in n.
   
   
   ------------------------------------------------
   Things yet to be computed:
   
           For both the + and - signs of F(n),
   
   	Tr( F(n) G(n) F(n) )  =  Tr( F^2(n) ) - n <n, n-1| F^2(n) |n, n-1>
                                 =  n(n-1)(2n-1)/6 - n(n-1)(2n-1)/6
                                 =  0
   				See Corollary 8.5.3
   
   	Tr( F(n) G(n) )  =  Tr( F(n) ) - n <n, n-1| F(n) |n, n-1>
                            =  n(n-1)/2 - n(n-1)/2
                            =  0
                              See Theorem 8.5
   
   	Tr( F(n) N(n) F(n) )  =  ?
   	Tr( F(n) N(n) )  =  ?
   	Tr( F(n) G(n) Q(n) )  =  ?           Important
   	Tr( F(n) Q(n) )  =  ?                Important
   	Tr( F(n) G(n) P(n) )  =  ?
   	Tr( F(n) P(n) )  =  ?
   	Tr( F(n) [ N(n) + (1/2)G(n) ] )  =  ?
   
   For n=2 using the explicit matrix representations in section XVI
   
   	Tr( Q(2) G(2) F(2) )  =  0,    [Q(2), F(2)]  =  0
   	Tr( P(2) G(2) F(2) )  =  -isqrt(2)
   	Tr( P(2) G(2) N(2) )  =  0
   	Tr( P(2) G(2) [N(2) + (1/2)G(2)] )  =  0
   



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Created: August 1997
Last Updated: May 28, 2000
Last Updated: July 21, 2002