Operator Trace Formulae

```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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