Natural Science Forum / Physics / Research / March 2008

Since I've seen no response to this, I'll make some simple observations.

I'm not sure how well-defined the probability space of unitary operators
over an infinite dimensional vector space is. For simplicity, let's look at
the finite dimensional case and hope that the results generalise in a nice
way.

If you start with a set |Phi(i)> of unit vectors, the effect of a unitary
operator U on these is essentially that of mapping them to any set of unit
vectors |u(i)> that have the same inner products:
 (1)  <u(i)|u(j)> = <Phi(i)|Phi(j)> for all i,j.
Thus, if you pick an orthonormal basis |w(k)> of the space spanned by the

 (2)  |Phi(i)> = sum_k  P_{ik} |w(k)>,
since the unitary matrix maps |w(k)> to other orthonormal bases (with
uniform probability...if this is well-defined for the infinite dimensional
case), picking a random U mapping |Phi(i)> to |u(i)> corresponds to picking
a random orthonormal basis |v(k)> = U |w(k)> and rewriting your expression
in terms of
 (3)  U|Phi(i) = |u(i)> = sum_k P_{ik} |v(k)>.
For <A>, if |u(0)>=|v(0>, |u(1)>=a|v(0)>+b|v(1)>,
 (4a) <A> = E[<u(0)|A|u(1)>] = E[a<v(0)|A|v(0)>+b<v(0)|A|v(1)>]
which, as the b-term gives expected value zero since the expected value of

 (4b) <A> = E[<v(0)|A|v(0>] = tr(A)/N
where N is the dimension of the space: for finite dimensional spaces, this
is no problem, but for infinite dimensional it is not clear to me that this
need be well-defined.

This approach should work for many of the other cases as well, of course up
until you get to the infinity problem.

The case where [U,H], if H has distinct eigenvalues, in the basis of the
eigenvalues of H such a unitary operator will simply be one which multiplies
each eigenvector with a constant: i.e. if |h(k)> are the eigenvalues of H,
U|h(k)>=c(k)|h(k)> for any set of c(k) with |c(k)|=1. Express your vectors
in this basis and write out the expression, and you should get somewhere. If
H has multiple eigenvalues, this only produces a subset of the unitary
operators, so more terms may possible cancel when taking the mean.

Einar

I'm not quite sure of all the difficulties on constructing an invariant
measure in the infinite dimensional unitary group and identifying
integrable functions on the group. But, suppose that were possible and
that the averages you propose in the definition of <A> were finite.
Then this average would have to satisfy certain algebraic properties.

First, let me write <A> in slightly different notation that emphasizes
its argument structure:

 <A> = [ <Phi0| [ A ] |Phi1> ]

Given your definition, this quantity would be linear in A and |Phi1>, and
antilinear in |Phi0>. Moreover:

 <A> = [ <Phi0| [ V^+ A V ] | |Phi1> ], for V unitary and
 [ <Phi0|V [ A ] |Phi1> ] = [ <Phi0| [ A ] V|Phi1>, for V unitary.

In other words, for fixed A, <A> has to be proportional to <Phi0|Phi1>,
and also, for fixed |Phi0> and |Phi1>, it has to depend only on the
spectrum of A. I belive, the only linear spectral invariant of A is
tr(A), provided it is trace-class. So, the only possibility appears to
be

 <A> = C tr(A) <Phi0|Phi1>,

where C is a normalization constant to be fixed for your convenience.
This certainly holds for the finite dimensional group U(N).

This average can (almost) be reduced to the same computation as <A>.
Note that integrating U^+ A U over all unitary U is a well known way of
producing an intertwiner (an operator that commutes with all U). In the
case of <A,B>, you are also constructing an intertwiner, but now it has
the form rho(U)^+ A(x)B rho(U) integrated over all unitary U, and you've
replaced |Phi0> by |Phi0>(x)|Phi0>, similarly for |Phi1>. Here (x)
denotes the tensor product and rho(U) = U(x)U. That is, you are no
longer working with the fundamental representation of the unitary group,
instead you are working with the product of two of its fundamental
representations. Unlike the fundamental one, this product representation
is reducible. In the general finite dimensional case, it splits into
three irreducible representations: the diagonal one, the antisymmetric
one, and the symmetric trace-free one.

So, using the fact that group integration projects A(x)B onto the space
of intertwiners defined on this product representation and invariance
properties similar to those of <A>, we can write:

 <A,B> = C tr(A) tr(B) (<Phi0|Phi1>^2 + D <Phi0|Phi0><Phi1|Phi1>)
           + E tr(A^+ B) <Phi0|Phi0><Phi1|Phi1>.

Again, C is a normalization coefficient and D and E depend on N (if we
are dealing with the finite dimensional U(N) group) in such a way that
they both go to 0 as N goes to infinity. However, if you are only
interested in the algebraic properties of this average, then you may
want to keep these coefficients and define them independently in some
convenient way.

Here you can always write A in block form, where each block maps one
eigenspace of H into another one. The above average can be done block by
block. If A_{EE'} is a block that maps the E'-eigenspace into the
E-eigenspace, then <A_{EE'}>_H must be 0. At the same time <A_{EE}>_H
reduces to the previous problem, where the averaging is now done over
the unitary group on the E-eigenspace.

Again, A and B can be decomposed into blocks and the average can be
taken on a block by block basis.

Hope this helps.

Igor