Isometric operators- spectrum preserving?

[muppet]

Hi all,
I'm working on Taylor's text on scattering (a reference from Peskin and Schroeder). They define the Moller operators \Omega which are isometric, satisfying
\Omega^{\dagger}\Omega=1
This is not necessarily the same as unitary in an infinite dimensional space, the difference being that an infinite dimensional space can be mapped isometrically onto a proper subset of itself, so that state vectors exist for which \Omega^{-1} is undefined. On states in the range of \Omega
\Omega^{\dagger}=\Omega^{-1};
for states outside of this range, \Omega^{\dagger}=0.

They show that
\Omega^{\dagger}H\Omega=H_0
where H is an interacting hamiltonian and H_0 its kinetic term. They argue that if Omega were unitary then H and H_0 would have to have the same spectra; only in the case where H has no bound states is Omega unitary, and the spectra coincide.

My question is: is it true that the spectrum of H_0 coincides with the energies of the non-bound states of H? This assumption is usually made in other treatments of scattering I've seen, and it seems a reasonable conclusion, but it's striking that the author belabours the non-unitarity of Omega rather than explains the reasoning behind what's taken as a given in other accounts, and the book is very careful to point out that other plausible statements can be wrong (for example, the Moller operators are non-unitary even though they are defined as the limit of a product of unitary operators, and the calculus of limits theorem doesn't hold for operators apparently, which is also something I'd like to understand better).

Thanks in advance.

 

[Bill_K]

is it true that the spectrum of H_0 coincides with the energies of the non-bound states of H?

I think you're right that this is a reasonable assumption but definitely an assumption. For example in a field theory in which the mass renormalization is finite, the space of continuum states of H and H₀ would be different.

 

[muppet]

Thanks for your reply.
I think it may actually be possible to demonstrate the partial equivalence of the spectra,
Let
H|\psi\rangle =E|\psi\rangle
and
\Omega|\phi\rangle=|\psi\rangle
Then
H|\psi\rangle=H\Omega|\phi\rangle=E\Omega|\phi \rangle
Acting on the left with \Omega^{\dagger} we then have
\Omega^{\dagger} H \Omega|\phi\rangle =H_0|\phi\rangle=E\Omega^{\dagger}\Omega|\phi \rangle =E|\phi\rangle
Therefore
H_0|\phi\rangle =E|\phi\rangle
i.e. the image of an eigenstate of the free hamiltonian under a Moller operator is an eigenstate of the full hamiltonian with the same eigenvalue.

Thinking about this setup a little more, however, it seems that taking the adjoint of
\Omega^{\dagger}H\Omega=H_0
implies that the free hamiltonian should annihilate the bound states, which seems highly suspect... any thoughts?