- #1
muppet
- 608
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Hi all,
I'm working on Taylor's text on scattering (a reference from Peskin and Schroeder). They define the Moller operators [tex]\Omega[/tex] which are isometric, satisfying
[tex]\Omega^{\dagger}\Omega=1[/tex]
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 [tex]\Omega^{-1}[/tex] is undefined. On states in the range of \Omega
[tex]\Omega^{\dagger}=\Omega^{-1}[/tex];
for states outside of this range, [tex]\Omega^{\dagger}=0[/tex].
They show that
[tex]\Omega^{\dagger}H\Omega=H_0[/tex]
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.
I'm working on Taylor's text on scattering (a reference from Peskin and Schroeder). They define the Moller operators [tex]\Omega[/tex] which are isometric, satisfying
[tex]\Omega^{\dagger}\Omega=1[/tex]
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 [tex]\Omega^{-1}[/tex] is undefined. On states in the range of \Omega
[tex]\Omega^{\dagger}=\Omega^{-1}[/tex];
for states outside of this range, [tex]\Omega^{\dagger}=0[/tex].
They show that
[tex]\Omega^{\dagger}H\Omega=H_0[/tex]
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.