# [time 908] Re: [time 907] A proof of unitarity

Matti Pitkanen (matpitka@pcu.helsinki.fi)
Fri, 8 Oct 1999 09:56:38 +0300 (EET DST)

I will print the text, read it and respond later.

Best,

MP

On Fri, 8 Oct 1999, Hitoshi Kitada wrote:

> Dear Matti,
>
> Rethinking about your proof, I found an alternative simple proof of unitarity
> of S-matrix. Of course this is a proof on a formal level.
>
> The proof:
>
> Set
>
> R_0(z)=(H_0-z)^{-1}, z: non-real, H_0=L_0(free), V=L_0(int),
>
> P = projection onto the eigenspace of H_0 with eigenvalue 0.
>
> Let m_0 in P\HH (=eigenspace of H_0 with eigenvalue 0). Then the scattering
> state m(z) satisfies (I omit the bracket notation)
>
> m(z)=m_0+m_1(z)=m_0-R_0(z)Vm(z).
>
> Thus
>
> Pm(z)=Pm_0+Pm_1(z)=m_0+z^{-1}PVm(z). (1)
>
> Therefore
>
> PVm(z)=zPm_1(z). (2)
>
> Your assumption is
>
> VPm_1(z)=0. (A)
>
> Thus (2) and (A) yield
>
> VPVm(z)=zVPm_1(z)=0, which means
>
> <m(z)|VPV|m(z)>=0,
>
> hence
>
> PVm(z)=0.
>
> This and (2) imply
>
> Pm_1(z)=z^{-1}PVm(z)=0.
>
> Thus by (1) we have
>
> Pm(z)=m_0.
>
> Therefore
>
> <m(z)|P|n(z)>=<m_0|n_0>: unitarity. (U)
>
>
> Comment 1. As Im z -> 0,
>
> m(z) = m_0 - R_0(z)Vm(z) = (1+R_0(z)V)^{-1}m_0
>
> would be outside the Hilbert space \HH. Then (U) might lose its meaning as Im
> z -> 0. This would require the introduction of some larger space \HH_-.
>
> Comment 2. The present formulation of yours uses P explicitly. Namely m=m(z)
> may be outside of P\HH. Thus it is free of my criticism in [time 882].
>
>
> Best wishes,
> Hitoshi
>
>

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