Fri, 8 Oct 1999 23:24:15 +0900

Dear Matti,

Matti Pitkanen <matpitka@pcu.helsinki.fi> 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.
>
> [MP] I hardly dared to open your posting(;-) and only wondered what
> kind of new devilish trick proving my poor S-matrix trivial you
> Escher, Bach'? If not you should do it.

I am a specialist of Goedel's proof.

> There are hilarious stories whose
> basic theme is Goedel's theorem: every complete axiom system with
> sufficient complexity contains inconsistency. In one story
> the great dream of Achilles is to build a record player which is complete.
> When N:th generation record player is constructed,
> Tortoise gives for poor Achilles as a gift a record whose name happens
> to be "I am record not playable by record player of generation N". Every
> time Achilles manages to construct a new version
> of his record player the record brought by Tortoise manages to smash it
> into pieces. So I will look at your proof with horror in my heart(;-).
>
>
> 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)
>
>
> [MP]
> Isn't it dangerous to talk about P|m(z)> since the action
> of P itself is defined as integral over infinitesimal circle around z=0?

No, your definition of P by integral does not work.

> You now how extremely tricky creature residue calculus is.
>
> Since I cannot afford too many record players, I am only willing
> to talk about P|m> defined as
>
> Int_C dz p(z)|m(z)
>
> with p(z)= (1/2pi)* (1/L_0-iz).
> **************************
>
>
> [Hitoshi]
> Therefore
>
> PVm(z)=zPm_1(z). (2)
> ***************
>
>
> [MP] This looks ok but I am a somewhat worried about applying P to
> |m_1(z)>.
>
> [MP]
>
> VPm_1(z)=0. (A)
>
> [MP] My assumption is actually weaker. P involves integration
> around small circle.

>
> [Hitoshi]
> Thus (2) and (A) yield
>
> VPVm(z)=zVPm_1(z)=0, which means
>
> <m(z)|VPV|m(z)>=0,
>
> hence
>
> PVm(z)=0.
>
> [MP] This states that state PV|m(z)> has zero norm. In real
> context this implies PV|m(z)> =0.

No, P and V does not commute.

> fact that state has vanishing p-adic norm does not
> however imply that state vanishes. In fact that rows of T
> matrix have zero norm p-adically. So that this step
> *******************
>
> [Hitoshi]
> This and (2) imply
>
> Pm_1(z)=z^{-1}PVm(z)=0.
>
>
> [MP] Here there is dangerous 1/z which goes to zero and the
> conclusion should be taken with grain of salt. In any case
> Pm_1(z)=0 leads to trivial S-matrix.

You need to treat 1/z and its limit Im z->0. You seem to assume the existence
of

lim_{Im z->0} (1-R_0(z)V)^{-1}.

But this is the main issue of the problem. The rest that you gave and I gave
are only miscellaneous arguments.

>
> [Hitoshi]
> Thus by (1) we have
>
> Pm(z)=m_0.
>
> Therefore
>
> <m(z)|P|n(z)>=<m_0|n_0>: unitarity. (U)
> **********
>
>
> [MP] By the same proof one has
>
> <m_0|n(z)> = <m_0|P|n(z)>= <m_0|n_0> so that S-matrix is trivial!
> And my record player would be smashed into pieces again!

Maybe you have to begin your proof on the basis of sound mathematics.
The limit Im z->0 is the point. Others are just miscellaneous and trivial
things for mathematicians.

>
>
> I think that the problem is that one should not include P|n(z)>
> into the vocabulary but talk only about P|n> involving integration
> over z.

Your formulae including n(z) in the integrand is different from what you want
to mean.

> zero norm for state does not imply vanishing of state in
> ******************
>
>
> 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,
>
> MP
>
>
>
>

Physicists look like to talk much. Mathematicians just talk about what they