Matti Pitkanen (matpitka@pcu.helsinki.fi)
Fri, 8 Oct 1999 20:32:47 +0300 (EET DST)

On Fri, 8 Oct 1999, Hitoshi Kitada wrote:

> 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
> > have discovered this time! Have you read Hofstadter's book 'Goedel,
> > 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.

This is not essential: one can just assume that P exists.

>
>
> > 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.
>
> Your integral is not well-defined.

One can just assume that P just exists. It does not
change my proof of unitarity.

>
> >
> > [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.

By adding one P (P^2=P) you get <m(z)|VPPV|m(z)=0> stating that
PV|m(z)> is zero norm state. This has nothing to do with commutation
of P and V.

And here comes the quintessential thing: in p-adic context zero norm
does not imply the vanishing of state. If you read the construction
of the nilpotent hermitian matrices this becomes clear.
My condition implies vanishing of T in real context: debate is not

It would be more constructive to look the consequences rather than
trying to destroy the proof by arguments which are doomed to
involve always some element of "real" thinking. One can
invent infinite number of arguments showing T=0 in real context:
T is indeed zero in real context! And I am extremely happy with it
since I have no deep justification for p-adics and beatiful cohomological
interpretation of S-matrix.

Every argument contains "<m|m>=0 --> |m>=0" somewhere not holding true
in p-adic context. The simplest manner to see that T=0 in real
case is based on positive definiteness of T^daggerT expectation
values.

>
> > In p-adic context the
> > 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
> > is not allowed p-adically.
> > *******************
> >
> > [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.

The essential step in your proof is the step above which in
real case is ok but not in p-adic case.

>
> >
> > [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.

Certainly it is and therefore it is dangerous to drop the integration
over a small curve and replace it with |n(z)>. This approach is from
which I started: formal description of scattering theory using iepsilon.
The description using residy calculus is much more elegant
and mor rigorous.

> >
> >
> > 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.

I fail to see what you mean with this. In any case, your deduction
does not hold in p-adic context for the reason that vanishing norm
does not imply vanishing state in p-adic context.

>
> > 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
> knows without making redundant comments.
>

Physicists are forced to generalize existing theories. The requirement
that everything is mathematically rigorous and sound restricts the
consideration to systems which are rather uninteresting physically.
The road from N-particle Schrodinger equation to configuration space
spinors is horribly long. This is the problem. What helps in this kind of
situation is physical and philosophical intuition making possible to
see the architecture.

Best,

MP

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