Matti Pitkanen (matpitka@pcu.helsinki.fi)
Fri, 8 Oct 1999 16:22:28 +0300 (EET DST)

Dear Matti,

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. 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?
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. 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
*******************

[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

[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!

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
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

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