Matti Pitkanen (matpitka@pcu.helsinki.fi)
Fri, 8 Oct 1999 16:22:28 +0300 (EET DST)
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. 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]
Your assumption is
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
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.
[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
over z. The p-adically weak point of your proof is that
zero norm for state does not imply vanishing of state in
p-adic context.
******************
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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