[time 927] Re: [time 923] Unitarity

Hitoshi Kitada (hitoshi@kitada.com)
Sun, 10 Oct 1999 04:31:26 +0900

Dear Matti,

I looked through your "formal" proof. I agree it is correct as far as
epsilon>0. And it is free of the criticism that you did not treat the inner
product <m|n> of the full states <m| and |n> before (i.e. you considered
<m|P|n> before).

I think physicists might say OK since many of them seem not aware of

My mathematical interest is in how the states <m| and |n> behave as
epsilon ->0. This is necessary to be considered because <m|=<m(z)| and
|n>=|n(z)> actually depend on z=i\epsilon and |m(z)> satisfies
(H+z)|m(z)>=(H_0+z)|m_0>=z|m_0> by your equation (2), hence
H|m(z)>=-z(|m(z)>-|m_0>) which is not necessarily 0 when epsilon>0. If one
wants to have equation H|m(i\epsilon)>=0, one needs to let epsilon ->0. If one
goes to the limit epsilon ->0, they would go outside the Hilbert space \HH.
For if the scattering states |m(i0)> and |n(i0)> remain in \HH, they are
genuine eigenstates of H and scattering does NOT occur. Thus it is necessary
for |m(i0)> and |n(i0)> to be outside \HH. Then the inner product <m(z)|n(z)>
goes to infinity as epsilon ->0 and S-matrix diverges. Thus one has to
consider "generalized" eigenfunctions or eigenvectors |m(i0> and |n(i0)> which
are characterized by the conditions

H|m(i0)>=0, and |m(i0)> does not belong to the Hilbert space \HH.

"Generalized" means that they are outside \HH.

In this respect, formal proof does not give the unitarity.

There will be a way to avoid this difficulty. A physicist Prugovecky whom
Stephen referred to was researching scattering theory when I was a graduate
student (probably he might have been a graduate student too or not long later
than that). As someone whom Stephen referred to said, Prugovecky is aware of
the difficulty of this sort, and uses Gel'fand triple to avoid it. This is
understandable recalling his career in scattering theory. As this example
suggests, physicists could be more careful in their treatment of divergences.
Renormalization technique would not give logical solutions but it could be
replaced by more mathematical arguments, which could be the shortest way to
get physicists' dreams. "Blind mathematicians" could give better help to
physicists than mathematical dreamers could, although personally I like to
remain a dreamer myself. This time I made criticisms on your proofs with a
dare to be away from the standpoint of a dreamer. This was because there
seemed a possibility that a rigorous proof without divergence might have been
gotten if some push was given. I hope you would not take my criticisms as
something like attempts to destroy your proofs.

Best wishes,

----- Original Message -----
From: Matti Pitkanen <matpitka@pcu.helsinki.fi>
To: <time@kitada.com>
Sent: Sunday, October 10, 1999 12:07 AM
Subject: [time 923] Unitarity

> Dear Hitoshi,
> I began to ponder your comment and looked formal scattering theory again
> and realized that unitarity proof for S-matrix formally generalizes to
> TGD case.
> Denote H_0== L_0(free), H==L_0(tot)= L_0(free)+ L_0(int) and
> V=L_0(int).
> a) One has the basic equation
> |m> = |m_0> - 1/(H_0+iepsilon) V |m> (1)
> b) One can multiply this equation by H_0+iepsilon and move terms
> proportional to |m> to the left hand side to
> get (H+iepsilon)|m> right hand side. Left hand side gives
> (H_0 +V)|m_0> -V|m_0> by adding and subtracting V|m_0>.
> Solving |m> one obtains
> |m> = |m_0> -1/(H+ iepsilon) V|m_0> (2)
> c) One can also solve |m_0> from the first equation
> |m_0> = |m> + 1/(H_0+iepsilon) V|m> (3)
> ******************
> Consider now the matrix element <m|n>: one must show that this
> is <m_0|n_0> in order to prove unitarity.
> a) Express first <m| in terms of <m_0| using (2)
> <m|n> = <m_0|n> +<m_0|V*1/(-H-iepsilon)|n> (4)
> b) One can use the fact that H annihilates |n>
> to remove 1/(L_0(tot).. term in front of V and replace
> the H=0 by -H_0=0 (due to inner product with <m_0|)
> to get
> <m|n> = <m_0|n> -<m_0|1/(H_0-iepsilon)V|n>

Here not -iepsilon, but +iepsilon.

> c) But by equation (3) the state proportional to |n> is in fact |n_0>
> and one has
> <m|n> =<m_0|n_0>.
> Thus one has formal unitarity. The calculation is extremely tricky.
> What do you think?
> Best,
> MP
> P.S
> I tend to believe that the condition V|m_1>=0 is correct condition
> since it leads to p-adics and is consistent with quantum criticality
> even if it would not be needed for unitarity.

This archive was generated by hypermail 2.0b3 on Sun Oct 17 1999 - 22:40:47 JST