Hitoshi Kitada (hitoshi@kitada.com)
Sun, 3 Oct 1999 13:26:04 +0900
Dear Matti,
My question in the following is that:
You stated the scattering space \HH_s is the same as the free space P\HH. This
means
\HH_s = P\HH,
hence
any free state u=|n_0> in P\HH=\HH_s satisfies
u = Pu.
Thus
1/(1+X) |n_0> = 1/(1+X) P|n_0> = |n_0>.
The last equality here follows from
1/(1+X) = 1-\sum_{n=1}^\infty (-X)^n
and
X P = 0
by
X = X= 1/(L_0 +ie*epsilon))*L_0(int)
and L_0(int) P|n_0> = 0.
Best wishes,
Hitoshi
----- Original Message -----
From: Hitoshi Kitada <hitoshi@kitada.com>
To: Matti Pitkanen <matpitka@pcu.helsinki.fi>
Cc: Time List <time@kitada.com>; Paul Hanna <phanna@ghs.org>
Sent: Sunday, October 03, 1999 1:04 PM
Subject: [time 861] Re: [time 860] Re: [time 855] Re: [time 847] Unitarity of
S-matrix
> Matti Pitkanen <matpitka@pcu.helsinki.fi> wrote:
>
> Subject: [time 860] Re: [time 855] Re: [time 847] Unitarity of S-matrix
>
>
> >
> >
> > On Sun, 3 Oct 1999, Hitoshi Kitada wrote:
> >
> > > Dear Matti,
> > >
> > > I considered your proof and previous notes. I have a following question
on
> the
> > > present proof:
> > >
> > > If u = Pu for any scattering states u, u satisfies
> > >
> > > Vu = 0
> > >
> > > by your assumption: L_0(int)P|n> = 0. (Here V = L_0(int) and u = |n>.)
> Then
> > >
> > > (I+X)^{-1}|n> = (I - R(z)V) u = u = |n>.
> > >
> > > (Here R(z) = (H-z)^{-1}, H=L_0(tot), z= i\epsilon in your notation.)
> > >
> > > This means there is no scattering: S = I.
> >
> >
> > I think that this is not the case.
> >
> >
> > 1/(1+X), X= 1/(L_0 +ie*epsilon))*L_0(int)
> >
> > operates on *"free"* state |n_0> in matrix element of the scattering
> > operator and L_0(int) does not annhilate it.
>
> What is the difference of |n_0> from |n>?
>
>
> 1/(1+X) acts like unity
> > only when acts on *scattering state* |n>.
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