Matti Pitkanen (matpitka@pcu.helsinki.fi)
Sun, 3 Oct 1999 08:47:18 +0300 (EET DST)
On Sun, 3 Oct 1999, Hitoshi Kitada wrote:
> 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 +i*epsilon)*L_0(int)
>
> and L_0(int) P|n_0> = 0.
Yes. Suppose that every state L_0(int)P|n> vanishes and states P|n> span
the space spanned by |n_0>. Then one can express |n_0> as superposition
of P|n>:s and conclude that L_0(int)|n_0> vanishes and S is trivial.
I am really amazed! How it is possible to obtain nontrivial S-matrix
at all in QM? Same kind proof for unitarity should hold also there!
The point is that I "know" that the expansion yields S-matrix. There
is now doubt about that. But how on Earth can I demonstrate the unitarity?
There is something which I do not understand but what is it?
Could this paradox be related to the taking of epsilon-->0 limit?
Condition
L_0(int)* P*(1/(1+X)) |m_0> =0
holds true only at in the sense of limit epsilon-->0 .
Limit of this equation would not be same as equation
obtained putting epsilon=0 from the beginning to get L_0(int|m_0>=0?
This would not be surprising since epsilon prescription can be seen
as a manner to make propagators well defined. I do not know.
Best,
MP
>
>
> 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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