# [time 865] Re: [time 861] Re: [time 860] Re: [time 855] Re: [time 847] Unitarity of S-matrix

Matti Pitkanen (matpitka@pcu.helsinki.fi)
Sun, 3 Oct 1999 10:05:18 +0300 (EET DST)

I noticed what might be the reason for the paradoxal conclusion
about the triviality of S-matrix.

The expression of S-matrix is

<m_0|Sn> = <m_0| P*(1/(1+X)|n_0>

Expand this to geometric series to get

...= delta (m,n) + sum_n <m_0| X^n|n_0>

= delta (m,n) + (1/i*epsilon) sum_n <m_0| L_0(int) X^(n-1)|n_0>

Here I have used X= (1/L_0(free)+iepsilon)L_0(int) to the first
X in the expansion in powers of X.

The point is that formula contains 1/epsilon factor!!

Thus the limit is extremely delicate. S-matrix is notrivial
if L_0(int)|m_0> is of order epsilon and goes to zero at
the limit epsilon->0.

This is dangerously delicate but I think that similar problems
must be encountered with ordinary time dependent scattering theory
when one restricts to 'energy shell' E=constant. The task would
be to find proper formulation or possibly understand why p-adics
save the situation.

On Sun, 3 Oct 1999, Matti Pitkanen wrote:

>
>
> 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>.
> >
> >
> >
>
>

This archive was generated by hypermail 2.0b3 on Sun Oct 17 1999 - 22:40:46 JST