Hitoshi Kitada (hitoshi@kitada.com)
Sun, 3 Oct 1999 16:06:37 +0900
Dear Matti,
Matti Pitkanen <matpitka@pcu.helsinki.fi> wrote:
Subject: [time 864] Re: [time 861] Re: [time 860] Re: [time 855] Re: [time
847]Unitarity of S-matrix
>
>
> 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 .
Yes, your guess is correct. The limit as epsilon -> 0 corresponds to taking
the limit t -> \infty in time dependent expression. Originally scattering
theory started with stationary theory, just from the equation you proposed:
\Psi = \Psi_0 -R_0(z)V \Psi.
(Sommerfeld, etc.) Time dependent method is later introduced, and it is only
recently that the method is recognized powerful.
When taking the limit of e.g. R_0(z) = R_0(E+i\epsilon) = (H_0 - E -
i\epsilon)^{-1} as \epsilon -> 0, R_0(z) does not remain a bounded operator
from \HH into itself anymore (although this is the case when \epsilon > 0),
and lim_{\epsilon->0}R_0(z) must be considered a (bounded) operator from \HH_+
to \HH_-, where \HH_+ \subset \HH \subset \HH_-, a Gel'fand triple. This is
because H_0 (or H) has continuous spectra (we consider general case without
super Virasoro condition). There are three cases:
1) If E (real) is not a spectrum of H_0 then R_0(E) = (H_0-E)^{-1} exists as a
bounded operator from \HH into itself.
2) If E is a point spectrum (of finite degree) then
-(2i\pi)^{-1}\int R_(z) dz
(integration is on a circle rounding E counter clockwise)
gives the projection P(E) onto the eigenspace corresponding to eigenvalue E.
3) But when a closed interval [F,E] (F<E) is included in the continuous
spectrum of H_0, none of the above holds, and we have
E_0(E) - E_0(F) = -(2i\pi)^{-1} \int R(z) dz
= -(2i\pi)^{-1} lim_{\epsilon->0}\int R(z) dz (*)
(z=\lam+i\epsilon or \lam-i\epsilon)
(integration is around a path which passes point E and F counter clockwise).
Here E_0(E) is called "resolution of the identity" that expresses H_0 as
H_0 = \int_{-\infty}^\infty \lam dE_0(\lam).
E_0(E) is a kind of operator-valued measure.
In this way, in the continuos spectra, one cannot take the limit
lim_{\epsilon->0} R_0(E+i\epsilon) directly but the limit has meaning only in
the sense of mean as (*) above. If one wants to get a pointwise limit
lim_{\epsilon->0} R_0(E+i\epsilon), one has to take a smaller space \HH_+ as
its domain and a larger space \HH_- as its range. This is the cause that we
have to consider Gel'fand triple (\HH_+,\HH,\HH_-), \HH_+ \subset \HH \subset
\HH_-.
Resolvent equation
R(z) - R_0(z) = -R_0(z)VR(z) = -R(z)VR_0(z)
and
R_0(z): \HH_+ -> \HH_-
R(z): \HH_+ -> \HH_-
say
R_0(z)VR(z): \HH_+ -> \HH_-.
But the range of R(z) is included in \HH_- and the domain of R_0(z) is \HH_+
that is smaller than \HH_-. If the resolvent equation holds at the limit
\epsilon->0, V must be an operator
V: \HH_- -> \HH_+.
This means that V needs to "decay" in some sense, and if this is satisfied the
resolvent equation holds at the limit \epsilon -> 0.
Equations in the previous TeX file can be justified if V satisfies this type
of assumption. Here is a possibility that S-matrix S(E) is well-defined with
super Virasoro conditions: (H-E)\Psi=0 and (H_0-E)\Psi_0=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.
Your thought is on the right track!
>
>
> Best,
>
> MP
Best wishes,
Hitoshi
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