# [time 868] Re: [time 864] 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 10:56:14 +0300 (EET DST)

On Sun, 3 Oct 1999, Hitoshi Kitada wrote:

> 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 .
>
[Hitoshi]
> 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).

Let me think: in TGD framework iepsilon is added to the propagators
1/p^2-m^2 +iepsilon to define momentum space integrals. Masses
get tiny imagy component and planewaves get small exponential factor
making them divergence at t-->infity. Therefore one goes outside the state
space. This would mean in TGD context that one consider eigenstates
of Diff^4 invariant momentum generator with slightly imaginary eigen
values: they are not normalizable configuration space spinor fields.

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.
>
OK

> 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.
>

I think I understand this.

> 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

You
>
> 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.

I think I understand this. You have continuous spectrum
and projector E_0 is replaced by dE_0/dE and integration yields
Int dEdE_0/dE = E_(E)-E_0(F).

>
> 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_-.
>

Can one explain verbally HH_+, HH and HH_-? Or actually 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.
>
OK. I think that I understand this. R_0(z) is defined in
HH_+. Therefore V must map HH_- to HH_+ in order that everything is
well defined.

> 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.
>
>
This means that one gets S-matrix without any reference to
unitary time evolution if this kind of condition holds true?

> >
> > 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!
>

Glad to hear that. I should be familiar with these delicacies but I am
not. It took 21 years before I ended at the concrete level
of worrying details of this kind(;-)!
I remember had a course on Hilbert spaces and resolutions
of identity something like 25 years ago: it seems that wheel must
be reinvented again and again(;-).

Best,
MP

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