# [time 870] Re: [time 868] Re: [time 864] Re: [time 861] Re: [time 860] Re: [time 855] Re:[time 847]Unitarity of S-matrix

Hitoshi Kitada (hitoshi@kitada.com)
Sun, 3 Oct 1999 18:08:28 +0900

Dear Matti,

Matti Pitkanen <matpitka@pcu.helsinki.fi> wrote:

Subject: [time 868] Re: [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,
> >
> > 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.

Yes, probably. BTW could I ask if your theory include weak and strong forces,
which Ben thinks a necessary condition for physical theory?

>
>
>
>
> 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?

In the example of usual Schroedinger operator:

H_0 = -Laplacian, H = H_0 + V

with

V = V(x) =O((1+|x|)^{-1-d}) as |x| -> \infty (d>0)

(short-range condition),

one can take

\HH = L^2(R^3),

\HH_+ = L^2_{+(1+d)/2}(R^3),

\HH_- = L^2_{-(1+d)/2}(R^3).

Here L^2_a(R^3) (a: real) is a Hilbert space with inner product

(f, g) = \int_{R^3} f(x) g(x)^* (1+|x|^2)^{a} dx

and norm squared

||f||^2 = \int_{R^3} |f(x)(1+|x|^2)^{a/2}|^2 dx < \infty.

Then it is known that

R_0(z) = (H_0-z)^{-1} and R(z) = (H-z)^{-1} (*)

have limits as Im z -> +0 or -0 as bounded operators from \HH_+ into \HH_-
(except for Re z = 0 and eigenvalues).

Note

R_0(z)VR(z) (= R_0(z) - R(z))

= R_0(z)(1+|x|^2)^{-(1+d)/4} \times

(1+|x|^2)^{(1+d)/4}V(1+|x|^2)^{(1+d)/4} \times

(1+|x|^2)^{-(1+d)/4} R(z).

By the decay assumption on V(x) the middle factor is bounded in \HH. The first
and third factors are bounded from \HH into \HH_- and from \HH_+ into \HH by
(*) above. Thus the limit of this product exists as a bounded operator from
\HH_+ into \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?

Yes partly. The method required depends on each concrete case. If V satisfies
the conditions good enough as in the above case of Schroedinger operator, no
time-dependent method is required and time dependent results follow from
stationary results. However to deal with many body case, time dependent
expression was required. It is not yet known if one can treat it without time
dependent expression.

>
>
> > >
> > > 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(;-).

My impression is that the so-called "mathematical scattering theory" that
treats these things has not been taken seriously by usual physicists. If they
were aware of these results/methods, the history might have been different.

>
> Best,
> MP
>
>

Best wishes,
Hitoshi

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