# [time 816] Re: [time 815] A summary on [time 814] Still about construction of U

Matti Pitkanen (matpitka@pcu.helsinki.fi)
Sun, 26 Sep 1999 21:23:52 +0300 (EET DST)

Thank you for good posting. Your are right in that Hilbert space
is extended. One however obtains S-matrix for which other half
of unitary condition with summation over intermediate states of
extended Hilbert space is satisfied and this makes
S-matrix physical. Other half of unitarity conditions
involving sum over the intermediate states in smaller Hilbert space is
lost.

See below.

On Mon, 27 Sep 1999, Hitoshi Kitada wrote:

> Dear Matti,
>
> Thank you for your answers. Rather sufficient data about your theory are
> collected. I understand your theory as follows. I omit the detailed structure of
> TGD, and summarize an abstract structure to see the essence:
>
> You define the following operators whose domains are included in a Hilbert space
> \HH:
>
> H = L_0(tot),
>
> H_0 = L_0(free) = \sum_n L_0(n),
>
> L_0(n) = p^2(n) - L_0(vib,n),
>
> V = H - H_0 = L_0(int).
>
> You expect that these operators can be extended to selfadjoint operators defined
> in \HH.
>
> (Note: The difference between symmetric (or Hermitian) operators and selfadjoint
> operators is if their adjoint operators are the same as the original operators.
> If the adjoint operator H^* is the same as H, H is called selfadjoint. There are
> examples that are not selfadjoint, but symmetric. E.g., let T be a Laplacian
> with domain C consisting of infinitely differentiable functions on R^n with
> compact supports, and consider T in the usual L^2(R^n) space with inner product:
> (f,g) = \int_{R^n} f(x) g(x)^* dx (g(x)^* is complex conjugate).
>
> Then clearly T is symmetric: (Tf,g) = (f,Tg) for any f, g in C. But the adjoint
> T^* is also defined for distributions h satisfying the condition:
>
> ||h|| = (\int_{R^n} \sum_{0=<|k|=<2} |D^k h(x)|^2 dx)^{1/2} < \infty.
>
> Here D = (d/dx_1, ..., d/dx_n), k = (k_1, ..., k_n) with k_j being nonnegative
> integer, D^k =(d/dx_1)^{k_1}...(d/dx_n)^{k_n}. The space of h above is called
> Sobolev space of order 2, and is denoted H^2(R^n). This space is larger than C.
> Thus T^* is a true extension of T, and T is symmetric but not selfadjoint.
> However T^* satisfies (T^*)^* = T^*, thus is selfadjoint. Usually one considers
> the maximal extension of T, and it is usually a unique closed extension of T. In
> this case T is called essentially selfadjoint. Once one knows T can be extended
> to a selfadjoint operator, one usually uses the same notation T to denote its
> extension.)
>
> Returning to your operators, I proceed without assuming super Virasoro condition
> for the time being. The reason will be clear in the following.
>
> Let z=E+ie, e>0, E: real. Since H and H_0 are selfadjoint, their spectra are
> confined in the real line. Thus the following definition makes sense:
>
> R(z)=(H-z)^{-1}, R_0(z)=(H_0-z)^{-1}.
>
> These are called resolvents, and are continuous (i.e. bounded) operators with
> domain \HH.
>
> (Note that for symmetric operators, the domains of their resolvents are not
> necessarily equal to \HH.)
>
> Resolvents satisfy the resolvent equation:
>
> R(z) - R_0(z) = - R_0(z)VR(z) = -R(z)VR_0(z)
>
> for any non-real number z. Set for f in \HH
>
> \Psi = R(z)f, \Psi_0 = R_0(z)f.
>
> Then resolvent equation gives
>
> \Psi = \Psi_0 - R_0(z)V\Psi (1)
>
> and
>
> \Psi = \Psi_0 - R(z)V\Psi_0. (2)
>
> The former equation is your equation (1) in [time 798] and is equivalent to (2).
> Namely (2) gives a solution of (1).
>
> If we assume super Virasoro condition on \Psi and \Psi_0:
>
> H\Psi = 0, H_0\Psi_0 = 0,
>
> we have from
>
> (H-z)\Psi = f = (H_0-z)\Psi_0,
>
> that
>
> -z\Psi = -z\Psi_0.
>
> Thus
>
> \Psi = \Psi_0
>
> if z not = 0. Thus the scattering operator U in your notation satisfies
>
> \Psi = U\Psi_0 = \Psi_0
>
> and
>
> U = I.
>
> This is not your expectation. Why this happened? There are two possible reasons:
>
> 1) The first is that we have assumed that both of \Psi and \Psi_0 are in the
> Hilbert space \HH. If we assume \Psi_0 is in \HH, then \Psi must be outside \HH.

This is certainly the case since Psi contains superposition of
off mass shell states. p^2-L_0(vib)=0 is not satisfied for Psi.
If this were not the case, the entire equation would be nonsensical
since right hand side would be of form (L_0(int)/+ie)Psi.
Thus we have Hilbert spaces which we could call Hilb_0 and Hilb.

One the other hand. Psi is image of on mass shell state under Psi_0-->Psi
and S-matrix is defined as matrix elements

SmM== <Psi_0(m),Psi (M)>.

One restricts outgoing momenta to on mass shell momenta in inner product.
This means projection of Psi (m) to the space Hilb_0 spanned by Psi_0:s
when one calculates inner products defining S-matrix.

One obtains unitarity relations

sum_N SmN (SnN)^* = delta (m,n)

from completeness in Hilb: sum_N |N> <N|=1

but NOT

sum_m smM (SmN)*.

since Hilb_0 completeness relation sum_m |m><m|=1 are not true in Hilb
but become sum_m |m><m>= P, P projector to Hilb_0.

But this seems to be enough! One obtains S-matrix with orthogonal
rows: this gives probability conservation plus additional conditions.
Colums are however not orthogonal.

>
> 2) The second is that we assumed z not = 0. If z = 0, then both of \Psi and
> \Psi_0 might be expected to be in \HH.
>
> The first case contains the problem to which space \Psi belongs. The expected
> space is a Hilbert space larger than \HH, whose norm is smaller than that of
> \HH.
>
> The second case must be considered as a result of taking the limit e->+0, so it
> would contain some difficult topological problems; contrary to the expectation
> above, \Psi would also be outside \HH.
>
> 1) corresponds to the case E not = 0, and 2) to E = 0.

>
> Now a brief comment based on ordinary quantum scattering theory:
>
> 1) If we consider general E, we would also have to consider \Psi_0 moving in a
> Hilbert space larger than \HH. (This is suggested by (4) below.)
>
> 2) is anticipated to be more difficult than 1). The study of this case would
> tell that your universe is in a resonance state.
>
>
> I conclude with a comment on a relation between (1) and the equation in [time
> 804]:
>
> S = (W_+)^* (W_-) = lim_{t->\infty} exp(itH_0) exp(-2itH) exp(itH_0)
>
> = I+2i\pi \int_{-\infty}^\infty E'_0(\lam)V R(\lam+i0)V E'_0(\lam)d\lam.
>
> -2i\pi \int_{-\infty}^\infty E'_0(\lam)VE'_0(\lam)d\lam
>
> An iteration of resolvent equation gives
>
> R(z)V = - \sum_{n=1}^\infty (-R_0(z)V)^n.
>
> This and (2) give a solution of (1):
>
> U(z) = I + \sum_{n=1}^\infty (-R_0(z)V)^n. (3)
>
> (Here the variable z is attached as we consider general case without super
> Virasoro condition.)
>
> Similarly, S becomes
>
> S
>
> = I + 2i\pi \int_{-\infty}^\infty E'_0(\lam)V (R(\lam+i0)V-I) E'_0(\lam)d\lam
>
> = I + 2i\pi \int_{-\infty}^\infty E'_0(\lam) VU(\lam+i0) E'_0(\lam)d\lam. (4)
>
> This gives a relation of your scattering operator U(z) with time dependent
> method.
>
> In the treatment of the equation (1), it is helpful to consider general context
> allowing z to take general values. Let me explain.
>
> The factor V in front of U(\lam+i0) in (4) is interaction term L_0(int). This
> seems to decay as a -> \infty, as you stated in [time 808]:

>
> [MP] [time 808]
> > Diff^4 invariant momentum generators are defined in the following manner.
> > Consider Y^3 belonging to delta M^4_+xCP_2 ("lightcone boundary").
> > There is unique spacetime surface X^4(Y^3) defined as absolute minimum
> > of Kaehler action.
> >
> > Take 3-surface X^3(a) defined by the intersection of lightcone
> > proper time a =constant hyperboloidxCP_2 with X^4(Y^3). Translate it
> > infinitesimal amount to X^3(a,new)and find the new absolute minimum
> > spacetime surface goinb through X^3(a,new). It intersectors
> > lightcone at Y^3(new). Y^3(new) is infinitesimal translate
> > of Y^3: it is not simple translate but slightly deformed surface.
> >
> > In this manner one obtains what I called Diff^4 invariant infinitesimal
> > representation of Poincare algebra when one considers also infinitesimal
> > Lorentz transformations. These infinitesimal transformations need
> > *not* form closed Lie-algebra for finite value a of lightcone proper time
> > but at the limit a--> the breaking of Poincare invariance is expected
> > to go to zero and one obtains Poincare algebra since the distance to
> > the lightcone boundary causing breaking of global Poincare invariance
> > becomes infinite. The Diff^4 invariant Poincare algebra p_k(a--> infty)
> > defines momentum generators appearing in Virasoro algebra.
>
> If this is the case, V would work to damp the behavior of U(\lam+i0)\Psi_0,
> which would result in that
>
> VU(\lam+i0)\Psi_0 belongs to a good Hilbert space, possibly to a space smaller
> than \HH (as experienced in ordinary quantum scattering).
>
> This would make the treatment of scattering operator and S-matrix possible:
> \Psi= U\Psi_0 with super Virasoro condition would be understood by considering
> the neighborhood of the desired eigenvalue 0 or E.

Best,
MP

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