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

Mon, 27 Sep 1999 02:04:58 +0900

Dear Matti,

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.

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 wishes,
Hitoshi

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