**Hitoshi Kitada** (*hitoshi@kitada.com*)

*Sun, 3 Oct 1999 06:24:38 +0900*

**Messages sorted by:**[ date ] [ thread ] [ subject ] [ author ]**Next message:**Hitoshi Kitada: "[time 859] Re: [time 858] Re: [time 855] Re: [time 847] Unitarity of S-matrix"**Previous message:**Stephen P. King: "[time 857] Re: [time 856] RE: [time 853] Re: [time 852] Another expresion of my weird idea..."**In reply to:**Ben Goertzel: "[time 856] RE: [time 853] Re: [time 852] Another expresion of my weird idea..."**Next in thread:**Matti Pitkanen: "[time 860] Re: [time 855] Re: [time 847] Unitarity of S-matrix"

Dear Matti,

I considered your proof and previous notes. I have a following question on the

present proof:

If u = Pu for any scattering states u, u satisfies

Vu = 0

by your assumption: L_0(int)P|n> = 0. (Here V = L_0(int) and u = |n>.) Then

(I+X)^{-1}|n> = (I - R(z)V) u = u = |n>.

(Here R(z) = (H-z)^{-1}, H=L_0(tot), z= i\epsilon in your notation.)

This means there is no scattering: S = I.

I reflect on the previous communication, and summarized my thought as a LaTeX

file at the bottom. This is partially a repetition of "[time 815] A summary on

[time 814] Still about construction of U," but it contains important changes

and new information about Gel'fand triple.

The problem is related with whether we take the subspace P\HH of the total

Hilbert space \HH before or later than constructing the solution \Psi. This is

equivalent to whether P commutes with H = L_0(tot). If commutes, your argument

works but the result reduces to the trivial one: S = I. So we have to

construct \Psi before taking the subspace P\HH. Then we need to think the

general problem without assuming super Virasoro conditions. The difficulty is

equivalent to the general problem that arises in usual quantization of GR: If

we should make the constraints first or later than quantization.

Best wishes,

Hitoshi

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\begin{document}

\F

Matti [time 814 and former] defines the following operators

whose domains and ranges are included in a Hilbert space

$\HH$:

\beq

&&H = L_0(tot),\nonumber\\

&&H_0 = L_0(free) = \sum_n L_0(n),\nonumber\\

&&L_0(n) = p^2(n) - L_0(vib,n),\nonumber\\

&&V = H - H_0 = L_0(int).\nonumber

\ene

He expects that these operators are extended to selfadjoint

operators defined in $\HH$.

\MP

\F

We proceed without assuming super Virasoro condition for the

time being. The reason will be clear in the following.

\MP

\F

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:

\beq

R(z)=(H-z)^{-1},\q R_0(z)=(H_0-z)^{-1}.\nonumber

\ene

These are called resolvents, and are continuous (i.e. bounded)

operators with domain $\HH$.

\F

Resolvents satisfy the resolvent equation:

\beq

R(z) - R_0(z) = - R_0(z)VR(z) = -R(z)VR_0(z)\nonumber

\ene

for any non-real number $z$.

\MP

\F

Set for $\Psi_0$ in $\HH$

\beq

U(z)\Psi_0=\Psi = R(z)(H_0-z)\Psi_0.\label{1}

\ene

Then

\beq

U(z)\Psi_0=\Psi &=&R(z)(H-z-V)\Psi_0\nonumber\\

&=&\Psi_0 - R(z)V\Psi_0.\nonumber

\ene

I.e.

\beq

U(z)=I-R(z)V.\label{2}

\ene

Resolvent equation reduces this to

\beq

U(z)\Psi_0=\Psi&=&\Psi_0 - R(z)VR_0(z)(H_0-z)\Psi_0\nonumber\\

&=&\Psi_0 - R_0(z)VR(z)(H_0-z)\Psi_0\nonumber\\

&=&\Psi_0 - R_0(z)V\Psi.\label{3}

\ene

This is Matti's equation \eq{1} in [time 798], whose solution

is given by \eq{1} or \eq{2} above. Another form of the solution

is obtained by iteration of resolvent equation:

$$

R(z)V = - \sum_{n=1}^\infty (-R_0(z)V)^n.

$$

This and (2) give a solution of (3):

\beq

U(z) = I + \sum_{n=1}^\infty (-R_0(z)V)^n.\label{4}

\ene

\MP

\F

If we assume super Virasoro condition on $\Psi$ and $\Psi_0$:

$$

H\Psi = 0, \q H_0\Psi_0 = 0,

$$

we have from \eq{1}

$$

U(z)\Psi_0=\Psi = R(z) (H_0-z)\Psi_0=-zR(z)\Psi_0.

$$

Multiplying $H-z$ to both sides gives

$$

-z\Psi = -z\Psi_0.

$$

Since $z=E+ie\ne 0$, this implies

$$

\Psi = \Psi_0.

$$

Thus Matti's operator $U(z)$ satisfies

$$

\Psi = U(z)\Psi_0 = \Psi_0

$$

and

$$

U(z) = I.

$$

This is not his expectation, but as far as we start from

$\Psi_0$ in $\HH$ we arrive at this conclusion. There seems

to be two possible ways to evade this:

\MP

\F

1) We have assumed that $\Psi_0$ is in the Hilbert space $\HH$.

If we assume $\Psi_0$ is outside $\HH$, then $\Psi$ might be

different from $\Psi_0$.

\MP

\F

2) We assumed $z \ne 0$. If $z = 0$,

then both of $\Psi$ and $\Psi_0$ might be expected to be

in $\HH$ without being zero vectors.

\MP

\F

The second possibility is expected to be true only when z=0

from the beginning, in which case we have by \eq{1}

$$

U(0)\Psi_0=\Psi=H^{-1}H_0\Psi_0.

$$

This is impossible because $H$ has a nonvanishing null space

as expected by super Virasoro condition:

$$

H\Psi=0.

$$

There remains only the first possibility.

This case contains the problem to which space $\Psi_0$ belongs.

The expected space is a Hilbert space $\HH_-$ larger than $\HH$,

whose norm is smaller than that of

$\HH$.

\MP

\F

A relation between \eq{1} and scattering operator in [time 804]

(H. Isozaki and H.Kitada: {\it Scattering matrices for two-body

Schroedinger operators}, Sci. Papers of the Coll. Arts \& Sci.,

The Univ. Tokyo, {\bf 35} (1986), 81-107)

\beq

S &=& (W_+)^* (W_-) = \lim_{t\to\infty}

\exp(itH_0) \exp(-2itH) \exp(itH_0)\nonumber\\

&=& 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\label{5}

\ene

is given by using \eq{2}

\beq

S&=& I + 2i\pi \int_{-\infty}^\infty E'_0(\lam)V

(R(\lam+i0)V-I) E'_0(\lam)d\lam\nonumber\\

&=& I - 2i\pi \int_{-\infty}^\infty E'_0(\lam)

VU(\lam+i0) E'_0(\lam)d\lam.\label{6}

\ene

This gives a relation of Matti's operator $U(z)$ with time dependent

method, when super Virasoro conditions are not assumed.

\BP

\F

{\bf Remark.} Time dependent expression in \eq{5} does not

imply the unitarity of $S$ because the unitarity at a finite

time $t$ does not imply the unitarity at the limit $t\to\infty$.

The proof of the unitarity of $S$ requires to prove the

{\it completeness} of wave operators $W_\pm$: the ranges of

$W_+$ and $W_-$ coincide. Then the unitarity of $S=(W_+)^*W_-$

follows.

\BP

\F

In the treatment of the equation \eq{3}, it would be helpful

to consider general context

allowing $z$ to take general values.

\MP

\F

The factor $V$ in front of $U(\lam+i0)$ in \eq{6} is interaction

term $L_0(int)$. This

seems to decay as $a \to \infty$, as Matti [time 808] stated.

\MP

\F

{\it If this is the case in some appropriate sense}, $V$ could work

to damp the behavior of $U(\lam+i0)\Psi_0$,

which would result in:

\begin{quotation}

\F

$VU(\lam+i0)\Psi_0=-V (R(\lam+i0)V-I)\Psi_0$ belongs to

a good Hilbert space, possibly to a space $\HH_+$ dual

to $\HH_-$ with respect to the inner product of $\HH$.

Thus we have a Gel'fand triple:

$$

(\HH_+,\HH,\HH_-)\q\mbox{with}\q \HH_+\subset \HH\subset\HH_-

$$

with $\Psi_0\in\HH_-$ and

$V (R(\lam+i0)V-I):\HH_-\longrightarrow\HH_+$ being bounded.

The operator $E'_0(\lam)$ is decomposed $E'_0(\lam)=K(\lam)^*K(\lam)$

for some bounded operator $K(\lam):\HH_+\longrightarrow \mbox{\bf h}$

with {\bf h} being a Hilbert space, thus its dual

$K(\lam)^*: \mbox{\bf h}\longrightarrow\HH_-=\HH_+^*$ and

$E'_0(\lam): \HH_+\longrightarrow \HH_-$ are bounded. \eq{6}

then becomes

\beq

S&=& I + 2i\pi \int_{-\infty}^\infty

K(\lam)^*[K(\lam)V (R(\lam+i0)V-I)K(\lam)^*]K(\lam)d\lam,\label{7}

\ene

where

$K(\lam)V (R(\lam+i0)V-I)K(\lam)^*=-K(\lam)VU(\lam+i0)K(\lam)^*$

is a bounded operator in {\bf h}. This operator is called transition

matrix and denoted $\TT(\lam)$. Scattering matrix $\SS(\lam)$

is defined as a bounded operator $\SS(\lam)=I+\TT(\lam)$ in {\bf h}.

\end{quotation}

These would make it possible to treat scattering operator and

S-matrix with super Virasoro conditions. I.e.

$\Psi= U(E+i0)\Psi_0=(I-R(E+i0)V)\Psi_0$ with general super

Virasoro condition $(H-E)\Psi=(H_0-E)\Psi_0=0$ would be understood

by considering

a neighborhood of the desired eigenvalue (spectrum) $E$.

\end{document}

**Next message:**Hitoshi Kitada: "[time 859] Re: [time 858] Re: [time 855] Re: [time 847] Unitarity of S-matrix"**Previous message:**Stephen P. King: "[time 857] Re: [time 856] RE: [time 853] Re: [time 852] Another expresion of my weird idea..."**In reply to:**Ben Goertzel: "[time 856] RE: [time 853] Re: [time 852] Another expresion of my weird idea..."**Next in thread:**Matti Pitkanen: "[time 860] Re: [time 855] Re: [time 847] Unitarity of S-matrix"

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