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

Sun, 3 Oct 1999 13:04:27 +0900

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

Subject: [time 860] Re: [time 855] Re: [time 847] Unitarity of S-matrix

>
>
> On Sun, 3 Oct 1999, Hitoshi Kitada wrote:
>
> > 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 think that this is not the case.
>
>
> 1/(1+X), X= 1/(L_0 +ie*epsilon))*L_0(int)
>
> operates on *"free"* state |n_0> in matrix element of the scattering
> operator and L_0(int) does not annhilate it.

What is the difference of |n_0> from |n>?

1/(1+X) acts like unity
> only when acts on *scattering state* |n>.
>
>
> >
> > 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.
> >
> I think that S is not trivial for the reason which I gave already above.
>
>
>
> Best,
> MP
>
>
> > \documentstyle[12pt]{article}
> >
> >
> > \oddsidemargin 0pt
> > \evensidemargin 0pt
> > \topmargin 0pt
> > \textwidth 16cm
> > \textheight 23cm
> >
> > \newcommand{\F}{\noindent}
> > \newcommand{\SP}{\smallskip}
> > \newcommand{\MP}{\medskip}
> > \newcommand{\BP}{\bigskip}
> >
> > \newcommand{\beq}{\begin{eqnarray}}
> > \newcommand{\ene}{\end{eqnarray}}
> >
> >
> > \newcommand{\HH}{{\cal H}}
> > \newcommand{\UU}{{\cal U}}
> > \newcommand{\OO}{{\cal O}}
> > \newcommand{\SS}{{\cal S}}
> > \newcommand{\TT}{{\cal T}}
> > \newcommand{\ep}{{\epsilon}}
> > \newcommand{\lam}{{\lambda}}
> > \newcommand{\Ltn}{{L^2(R^\nu)}}
> > \newcommand{\Ltnn}{{L^2(R^{3n})}}
> > \newcommand{\tX}{{\tilde X}}
> > \newcommand{\tP}{{\tilde P}}
> > \newcommand{\ve}{\vert}
> > \newcommand{\V}{\Vert}
> > \newcommand{\tx}{(x_0,x_1,x_2,x_3,x_4)}
> >
\newcommand{\txp}{(x_0^\prime,x_1^\prime,x_2^\prime,x_3^\prime,x_4^\prime)}
> >
> > \newcommand{\eq}[1]{(\ref{#1})}
> > \newcommand{\eqs}[2]{(\ref{#1}--\ref{#2})}
> >
> > \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}
> >
> >
> >
>

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