**Matti Pitkanen** (*matpitka@pcu.helsinki.fi*)

*Sun, 3 Oct 1999 07:56:38 +0300 (EET DST)*

**Messages sorted by:**[ date ] [ thread ] [ subject ] [ author ]**Next message:**Matti Pitkanen: "[time 864] Re: [time 861] Re: [time 860] Re: [time 855] Re: [time 847] Unitarity of S-matrix"**Previous message:**Hitoshi Kitada: "[time 862] Re: [time 861] Re: [time 860] Re: [time 855] Re: [time 847] Unitarity of S-matrix"**In reply to:**Matti Pitkanen: "[time 860] Re: [time 855] Re: [time 847] Unitarity of S-matrix"**Next in thread:**Matti Pitkanen: "[time 875] Still about the unitarity of S-matrix"

On Sun, 3 Oct 1999, Hitoshi Kitada wrote:

*> 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>?
*

|n_0> are the incoming "free states" or on mass shell states and |n> are

scattered states. P|n> are their projections to the space

of "free" states.

*>
*

*>
*

*> 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{\qqq}{\qquad\qquad}
*

*> > > \newcommand{\q}{\qquad}
*

*> > > \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}
*

*> > >
*

*> > >
*

*> > >
*

*> >
*

*>
*

*>
*

**Next message:**Matti Pitkanen: "[time 864] Re: [time 861] Re: [time 860] Re: [time 855] Re: [time 847] Unitarity of S-matrix"**Previous message:**Hitoshi Kitada: "[time 862] Re: [time 861] Re: [time 860] Re: [time 855] Re: [time 847] Unitarity of S-matrix"**In reply to:**Matti Pitkanen: "[time 860] Re: [time 855] Re: [time 847] Unitarity of S-matrix"**Next in thread:**Matti Pitkanen: "[time 875] Still about the unitarity of S-matrix"

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