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

*Sun, 3 Oct 1999 06:50:51 +0300 (EET DST)*

**Messages sorted by:**[ date ] [ thread ] [ subject ] [ author ]**Next message:**Hitoshi Kitada: "[time 861] Re: [time 860] Re: [time 855] Re: [time 847] Unitarity of S-matrix"**Previous message:**Hitoshi Kitada: "[time 859] Re: [time 858] Re: [time 855] Re: [time 847] Unitarity of S-matrix"**In reply to:**Matti Pitkanen: "[time 855] Re: [time 847] Unitarity of S-matrix"**Next in thread:**Hitoshi Kitada: "[time 861] Re: [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. 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:**Hitoshi Kitada: "[time 861] Re: [time 860] Re: [time 855] Re: [time 847] Unitarity of S-matrix"**Previous message:**Hitoshi Kitada: "[time 859] Re: [time 858] Re: [time 855] Re: [time 847] Unitarity of S-matrix"**In reply to:**Matti Pitkanen: "[time 855] Re: [time 847] Unitarity of S-matrix"**Next in thread:**Hitoshi Kitada: "[time 861] Re: [time 860] Re: [time 855] Re: [time 847] Unitarity of S-matrix"

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