Matti Pitkanen (matpitka@pcu.helsinki.fi)
Sun, 3 Oct 1999 06:50:51 +0300 (EET DST)
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}
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> \newcommand{\F}{\noindent}
> \newcommand{\qqq}{\qquad\qquad}
> \newcommand{\q}{\qquad}
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> \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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