Hitoshi Kitada (hitoshi@kitada.com)
Sun, 3 Oct 1999 06:24:38 +0900
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}
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