Hitoshi Kitada (hitoshi@kitada.com)
Sun, 3 Oct 1999 11:10:00 +0900
Dear Matti,
To former LaTeX file in [time 858], I should have added the following to the
last of a quotation in page 3 (this is not a quotation actually, but it is
used to make the page easy to read):
If $S$ is shown to be unitary and $\TT(\lam)$ is continuous in $\lam$
in some sense, then it follows that
$\SS(\lam):\mbox{\bf h}\longrightarrow\mbox{\bf h}$ is unitary
for any real $\lam$.
i.e. the part parenthesized by quotation in page 3 should be
\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}.
If $S$ is shown to be unitary and $\TT(\lam)$ is continuous in $\lam$
in some sense, then it follows that
$\SS(\lam):\mbox{\bf h}\longrightarrow\mbox{\bf h}$ is unitary
for any real $\lam$.
\end{quotation}
Best wishes,
Hitoshi
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