**Hitoshi Kitada** (*hitoshi@kitada.com*)

*Sun, 3 Oct 1999 11:10:00 +0900*

**Messages sorted by:**[ date ] [ thread ] [ subject ] [ author ]**Next message:**Matti Pitkanen: "[time 860] Re: [time 855] Re: [time 847] Unitarity of S-matrix"**Previous message:**Hitoshi Kitada: "[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"

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

**Next message:**Matti Pitkanen: "[time 860] Re: [time 855] Re: [time 847] Unitarity of S-matrix"**Previous message:**Hitoshi Kitada: "[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"

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