# [time 889] Re: [time 888] Re: [time 886] Unitarity finally understood!

Thu, 7 Oct 1999 13:20:44 +0900

Dear Matti,

Let me make a question at each posting for the time being.

Matti Pitkanen <matpitka@pcu.helsinki.fi> wrote:

Subject: [time 888] Re: [time 886] Unitarity finally understood!

>
>
> Dear Hitoshi,
>
> The posting did not containg any proof of unitarity. I attach a latex file
> with proof.
>
> Best,
> MP
>
>
> \documentstyle [10pt]{article}
> \begin{document}
> \newcommand{\vm}{\vspace{0.2cm}}
> \newcommand{\vl}{\vspace{0.4cm}}
> \newcommand{\per}{\hspace{.2cm}}
> %matti
>
>
> \subsection{Formal proof of unitarity}
>
>
> Consider now the formal proof of the unitarity.
> Orthogonality condition guaranteing
> unitarity can be expressed also as the condition
>
>
> \begin{eqnarray}
> \frac{1}{1+X^{\dagger}} P\frac{1}{1+X}&=&G \per ,\nonumber\\
> \nonumber\\
> G(m,n) &=&\delta (m,n) \langle m\vert m\rangle\per .
> \end{eqnarray}

>From where the projection P comes?

My understanding:

|m> = |m_0> + |m_1> = |m_0> - X|m>,

X=(L_0+iz)^{-1}V, V=L_0(int),

thus

|m> = (1+X)^{-1}|m_0>.

So unitarity

<m|n> = <m_0|(1+X^\dagger)^{-1}(1+X)^{-1}|n_0> = <m_0|n_0>

means

(1+X^\dagger)^{-1}(1+X)^{-1} = G.

Here P in

|m_0> = P|m_0>

cannot come to the center between (1+X^\dagger)^{-1} and (1+X)^{-1}, since P
does not commute with L_0 and V.

Best wishes,
Hitoshi

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