**Matti Pitkanen** (*matpitka@pcu.helsinki.fi*)

*Thu, 7 Oct 1999 06:53:57 +0300 (EET DST)*

**Messages sorted by:**[ date ] [ thread ] [ subject ] [ author ]**Next message:**Hitoshi Kitada: "[time 889] Re: [time 888] Re: [time 886] Unitarity finally understood!"**Previous message:**Hitoshi Kitada: "[time 887] Re: [time 886] Unitarity finally understood!"**In reply to:**Matti Pitkanen: "[time 886] Unitarity finally understood!"**Next in thread:**Hitoshi Kitada: "[time 889] Re: [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

On Thu, 7 Oct 1999, Hitoshi Kitada wrote:

*> Dear Matti,
*

*>
*

*> I am still stumbling on the first step of your proof in "1. The condition
*

*> guarateing unitarity." Could you explain proof of this section step by step
*

*> without omitting any steps?
*

*>
*

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

*>
*

*> Subject: [time 886] Unitarity finally understood!
*

*>
*

*>
*

*> >
*

*> >
*

*> >
*

*> >
*

*> > Dear Hitoshi,
*

*> >
*

*> > I believe that I understand the unitarity condition now.
*

*> > To streamline the notation I just denote
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*> >
*

*> > L_0(int)==V ,
*

*> >
*

*> > and
*

*> >
*

*> > L_0(free)==L_0.
*

*> >
*

*> > |m>= |m_0> +|m_1>
*

*> >
*

*> > is the decomposition of scattering state to free or bare state
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*> > and genuine scattering contribution (cloud of virtual particles.
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*> >
*

*> > S=1+T
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*> > ******************************************************
*

*> >
*

*> >
*

*> > 1. The condition guarateing unitarity
*

*> >
*

*> > The condition guaranteing unitarity is
*

*> >
*

*> > V|m>=V|m_0>
*

*>
*

*> Is this the only assumption in proving unitarity? Could you explain proof so
*

*> that a blind mathematician can understand it?
*

*>
*

*> >
*

*> > The action of interaction operator to dressed
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*> > state is same as to bare state. The contribution
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*> > of virtual cloud of particles to interaction vertex
*

*> > vanishes.
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*> >
*

*> > Vanishing coupling constant and wave function renormalization
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*> > are immediate consequences. This is just what is
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*> > expected on basis of quantum criticality.
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*> > Kahler coupling strength is fixed point of coupling constant
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*> > evolution and since it is the fundamental coupling, all
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*> > vertices are RG invariant.
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*> >
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*> > p-Adic coupling constant evolution means that Kahler coupling depends
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*> > on p-adic prime labelling configuration space sectors
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*> > so that continuous coupling constant evolution
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*> > is effectively replaced by a discrete one.
*

*>
*

*>
*

*> Best wishes,
*

*> Hitoshi
*

*>
*

*>
*

*>
*

\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}

\noindent This condition can be written in the form

\begin{eqnarray}

\langle m_0\vert n_0\rangle + \langle m_0\vert P \vert n_1\rangle

+\langle m_1\vert P \vert n_0\rangle + \langle m_1\vert P \vert n_1\rangle

= G(m,n)\per .

\end{eqnarray}

The proof of unitarity splits in two basic steps.

\vm

a) Consider first the last term appearing at the

left hand side:

\begin{eqnarray}

\langle m_1\vert P \vert n_1\rangle &=&

\oint_C d\bar{z} \langle m_0\vert

\sum_{k>0}\left[L_0(int)\frac{1}{L_0(free)-i\bar{z}}\right]^k

\frac{1}{L_0(free)-i\bar{z}} P \vert n_1\rangle \per .

\nonumber\\

\end{eqnarray}

\noindent The first thing to observe is that

$\langle m_1\vert$ has operator

$L_0(int)\frac{1}{L_0(free) -i\bar{z}}P$

outmost to the right. Since projection operator

effectively forces

$L_0$ to zero, one can commute $L_0(int)$ past

the operators $1/(L_0(free)-i\bar{z})$

so that it acts directly to $P\vert n_1\rangle$. But

by the proposed condition $L_0(int)P\vert n_1\rangle=0$

vanishes!

\vm

b) Consider next second and third terms at the left hand side

of the unitarity condition. The sum of these terms can

be written as

\begin{eqnarray}

\begin{array}{l}

\langle m_0\vert P \vert n_1\rangle+ \langle m_1\vert P \vert n_0\rangle

\\

\\

= \frac{1}{2\pi}\oint_C dz

\langle m_0 \vert \frac{1}{L_0(free)+iz} \sum_{k>0} X^k

\vert n_0\rangle \\

\\

+ \frac{1}{2\pi}\oint_C d\bar{z} \langle m_0

\vert \sum_{k>0}

(X^{\dagger})^k L_0(int)\vert \frac{1}{L_0(free)-i\bar{z}}

\vert n_0\rangle \per .\\

\end{array}

\end{eqnarray}

\noindent One might naively conclude that the sum of

these terms is zero since the overall sign factors are different

(this looks especially obvious in the dirty $1/i\epsilon$-approach).

This is however the case only if on mass shell states do not

appear as intermediate states in terms $X^k$. Unless this

is the case one encounters difficulties.

\vm

c) One can

project out on mass shell contribution to see what kind

of contributions one obtains: what happens that the conditions

$L_0(int)P\vert m_1\rangle=$ guarantees that these contributions

vanish! Consider the second term in the sum to see how this happens.

The on mass shell contributions from

terms $X^k\vert n_0\rangle$ can be grouped by the following

criterion. Each on mass shell contribution

can be characterized by an integer $r$

telling how many genuinely off mass shell powers of $X$ appear

before it. The on mass shell contributions which

come after r:th X can be written in the form $X^r PX^{k-r}$

The sum over all these terms coming from $\sum_{n>0} X^n$

is obviously given by

$$ X^r P\sum_{k>r} X^{k-r}\vert m_0\rangle = X^r P\vert m_1\rangle

=0

$$

\noindent and vanishes

sinces $X^r$ is of form $...L_0(int)$ and hence

annihilates $\vert m_1\rangle$.

Thus the condition implying unitarity also implies that

on mass shell states do not contribute to the perturbative

expansion.

\vm

The condition implies not only the unitarity

of the S-matrix but also that wave function renormalization

constants are equal to one so that these cannot serve

as sources of divergences.

\end{document}

**Next message:**Hitoshi Kitada: "[time 889] Re: [time 888] Re: [time 886] Unitarity finally understood!"**Previous message:**Hitoshi Kitada: "[time 887] Re: [time 886] Unitarity finally understood!"**In reply to:**Matti Pitkanen: "[time 886] Unitarity finally understood!"**Next in thread:**Hitoshi Kitada: "[time 889] Re: [time 888] Re: [time 886] Unitarity finally understood!"

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