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

Matti Pitkanen (matpitka@pcu.helsinki.fi)
Thu, 7 Oct 1999 06:53:57 +0300 (EET DST)

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
> >
> > 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
> > and genuine scattering contribution (cloud of virtual particles.
> >
> > S=1+T
> > ******************************************************
> >
> >
> > 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
> > state is same as to bare state. The contribution
> > of virtual cloud of particles to interaction vertex
> > vanishes.
> >
> > Vanishing coupling constant and wave function renormalization
> > are immediate consequences. This is just what is
> > expected on basis of quantum criticality.
> > Kahler coupling strength is fixed point of coupling constant
> > evolution and since it is the fundamental coupling, all
> > vertices are RG invariant.
> >
> > p-Adic coupling constant evolution means that Kahler coupling depends
> > on p-adic prime labelling configuration space sectors
> > so that continuous coupling constant evolution
> > 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}

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