[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.


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}

\subsection{Formal proof of unitarity}

Consider now the formal proof of the unitarity.
Orthogonality condition guaranteing
unitarity can be expressed also as the condition

\frac{1}{1+X^{\dagger}} P\frac{1}{1+X}&=&G \per ,\nonumber\\
G(m,n) &=&\delta (m,n) \langle m\vert m\rangle\per .

\noindent This condition can be written in the form

\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 .

 The proof of unitarity splits in two basic steps.


a) Consider first the last term appearing at the
left hand side:

 \langle m_1\vert P \vert n_1\rangle &=&
\oint_C d\bar{z} \langle m_0\vert
\frac{1}{L_0(free)-i\bar{z}} P \vert n_1\rangle \per .

\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 $L_0(int)P\vert n_1\rangle=0$


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

 \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 .\\

\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.


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

\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


 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.


This archive was generated by hypermail 2.0b3 on Sun Oct 17 1999 - 22:40:46 JST