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

*Fri, 1 Oct 1999 11:25:22 +0300 (EET DST)*

**Messages sorted by:**[ date ] [ thread ] [ subject ] [ author ]**Next message:**Hitoshi Kitada: "[time 848] Re: [time 847] Unitarity of S-matrix"**Previous message:**Hitoshi Kitada: "[time 846] Re: [time 845] Re: [time 844] Re: [time 841] Re: [time 839] What does an LS observe?"**Next in thread:**Hitoshi Kitada: "[time 848] Re: [time 847] Unitarity of S-matrix"

Hi all,

we had somewhat emotional discussions with Hitoshi about

unitary of the TGD S-matrix. I decided to look the problem

again and realized that unitarity is indeed guaranteed formally

by the mere algebraic structure of the scattering solution

and the fact that outgoing states are projections

of scattering state to the space of on mass shell

free states and indeed satisfy single particle

Virasoro conditions. Off mass shell states appear only

as intermediate states.

p-Adics might be forced by the divergence of

normalization constants: in quantum field theories

normalization constants indeed diverge.

What is remarkable that the mere algebraic structure

of scattering states of time dependent perturbation theory

guarantees unitarity: time development operator is not needed.

The possible delicacies related to the Hilbert space

extensions are problems about which one cannot say must

at this stage: personally I am satisfied with

the formal proof of unitarity.

I include the latex field providing the proof of formal

unitarity treating L_0(int) as perturbation.

Best,

MP

\documentstyle [10pt]{article}

\begin{document}

\newcommand{\vm}{\vspace{0.2cm}}

\newcommand{\vl}{\vspace{0.4cm}}

\newcommand{\per}{\hspace{.2cm}}

{\it 5. Proof for the unitary of S-matrix}

\vm

S-matrix is defined between the projections $P\Psi$ of scattering

states to "free"

states satisfying free Virasoro conditions.

Therefore the Hilbert spaces of "free" and projected

scattering states are at least

formally identical.

This means that off-mass-shell states appear only as intermediate

states in the perturbative expansion of the S-matrix

just as they do in the standard quantum field theory.

\vm

S-matrix is unitary if

outgoing states are orthogonal to each other. This follows from

the definition of S-matrix as

\begin{eqnarray}

S_{n,m}&=& \langle m_0\vert n\rangle\per ,\nonumber\\

\end{eqnarray}

\noindent where $m_0$ is incoming state and $n$ is scattering state

normalized to unity.

Unitary condition reads as

\begin{eqnarray}

\sum_r S_{m,r}(S_{n,r})^*&=& \delta (n,m)\per .

\end{eqnarray}

\noindent where summation is over the "free" states $\vert r_0\rangle$

to which quantum jump occurs.

\noindent Unitarity condition reads explicitely as

\begin{eqnarray}

\sum_r S_{m,r}(S_{n,r})^*&=&\sum_{r} \langle n\vert r_0 \rangle\langle

r_0\vert m\rangle = \langle n\vert m\rangle\per .

\end{eqnarray}

\noindent Here the completeness of the "free" state basis has been used.

Hence unitarity holds true if one has

\begin{eqnarray}

\langle m\vert n\rangle \propto \delta (m,n)\per .

\end{eqnarray}

\noindent provided that the normalization constant

for the outgoing states are finite. In quantum field theories

this is not usually the case and this could be the reason

for why p-adics are necessarily needed.

\vm

In case of Schr\"odinger equation one can prove orthogonality

of the scattering states

by noticing that "free" and scattering state basis are related

by a unitary time development operator, which preserves

the orthonormality

of the incoming states. Now the situation is different. The

combinatorial structure is same as in wave mechanics but

genuine time development operator need not exist and one

must resort to the hermiticity of $L_0(free)$ and $L_0(int)$

plus the general algebraic structure of the scattering states

in order to prove the unitarity.

\vm

One can treat $L_0(int)$ perturbatively by assuming

it to be proportional to coupling constant $g$, which

is put to $g=1$ at the end. In this manner

one can systematically study the perturbative expansion for

the inner product $\langle m\vert n \rangle$ for scattering

states by using the explicit expression definition of $\vert m\rangle$

as a scattering state to show $\langle m\vert n \rangle\propto \delta

(m,n)$.

The result can be seen to hold order by order by studying the explicit

expression of $\langle m\vert n \rangle \propto \delta (m,n)$ given

by:

\begin{eqnarray}

\langle m\vert n \rangle &=& \delta (m_0,n_0)\nonumber\\

&+& \langle m_0\vert \frac{1}{L_0(free)+i\epsilon}g L_0(int)\vert

n\rangle

+\langle m \vert gL_0(int) \frac{1}{L_0(free)-i\epsilon}\vert n_0\rangle

\nonumber\\

&+&\langle m \vert gL_0(int) \frac{1}{L_0(free)-i\epsilon}P

\frac{1}{L_0(free)+i\epsilon} gL_0(int)\vert n\rangle\per ,\nonumber\\

\vert n \rangle &=&\vert n_0\rangle +

\frac{1}{L_0(free)+i\epsilon}gL_0(int)

\vert n \rangle\per ,\nonumber\\

\langle m \vert &=&\langle m_0 \vert+ \langle m \vert

gL_0(int)\frac{1}{L_0(free)-i\epsilon} \per .

\end{eqnarray}

\noindent

In the equations above $P$ denotes the projector to the space of the

"free" states.

These equations are understood as powers series with respect

to $g$ at $g=1$.

\vm

It is easy to directly check that orthogonality conditions

hold true in the lowest orders. The proof that orthogonality holds true

in order $k+1$ is obtained by simply replacing in the inner product

the terms linear in $g$ with $k+1$:th order approximation

\begin{eqnarray}

\vert n \rangle_{k+1} &=&\vert n_0\rangle +

\frac{1}{L_0(free)+i\epsilon}gL_0(int)

\vert n \rangle_k \per .\nonumber\\

\langle m \vert_{k+1} &=&\langle m_0 \vert+ \langle m \vert_k

gL_0(int)\frac{1}{L_0(free)-i\epsilon} \per ,

\end{eqnarray}

\noindent and substituting to the expression of

$\langle m\vert n\rangle$ whereas

in the term quadratic in $g$ $k$:th order approximation is used.

If the expansion in powers of $g$ converges, orthogonality conditions

hold true.

\end{document}

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