# [time 847] Unitarity of S-matrix

Matti Pitkanen (matpitka@pcu.helsinki.fi)
Fri, 1 Oct 1999 11:25:22 +0300 (EET DST)

Hi all,

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.

\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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