# [time 855] Re: [time 847] Unitarity of S-matrix

Matti Pitkanen (matpitka@pcu.helsinki.fi)
Sat, 2 Oct 1999 09:05:13 +0300 (EET DST)

Dear Hitoshi,

Your comment about proof was quite to the point. Neither
had I really undestood it! There was still one piece
lacking and this was the *assumption* L_0(tot)P|m>=0, not only
L_0(tot|m>=0. With this physically natural assumption
stating that off mass shell contributions to physical
states are negligible, everything works
nicely. See for details below.

On Sat, 2 Oct 1999, Hitoshi Kitada wrote:

> Dear Matti,
>
> I do not understand your proof. Comments are below.
>
> Matti Pitkanen <matpitka@pcu.helsinki.fi> wrote:
>
> Subject: [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.
>
>
> Do you mean a replacement of |n> by |n>_{k+1}? If so we have as the first term
> linear in g:
>
> g<m_0|(L_0(free)+ie)^{-1}L_0(int)+L_0(int)(L_0(free)-ie)^{-1}|n_0>.
>
> How this term can be zero when n_0 not = m_0? Do you assume that the
> projection P commutes with the interaction L_0(int)? If so how this
> assumption is justified?

a) This assumption can be justified by noticing that m_0 and n_0 are
annihilated by L_0(free) and one gets just 1/iepsilon factors
and two terms cancel each other. But *in higher orders*
one obtains problematic terms as I realized!

b) Your comment about *commutativity* of P and L_0(int) is to the
point. I reformulated the attempt to prove unitarity using
expansion of the scattering operator as geometric series.

c) Very essential element is the *assumption* that L_0(int) annhilates
projections of the scattering states to the space of "free" states.

(L_0(free)+L_0(int))|m> =0 does not necessarily imply

L_(tot)P\vert m> = (L_0(free)+ L_0(int))P|m>=0 which in turn implies
L_0(int)P|m>=0.

*IF* latter assumption is made everything follows neatly and as far
as I can see S-matrix *IS* nontrivial.

d) This assumption implies that wave function
renormalization is trivial and certainly does not diverge therefore! In
QFT:s situation is different.

I attach piece of text showing unitarity. Certainly essentially similar
proof must appear also in ordinary time dependent perturbation
theory so that there is nothing new with it. Except for me, of
course(;-)!
*****************************

An explicit expression for the scattering solution is as geometric series

\begin{eqnarray}
\vert m\rangle &=& \frac{1}{1+X}\vert m_0\rangle\per ,\nonumber\\
\langle m\vert &=& \langle m_0 \vert \frac{1}{1+X^{\dagger}}\per
,\nonumber\\
X&=& \frac{g}{L_0+i\epsilon}L_0(int)\per ,\nonumber\\
X^{\dagger}&=& gL_0(int)\frac{1}{L_0-i\epsilon}\per ,\nonumber\\
\end{eqnarray}

\noindent Essential assumption
in the proof is that $L_0(int)$ annihilates
projections of the scattering states to the space of
"free" states:

\begin{eqnarray}
L_0(int)P\vert n\rangle &=&0\per .
\end{eqnarray}

\noindent This condition follows immediately if one assumes that
$L_0(tot)\vert m\rangle=0$ implies

\begin{eqnarray}
L_0(tot)P\vert n\rangle&=&0\per .
\end{eqnarray}

\noindent since $L_0(free)P\vert n\rangle=0$. Thus $L_0(tot)$
and $P$ must effectively commute. If $L_0(tot)$ annihilates
both $\vert n\rangle$ and $P\vert n\rangle$, it seems that scattering
states span the same space as "free" states. Physically this condition
means that the presence of off-mass shell components
in the state $\vert n\rangle$ has no physical consequences
so that one
can {\it define} scattering states as states $P\vert n\rangle$.
The highly nontrivial consequence is that the states $P\vert n\rangle$
have unit norm: wave function normalization is trivial.
This would mean that quantum TGD is free of divergences related
to wave function renormalization. The fact that wave function
renormalization constant is nontrivial in quantum field theories,
suggests that the condition $L_0(tot)P\vert n\rangle =0$
is too strong and raises the question whether one
could somehow cope without this condition.

\vm

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

\begin{eqnarray}
\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 .
\end{eqnarray}

\noindent Using the fact that $X^{\dagger}$ annihilates
$P\vert m\rangle$ from right and $X$ annilates $\langle m\vert P$
from left, one can reduce this condition to

\begin{eqnarray}
\frac{1}{1+X^{\dagger}}P + P\frac{1}{1+X}&=&G \per .
\end{eqnarray}

\noindent This equation gives infinite
number of conditions by using geometric
power series expansions with respect to $g$:

\begin{eqnarray}
(X^{\dagger})^nP + P X^{n}= 0\per , \per n >0 \per .
\end{eqnarray}

\noindent Writing this explicitely and noticing
that one can drop $\i\epsilon$ from denominators
when $\frac{1}{L_0-i\epsilon}$ acts on off mass shell state,
one finds that
the two terms are apart from sign factor equal
to $\frac{1}{i\epsilon}(L_0(int)\frac{1}{L_0})^{n-1}L_0(int)$
but have different signs so that they cancel.

>
> By the way, the operator you define by
>
> S Psi_0 = Psi,
>
> where Psi satisfies
>
> Psi = Psi_0 -(L_0(free)+ie)^{-1}L_0(int) Psi,
>
> is known equal to the wave operator expressed on the energy shell. This is a
> common sense in scattering theory, known since around 1950 or before. Wave
> operator W = W_+ or W_- is defined by
>
> W_+ = lim_{t->\infty} exp(itL_0(tot)) exp(-itL_0(free)),
>
> in time dependent expression. There is also a stationary definition:
>
> (W_+ f, g) = (f, g)
>
> - (2i\pi)^{-1}\int_{-\infty}^\infty
>
> (V(H_0-\lam-i0)^{-1}f, (H-\lam-i0)^{-1}g) d\lam.
>
> Scattering operator S is defined by
>
> S = (W_+)^* W_-.
>
> But if you can prove your assertion for the wave operator W (which is equal to
> your S), i.e.
>
> W W^* = I and W^* W = I (S S^* = I, etc. in your case),
>
> then it follows from this that scattering operator S is unitary:
>
> S S^* = (W_+)^* (W_-) (W_-)^* (W_+) = (W_+)^* (W_+) = I and S^* S = I.
>

Yes. I have proven unitarity but not using time development
picture which would make proof unnecessary but just the general structure
of scattering solution. Similar algebraic proof must exist
also in wave mechanics and it would be interesting to know
whether the is counterpart for the assumption

*L_0(int) annihilates projections of the scattering states on space of
"free" states.*

Physically this means that all vertices describing
decay of on mass shell states to on mass shell states vanish: rather
reasonable requirement physically.

Since this condition follows from L_tot|m> =L_(tot)P\vert m>
it *very nearly* states Hilb=Hilb_0 "effectively". At least
*formally* this is the case since the projections of scattering states to
Hilb_0 have also *unit* norm. In finite-dimensional case this certainly
would imply Hilb=Hilb_0.

> Thus if your present proof works, it would give a "formal" proof. There is,
> however, usually subtlety that requires quite hard analyses in relation with
> topologies of Hilbert spaces of Gelfand triple even in the single particle
> case. If your present proof could be justified at a formal level, it might
> not be easy to get a rigorous logical proof.
>

You are certainly right. The assumption that projections of scattering
states "effectively" span space of "free" states is essential and could
follow from L_0(tot)P|m>=0. Perhaps this has something to do
with Gelfand triples. Could the space in the middle of Gelfand triple
correspond to states P|m>?

Remembering that states are spinor fields in infinite-dimensional space
of 3-surface suggests that this piece of proof belongs to functional
theorist rather than poor theoretical physicists(;-).

Best,
MP

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