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

*Sat, 2 Oct 1999 09:05:13 +0300 (EET DST)*

**Messages sorted by:**[ date ] [ thread ] [ subject ] [ author ]**Next message:**Ben Goertzel: "[time 856] RE: [time 853] Re: [time 852] Another expresion of my weird idea..."**Previous message:**Stephen P. King: "[time 854] Re: [time 852] Another expresion of my weird idea..."**In reply to:**Ben Goertzel: "[time 852] RE: [time 851] Re: [time 850] RE: [time 849] Another expresion of my weird idea:"**Next in thread:**Hitoshi Kitada: "[time 858] Re: [time 855] Re: [time 847] Unitarity of S-matrix"

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

**Next message:**Ben Goertzel: "[time 856] RE: [time 853] Re: [time 852] Another expresion of my weird idea..."**Previous message:**Stephen P. King: "[time 854] Re: [time 852] Another expresion of my weird idea..."**In reply to:**Ben Goertzel: "[time 852] RE: [time 851] Re: [time 850] RE: [time 849] Another expresion of my weird idea:"**Next in thread:**Hitoshi Kitada: "[time 858] Re: [time 855] Re: [time 847] Unitarity of S-matrix"

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