# [time 848] Re: [time 847] Unitarity of S-matrix

Hitoshi Kitada (hitoshi@kitada.com)
Sat, 2 Oct 1999 01:53:31 +0900

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?

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.

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.

> If the expansion in powers of $g$ converges, orthogonality conditions
> hold true.
>
>
> \end{document}
>

Best wishes,
Hitoshi

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