[time 803] Re: [time 801] Re: [time 799] Still about construction of U


Matti Pitkanen (matpitka@pcu.helsinki.fi)
Sat, 25 Sep 1999 11:19:21 +0300 (EET DST)


On Sat, 25 Sep 1999, Hitoshi Kitada wrote:

> Dear Matti,
>
> I have several questions on your construction of S-matrix.
>
> 1.
>
> > Contrary to earlier expectations, it seems that one cannot assign
> >explicit Schr\"odinger equation with S-matrix although the
> >general structure of the solutions of the Virasoro conditions
> >is same as that associated with time dependent perturbation theory
> >and S-matrix is completely analogous to that obtained as
> >time evolution operator $U(-t,t)$, $t\rightarrow \infty$ in
> >the perturbation theory for Schr\"odinger equation.
>
> Does this mean that your former equation
>
> >\begin{eqnarray}
> >i\frac{d}{dt}\Psi&=& H\Psi\per , \nonumber\\
> >H &\equiv& k\left[-P_T^2 - L_0(vib)-L_0(int)\right]\Psi\per .
> >\end{eqnarray}
>
> is wrong or cannot be derived by the former note?

As such the equation is probably wrong. The point is
that I was forced to make *ad hoc* replacement
of Diff^4 invariant translation generator p_+ with id/dt

p_+--> id/dt

in order to obtain Schrodinger equation. The introduction
of time t leads to potential problems with Poincare invariance,
which however could be avoided. But this is the main
ad hoc element of construction.

Starting directly from Super Virasoro conditions and just
writing "scattering solution" for them one avoids
all ad hoc hypothesis and manifest Poincare invariance is achieved.
One however loses Schrodinger equation but this is not needed
since informational "time evolution" is
totally characterized by S-matrix. Thus I am tending to believe that
Heisenberg was right: S-matrix has nothing to do with time evolution with
respect to geometric time.

>
> 2.
>
> >\begin{eqnarray}
> >\Psi&=&\Psi_0 + \frac{V}{E-H_0-V+i\epsilon} \Psi \per .
> >\end{eqnarray}
>
> (This is equation (1) of your note.)
>
Sorry: this equation is mistyped:

\Psi&=&\Psi_0 + \frac{V}{E-H_0+i\epsilon} \Psi \per .

The presence of V in denominator would make it to diverge.
The ordering is also important: V is to the right.

>
> >Since ordinary Schr\"odinger equation is consistent with the scattering
> >matrix formalism avoiding elegantly the difficulties with the
> >definition of the limit $U(-t,t)$, $t\rightarrow \infty$, it
> >is natural to take this form of Schr\"odinger equation as starting
> >point when trying to find Schr\"odinger equation for the 'time' evolution
> >operator $U$. One can even forget the assumption
> >about time evolution and require only
> >that the basic algebraic information guaranteing
> >unitarity is preserved. This information boils down to the Hermiticity
> > of free and interacting Hamiltonians and
> >to the assumption that the spectra
> >non-bound states for free and interacting Hamiltonians
> >are identical.
>
>
> It is known that to consider the limit as \epsilon -> 0 in the Schrodinger
> equation (1) of your note is equivalent to considering the time limit as t ->
> \infty of exp(-itH). So you cannot avoid the difficulty: Below I will try to
> show this.

In TGD framework single particle Virasoro generators L_0(n) define
propagators

1/(p^2-L_0(vib)+i*epsilon)

appearing in stringy diagrams. L_0(vib) is integer valued and gives rise
to the universal non-negative integer valued mass squared spectrum of
string models (in suitable units).

In present case i*epsilon is completely equivalent with
the presence of i*epsilon in the propagators of relativistic quantum field
theory: epsilon term guarantees that momentum spacetime integration
over virtual momenta is performed correctly in case that one
is forced to integrate over pole of propagator.

As far as I can understand this has nothing to do with time but I
could of course be wrong.

[In string models similar expansion is derived from
stringy perturbation theory in which time parameter corresponds
to string coordinate. The problem in string models is
that this gives only first quantized theory and one should
second quantize. In TGD this problem is not encountered. Super
Virasoro generators are of second quantized form from
beginning. Also topological description of particle reactions
emerges automatically via the decomposition of L_0(tot) to
a sum of single particle Virasoro generators L_0(n). This
decomposition was described in later posting.]

>
> The Schrodinger equation (1) in your note should be written as:
>
> \Psi = \Psi_0 - V ((E+i \epsilon)-H)^{-1} \Psi, (1)

Yes with H replaced with free Hamiltonian: I am sorry for my typo.
I hope that it does not affect you conclusions.

>
> or
>
> \Psi = \Psi_0 - ((E+i \epsilon)-H)^{-1} V \Psi. (2)
>
> The reason we consider two equations is that the fractional expression in
> your note has two interpretations, since the relevant factors are operators,
> hence they are noncommutative in general.

Yes. In my equation 1/(E-H_0 +iepsilon)*V is meant to be the
correct ordering. This is easy to see: by operating with E-H_0 to both
sides of the equation one obtains (E-H_0) Psi= (E-H_0)Psi+ VPsi =VPsi
which is just Schrodinger equation. Provided that order is
correct.

> Also the sign on the RHS (right
> hand side) should be minus if the definition of H is H = H_0 + V.
>

> We write z = E+i\ep, where \ep = \epsilon > 0 and E is any real number.
>
> In case (1), the equation is equivalent to
>
> (I - V (H-z)^{-1} ) \Psi = \Psi_0.
>
> This is rewritten as
>
> (H - z - V) (H - z)^{-1} \Psi = \Psi_0,
>
> which is equivalent to
>
> (H_0-z) (H-z)^{-1}\Psi = \Psi_0 (1)'
>
> or
>
> (H-z)^{-1}\Psi = (H_0-z)^{-1}\Psi_0 (1)"
>
> or
>
> \Psi = (H-z)(H_0-z)^{-1}\Psi_0
>
> = \Psi_0 + V(H_0-z)^{-1}\Psi_0 (1)'''
>
> by H = H_0 + V.
>
> Thus if \Psi_0 is a given fixed state function, \Psi depends on z = E+i\ep,
> and it should be written as a function of z:
>
> \Psi = \Psi(z).
>
> Note that these hold only when \ep > 0, NOT for an infinitesimal number \ep
> because the inverse (H-(E+i\ep))^{-1} does not exist for \ep = 0 in general.
>
> In case (2), the equation is equivalent to
>
> \Psi_0
>
> = (I - (H-z)^{-1} V)\Psi (2)'
>
> = (H-z)^{-1} (H-z-V)\Psi
>
> = (H-z)^{-1} (H_0 - z)\Psi.
>
> This is rewritten:
>
> \Psi(z) = (H_0-z)^{-1} (H-z) \Psi_0. (2)"
>
> If we transform \Psi_0 and \Psi to
>
> \Phi_0 = (H-z) \Psi_0,
>
> \Phi = (H-z) \Psi,
>
> the equation (2)" becomes
>
> \Phi(z) = (H-z) (H_0-z)^{-1} \Phi_0. (2)'''
>

Certainly one cannot get rid of time in ordinary wave mechanics.
If you have energy, you have also time!

In TGD approach one has

L_0(tot) Psi=0 rather than HPsi = EPsi! No energy, no time!!

By the way, this condition is analogous to your condition
that entire universe has vanishing energy

HPsi=0

Thus there is something common between our approaches!

Also in general Relativity Hamiltonian vanishes as a constraint.
In TGD However General Coordinate invariant
as gauge invariance is replaced by Super Virasoro invariance
which operates at lightcone boundary and H is replaced with L_0.

Best,

MP



This archive was generated by hypermail 2.0b3 on Sat Oct 16 1999 - 00:36:41 JST