[time 680] Re: [time 678] Re: [time 676] Reply to NOW/PAST question


Hitoshi Kitada (hitoshi@kitada.com)
Mon, 6 Sep 1999 16:00:03 +0900


Dear Stephen,

Stephen P. King <stephenk1@home.com> wrote:

Subject: [time 678] Re: [time 676] Reply to NOW/PAST question

> Hi All,
>
> Hitoshi Kitada wrote:
>
> > If you set like this, this x is equal to
> >
> > x = - i Deltat H.
> >
> > (You forgot minus sign in the above).
> >
> > In this setting, we have an exact identity
> >
> > Psi(t+Deltat) = exp(-i Deltat H) Psi(t) = exp (x) Psi(t)
> >
> > according to the Schroedinger equation. This equals
> >
> > Psi(t+Deltat) = (1+ x + x^2/2! + x^3/3! + x^4/4! + ...) Psi(t),
> >
> > which seems different from your calculation:
> >
> > > Psi(t+Deltat)/Psi(t) = [ 1/(1 - x) ],
> >
> > i.e.
> >
> > Psi(t+Deltat)=(1+x+x^2+x^3+x^4+...)Psi(t)
> >
> > Do you mean to imply what we actually observe is different from the exact
> > physical process to this amount? If so, then why/how?
>
>
> What is the difference? Does it diverge or converge or neither as x ->
> \infinity?

1/(1-x) = 1+x+x^2+x^3+x^4+...

converges for |x|<1, but diverges for |x|>1, while

exp(x) = 1+ x + x^2/2! + x^3/3! + x^4/4! + ...

converges for all (complex (or operator)) x. No need to say that they have
different values for the same value of x ... ?

>
> Bill, what if every observer related to the values v and c in (v^2/c^2)
> could only interfere with a finite number of other observers, but there
> exist at least an infinity of them?
>
> Puzzled!
>
> Stephen
>

Best wishes,
Hitoshi



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