# Re: Quantum Time Travel

--- Russell Standish <[EMAIL PROTECTED]> wrote:
> An observer moment is not devoid of information. The
> mere fact that it is an observer moment, implies
> that the observer can observe itself. Following the
> logic of my Occam paper, one can conclude that
> measure of an observer moment - i.e. _given_ that I
> am an observer is highly non-uniform, with greatest
> measure given for systems indestinguishable from
> lawlike (hence no WRs).

Sounds OK.

> Now with the multiverse, for which there is an
> objective measure uniform measures can exist as
> (rather unnatural) solutions to the
> SE. eg a superposition of all plane waves
> \int_{-\infty}^\infty
> exp(-ipx/\hbar)dp. This beast is clearly
> unnormalisable, but that is not a problem in itself.

That's also known as a position eigenstate.

> However, observers will constrain the form of the
> universal wavefunction such that the measure is
> nonuniform, effectively giving it a value.

The above sentence doesn't make sense.  First, the
term 'measure' is only defined with respect to
observer-moments.  Second, obsevers do not change the
universal wavefunction, though they do for many
practical purposes have to deal only with the relative
state.

> The more information one has, the more non-uniform
> will be the measure.

The objective measure distribution itself remains
the same, but one can define a conditional measure
distribution to reflect the information, useful for
Bayesian purposes.  (One can work with the origianl
objective measure and carefully apply Bayesian
analysis.)

> But it is still the observer selecting a quantum
> history that defines the measure.

What's a "quantum history"?
An observer-moment doesn't see a history.

=====
- - - - - - -
Jacques Mallah ([EMAIL PROTECTED])
Physicist  /  Many Worlder  /  Devil's Advocate
"I know what no one else knows" - 'Runaway Train', Soul Asylum
My URL: http://hammer.prohosting.com/~mathmind/

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