Re: Many Pasts? Not according to QM...

From: Russell Standish <r.standish.domain.name.hidden>
Date: Sun, 12 Jun 2005 18:29:39 +1000

On Sat, Jun 11, 2005 at 07:43:30PM -0700, "Hal Finney" wrote:
> Jesse Mazer writes:
> > But I explained in my last post how the ASSA could also apply to an
> > arbitrary "next" observer-moment as opposed to an arbitrary "current"
> > one--if you impose the condition I mentioned about the relation between
> > conditional probability and absolute probability, which is basically
> > equivalent to the condition that each tank is taking in water from other
> > tanks at the same rate it's pumping water to other tanks, then the
> > probabilities will be unchanged.
>
> One thing I didn't understand about this example: how do you calculate
> the probabilities which relate one observer-moment to a potential
> successor observer-moment? And do they have to satisfy the rule that
>
> p(x) = sum over all possible predecessor OM's y of (p(y) * p(x|y))
>
> where p(x|y) is the transition probability from predecessor OM y to
> successor OM x? In other words, is probability conserved much as fluid
> flow would be in tanks which had constant fluid levels?
>

In this notation, the ASSA gives a measure p(x)=\sum_y p(y) p(x|y),
and the RSSA gives p(x|y) for the same OM.

Also the ASSA gives a non-zero measure p(z), where z weas a
predecessor of y. However, the RSSA gives p(z|y) = 0.

The event on the left hand side should be read "next observer moment".

The two sampling assumptions do not add up!

> I'd be interested in any ideas for how one might calculate a priori the
> p(x|y) probabilities.

Use the Born rule. Using bra-ket notation, p(x|y), where x is at time
t, and y is at time 0,

   p(x|y) = <x|exp(-i\hbar Ht) |y>/\sqrt{<x|x>}\sqrt{<y|y>}

For the gory details, read section 4 of "Why Occam's Razor".


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Received on Sun Jun 12 2005 - 04:45:12 PDT

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