Re: [ Jacques Mallah]

From: russell standish <>
Date: Mon, 9 Feb 2009 13:02:31 +1100

On Sun, Feb 08, 2009 at 08:33:52PM -0500, Jesse Mazer wrote:
> > Date: Mon, 9 Feb 2009 11:47:02 +1100
> > From:
> > To:
> > Subject: Re: [ Jacques Mallah]
> >
> >
> > Jesse, you need to fix up your email client to follow the usual
> > quoting conventions, wrap lines etc.
> I'm using hotmail, any idea what I'd need to do to make it work
correctly? Looking at how the posts show up on google groups, it looks
like my previous posts didn't have this problem, I'm not sure what
went wrong with the last one.

This time it looks better, although you might want to enable word wrap
on 80 columns for the text only mode.

I've never used Hotmail, so can't comment on specifics. I would hope
that it doesn't trash email conventions built up over a number of
years, but given that its Microsoft owned, and they did precisely that
with the abomination called Outlook, one can't be too optimistic.

> >
> > By working *effectively*, I think you mean the subjective experience (or
> > 1st person experience) should be as though \psi is transformed to
> > \psi_i, where \psi_i is an eigenvector of the relevant observable.
> >
> > Of course no such thing happens in the MWI, ie the 3rd person
> > description. \psi remains unaltered in this case, and is undergoing
> > regular unitary evolution.
> Yes, my understanding is that decoherence might be of some use in explaining the appearance of collapse even if we assume that an experimenter measuring a quantum system merely becomes entangled with it, with the wavefunction of the combined experimenter/quantum system continuing to evolve in a unitary way. There's a discussion of this on Greg Egan's page at

I think the einselection issue should be orthogonal to issues of
observer measure, but suspect that Mallah may be using
it. Einselection does have its problems, though.

> >
> > In any case, there is no requirement for to be given
> > by the product of the Born rule probability with .
> I don't understand, why is this implied by what Jacques or I said? My comment was that the "Born rule probability" is equal to , and since is the amplitude, it's accurate to say the probability is the amplitude-squared (although after editing the wikipedia article to say this, I was told that this only works for operators where the eigenstates are 1D eigenvectors, and that there can be cases where the eigenstate is an 'eigenspace' with more than one dimension in which case the Born rule probability can't be written this way). I don't see how this implies that = (Born rule probability)*, i.e. = , which is what you seem to be accusing Jacques of claiming above. (By the way, this would probably be easier to read if a different symbol than \psi was used for eigenstates, as I did in my previous post)

All I have ever said was that effective probability given by the
squared norm of the projected eigenvector does not follow from Born's
rule. It can't follow, because Born's rule says nothing about what the
normalisation of the state vector after observation should be. It is a
conditional probability only.

Now Jacques must have a model for what happens to the state vector
after observation - I made an interpretation that the normalisation
might be given by chaining Born rule probabilities with the squared
norm of the original state just to try an advance understandinf, but
regardless of whether that is correct, Jacques needs to come clean
about what model he is using. What does "total squared amplitude of a
branch" actually mean?

> Jesse
A/Prof Russell Standish                  Phone 0425 253119 (mobile)
UNSW SYDNEY 2052         
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Received on Sun Feb 08 2009 - 21:02:48 PST

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