> Date: Mon, 9 Feb 2009 11:47:02 +1100
> From: [email protected]
> To: [email protected]
> Subject: Re: [[email protected]: 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.
>
> 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
http://gregegan.customer.netspace.net.au/SCHILD/Decoherence/Decoherence.html
>
> 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)
Jesse
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