RE: [ Jacques Mallah]

From: Jesse Mazer <>
Date: Sun, 8 Feb 2009 20:33:52 -0500

> 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.

> 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

> 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)
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Received on Sun Feb 08 2009 - 20:34:31 PST

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