----------------------------------------> From: lasermazer.domain.name.hidden> To: everything-list.domain.name.hidden> Subject: RE: [KevinTryon.domain.name.hidden.com: Jacques Mallah]> Date: Sun, 8 Feb 2009 20:33:52 -0500>> 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)Arrrgh, now it looks like it snipped out my bra-ket notation for no discernable reason. Trying again:My comment was that the "Born rule probability" is equal to <\psi|\psi_i><\psi_i|\psi>, and since <\psi_i|\psi> 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 <\psi_i|\psi_i> = (Born rule probability)*<\psi|\psi>, i.e. <\psi_i|\psi_i> = <\psi|\psi_i><\psi_i|\psi><\psi|\psi>, 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 - 20:42:52 PST