# RE: [KevinTryon.domain.name.hidden: Jacques Mallah]

From: Jesse Mazer <lasermazer.domain.name.hidden>
Date: Sun, 8 Feb 2009 06:46:04 -0500

Russell Standish wrote:> > According to Wikipedia, Born's rule is that the probability of an> observed result \lambda_i is given by <\psi|P_i|\psi>, where P_i is the> projection onto the eigenspace corresponding to \lambda_i of the> observable. > > This formula is only correct if \psi is normalised. More correctly,> the above formula should be divided by <\psi|\psi>.> > This probability can be interpreted as a conditional probability - the> probability of observing outcome \lambda_i for some observation A,> _given_ a pre-measurment state \psi.> > What is important here is that it says nothing about what the state> vector is after the measurement occurs. There is a (von Neumann)> projection postulate, which says that after measurement, the system> will be found in the state P_i|\psi>, but as I said before, this is> independent of the Born rule, and also it does not state what the> "amplitude" (ie magnitude) of the state is. The v-N PP is also distinctly not> a feature of the MWI (it is basically the Copenhagen collapse).The projection postulate needs to work *effectively* in the MWI though, in order for it to make the same predictions about actual experimental results as the Copenhagen interpretation. If an observer makes one measurement and then makes a later measurement of the same system, they can collapse the system's quantum state onto an eigenstate at the moment of the first measurement and evolve that new state forward, and this will give correct predictions about the probabilities of different outcomes when the second measurement is made. I think part of the difficulty with connecting the MWI to actual observed probabilities is explaining *why* this rule should work in an effective sense.> > > I think the quote I was responding to was the following:> > "In an ordinary quantum mechanical situation (without deaths), and> assuming the Born Rule holds, the effective probability is proportional> to the total squared amplitude of a branch."> > If you compare it with the description of the Born rule above (which> computes a conditional probability), there is no sense in which one> can say that "the effective probability is proportional to the total> squared amplitude of a branch" follows directly from the Born> rule. Jacques is assuming something else entirely - perhaps> einselection?The notion that probability is proportional to squared amplitude is equivalent to the wikipedia definition you posted above. The projection operator P_i onto a given eigenstate |\lambda_i> is really just |\lambda_i><\lambda_i| (see the text immediately above 'Postulate 5' near the bottom of the page at http://xbeams.chem.yale.edu/~batista/vvv/node2.html ), which means that when this operator acts on a given state vector |\psi>, it gives (|\lambda_i><\lambda_i|)|\psi> = |\lambda_i>(<\lambda_i|\psi>), and (<\lambda_i|\psi>) is just a scalar c_i, so this becomes c_i * |\lambda_i>. This c_i is referred to as the "amplitude" that the original state |\psi> assigns to the eigenstate |\lambda_i> (by something called the 'expansion postulate' it's possible to write |psi> as a weighted sum of all the eigenstates of a measurement operator, and the weights on each eigenstate are just the amplitudes for each eigenstate). So, if the probability is <\psi|P_i|\psi> (which I think assumes that |\psi> has been normalized), then since P_i = |\lambda_i><\lambda_i|, this is the same as saying the probability is <\psi|\lambda_i><\lambda_i|\psi>. If <\lambda_i|\psi> is the amplitude c_i, then <\psi|\lambda_i> is just the complex conjugate (c_i)*, so the probability is just the amplitude times its own complex conjugate, often referred to as the "amplitude-squared".Jesse
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Received on Sun Feb 08 2009 - 06:46:17 PST

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