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

From: Jesse Mazer <lasermazer.domain.name.hidden>
Date: Fri, 6 Feb 2009 08:16:20 -0500

> His discussion of the Born rule is incorrect. The probability given by
> the Born rule is not the square of the state vector, but rather the square
> modulus of the inner product of some eigenvector with the original
> state, appropriately normalised to make it a probability. After
> observation, the state vector describing the new will be proportional
> to the eigenvector corresponding the measured eigenvalue, but nothing
> in QM says anything about its amplitude.

I may be misunderstanding you, but I don't think it's correct to say that "nothing in QM says anything about its amplitude"--in QM every state vector can be expressed as a weighted *sum* of the eigenvectors for any measurement operator (vaguely similar to Fourier analysis), and the Born rule says the probability the system will be measured in a given eigenstate should be given by the square of the amplitude assigned to that eigenvector in the sum which corresponds to the state vector at the moment before the measurement (technically the probability the amplitude multiplied by its own complex conjugate rather than the amplitude squared, so if the amplitude is x + iy you multiply by x - iy to get a probability of x^2 + y^2, but it's common to just say 'amplitude squared' as shorthand).

Jesse

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Received on Fri Feb 06 2009 - 08:16:37 PST

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