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

From: Jesse Mazer <lasermazer.domain.name.hidden>
Date: Fri, 6 Feb 2009 08:59:44 -0500

Ah, never mind, rereading your post I think I see where I misunderstood you--you weren't saying "nothing in QM says anything about" the amplitude of an eigenvector that you square to get the probability of measuring that eigenvector's eigenvalue, you were saying "nothing in QM says anything about" how the length of the state vector immediately after the measurement "collapses" the system's quantum state is related to the length of the eigenvector it collapses onto (since the probabilities given by squaring the amplitudes of the eignevectors always get normalized I think it doesn't matter, the 'direction' of the state vector is all that's important).

Still, I don't quite see where Mallah makes the mistake about the Born rule you accuse him of making, what specific quote are you referring to?

Jesse

From: lasermazer.domain.name.hidden
To: everything-list.domain.name.hidden
Subject: RE: [KevinTryon.domain.name.hidden: Jacques Mallah]
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

--~--~---------~--~----~------------~-------~--~----~
You received this message because you are subscribed to the Google Groups "Everything List" group.
To post to this group, send email to everything-list.domain.name.hidden
To unsubscribe from this group, send email to everything-list+unsubscribe.domain.name.hidden
For more options, visit this group at http://groups.google.com/group/everything-list?hl=en
-~----------~----~----~----~------~----~------~--~---
Received on Fri Feb 06 2009 - 08:59:57 PST

This archive was generated by hypermail 2.3.0 : Fri Feb 16 2018 - 13:20:15 PST