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Quantum expectation value (griffiths)

  1. Sep 2, 2011 #1
    1. The problem statement, all variables and given/known data

    If x is the position of a particle then the expectation value of x, <x> is :

    4dfcd01ed4ea36ccfe33487d3565fb05.png

    (I got lazy and just copied an image from Wiki, just pretend <x> is on the lhs of the eqn)

    When Griffith derives an expression for d<x> / dt, he uses the fact that dx/dt is zero, since "the wave function collapses after the first measurement and subsequent measurements will repeat the same result, if they're performed quickly."

    What does he mean by this? Surely the wave function evolves according to the schrodinger equation after the first measurement... so how fast must the subsequent measurements be taken to ensure that dx/dt = 0?
     
  2. jcsd
  3. Sep 2, 2011 #2
    What he means by 'collapsing wave function' is the fact that the wave function just represents a probability of where the particle is most likely to be. When you take a measurement, you now know where the particle is, so there is a 100% chance of finding the particle of where you have measured it, and a 0% chance of finding it anywhere else. Because of your measurement, you've disturbed the system, and unless you reset the system, nothing is going to change. The whole 'if they're performed quickly' thing is just a statement of the fact that your particle is going to move from where you measured it eventually, but to all intents and purposes, you can say that it won't, and thus the same result will be repeated each time you make a measurement.

    Unfortunately you're going to have to get used to statements such as 'if they're performed quickly'; he's basically covering his back from someone saying 'well what if you wait 1 million years...' etc.
     
  4. Sep 2, 2011 #3
    So the wave-form of that particular particle only begins again, and evolves according to the schrodinger equation, right after a measurement "ends"?
     
  5. Sep 2, 2011 #4

    xts

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    I don't know Griffith's book, but I may imagine what he wanted to say (skipping math and derivations - find them at Griffith's book):

    We have some quantum process. E.g. an radioactive atom decaying with emission of alpha particle. This particle may go anywhere. Till now the process is described by Schrödinger's equation. But the alpha particle goes into our favourite Wilson chamber. It ionizies some molecule (we'll see a drop of mist here later). That is our first measurement. It may occur anywhere - we can't say what direction alpha particle takes after decay.
    But as the particle got measured, its further evolution must be calculated using Schrödinger's equation with new starting conditions: not as it origines from original atom, but taking as a starting point the place where it got measured. A very short while later it ionizies another molecule - very close to previous one. And so on. Thus we may see the track of the particle - even, if it is a quantum particle. Every measurement is very unprecise (the drop in Wilson chamber has one micrometer or so in diameter - that's large span from atomic perspective) - so it don't affect by Heisenberg Principle particle's momentum significantly.
     
  6. Sep 2, 2011 #5
    Yes, basically!
     
  7. Sep 3, 2011 #6
    Thanks to both of you, makes sense.
     
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