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Precise measurements in Quantum Mechanics

  1. Apr 26, 2013 #1
    It follows from the Expectation value postulate that an observable A, associated with the operator A^, can be precisely measured only if the wave function ψ of the system is an eigenfunction of A^ .

    Accordingly, the position and momentum of a particle can never be precisely measured because the wave function (energy eigenfunction) is not an eigenfunction of the operators of these properties.

    In the popular statement of the uncertainty principle, viz., the position and momentum of a particle cannot be measured precisely and simultaneously, it appears that one of these properties can be precisely measured sacrificing the other. Does it not contradict the above deduction from the expectation value postulate, that the wave function should be an eigenfunction of the operator of the property in order to measure the property precisely?
     
  2. jcsd
  3. Apr 26, 2013 #2
    A quantum state cannot be a simultaneous eigenfunction of position and momentum. Position eigenstates do exist, as do momentum eigenstates. You just can't be an eigenstate of both.

    You may take an arbitrary state and measure its position, so that it becomes (collapses into) a position eigenstate. The state now has a determinate, known position. The probability that is a particular value is given by square of the projection of the original state onto the position basis (ie this is probably what you mean by the expectation value postulate). Note that, unless the original state was a position eigenstate, your new state is different; the act of measurement (the projection onto the new basis) has altered the state!

    You may then take this position eigenstate and measure its momentum, so that it becomes a momentum eigenstate. The state has changed again; now it has determinate, known position.

    If you had two commuting observables A and B (position and momentum are NOT commuting), you could take an eigenstate of A and then measure its B without changing the state. This is because commuting observables share an eigenbasis: the projection (the measurement) has no effect.
     
  4. Apr 26, 2013 #3

    kith

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    Why should this follow? If I throw a standard dice, my expectation value is 3.5 but my measurement outcome is always "precisely" an integer between 1 and 6.
     
  5. Apr 27, 2013 #4
    Precise measurements in QM

    But, you cannot predict whether you get 1 or 2 or ....6 in each measurement you make. This leads to uncertainty.
     
  6. Apr 27, 2013 #5
    I think that, by "precise", mpkannan means "determinate". If ψ is an eigenstate to begin with, we are guaranteed that its measurement will yield a particular value (the corresponding eigenvalue). Otherwise, we can't predict with certainty the outcome of the measurement.

    "Precise" probably isn't the most appropriate term since it is typically used as kith used it.
     
  7. Apr 27, 2013 #6

    Nugatory

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    In a VERY hand-wavy way (hand-wavy because I'm using ordinary uncertainty about the state of the die, a very different mathematical proposition than superposition of eigenfunctions)....

    WARNING - hand-waving follows. Use at your own risk!

    While the die is in the air the expectation is indeed 3.5. Once it lands on a flat surface and stops (that is, interacts with the measuring apparatus) it will settle into one of six states defined by which face is up; all six of these are very different from the state while it's in the air. When the die is in the in-the-air state, I cannot claim to have made any measurement of its value; all I can say is that if I made such a measurement the expectation value would be 3.5. After I've forced the die into one of the six resting-on-a-flat-surface states, I can read its value off the top face with absolute precision, and (because the die is in a different state) the expectation value of the next reading is not 3.5.

    We could push the analogy to the breaking point (Hey - it's your analogy not mine! - don't blame me!) by saying that the in-the-air state is analogous to a superposition of the six on-a-flat-surface states which are analogous to eigenstates of the "value operator".

    However, we can't push the analogy far enough to answer OP's question (How do we reconcile the uncertainty principle with the fact of precise values of observables in some eigenstates) because we'd need a second non-commuting observable, and dice don't have that.
     
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