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Position & Momentum

  1. Apr 13, 2010 #1

    jaketodd

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    If you have an array of momentum detectors, if there is such a thing, couldn't the detector that is hit in the array simply report its momentum measurement along with its identifier in the array, which would reveal the position?

    Thanks,

    Jake
     
  2. jcsd
  3. Apr 13, 2010 #2

    jtbell

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    Sure. The uncertainty principle isn't about the limitations of detectors. You can measure the position and momentum of a single particle to any precision your detectors are capable of.

    If you now repeat the experiment many times, using identically prepared particles, in general you get different results for the momentum and position each time. The momentum values have a mean (average) and a standard deviation [itex]\Delta p[/itex]. The position values also have a mean and a standard deviation [itex]\Delta x[/itex]. These standard deviations are what appear in the HUP:

    [tex]\Delta x \Delta p \ge \hbar / 2[/tex]
     
  4. Apr 13, 2010 #3

    jaketodd

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    Why is there a minimum uncertainty as expressed in the equation you provided? Couldn't preparation get more exact and detectors more accurate to where the product of the standard deviations of momentum and position are less than hbar/2?

    Thanks,

    Jake
     
  5. Apr 13, 2010 #4
    No, and that is what the principle exactly states.

    You can measure precisely one number, with the cost of an uncertain measurement of the second.
     
  6. Apr 13, 2010 #5

    DrChinese

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    You may find it easier to think of this way: You have 2 non-commuting properties, p and q, and choose to measure them to unlimited precision in successive experiments. The question: does that mean it had those values simultaneously? No, the uncertainty principle expresses a different conclusion.

    Now, this has nothing to do with the idea that the act of measurement imparts some specific change to the particle as a deterministic interaction (like billiard balls). If it did, you could consider the change to the measurement apparatus and deduce what the change was. But that won't work.

    In fact, pairs of entangled particles respect the HUP. Measure p on Alice and q on Bob. What then happens when you measure q on Alice and p on Bob? You see the HUP in action. That would not be true if you could measure Alice and Bob without consideration of the HUP.
     
  7. Apr 13, 2010 #6

    dx

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    It may be worth mentioning that when one says that "the position and momentum of a particle cannot be measured simultaneously", we are not imagining a picture of a classical particle which has an objective momentum and position in the classical sense which cannot be simultaneously "known", but are in fact saying that the ordinary classical picture has a "latitude" where in any given experimental situation the use of certian space-time relations in effect excludes the simultaneous application of energy-momentum relations, and therefore makes room for the existance of the non-classical quantum-mechanical regularities, which cannot be visualized in the customary manner without leading to contradictions.
     
  8. Apr 13, 2010 #7
    Although the quantum picture looks very counter intuitive, you can still see some problems with the classical picture from a purely intuitive point of view.

    If there were no uncertainty relation then you could take a particle in a box and store an infinite amount of information in that system with the particle having only a finite amount of energy.

    It can also be shown that the laws of classical mechanics allow you to build a computer whose clock cycle halves every cycle, so you can do an infinite amount of computations in a finite time. This together with the fact that you have an infinite amount of memory space available, makes it possible to build a computer that can settle whether or not certain mathematical theorems are true or false, even if they are unprovable, using brute force alone. It would even be able to check whether or not a proof exists.
     
  9. Apr 13, 2010 #8
    Was that the motivation for Beckenstein re the Beckenstein Bound?
     
  10. Apr 13, 2010 #9
    The Bekenstein bound is obtained by going a step further. So, quantum mechanics implies that in some volume containing one particle, you can store only a limited amount of information using less than some given amount of energy. Also this amount of information depends on the mass of the particle you use. So, you can still store an arbitrary amount of information if only you can use arbitrary large amounts of energy. However, there is then still a limit due to General Relativity. Above a certain limit, the volume will collapse into a black hole.

    It turns out that the maximum amount of information you can store in a volume if the amount of energy you can use is unrestricted, is proportional to the surface area bounding the volume. This is the Bekenstein bound.
     
  11. Apr 13, 2010 #10
    Ahhhh, and this is why you can infer that a BH doesn't have a "bound breaking" remnant, based on the area of its event horizon?
     
  12. Apr 13, 2010 #11

    I hope you don't mean to say you can measure the position and momentum of a single particle to any precision simultaneously.
     
  13. Apr 13, 2010 #12

    Matterwave

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    Are simultaneous measurements even allowed in QM? Can you in fact measure 2 quantities at the exact same time? I always thought in QM you had to measure 1 first and then measure the other one - hence the operators representing measurements kind of thing.
     
  14. Apr 13, 2010 #13
    yes you can in Principle, the HUP is about statistics, i.e many measurements.

    Just have a look at the derivation of HUP and the definition of standard deviation which is what the delta x and delta p are referring to.
     
  15. Apr 13, 2010 #14
    You should also just go through the derivation of HUP and the definition of standard deviation again.
     
  16. Apr 13, 2010 #15
    I see what your saying I think. But explain this to me...

    If you make both measurements on position and momentum, at the same time... one of the postulates of quantum mechanics says that measurement leaves the particle in an eigenstate it was observed to be in, but the position and momentum eigenstates are not the same. Are you telling me the particle can be observed to be in both a position eigenstate and a momentum eigenstate at the same time? What exactly are you saying?
     
  17. Apr 13, 2010 #16

    Matterwave

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    My question is less to do with the HUP than the simple question of whether one can in fact make simultaneous measurements of 2 observables. I suppose one technically could, since measuring x will allow the measurement of x^2 or some such...but I'm thinking more in terms of "independent" (at least explicitly, perhaps not implicitly) operators.
     
  18. Apr 14, 2010 #17
    But as been told, it is possible to have wavefunctions which have such behaviour under such application of operators.

    Can you in classical dynamics measure two things at the same time? the position and the momentum? No you need two position measurments to determine the momentum.
     
  19. Apr 14, 2010 #18
    To see the difference between exp[i k x] and exp[i (k + delta k) x] requires one to consider a range of length 1/delta k. So, to get more and more precise momentum measurements, you need to use larger and larger measurement devices.
     
  20. Apr 14, 2010 #19
    ansgar,

    Ah yes, I think I see now. Let's say the position of a particle is measured to high accuracy. This leaves the particle in an eigenstate such that the value obtained in a successive measurement of the momentum is uncertain to a degree in accord with HUP. However, one can still measure to arbitrary accuracy what this momentum is. In an ensemble the values measured of position and momentum will have a distribution with a standard deviation given by the HUP.

    And just to be clear for myself, if the operators of two observables, p and q, commute , then they share common eigenstates and it is possible to measure p and know with complete confidence what successive measurement of q will be.

    Tell me if this is the correct way of looking at it. Thanks.
     
  21. Apr 14, 2010 #20
    This is a great point. If momentum is m(dx/dt) and the object's position is measured to an accuracy so small that even the distance dx goes to 0 the term momentum no longer has any meaning.

    This is also a good quote:

    "In much the same way that the existence of perpetual motion
    machines is ruled out by the Second Law of Thermodynamics, the existence of an
    apparatus which would give a precise determination of position and momentum is in
    conflict with the nature of physical states in quantum mechanics."

    I would also like to point out that this was always a consequence in wave mechanics equations far before QM came about.
     
  22. Apr 14, 2010 #21

    DrChinese

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    It is possible to know 2 COMMUTING observables simultaneously, but not 2 NON-commutung observables.

    I guess a good example would be photon polarization and frequency, which commute. You can know these to unlimited precision simultaneously.
     
  23. Apr 14, 2010 #22
    Is that simply because they share eigenvectors?
     
  24. Apr 14, 2010 #23
    on the question of simultaneous measurements...

    on the American Institute of Physics webpage... http://www.aip.org/history/heisenberg/p08a.htm ,

    they say ...

    "The uncertainty relations may be expressed in words as follows.

    The simultaneous measurement of two conjugate variables (such as the momentum and position or the energy and time for a moving particle) entails a limitation on the precision (standard deviation) of each measurement. Namely: the more precise the measurement of position, the more imprecise the measurement of momentum, and vice versa. In the most extreme case, absolute precision of one variable would entail absolute imprecision regarding the other."

    i'm thinking the statistical interpretation is just another way of looking at HUP

    Basically..but there is a subtlety. Even if they commute, the particle must be in an eigenstate of both observables if you are to measure simultaneously the values to arbitrary precision. If they don't commute, then this can never be the case.
     
    Last edited: Apr 14, 2010
  25. Apr 14, 2010 #24
    We are still confused about the difference between a single measurement and repeated measurements. A single measurement has nothing to do with uncertainty! There is no way to determine the uncertainty when we have only a single measurement.

    Many repeated measurements give many different results. If we repeat the experiment enough times, then we obtain the entire eigenvalue spectrum of the observable being measured. The probability distribution of all those values can be used to determine the uncertainty in the observable being measured.

    Position and momentum can be measured simultaneously with any precision allowed by the measuring apparatus. The uncertainty principle does not say that [tex]p_y y \ge \frac{\hbar }{2}[/tex], where [tex]p_y [/tex] is the measured value of momentum and y is the measured value of position Rather, it does say that [tex]\Delta p_y \Delta y \ge \frac{\hbar }{2}[/tex], where [tex]\Delta p_y[/tex] is the uncertainty in momentum and [tex]\Delta y[/tex] is the uncertainty in the position.

    Consider a slit experiment where a photon passes through slits and is detected on a distant screen. We see a dot on the screen at the position of the photon. Thus, we know the photon's position at the instant it hit the screen. But, we also know the momentum [tex]p_y = p\sin \theta[/tex] by measuring the scattering angle [tex]\theta [/tex]. But these single values of position and momentum have nothing to do with uncertainty or the uncertainty principle.

    Now, if we repeat the same slit experiment many times we get many different values for [tex]p_y[/tex]. There is a statistical distribution of those different values and the standard deviation of that statistical distribution is what we call the uncertainty in momentum for this experiment. Likewise for the position.

    Many reject the projection postulate. It is more of an interpretation than a postulate. There is no need to know the state of the photon after it is detected. Usually the photon is lost in the detection material after detection and we have no idea what its quantum state is.

    What we should say is, "there is no experiment with a state function that is an eigenfunction of both momentum and position". There is no preparation procedure for which repeated measurements always yield the same momentum value and the same position value. For such a case [tex]\Delta p_y = 0[/tex], [tex]\Delta y = 0[/tex], and the Heisenberg uncertainty principle would be violated. We believe this to be impossible and, up till now, all experiments agree.

    Finally, a specific answer to your question:
    No, we cannot observe a particle to be in an eigenstate of momentum and in an eigenstate of position at the same time.
    (But, then you must reject the projection postulate.)
     
  26. Apr 15, 2010 #25
    The measurement process must leave the system in one of the eigenstates of the observable operator. States with no uncertainty are eigenstates of Hermition operators.

    If you measured two observables simultanously the state would be left with zero uncertainty in both observables. The physical state must then be in an eigenstate of both observables simulatanously following the measurement.

    In the case of position and momentum, there are no physical states that are eigenstates for both x and p.
     
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