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Entanglement problem

  1. Feb 9, 2009 #1
    I'm new to this, so I can only express and understand simple language (as far as quantum mechanics is concerned). Please try to keep the replies as less technical as possible.
    Consider a simple example: Two particles ( A and B ) are ejected in opposite directions, entangled in a way such that their spins along any axis are along opposite directions. Consider only spins along x and y axes for this.

    When the x-spin of A is measured, according to the uncertainty principle, info about the y-spin of B is lost, as the x-spin of B is now known. So if the y-spin is measured after, it can have ANY possible value.

    NOW THE PROBLEM : If the x-spin of A, and y-spin of B are SIMULTANEOUSLY measured, wouldn't we get to know both the x and y spins of both particles, thereby violating the uncertainty principle?

    (And if this kind of simultaneous measurement is impossible, how is it so?)
  2. jcsd
  3. Feb 10, 2009 #2
    No, can know for 2 particles - not a problem. Problem for 1 particle - no good.
  4. Feb 10, 2009 #3
    Hello, acxler8

    In quantum physics, simultaneous measurments on entangled systems are modelized as if the measurments took place immediately one after another.

    You can suppose that the x-spin of A is measured fisrt. The x-spin of B is then known. Now, the two particles are no more entagled. Thus the y-spin measurment on B takes place, but gives no information on the y-spin of A. It also destroys the information that A had previously got about the x-spin of B. After both measurments, the operators A and B have information about their own particule, and no information about the other.

    We can also suppose that B performs his measurment just before A. The intermediate steps are reversed in regard to A and B, but the result is the same. A knows the x-spin of her particle, B knows the y-spin of his particle, but both don't know the spin of the other.

    Both descriptions of the intermediate steps lead to the same predictions, therefore they are equivalent, as long as they lead to no observable difference.
  5. Feb 10, 2009 #4


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    I know this is tricky but you are not violating the uncertainty principle.

    First be careful about thinking of the spin of the particle as being an encoded state. Stick to only the language of describing what you know about how a certain outcome of a given measurement will go.

    Measuring y-spin of A tells you that IF you measure the y-spin of B you will know the outcome. If you don't measure the y-spin of B then there is no outcome to know.

    If you measure the x-spin of B then you no longer know what a subsequent y-spin measurement of B will be (satisfying the uncertainty principle).

    If it helps you can also take the A vs. B variable and make it an observable, then mix things up. You can describe "the y-spin + particle" vs "the y-spin - particle" and speak of each as being in dual anti-correlated superpositions of going in the A vs. B directions.
    You thereby factor the two particle system into different "this particle" and "that particle" parts.

    What this shows us is that it is not quite proper to think of "the B particle having given (unknown) spin state" as we can also think of "the y-spin + particle" as being in a superposition of A and B position "states". Avoid the "state" mindset and stick to what will happen when a given observation is made.

    Then remember that assuming a counterfactual i.e. "if instead of measuring x-spin for B we had measured y-spin" you must also reset the assumption of a given outcome for the measurement of particle A. Holding the actual outcome of the A measurement fixed and assuming you are talking about the very same experiment while changing facts about what you measure in the B case is what leads to the seeming conflict. Its subtle I know but vital to correctly understanding the logic of "what we know" and the implications of the uncertainty principle.

    This subtle error comes from our thinking in objective terms. As long as we are working with objective states of a classical system we can play the counterfactual game because the systems could have been measured completely prior to the experiment and the outcomes predetermined and independent of changes in assumption about what is measured.

    But in the quantum case one is not describing objective states but rather the state of knowledge about the system. With this higher level of abstraction changing assumptions does change our assumed knowledge which is the subject of the logical inferences.

    It is like the paradox "If we know that 'we know nothing' then we know something!" Since the logic is applied to our knowledge and not classically objective reality we must be careful to stay within the confines of the epistemological definition of "knowledge" to avoid semantic tangles. In the case of QM "what we know" is only "what we measure" and we must avoid assumptions about knowledge independent of acts of measurement. This is what is meant in the Copenhagen interpretation about "no underlying reality" one must completely excise "reality" language from the description of the quantum system.

    Unfortunately the term "state vector" is still used in texts. If you parse this as "state of our knowledge"-vector then you do OK. My thesis advisor was careful to always refer to it as a "mode vector" as in mode of production for the system (and dually mode of registration). It helps if you go ahead and work in the density operator formulation where (in addition to a more general description) it is easier to see the density operator as the quantum version of a probability distribution for the system i.e. obviously an encoding of knowledge about outcomes instead of a representation of the physical system's state of reality.

    I.M.N.S.H.O. the various other interpretations of QM e.g. Bhom's pilot waves and Everett's many-worlds are attempts to reconcile the predictions of QM with the mistake of reification of the wave-function/"state"-vector/mode vector and the improper application of the logic of our knowledge about system outcomes.
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