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Quantum Entanglement: What would happen if

  1. May 28, 2010 #1
    What would happen if one had two entangled particles and performed a position measurement on one and a momentum measurement on the other? If one kept performing these measurements, perhaps first the position measurement on a particle A, then the momentum measurement on particle B, would the momentum measurement create an ambiguous position for particle A whose position would therefore potientially be different upon subsequent measurement? If not, it seems as though one could measure indirectly both the position and momentum of a particle. Someone please help me with this.
  2. jcsd
  3. May 29, 2010 #2


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    You have just rediscovered the famous Einstein-Podolsky-Rosen paradox.
  4. May 29, 2010 #3
    If the particle is moving of course its position will change on subsequent measurements, If the particle isn't moving its momentum is ~zero (actually ~h), your measurement of the position will be no more precise than ~h each measurement.
  5. May 29, 2010 #4
    And of course, the first measurement process would break the entangled state.
  6. May 29, 2010 #5
    So after entangled particles are measured, they cease to be entangled and subsequent measurements have no bearing on each other? Am I understanding you correctly?
  7. May 29, 2010 #6


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    Yes, that is correct. A position measurement will break the entanglement on position/momentum. It is interesting to note that particles can be entangled on several bases, and you can break the entanglement on one basis without necessarily breaking it on another. Spin for example.
  8. May 29, 2010 #7
    So if I were to measure the position on one particle and the momentum of the other, the measurement of the momentum at that point has no bearing on anything because the particles are no longer entangled?

    Why then, if I were to measure say the spin on one particle which was entangled, would I know the spin of the other particle since in my measuring the spin, I would be breaking the entanglement?
  9. May 29, 2010 #8

    yes, because In measuring the position you execute a nonunitary transformation of the entangled wave-function component governing both particles' position-momentum state. Nonunitary means irreversible (sometimes called "wave-function collapse"), so a further measurement of either property would only apply to each individual particle and not the entangled state.

    As DrChinese mentioned, the partlcles may still be entangled with respect to other conjugate properties, time/energy or angular momentum wrt to orthogonal axes, but I guess you'd need a clever way to do the measurement to ensure that.

    When measuring spin of one entangled particle you fix the spin value of the other particle "instantly" (or at FTL speed), this has been experimentally verified dozens of times. So it doesn't matter if they are no longer entangled, both their spin states are determined.
  10. May 29, 2010 #9
    Is there a way to alter the spin of one particle once you have measured it? If so, would this in turn alter the spin of the corresponding entangled particle?
  11. May 29, 2010 #10
    yes and, no
  12. May 29, 2010 #11
    what do you mean? Yes to the first and no to the second? or yes and no as in this is a fuzzy area or my question is not specific enough?
  13. May 29, 2010 #12


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    Yes, it would. But you cannot change it to a specific value. It will be purely random as far as anyone can tell. That is why an earlier answer said yes and no.
  14. May 30, 2010 #13
    Thank you both for all your help. If I wanted to compose a table of cause and effect for quantum entanglement as in 'if I do this to A, that happens to B' in order to help myself better understand it, do you know some good papers by which these realtions are shown or by which I could infer these properities?
  15. May 30, 2010 #14
    Without classical means of confirmation, you do something to 'A', and what happens to 'B' is not predictable.
  16. May 30, 2010 #15
    By "yes and, no" (notice the comma) I meant that,

    "yes" you can alter the particle's spin after the first measurement, eg by having it interact with a third particle (but you can't predict the outcome, it may be the same afterwards or it may change, as DrC stated)

    and, "no" it would not have any effect on the previously entangled partlcle.

    sorry if that wasn't clear.

    Remember that in QM the outcome of a measurement is a probabilistic result not a deterministic one, at least according to our current model of QM. But by using conservation laws and entangled particles you can know about the unmeasured value of one of the entangled particles by just measuring the other one.
  17. Jun 5, 2010 #16
    Entangled or not entangled

    Suppose a staionary molecule of hydrogen gas dissociates into two hydrogen atoms. Aren't the linear momenta of the two atoms moving in the opposite directions correlated in that measuring the momentum of one of the atoms instantaneously tells you the momentum of the other (non-measured) atom.

    Isn't this situation the analogous as the measurement of the two correlated photons emitted in opposite directions by an excited Calcium atom (the Alan Aspect experiment).

    Yet previous posts on this subject say that once a measurement is made, enanglement is destroyed. I don't see any difference between the two experiments. Measurement of one property: polarization for the photon or momentum for the hydrogen atom, both "disturb "the system and instantly give you information about the non-measured species. Both are entangled systems, aren't they?

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