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I Is entanglement severed by wave function collapse?

  1. Jan 25, 2017 #1
    What has confused me for a long time is the interaction between superposition and entanglement. That is, what happens when one member of a pair of entangled particles passes through a filter that selects for an observable that is incompatible to the observable in which the pair is entangled?

    Say we have a pair of particles entangled in some observable A, so that one is definitely in the state A+ and the other is definitely in the state A-, we just don't know yet which is which. Let one of those particles pass through a filter that collapses the wave function of the particle to a definite state of B, an observable incompatible to A, and that resulting B particle is then passed through an A filter that collapses the wave function to a definite state of A which we also do not yet know. Now we look at both particles, the one original and the one passed through the filters. What state pairs will we see for this pair of particles? Will we ever see ++ or --? If we do, then it would seem that collapsing the wave function to a definite state of B severs the entanglement in A between the first two, original particles.

    My question is: Do the results of actual experiments indicate that the entanglement is severed in this situation or does it survive so that we still will only see +- & -+ state pairs?

    Based on posts I've read, it seems that the entanglement will be severed but I'm just not sure I've correctly interpreted what I've read.
  2. jcsd
  3. Jan 25, 2017 #2


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    One technical meaning of entanglement is this: A composite system in a pure state is entangled if the state cannot be written as the product of the states of the components. The prototypical case is an electron/positron pair with total spin zero. The composite state (only considering spin) can be written as: [itex]\frac{1}{\sqrt{2}}(|u\rangle |d\rangle - |d\rangle |u\rangle)[/itex], where [itex]|u\rangle |d\rangle[/itex] is the state in which the electron is spin-up in the z-direction, and the positron is spin-down, and [itex]|d\rangle |u\rangle[/itex] is the state in which the electron is spin-down and the positron is spin-up. The composite system is in a superposition of these two states. So it doesn't make any sense to talk about the state of the electron--it doesn't have a (pure) state, or the state of the positron. When two systems are entangled, there is a composite state, but there is no state for either system, separately.

    Now, if you measure the spin of the electron, and find it to be spin-up, then you immediately know that the positron was spin-down. So after the measurement (in the collapse interpretation, anyway), the system would be described by the state [itex]|u\rangle |d\rangle[/itex]. That state is not entangled, because the electron and positron both have states.

    It might not always be the case, but under the collapse interpretation of QM, measurement can destroy entanglement.
  4. Jan 25, 2017 #3


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    This description of things leaves out the state of the device (or person) who did the measurement. If you treat the observer/measuring device quantum-mechanically, as well, then a measurement involves even more entanglement: the state of the observer/measuring device becomes entangled with that of the system being measured. If you assume collapse, then the collapse destroys the entanglement (some people believe that there is no physical collapse, which means that in general, things just keep getting more and more entangled, until eventually everything is entangled with everything else).
  5. Jan 25, 2017 #4


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    Sure. If you performed a non-commuting spin measurement, for example, there would no longer be a entangled state for the initial observable between the 2 original particles.
  6. Jan 25, 2017 #5
    Thank you both very much for your responses. They are very helpful. I still have a lot to learn, but this seems to have cleared a few clouds out of the way. Thanks.
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