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I Condition for quantum entanglement?

  1. Mar 23, 2016 #1
    Somebody told me that the condition that must be met for Quantum Entanglement in a system, is that the sum of the wavefunctions of the individual particles must equal the overall wavefunction of the system. But isn't this the case anyways with any system of two particles whether they are entangled or not?
  2. jcsd
  3. Mar 23, 2016 #2
    It is the correlations between subsystems of a quantum system (a particle in a system of two particles)
    that can give rise to distinction from classical systems(correlations in classical systems can always be described in terms of classical probabilities),
    whereas ;
    this is not always true in quantum systems.
    Such non-classical correlations lead to entanglement
    Let us take a two particle system whose individual particle states are pure states and the measurement done on each particle is independent of the such measurements done on the other- meaning the states are not correlated - no entanglement.
    But if the two particle states are entangled in a manner such that a local measurement of particular observable done on particle 1 -predefines the result of such measurements on particle 2-
    then we get the ' entangled states' of the system
  4. Mar 23, 2016 #3


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    That's a very bad start.

    That doesn't make sense. Have a look at the Wikipedia page on quantum entanglement for starters. The first sentence is very good:
    For two particles in a pure state, they will be entangled if the state can't be written as the product of the state of particle 1 with the state of particle 2.
  5. Mar 23, 2016 #4


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    Staff: Mentor

    "Somebody told me" is not an a good reference under the Physics Forums rules; if you can't tell us where you heard it we can't help you understand it. However....
    I'm not sure what this could mean - you can't just add vectors in different Hilbert spaces to get a sum. You might give the first few pages of http://www.pa.msu.edu/~mmoore/Lect24_TensorProduct.pdf a try.
  6. Mar 23, 2016 #5


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    No, that is not always the case. There are a couple of general conditions:

    1. Usually there is a conserved quantity; in that limited basis you could say that x1+x2 sum to something that is a constant.
    2. A pair of particles with known/definite property values will not be entangled on that basis. So x1 and x2 cannot be definite prior to measurement.

    So a typical system of 2 particles will, if prepared reasonably, meet condition 1. But it won't meet condition 2 unless you make non-commuting measurements on both. And then it will meet condition 2 but no longer will meet condition 1 (because the values are indefinite and the sum is not a constant).

    On the other hand: entangled pairs meet both requirements, and generally any pair of particles that meets those requirements are entangled on that basis. I am sure there are specific exceptions, hopefully you get the idea.
  7. Mar 23, 2016 #6
    Do the electrons that share any particular orbital or sub-orbital of an atom satisfy these two requirements?
  8. Mar 27, 2016 #7


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    If the joint quantum state comes from a pair of independent non-interacting (but otherwise arbitrary) particles, then the joint state is separable.
    If the joint state is an arbitrary mixture (not superposition) of such states, then it is also separable.
    All other states are entangled.
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