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Entanglement puzzle

  1. Mar 27, 2007 #1
    A friend of mine wrote a paper on entanglement, measurement, and simultaneity.

    the professor provided an interesting question, which he felt demonstrated a violation of special relativity. Of course, this is most likely due to a lack of understanding on the professor's part, but neither I nor my friend know enough of entanglement to say what is wrong exactly.

    Thought experiment:

    Suppose you have two particles (A and B), each with two observables X and Y. A and B are in entangled states of X, but in pure states of Y. Moreover, these observables, X and Y, are non commuting.

    Suppose an experimenter performs rapid and succesive measurements of observable Y on particle B. Then at some point the other observer makes a measurement of X on particle B.

    Since the X states of both particles are entangled, a measurement of X on A will determine the state of B.

    The professor argues that by measuring X on particle A this would create a change in the Y state of particle B (remember X and Y don't commute) (which is continually under measurement by the other experimenter) - and hence the observer of particle B will know that observer A has measured the state of his particle - a clear transmission of information instantenously.


    It seems to me this argument relies on a faulty understanding of entanglement. I suspect that it's not possible to create a system in which only two of the variables are entangled - however none of my texts mention anything useful about entanglement. Griffith's only discusses entanglement of spins, but doesn't say one way or another whether or not the total angular momentum state has to be entangled as well, for instance.

    Can someone guide me to some literature that will help illiuminate this argument?
     
  2. jcsd
  3. Mar 27, 2007 #2

    George Jones

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    This isn't possible. An entangled state is a state that can't be written as a (tensor) product of a state of A with a state of B. Either a state of the system is entangled or it isn't; the state can't be entangled with respect to some one-particle observables and not others.

    For example, considered an unentangled state of the two-particle system that is the product of a linear combination of the eigenvectors of Y_A with a linear combination of the eigenvectors of Y_B. By a simple change of basis, this same state is the product of a linear combination of the eigenvectors of X_A with a linear combination of the eigenvectors of X_B. The state is still unentangled.
     
  4. Mar 27, 2007 #3

    vanesch

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    Yes, I agree with George Jones here. When we talk about entangled systems (particles for instance), this means already that we have defined two distinct hilbert spaces for both, HA and HB. The total Hilbert space is then HA x HB (tensor product). For instance, if they represent two distinct particles, then HA will be the hilbert space spanned by the positionstates (or momenta, it is the same space) of the first particle, and HB will be the hilbert space spanned by the position states of the second particle.

    Now, an entangled system is a system whose state is an element of HA x HB, but not writable as a |a> x |b> state, but rather a superposition thereof.

    An observable of the first system, is an observable with the form: X = O1 x 1. O1 acts upon HA, and the unit operator acts on HB. An observable of the second system is of the form Y = 1 x O2.

    As such, if X is an observable "of the first system" and B is an observable "of the second system", they commute. No choice.

    Now, of course there are observables on the total space HA x HB which do not commute, but they will not be of the form O x 1 or 1 x O. They will be of the form O1 x O2 or even combinations thereof. As such, they represent *correlation measurements*, and not, individual measurements on system A or on B. These are necessarily commuting. It is easy to show:

    [ O1 x 1, 1 x O2 ] = (O1 x 1) (1 x O2) - (1 x O2) (O1 x 1) = O1 x O2 - O1 x O2 = 0.
     
  5. Mar 28, 2007 #4

    DrChinese

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    Gents,

    I am not disagreeing with what you are saying, but I wanted to be clear about it. You are saying that a pair of photons (say) cannot be entangled as to momentum (say) at the same time they are in a known state as to polarization... do I have it right?
     
  6. Mar 28, 2007 #5

    DrChinese

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    And the reason I am asking: I thought it WAS possible to have some observables entangled while others are not.

    Specifically: Indiviual Type I PDC crystals produce photon pairs with known polarization. Normally there are a pair of crystals and they are at 90 degree angles to each other. In this configuration, with a suitable phase shift for the source laser, they come out polarization entangled (the output streams of the crystals are mixed). But if one of the crystals is removed (so there is a single PDC crystal), then the output stream is not polarization entangled. But I had assumed it would be entangled as to the momentum observable still. Guess not?
     
  7. Mar 29, 2007 #6

    George Jones

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    In this case. each one-particle state space is itself a (tensor) product of polarization space with momentum space, which means that polariaztion entanglement is independent of momnetum entanglement. It also means that polariztion and momentum measurements, in principle, commute, which contradicts a condition in the original post.

    Independent entanglement means also that both momentum and polarization can be entangled - hyperentanglement - but you probably know more about this than do I.
     
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