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Cluster Decomposition in QFT

  1. Nov 5, 2011 #1
    Recently a student brought the following to my attention from Weinberg’s Quantum Theory of Fields, Volume I, from page 177, which I must admit that it stumped me. Here Weinberg introduces the concept of Cluster Decomposition: “It is one of the fundamental principles of physics (indeed, of all science) that experiments that are sufficiently separated in space have unrelated results…” The quantum entanglement a la EPR paradox, however, tells us that the measurement of the polarization of a photon from a [itex]{\pi}^o[/itex] decay measured a year later will determine the polarization of the other photon (two light years away from this measurement.) This is because the decay photons from a [itex]{\pi}^o[/itex] are entangled forever, no matter how far apart they may be. This sort of thing was discussed and dissected endlessly in the past fifty years, but today we simply accept this non-local entanglement as part of Quantum Mechanics.

    So back to Weinberg: what is he saying here then? If you were to restate his words a little more precisely, how would you restate the cluster decomposition?
  2. jcsd
  3. Nov 5, 2011 #2


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    Far be it from me to "restate Weinberg more precisely"! But what he's talking about is a situation in which all of the in states (α1, α2,...), (αjj+1,...) are known and independent. In your pion example the in states α1 and αj are correlated and dependent.
  4. Nov 5, 2011 #3


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    This is only true if they your system of entangled particles is isolated. I think in most practical scenarios, interactions with the environment rapidly destroy this entanglement by decoherence.

    /edit: deleted a scruffy analog
    Last edited: Nov 5, 2011
  5. Nov 7, 2011 #4


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  6. Nov 7, 2011 #5
    Thanks Demystifier! I was not aware of the previous thread of a similar discussion. I agree with you that Weinberg's statement is a little careless (at best). Indeed your restatement in post#7 is almost exactly what I believe the right wording may be.
  7. Nov 8, 2011 #6


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    I'm glad that I've been helpfull.
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