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GHZ entanglement

  1. Feb 21, 2012 #1
    Hi there,

    I was looking through Zeilinger's "Dance of the Photons" where I re-read the section of three-photon entanglement. If one photon is measured in H polarisation, the other two take on H - likewise with V polarisation.

    What I find interesting is if we measure one photon and its H polarisation, another (delayed) at 45 degrees, and the last at vertical - when the first photon takes on H polarisation, don't the other two? So when the last photon meets the V polarised filter, shouldn't it fail? But according to GHZ results, it should be vertical.

    So the other two photons are NOT in H polarisation when the first one meets the polariser? Or the first photon is NOT in H polarisation - evolves as a superposition in accord with Scrodinger equation (no collapse occurs).

    Unless I've mis-understood something?
  2. jcsd
  3. Feb 21, 2012 #2
    I see where I have mis-interpretated it. The author doesn't make it clear under the diagram he is measuring in another polaristion direction before the photon becomes V or H polarised. v' h' and all that stuff just makes it more confusing.
  4. Mar 6, 2012 #3
    If we have the GHZ state: |V>|H>|H>|V> + |H>|V>|V>|H>, if photons 1 and 4 take on polarisation 135, do photons 2 and 3 take on the opposite - 45 polarisation?
  5. Mar 6, 2012 #4
    To find the answer to this, all you need to do is write |H> and |V| in terms of |45> and |135>.
  6. Mar 6, 2012 #5
    So replace |H> with |45> and |V> with |135> ?

    Because expanding |H> to include |45> and |135>, I'm not sure if they have to match (i.e. with the other |H>)
  7. Mar 6, 2012 #6
    No, write |H> and |V> as linear combinations of |45> and |135>. Each of them will have an expression that's something like (1/√2)(|45>+|135>). I don't remember the exact expressions off the top of my head.
  8. Mar 6, 2012 #7
    So |45>|135>|135>|45> + |135>|45>|45>|135> ?
  9. Mar 6, 2012 #8
    No, that's not how it's done.
  10. Mar 6, 2012 #9
    How is it done? Do photons 1 and 4 share the same polarisation, likewise with 2 and 3? (e.g. in |H>|V>|V>|H>).

    Do we expand it as |45>|45>|45>|45>+|45>|135>|135>|45>+|135>|45>|135>|135> etc.?
  11. Mar 6, 2012 #10
    Yes, that's how it's done. We write |H> and |V> in terms of |45> and |135>, and so we expand the whole quantum state as a sum of products like |45>|45>|45>|45>.
  12. Mar 6, 2012 #11
    "Combinations with odd numbers of |45> do not occur".

    But you can write those from |V>|H>|H>|V>+|H>|V>|V>|H>?
  13. Mar 6, 2012 #12
    I told you, just write |H> and |V> in terms of |45> and |135>. Then just expand out the products using the distributive property, like |45>(|45>+|135>)=|45>|45>+|45>|135>
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