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Quantum Interference of atoms

  1. Apr 10, 2015 #1
  2. jcsd
  3. Apr 10, 2015 #2
    They travel in the same direction from the Bose-Einstein condensate source towards the Bragg mirror, but with variable spacing. They observe the interference effect when the spacing is very small.
     
  4. Apr 10, 2015 #3
    Thanks Craigi, it wasn't clear as it stated they are fired at opposite sides of the mirror. Does it mean one hits the front wall and one hits the back wall then?
     
  5. Apr 10, 2015 #4
    Perhaps I posted too hastily. This should clear it up:
    http://en.wikipedia.org/wiki/Hong–Ou–Mandel_effect

    This is the effect that they emulated with atoms.
     
  6. Apr 10, 2015 #5
    That's exactly what I was looking for! So is the narrative saying that diagrams 1 and 4 can happen and 2 and 3 cannot (for certain momenta/mirror thickness combinations)?
    Very cool!
     
  7. Apr 10, 2015 #6
    It appears to me with a 50/50 splitter any of those 4 possibilities are equally likely, just that 2 and 3 are indistinguishable from each other... whether both photons reflect or not the results are the same.

    What I am unclear about is how that is related to the dip caused by indistinguishable atoms?
     
  8. Apr 10, 2015 #7
    Indistinguishable atoms undergo interference in the same way as the photons in the original experiment. When the path integral is performed, for the particle pair, the paths for particles in seperate directions are cancelled.

    The paths for distinguisable particles don't interfere in this way.
     
    Last edited: Apr 10, 2015
  9. Apr 10, 2015 #8
    Instead of destructive interference the atoms come out in pairs, as in molecules?
     
  10. Apr 10, 2015 #9
    Molecular bonding isn't relevant to the experiment. The atoms are Helium atoms, so they don't form molecules under normal conditions.
     
  11. Apr 10, 2015 #10
    Ok I get it. The lack of bonding is relevant because they are Helium-4 they shouldn't interfere, and only when they are indistinguishable atoms they cause a "pairing" interference where one atom is reflected while the other passes.

    The atoms both just stop moving?
     
    Last edited: Apr 10, 2015
  12. Apr 11, 2015 #11
    So the situations 2 and 3 are identical except for a phase difference, so cannot be observed?
     
  13. Apr 11, 2015 #12
    Draw out wave packets for each case to convince yourself that 2 and 3 correspond to destructive interference and 1 and 4 correspond to constructive interference, but dont forget the phase shift.
     
  14. Apr 11, 2015 #13
    I am uncertain what you have in mind for me to do next. How would I draw a wave packet? What do they look like? I am not sure how cases 1 and 4 can interference as they are different.
     
  15. Apr 11, 2015 #14
    Have you studied Vibrations and Waves yet?

    Are you sure that you understand superposition of waves, constructive interference and destructive interference?

    If not, start here:
    http://en.wikipedia.org/wiki/Interference_(wave_propagation)

    We're not talking about case 1, interfering with case 4. We're talking about of both wavefunctions splitting and taking both outgoing routes. However, interference determines which of the 4 cases result in a non-zero probability of finding an outgoing particle.
     
    Last edited: Apr 11, 2015
  16. Apr 11, 2015 #15
    I must have got the wrong end of the stick as you mentioned constructive inference of 1 and 4 in post #12. Yes I have studied wave mechanics, but I still don't know what you mean about drawing a wave-packet.
     
  17. Apr 12, 2015 #16

    zonde

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    Gold Member

    Yes, 2 and 3 have the same destructive interference and are not observed (are observed considerably less frequently). This in experiments is observed as HOM dip in coincidence counts as you change length of one arm.
     
  18. Apr 12, 2015 #17
    Wave packet description here:
    http://en.wikipedia.org/wiki/Wave_packet

    The envelope overlap of the two packets determines the shape of the dip.
     
  19. Apr 12, 2015 #18
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