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Boson statistics

  1. Sep 29, 2012 #1
    It is said that is there already n bosons in a particular quantum state, the probability of another boson joining them is (n+1) times larger than it would have been otherwise. But if we apply this rule to calculate probability for one horizontally (H) polarized photon to join a bunch of n=99 diagonally (D) polarized photons, we'll get (99+1) * |< D|H>|^2 = 100 * (1/2) = 50. Why does the rule fail?
    Last edited: Sep 29, 2012
  2. jcsd
  3. Sep 29, 2012 #2


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    I think it doesn't fail, so I'm sure someone will explain to me why I'm wrong! :smile:

    To make it easier, let me switch Horizontal and Diagonal. The rule is a*|n> = √(n+1)|n+1>.

    Suppose we have a state with 99 Horizontally polarized and zero Vertically polarized photons, written |99,0>. And we add one Diagonally polarized photon to it. The creation operator for a diagonally polarized photon is aD* = (aH* + aV*)/√2. We get
    aD*|99,0> = (aH*|99,0> + aV*|99,0>)/√2
    = (100|100,0> + |99,1>)/√2.
    Last edited: Sep 29, 2012
  4. Sep 29, 2012 #3

    Ken G

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    How does that argument work? Sounds like some kind of stimulated emission process, but I would have thought of at as a physical process that is stimulated by the n photons, not a statistical probability effect.
  5. Sep 29, 2012 #4


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    This should be root of n+1.
  6. Sep 29, 2012 #5


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    What does "otherwise" mean exactly?
  7. Sep 29, 2012 #6
    "What does "otherwise" mean exactly?"
    This is a good question, and the one that I think lies at the root of the matter. If you read almost any textbook, they will simply say "the probability increases". My understanding is, what increases by (n+1) is the conditional probability of transition from |H> to |D> given that the same transition was made by n other bosons. Or formally,

    Prob(D | n photons in state H) = (n+1) Prob (D | zero photons in state H)

    (where the vertical bar means "conditioned on"). Is that the right way to put it? And if yes, would that mean the ONLY way to resolve the paradox is to say that conditional probability on the right is NOT equal 1/2?
    Last edited: Sep 29, 2012
  8. Sep 29, 2012 #7
    I replaced 100 with 10 as amplitude is given by sqrt 100 (not that it changes much). If the probabilities were to relate as 100:1, that would solve the problem. However, what happened to normalization? Looks like the amplitudes 10/√2 and 1/√2 must be attenuated by a normalizing constant?
  9. Sep 30, 2012 #8


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    It is not that easy. Bosonic final state stimulation results in the probability amplitude for processes which lead to bosons ending up in the same state and being indistinguishable to be proportional to n+1. The n is the stimulated emission/scattering/whatever effect, while the 1 is the spontaneous emission/scatteing/whatever.

    Photons do not just join and you cannot just add some photon to 99 photon state by will. You need to define some process which does. So you first need to know how you want to make your horizontally polarized photon interact with something that can turn it into being in the same state the photons with horizontal polarization are in. This might be absorption and reemission or something similar. Just superposing the photons or arranging them in some random fashion on a beamsplitter does not work.
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