Photons and decoherence

  1. I read it some where that there is very small decoherence for photons. The reason being that photons do not interact with each other (Is that because photons are chargeless, colorless and flavourless particles?) and hence the information that they contain tends to stay with them. They have a small cross-section. However if you consider the compton scattering, there the photon is interacting. Secondly, in vacuum, a photon interacts with the quantum fluctuations, right? So how is it that there is little or no decoherence for photons?
  2. jcsd
  3. mfb

    Staff: Mentor

    Decoherence is not a property of particles, that statement does not make sense.
    Their interaction cross-section is not zero, but tiny, for the reasons you noted.
    For photon-photon interactions.
    Sure, as an electron is charged.
    That does not lead to decoherence, independent of the setup.
  4. If I am considering a qubit make out of single photon and I say that decoherence is small for photons, that would not make sense?
  5. ZapperZ

    ZapperZ 29,763
    Staff Emeritus
    Science Advisor
    Gold Member

    It would only make sense if it is in relations to other photons! Coherence is a collective property! It isn't a property of individual particle! Look it up!

    Better yet, show me how I would measure the "coherence" of a single photon, isolated photon.

  6. You don't have to scream at me. I was considering photonS.
    I was confused about "Decoherence is not a property of particles". Electrons are elementary particles. They interact with each other and lose information. Photons are particles too, no?
  7. secondly, why doesn't the interaction of photons with vacuum fluctuations "... lead to decoherence..."?
    Also, in the presence of charge, there is vacuum polarization. Do these vacuum fluctuations happen only in the presence of charged particles?
  8. ZapperZ

    ZapperZ 29,763
    Staff Emeritus
    Science Advisor
    Gold Member

    That wasn't screaming. That was emphasizing since it appeared that you missed it. THIS IS SCREAMING!

    Yes, but you misunderstood what that meant. If you look at the Particle Data Book, you'll see that "coherence" is nowhere to be found in the description of each of these particles. Even your electrons! So what are you disputing here? That is what is meant by the fact that it is not a property of the particles.

    You will also notice that there is no description of its temperature, it's pressure, etc.. etc. Yet, they are particles and only in a collective state (and no, I'm still not screaming) would the properties of coherence, temperature, etc. make any sense. So again, what are you disputing here?

  9. I completely understand that coherence is experienced by a collective state only. I misunderstood what mfb had written.
    Please just confirm if the following claim of mine is correct or not: "Photons do not interact much with the matter because they do not have color, flavor and charge. The interaction of photons with themselves is also small. So, in presence of photons or any other matter, photons do not lose information (as their probability of interaction is small)."
  10. UltrafastPED

    UltrafastPED 1,919
    Science Advisor
    Gold Member

    This is incorrect ... photons do interact with matter because they interact with the electric charge.

    Where did you get this from?
  11. Okay, so I read the following:

    "Single Photons as Qubits
    Single photons are largely free of the noise or
    decoherence that plagues other systems..."


    I am trying to come up with reasons to justify the decoherence point.
  12. If I am not mistaken, (contrary to what has been expressed in previous posts) coherence very often involves a single particle. Coherence very often means a stable phase shift between two states of the same particle, which in turn implies indistinguishability of both states. In the case of an electron or neutron, after going through a very small double slit on a screen, its wave would start to spread and we would have a large uncertainty in its position. Up until this point, and assuming some conditions on the way the particle approached the double slit, there would be coherence between the states corresponding to having passed through each slit. As soon as the particle interacts with something macroscopic, in a way that path information can be recorded, coherence is lost. In the case of the bubble chamber, decoherence would happen very often and we observe a narrow path.
    In the case of photons, the interactions we normally think about are emission and absorption of the photon. So in this case we don't have a situation in which the photon bumps into something in such a way that we can detect where it is and the photon keeps going. Once we know where it is, it has been "destroyed" (absorbed). If we shoot the photon into a space where there are absorbing particles, then the photon will be absorved very quickly, but if we had the photon travelling through an optical fiber with low absorption for that wavelength, it would have a better chance to survive and mantain coherence until it is detected. Probably possible interactions that would not absorbe the photon, such as bending of the path along the fiber or passing through a beam splitter, don't record (in a permanent way) information such as path information or polarization, and therefore don't imply decoherence. So it seems that there is a close relationship between distinguishability of the states and coherence between them. It would be interesting to have some simple example showing how distinguishability implies dephasing.
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook

Have something to add?

Draft saved Draft deleted