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Is position not an observable of a photon?

  1. Jul 24, 2010 #1
    Is position not an observable of a photon?
  2. jcsd
  3. Jul 24, 2010 #2


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    It isn't. For massive particles in relativistic quantum field theories, you can construct a position operator called the Newton-Wigner position operator, but this doesn't work for massless particles. (I'm sure there's a good reason why you can't define a position operator for massless particles, but I don't know it).
  4. Jul 24, 2010 #3
    No, but you can detect the position of its interaction with another particle, the interaction destroys the photon.

    A photon has no rest frame wrt which you could specify its position.

    Theoretically it is not proven that photons move at exactly "the speed of light", so the possibility remains that a very small mass photon exists which does have a rest frame. Then we'd just need to rename the term "speed of light" to be the the speed invariant under a lorentz transformation.

    But in current models we talk about photon number and energy density of the em field rather than photon position.
  5. Jul 25, 2010 #4

    Here I have a question:

    Suppose I emit a light pulse whose length is very short, so we can consider the pulse contains only a few phtons, for example 2 photons.

    If this 2-photon pulse interact with an atom at point(x,t) and one phton is destroyed, can we say that the position of another photon at time t is x?
  6. Jul 25, 2010 #5
    No, you can only say that one photon interacted with an atom at position x at time t
  7. Jul 25, 2010 #6
    Thanks for your answer. But can you tell me why I can't talk about the position of another photon?
  8. Jul 30, 2010 #7
    Neutrino also move at speed c. Can we specify its position?
  9. Jul 30, 2010 #8
    Well you can talk about the position of another proton; actually you can talk about the position of all protons. Except the fact that you would have to be outside the space-time continuum to do that. It's like a 2D and a 3D plane. The 2D plane doesn't know that 3D plane exist and the 2D plane cannot see all the interactions of 2D plane because it is within the 2D plane itself. It is interacting with 2D plane, it within 2D plane so it's ability to see mupliple positioning of proton becomes limited, in fact imposible. In 3D plane, however, we can see the interactions of protons in a 2D plane. We can see them all. Why because we are no longer percieving reality from a 2D plane. However, not that we steped into 3D plane, we cannot see all the posibilities of 3D plane until we move into 4D plane and so on. think of it like a Russian doll!

  10. Jul 30, 2010 #9


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    The Newton-Wigner position operator is not the only way to construct a position operator in relativistic quantum physics.
  11. Jul 30, 2010 #10
    I am not pretty sure about it, but

    Suppose you have a one slit experiment. The picture bellow shows the situation. If the green dot in the ecran is a detected photon spot, at time t2, then it seems that ona can infer, using simple c velocity kinematics, that at a given time t1 < t2 the photon was inside the slit and therefore, has a certain position, with a certain reality content.

    View attachment quantum fig1.pdf

    Best Regards

  12. Aug 1, 2010 #11
    We are talking about the position of the photon, not proton.
  13. Aug 1, 2010 #12

    Can you show us some examples of other position operators?
  14. Aug 1, 2010 #13
    Hi All.
    We can say something about position of light, e.g. trace of beams, focusing light by rens in geometric optics.
    In these cases threshold or limit of position observation is h'/p =λ wavelength.
    Last edited: Aug 2, 2010
  15. Aug 2, 2010 #14


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    For example, the spacial part of the spacetime position operator in
    http://xxx.lanl.gov/abs/0811.1905 [Int. J. Quantum Inf. 7 (2009) 595]
  16. Aug 2, 2010 #15
    One good reason is that photons are masseless and move at the speed of light and have no rest frame! Then also they are bosons, so you can't tell which are which.

    Also, the position operator is a very fishy concept in relativistic quantum physics, since at Compton wave-length there are no single particles anymore, but particles pop in and out of existence. Relativistic quantum physics is a multiparticle theory, that's why the position operator is completely abandoned and instead becomes a parameter to a quantum field. (At least thta's the story in all the standard textbooks and lecture notes.)
  17. Aug 2, 2010 #16
    The position operator for a photon is a very big problem. A good review article on the subject is
    by Ole Keller

    Another person who has done a lot of work on the subject is Margaret Hawton.

    It's hard to count the problems with this idea. In general a non relativistic (schrodinger) approach is not good for photons (they travel too fast), and in relativistic quantum (field) theory there is no concept of single photons. In general position is a problem in relativistic quantum theory, but it's much more of a problem for photons. It turns out that even during the emission process the photon is smeared over a volume larger then the Bohr radius of the atom.

    Other problems are that attempts at a wigner like position operator give a position operator with non commuting parts (i.e X does not commute with Y). Another approach gives a well localized electric field if the magnetic field is non localized and vice versa. Remember that a photon is an electromagnetic wave.

    Hope this helps
    Last edited by a moderator: Apr 25, 2017
  18. Aug 3, 2010 #17


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    I think neither of these arguments is a good argument against position operator of a photon. Let me explain.

    A classical massless particle also moves with the velocity of light. Does it mean that its position is not well defined? Of course not.

    In nonrelativistic QM, the position operator is well defined even for symmetric (bosonic) wave functions.

    In relativistic quantum physics, particles pop in and out only if there are interactions. This does not happen for free particles (Klein-Gordon or Dirac equation). Even with interactions, the vacuum and the lowest-mass one particle states are stable.

    Relativistic quantum mechanics (Bjorken and Drell I) can be well interpreted as a single particle theory, or a theory with a fixed number of particles. It is INTERACTIONS, not relativity, which causes physical particle creation and destruction.

    (Yes, you are right that this is explained so in many introductory textbooks, but these explanations are wrong.)
  19. Aug 3, 2010 #18
    Regardless of interactions there is no position operator for photons. True localization is less of a problem for free photons, but the operator ( d/dp for example) does not work. Other choises can be constructed for spinless (massless) relativistic particles but a photon position operator is still a problem. The problem is however much reduced in the case of no interactions where some kinds of position operators can be defined (such as the one Hawton uses) . There is as far as I know no cannonical photon position operator.
  20. Aug 3, 2010 #19


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    This is simply wrong. A one-photon state is a well defined state in the Hilbert space of relativistic quantum field theory. Even if the number of photons change due to interactions, that only means that the state is a superposition of states with different numbers of particles. But each term in the superposition is well defined (otherwise the superposition itself would not make sense), which means that the states with definite number of particle are well defined.

    As I already mentioned in a previous post, there are also other definitions of the position operator.

    A 1-photon quantum state is NOT an eigenstate of the operator of electromagnetic field. That means that electromagnetic field is not well defined for a 1-photon state, but it does not mean that the position of such photon is not well defined. Indeed, there exists a 1-photon wave function which is well localized in space at a given time.
  21. Aug 3, 2010 #20
    Where can I find this?
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