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Does a Photons Energy (hv) spread out to all locations?

  1. May 5, 2004 #1
    A photons energy is 'hv', does this spread out to all locations or is it only contained at a specific radius?
     
  2. jcsd
  3. May 5, 2004 #2
    Looking at the photon from the "wave" point of view, you could ask if the wave can propagate in all directions from an emitting atom, forming an sphere that gets larger and larger as the wave travels away from the atom.
    I think the answer to this is NO. There are no spherical waves for photons. But you could have the wave associate with a photon propagating with a shape similar to that of radio waves from the radio station's antena. If I recall correctly, these waves look more like rings. Still in this case, the photon would appear to be "spread out" in a quite large region until you detect it, time at which there is what has been called the "collapse of the wave function" and the photon becomes localized.
    You could ask where was the energy of the photon before the wave collapses, but I think you would get contradictory answers to this question, with some people saying that the question itself doesn't make sense, (Copenhagen interpretation of Bohr, Heisenberg) and others explaining it thorugh other different interpretations, including the quite popular "many worlds interpretation" based on work by Everett and Brice DeWitt.
    These interpretations address the wave-particle duality, the collapse of the wave function, Schrodinger's cat, and other paradoxes of quantum mechanics.
    Now if you were looking for understanding of the distribution of the probability to find the particle, the uncertainty principle gives you a guide on this. A photon normally doesn't have a totally definite energy, but its energy is kind of fuzzy but close to a certain value. To each probable energy, there is associated a wavetrain of different wavelength. All the wavetrains superposed (wave packet) add up constructively in one place, which is the most likely place to find the photon. (constructive and destructive interference). So the wave packet gives you a picture of the distribution in space at a certain time to find the photon, but the energy itself can not be considered to be "spread out" through the whole packet, but only to materialize at a very small region of space when you detect the photon.
    If this doen't answer your question, please post again and I'll try to help (If I know the answer)
    --Alex Pascual--
     
    Last edited: May 5, 2004
  4. May 5, 2004 #3
    the answer is that you cannot say anything at all about a photon in between the time it is emitted and the time it is absorbed.
     
  5. May 5, 2004 #4
    Thanks for the reply, lets look at the 'wave' point of view

    Ok I will redifine my question, hopefully this will tempt -jnorman as well.

    How much of a 'hv' wave is contained with a volume of space, and what is the radius of the energy, in geometric terms?

    Photons are assigned a precise value, it is measured and catogorized as being hv, does this energy move in discrete bundles, ie an exact amount 'hv'. Does this energy remain within a definate volume when passing from one location to another?

    So we can now ask, how much of a Photons energy is spread out as it leaves one location (emission zone), and by how much does this change the value when it arrives at another location (absorbtion zone)?

    As I believe that all recorded values of photons leaving one location and arriving at another show the energy as being 'hv', so this should expel the myth that a photon's energy does spread out? The duality seems to be convienient, for instance from a location of measurment called (A) a photon is detected so it is leaving this location, and then at a later moment at location (B) it is detected as being absorbed into (B). Now surely the distance in space between A and B means that a Photon should be spreading outwards, if so then some of its 'wave' has to be at another location?
     
    Last edited: May 5, 2004
  6. May 5, 2004 #5
    Are you saying that I can assign any value I like to the space between atoms?..or are you stating a Photon has no precise value other than at locations of detection, ie emmision zones and Absorbtion zones.
     
  7. May 5, 2004 #6
    It appears you can only say that there is a probability of finding the photon in certain places. As the electron leaves the source, the probability wave starts spreading smomewhat (would spread very little laterally in a laser beam). Now even if the probability wave has spread, I don't think you can spread the hv energy to this larger value. If the energy was indeed distrubuted that way, when the wave collapses at detection, it would all have to suddenly run to the point of detection. I would say that you can assign to all points in space where the particle may be detected, the same energy hv, but with a partial probability. When the wave collapses, the sudden change is in the probabilities. It does not involve any instantaneous tranfer of energy.
    The different interpretations of quantum mechanics would describe this phenomenom differently.
    --Alex--
     
  8. May 6, 2004 #7
    olias - "a Photon has no precise value other than at locations of detection" - this is generally correct.

    alex - do not fall into the trap of assuming there is an understandable mechanism as to how the process operates. macro-logic is not applicable here. QM does not just indicate that the probability function applies only to our inability to measure or speak of a precise location - it states that there IS NO location.
     
  9. May 7, 2004 #8
    jnorman: I don't think I used macro-logic. On the other hand, I think you are getting into the grounds of interpretation (Copenhagen) . I tried to be interpretation-neutral.
    All I was saying is that there is indeed something you can say about the photon between emission and detection, and it is that there is a probability distribution of finding it at different places. Once the photon is detected, there is a collapse of the wave function at places other than the point of detection. This reflects the mathematical treatment, and I am not saying if the wavefunction is "real" or not. I don't even think your friend Bohr would disagree with what I just said.
    --Alex--
     
  10. May 8, 2004 #9
    If I recall correctly, these waves look more like rings.
    Wrong! All electromagnetic radiation unless otherwise modified propagates istropically i.e in spherical waves.
     
  11. May 8, 2004 #10
    McQueen,
    I am certainly not an expert in this and I don't have a detailed argument against your assertion. However, I recall reading about someone (it was someone famous such as Einstein or Bohr) saying that there are no spherical waves of electromagnetic radiation. On the other hand, if you look at pictures of the electromagnetic waves produced by radio antenas, (dipoles) they are certainly not spheres. I think if you start looking at the directions of the magnetic and electric field all around the sphere, you soon find that there is no configuration that can accomodate both fields, always perpendicular to each other, and with the magnetic lines of force closing on themselves. (I apologize for using the outdated concept of lines of force).
    Anyway, I think using Maxwell's equations and vector calculus it should be easy to prove. (nice exercise for an E&M class).
    But if you have convincing evidence or a quote from some text that proves me wrong, it would be nice to hear it.
    It would also be nice to hear from jnorman about this.
    --Alex--
     
    Last edited: May 8, 2004
  12. May 8, 2004 #11

    Gokul43201

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    I'm quite sure you can have both, depending on the geometry of the radiator. An antenna produces cylindrical waves, while a point source - like an accelerating charge -produces spherical waves.
     
  13. May 9, 2004 #12
    This is interesting, are I correct in saying that a far away luminous source such as say the Galaxy M87, there are no photons travelling in the Vacuum space from there (source) to here, disregarding the implications for Dark Matter?

    Of course this means that Gravitational lensing cannot occur? Are you sure about this? another thing how does one acertain that light travels in a straight path.

    I understand that looking at images from the cosmos one can definately see that visable light is not continueous, ie for M87 there is a discrete location around the Galaxy that is visible (source), but then one has to ask how much of the photons at source (leaving) , and how much are arriving at M87? from other far away Luminous source's, being that photons as you state this is a 'location of detection', but it is also a location where photons are arriving from elsewhere in the cosmos.
     
  14. May 10, 2004 #13
    It is not that the photon does not have a precise "value" (whatever that means) , but that it does not have a precise position. This doesn't mean that the photon can be "anywhere", it only means that there is uncertainty in it's position.
    The behavior of light can be described by ray optics or by physical optics. Ray optics is what you study in elementary courses and it is valid when dealing with lenses, mirrors, etc. . But the phenomenom of diffraction can not be explained by ray optics. (Look at the text "Optics" by Hecht for more detail) Things like gravitational lensing can be explained both in terms of waves as well as in terms of photons.
    There might be different approaches or interpretations with respect to the state of the photon between emission and detection. You might also want to look into Feynman's path integral.
    Gokul: I think I have read that light emmited by attoms is not emmited in the form of spherical waves.
    --Alex--
     
  15. May 10, 2004 #14
    alex - "there is a probability distribution of finding it at different places" - yes.

    "there are no spherical waves of electromagnetic radiation" - we cannot say this, but i agree with it in priciple as it is generally in line with CI.

    "there is no configuration that can accomodate both fields" - correct.

    gokul - "An antenna produces cylindrical waves" - this is a very general statement which is one of those standard oversimplifications which can assist in teaching laypersons about radio. it really has no connection to the reality of photon propagation. read QED again.

    olias - "there are no photons travelling" - i can agree with this much of what you said.

    alex - "This doesn't mean that the photon can be "anywhere", it only means that there is uncertainty in it's position" - nope, the photon CAN be anywhere or nowhere - we just cannot say (as you indicate, according to CI). "described" is very different from understanding fundamental nature. of course, here we move from the world of physics into, again, the world of interpretation. have you read "wholeness and the implicate order"? extraordinary application of thought, though i dont agree with it.
     
  16. May 10, 2004 #15
    I guess by CI you mean Copenhagen Interpretation right?
    Regardless any interpretation, if you have light emmited by a laser, I don't think you can find the photon behind the laser or even 1 foot to the side of the beam. There may be waves there but they cancel.
    --Alex--
     
  17. May 12, 2004 #16
    Alex


    Waves propagating along the surface of a medium , such as for instance in water , propagate in circles . That is , the totality of points of maximum displacement from equilibrium position form circles. All these points vibrate in-phase , and are known as plane waves. For waves propagating not along the surface but inside a medium , the totality of points vibrating in-phase constitutes a definite shape. In an isotropic medium , that is in a medium in which the phase velocity is independent of direction , the surfaces of equal phases propagating from a point source are spheres . Such waves are termed spherical . As regards the photon the following sets are normally used
    to describe the states of a photon :
    (a) k x , k y , k z , [tex]\alpha [/tex].
    And
    (b)E , M2 , M z , P. .
    Here k x , k y , k z are the projections of the wave vector of the radiation ; [tex]\alpha[/tex] is the polarization of the photon ; M2and Mz are the square of the momentum and the projection of the momentum of the photon respectively ; P is a quantum number called the spatial parity. The set (a) is used to describe the states of photons corresponding to plane classical waves And are called [tex] k\alpha [/tex] states . The set (b) is employed to describe the states of photons belonging to spherical classical waves. Just as a spherical wave may be represented as a super position of plane waves , the states of a photon described by set (a) may be represented as a “super position “ of states described by set (b). The converse statement regarding the representation of plane waves as a super position of spherical waves is also true. This is known as the principle of superposition of states.
     
  18. May 13, 2004 #17
    You might consider that the confusing issues surrounding the Copenhagen interpretation versus the contentions of Einstein and Schrodinger revolve around an issue of a non-dissapating photon energy. It does not at all require we consider a laser as our best physical production of a low dispersal beam. The experiments conducted on one (or few) photons fired individually gives results that appear to say that the photon does not in any way disperse as it travels. It makes no difference that we consider a tightly controlled beam. It lead Born, Bohr and Einstein to agree on one thing, there can be no physical reality of particle (or portion of particle) assigned to the probabilities along the wave front. Einstein said, the wave packets [and their associated probabilities] are merely symbolic and must be viewed as probability waves [not physical]. The sudden vanishing of the transmitted packet at one position and the appearance of it at another position is no more mysterious than the vanishing probability that tails may turn up when a coin is tossed and heads has been seen to fall.
     
  19. May 13, 2004 #18
    inre: "I don't think you can find the photon behind the laser or even 1 foot to the side of the beam."

    ahhh - but you certainly can. certainly the probability is very low, but it is not zero. i cant really point you to a better reference for this than QED by feynman. i am reminded of the quote - "things are not only stranger than we imagine, things are stranger than we CAN imagine."
     
  20. May 14, 2004 #19
    Can you make a clear statement as to the existence of travel by Photons?

    Simplistic question to your previous posts:Do Photons travel Distance?..again a Photon that is measured at one location and later at another location is process defined needed to MEASURE?

    At what does speed does a probability happen for a Un-Certainty of measure to be Imprecise?
     
  21. May 19, 2004 #20
    The energy of a Photon that is in motion 'propergating' = Kinetic waves, the energy of photons that are not propergating are Potential Waves.

    Is (hv) the sum of a Kinetic (Dynamical) or Potential (Static) Energy?

    When a Photon arrives at an Atom, the energy level is increased, so Electron is lifted into a another orbit/energy level?

    Does this mean the energy of Photons are 'Static' or Potential before emmited back away from Atoms?

    So can or do?.. Photons become entangled with Electrons during the Transition period of energy transfer?..what I guess I am asking is 'if' the Electron does a complete orbit before 'kicking' a photon back out from whence it came, do photons 'follow' the Electron around for a complete orbit?..or do Photons 'drag' an Electron to a different Energy Level then return the Electron to its original orbit before emerging from the Atom?

    I do not want to use the terminology from the film 2010?..you know where they only have a shortage of fuel so use the Gravitational pull of the Sun to reach the correct orbital velocity( similar to frame dragging?) :smile:

    So..does the Photon remain static until the Electron has gone around once to punch the photon back away from Atoms?..or is there a Dual dynamical process involved?

    Just a query. :rolleyes:
     
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