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Qualitative description of photon from faraway star

  1. Sep 30, 2012 #1
    A photon is (described by) an electromagnetic wave, meaning the oscillation of the E and the B field.

    Now there is this single photon that comes from a faraway star and hits the CCD detector of some astronomical satellite. As an approximation consider that it traveled through vacuum for the last several million years since it was send out by some start.

    Just before the photon shoves it energy into the CCD and is recorded, what is the qualitative description of the E and the B field. I would be interested in statements like

    - The fields are nearly/completely zero outside a radius of ... around the position just in front of the CCD.
    - 99% of the fields' energy is concentrated just in front of the CCD.

    Or, to the contrary something like.

    - The fields are spread far over the cosmos, roughly in a sphere around the star that emitted the photon.
    - Only when the photon is actually registered by the CCD, the collapse of the wave function instantly sucks the whole energy, just previously spread over this sphere, into the CCD.

    Or any other description that gives a qualitative idea of where this photons is "located" just before it hits the CCD.

    Thanks,
    Harald.
     
  2. jcsd
  3. Sep 30, 2012 #2
    I'd like to see some thoughts on this, too.

    If the photon is characterized by the surface of a sphere around its place/time of origin, it seems hard to imagine that this sphere does not encounter appropriate opportunities to absorb the photon rather early compared to millions of light years of expansion...

    For each radial distance of sphere expansion, would not every photon sphere certainly encounter all possible absorption opportunities on that radius it in all directions ... after a few million years this opportunity for the surface to have an absorption encounter would seem to be high.

    Photon spheres from the Sun would all encounter the planet Mercury - a whole planet of opportunities to absorb the photon. Eclipses and ordinary shadows seem to show that absorption is not a rare thing... so the expectation with an intervening planet might be that little or no photons make it as far as Earth; yet we know they do... and seem to make it a long way when coming from very distant sources, including cosmic ray background.

    And, with the sphere concept, the absorption is only allowed to happen once at one point on its surface, which would seem to have to prevent the absorption at all other locations on that surface even if they are separated by millions of light years. In a sense, all locations on the sphere would seem to need to have a kind of "entanglement" behavior to ensure a single absorption event for the entire life of the surface.
     
  4. Sep 30, 2012 #3

    Bill_K

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    But there is no difference in the odds, bahamagreen. Whether you regard the photon as a complete spherical wavefunction or as a small confined object, there is equal unlikelihood that it is aimed so precisely that it hits the Earth from 100 light years away. And yes, it only gets to hit one thing, and there are many other things out there waiting to be hit. But every point you made in your argument against spheres applies just as well to the perilous adventures of a tiny discrete photon.

    PS - Actually it's only a hemisphere. :wink: The photon presumably was emitted from the surface of the star, and only half the sky would be visible from there.
     
  5. Oct 1, 2012 #4
    How does Bill_K's argument - that the odds are the same whether the photon is considered to be a light sphere or considered to be a small confined object - resolve the issue that if the photon is a light sphere, then what stops the light sphere from collapsing at several points along its light sphere, once those surface points are separated by large distances?
     
  6. Oct 1, 2012 #5
    Bill_K,
    I'm not seeing how the odds are the same.

    If the photon is considered to be a small confined object, then it would need to be "...aimed so precisely that it hits the Earth from 100 light years away..."; but if it is regarded as an expanding sphere it will hit everything outside its point of origin in every direction without aiming until there is an absorption at one point on the surface of expansion.
    How would you aim an expanding sphere, or need too?

    And then that raises the question robinpike and I asked; how is the absorption restricted to a single event when the radius is very large?
     
  7. Oct 1, 2012 #6

    mfb

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    Two possible views:

    Quantum mechanical, with collapse interpretation: Every photon is emitted in a wide angle as a wave function. Each object which gets hit by the wave function acts as a measurement - the wave function collapses to "hit that object (and deposited all its energy there)" or "did not hit that object (and continues as wide-spread wave function)". The probability that a photon hits your CCD is tiny, but as we have so many photons, some of them do.
    The same can be done with all other interpretations of quantum mechanics.

    Alternative approach: Assume that all photons have some direction, some of them will hit the CCD. That gives good results if you do not try to perform double-slit experiments or similar things.
     
  8. Oct 1, 2012 #7

    Bill_K

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    I maintain that the second approach does not happen. The nature of the photons depends on the process that creates them. Stars produce black body radiation, and black body radiation comes from random collisions between atoms. Each collision is predominantly electric dipole, and the radiation from it is therefore dipole-distributed over solid angle. To get a photon that is "aimed", with a narrowly defined k vector, you would need a large-L process.
     
  9. Oct 1, 2012 #8

    mfb

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    It does not happen, and as I posted it will lead to wrong predictions in interference effects (like a double slit). However, it gives the correct result if you just want to know how many photons will hit your CCD pixels.
     
  10. Oct 1, 2012 #9
    It seems like I preselected the answers by posting in the Quantum Physics group. Maybe I should have gone to the classical physics, because I would really like to get a qualitative idea about the E and the B field.

    Bill_K says that one process to have created the photon is a dipole and
    So can someone explain that last one with slightly more detail. I take it that "angle" means actually an inverted funnel with that opening angle. I further guess that the photon (E-B/ field ripple) emanates at the narrow end and over time travels out of the funnel while in the form of the cut of a spheres surface matching the "angle". As for the "thickness" of the surface, i.e. the radial distance within which the E-B field ripple is clearly nonzero, can we roughly say how much it is in units of the wavelength?

    Or is my description completely wrong. How is it then?
     
  11. Oct 1, 2012 #10

    Bill_K

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    No, you came to the right place! It's important to remember that, first and foremost, a photon is a quantum mechanical object. We often get this question: "What does a photon really look like? What shape is it? How long? Is it just one cycle long, or N cycles long, etc." These are meaningful for classical wavetrains, but a single photon is described by a wavefunction that gives you a probability (amplitude) of finding a photon at a particular point. So the extent of the wavefunction is not the same as a region of E and B fields. Any more than the electron's orbit in a hydrogen atom tells you how big the electron is!
    See the picture of electric dipole radiation in Wikipedia under "dipole antenna".
     
  12. Oct 1, 2012 #11
    Well, but the E and B fields are what I am after. Are you saying that the Wikipedia picture qualitatively also describes a single photon?
     
  13. Oct 1, 2012 #12

    Bill_K

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    I give up. Sorry.
     
  14. Oct 2, 2012 #13
    birulami - your question is a good one and I understand your continued frustration. It may be a simple matter of certain respondents saying one thing and meaning another, but as it stands, one has the impression of a confusing and utterly nonsensical picture of a photon having on one presumably valid possible interpretation, a physically real extent of maybe billions of light-years in radius, magically undergoing instantaneous collapse down to a point at say some CCD detector. It needs to be clearly stated whether that possibly billions of light-years radial extent refers to a fantastically diluted spherical shell of physically real E & B fields, in keeping with at least roughly the field of a classical radiator, or it just refers to a point-like photon-as-particle, or at least spatially compact entity, whose probability of detection is described by said billions of light-years radius shell. Or something else gain.
     
  15. Oct 2, 2012 #14

    Bill_K

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    If you mean me, Q-reeus, I have said the same thing consistently throughout, and as clearly as I can. Sorry if you're confused.
    It is not "magical", just quantum mechanics. QM has other interpretations besides wavefunction collapse, but they amount to the same magic trick.
    I have said it's a wavefunction, which measures probability. I include wavefunctions in things which are "physically real." And yes, a photon does have E and B fields, and yes, they are fantastically diluted a billion miles away, but they have meaning only at the point of detection. Trying to visualize them as a sphere's worth of classical EM field is totally incorrect.
     
  16. Oct 2, 2012 #15
    Wasn't just you, but did have your comments in #7 in mind:
    Well that reads a lot like 'classical radiation field' to me. As I said last time, one needs to be very careful to clearly delineate if above does or does not imply a physically real classical or near-enough-to classical field distribution - and it sure seems to.
    I'm a complete layman when it comes to subtleties of QM, but perhaps this is where the proper understanding of this 'magic trick' needs to be spelled out - layman's language of course.
    This leaves me confused as ever. On the one hand, you say spherical shell represents a wavefunction (= probability calculational tool?), on the other that there are physically real and fantastically diluted E & B fields = classical EM prediction for say dipole oscillator. So what is it then? Physically real instantaneous collapse of a billions of light years radius spherical pulse of real E & B fields, or just detection at a point [STRIKE]by[/STRIKE] of a point-like bundle of energy that was in fact always point-like? Not trying to be argumentative here btw - I am genuinely baffled.
     
    Last edited: Oct 2, 2012
  17. Oct 2, 2012 #16
    It isn't (2) because the wavefunction was not always point-like.

    It also isn't (1) because it's not the EM fields which collapse, it's the photon wavefunction. They are crucially different because the EM field is a observable measurement, whereas the wavefunction is not. It's not as if the EM field is suddenly concentrated to a point where the photon happens to be measured- it's always concentrated around the photon's position, and this position could be at many different points in space until it is measured.


    That's my understanding anyway. Everything must be thought of via the wavefunction, which alone collapses due to measurement by a physical operator. Until that collapse occurs, there have, by definition, been no measured EM fields due to the photon anywhere.
     
  18. Oct 2, 2012 #17

    vanhees71

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    Well, then you should first define, what you understand under the "photon's position". There is no observable in the strict sense of quantum theory which can be interpreted as "position" for massless particles with spin greater or equal to 1.

    Also, there is no necessity whatsoever to believe in the collapse of a quantum state at the "instant" of measurement, which is anyway always a finite (uncertain) time interval.

    Besides, what we measure as "light" from distant sources (particularly in the extreme case of a very distant galaxy) is almost never a single photon but a coherent state and thus a very large extended "wave packet" with an indetermined photon number (the dimmer the light is you want to detect the more it's well approximated by a superposition of the vacuum and a single-photon state).

    You get a far better approximate intuitive idea about "light" from a distant source when thinking in terms of classical em. waves than in terms of thinking about "photons".
     
  19. Oct 2, 2012 #18
    Thanks for your input MikeyW, but I never claimed or implied the wavefunction was point-like. On the contrary that the 'wavefunction' (but hang on - isn't it a no-no for photon to even have a wavefunction?!) was 'spherical pulse-like' in predicting probability of detection of what was always a concentrated entity known as photon. For me one telling argument here is conservation of momentum. When a photon is absorbed say at that quintessential CCD detector, there is an an impulse dp = hf/c. It acts in a very specific direction - away from the source. If we wish that momentum conservation is preserved at all times, how can it be otherwise than that the source of that absorbed photon suffered an equal and opposite recoil momentum when emitting in the first place? Which is inconsistent with a spreading axially-symmetric classical dipole oscillator field, agreed?
    .
    I agree with this part - without getting into issues about nature of 'wavefunction collapse'.
     
  20. Oct 2, 2012 #19
    Is this just nit-picking? When I say position I mean the position of the CCD which records the photon, and when I say "instant", I mean at the time that we get the measurement signal. Do these minor alterations not remove your objections?

    I would think any correct explanation necessarily uses quantum mechanics, and the classical analogue may be more or less useful as a rough explanation depending on the circumstances. For a solar panel facing the sun, I'd say the classical explanation is fine. For a very distant star emitting black-body radiation, I would say the classical explanation can lead to some big problems, not least because the two fundamental processes (BB radiation and the CCD's photoelectric effect) are strictly non-classical.

    In this classical explanation, do we have a huge (hemi)sphere of EM waves which decrease with 1/r^2? What is the rough value of the EM field when a visible light from a yellow giant travels a billion light years? How can a CCD possibly detect such a small signal over the instrument noise?
     
  21. Oct 2, 2012 #20

    Cthugha

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    Typically it will be in a thermal state, but that will of course only extend the indeterminacy of the photon number.

    Well, BB radiation needs QM to describe the frequency distribution, but is that really important here? The photoelectric effect does not need a quantized em field. It works fine with a classical field. Therefore antibunching, not the photoelectric effect. is the unambiguous proof of single photons existing under some circumstances.

    Why not? some EMCCDs already have single photon sensitivity at a really low dark count. Besides, localizing the light field is not that easy. All photons inside one coherence volume are indistinguishable in principle. Far away from the source the coherence volume gets larger. Therefore the coherence volume can be quite large. Hanbury Brown and Twiss managed to measure the angular diameter of Sirius this by measuring the coherence length of its emission that way already in the fifties. Then, the coherence volume was at least large than two large telescopes.
     
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