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How do a photon's two fields look like

  1. May 14, 2007 #1
    How do a photon's two fields look like?

    The photon is a electromagnetic wave and the electric as well as the magnetic part of the wave can be written as follows (taken from http://en.wikipedia.org/wiki/Electromagnetic_wave_equation):

    [itex]E(\vec r, t) = g(\omega t - \vec k \vec r)[/itex]

    where [itex]\vec r[/itex] is any three dimensional vector, [itex]\vec k[/itex] is a fixed 3D vector that basically denotes the direction of propagation of the field in dependence of time, and [tex]\vec k \vec r[/tex] denotes the scalar product of the two vectors.

    To simplify the question to come, I select [itex]\vec k=(1,0,0)[/itex] and with [itex]\vec r=(x,y,z)[/itex] we get

    [itex]E(\vec r, t) = g(\omega t - x)[/itex]

    This field propagates in the direction of the x-axis and intuitively (and given figures in books) one assumes that the field propagates on the x-axis. This, however, is not true, because for the complete 2D plane [itex](x_0, y, z)[/itex] where [itex]x_0[/itex] is a fixed value and y and z vary over the real values, we get:

    [itex]E((x_0, y, z), t_0) = g(\omega t - x_0)[/itex]

    That means that the value of E is identical over the whole y,z-plane for each x_0.

    So when looking only at the values of E (not at the equation), nothing would tell you in relation to which x-axis position within the y,z-plane you have to compute it. In fact you can choose any position of the x-axis.

    Now, if a photon hits my eye and I see light, I would again have thought that it travelled on a line right into my eye. On the contrary, the reasoning above seems to indicate that a whole y,z-plane hit my eye but somehow decided that my eye is "in the middle" and therefore produces an interaction to make me see light.

    Is this description generally true or am I missing something? Do the photon's fields indeed not have a "center line" along which they travel but rather just a direction? How come I can actually see a photon then? How is it decided that it interacts in my eye and not a meter left of me with the wall?

    Harald.
     
    Last edited: May 14, 2007
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  3. May 14, 2007 #2

    Claude Bile

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    The classical field is analogous to the wave-function of the photon. Each individual photon does not have a well-defined position until it is "detected" through some localised interaction. Where the photon is detected is governed by Quantum Mechanics and is probabilistic in nature.

    Claude.
     
  4. May 15, 2007 #3
    So am I right then to derive that the value of the field is constant (identical, fixed) for a given point in time on each plane orthogonal to the direction of propagation?

    This is pretty amazing. If I look straight into a laser, I would assume that the line connecting my eye and the laser is somehow distinguished from all other lines in the universe that are parallel to it. And indeed despite the not well-defined position, the laser's photons all end up in my eye. At least there is usually no indication of light outside the main path of a laser's beam of light. So the connecting line between my eye and the laser is special, but the field equations don't seem to provide a hint thereto. What is wrong here?

    Harald.
     
  5. May 15, 2007 #4

    cepheid

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    Look,

    The problem is that the equations you are considering are for an ideal wave, which, as you have pointed out, exists with a non zero field amplitude throughout all space, and will have "planes of constant electric field amplitude" (which are the yz planes for each given value of x). Furthermore, if you make the further assumption that g is periodic and sinusoidal, i.e.:

    [tex] g(\omega t - \vec{k} \cdot \vec{r}) = \textrm{Re}[Ae^{i\vec{k} \cdot \vec{r}}] [/tex]

    then these planes of constant amplitude are also planes of constant phase, and we have what is often referred to as an ideal plane wave. Clearly this is a mathematical idealization. An ideal plane wave is a good model for the light coming from a distant point source (e.g. an astronomical object), because the spherical wavefronts emitted from this source are so enormous upon reaching earth that the section of them we're seeing can be considered essentially flat. (Using the ray model, we say that the light rays coming from this object are nearly parallel).

    The ideal plane wave is only one particular solution to the wave equation in optics and is most emphatically NOT a good model for the light from a laser beam. Although the light is very nice and straight, the field does not exist throughout all space with any significant amplitude, but is localized. To understand this, you have to look into an area of optics known as beam optics, and a specific solution to the wave equation known as a Gaussian beam. In beam optics, you solve the paraxial wave equation, which reveals that for propagation close to the optical axis, we can have solutions of the form of a modulated plane wave (the plane wave is like the carrier, and it is amplitude-modulated by a slowly varying (wrt the wavelength) envelope function. One such example is a Gaussian beam, an excellent model for a laser beam. Unlike in an ideal plane wave, if you look at the cross section of Gaussian beam in a plane perpenduclar to the direction of propagation, the intensity is NOT constant, but falls off *radially* like a Gaussian (hence the name). By intensity in this context I mean irradiance (I) (power per unit area in watts per metre^2 passing through the cross-sectional plane: [itex] I \propto |E|^2 [/itex]). As a result, most of the energy (some 85% I think) is contained within a circle centred on the optical axis called the beam waist that has some finite radius, which funnily is enough is not constant over length of the beam (even a laser beam diverges), but is nonetheless finite. The probabililty of finding a photon significantly outside this beam radius would be pretty low.

    Edit: btw, never look directly into a laser.
     
    Last edited: May 15, 2007
  6. May 16, 2007 #5
    Thanks a lot for the explanation. The radial Gaussian drop-off is immediately intelligible.

    There is one thing, however, that irritates me still. Your explanation seems to carefully avoid to state whether this all holds for a single photon too. I know that the question can easily go the wrong way due to particle/wave duality. I will nevertheless try to phrase it. You write
    and later
    My question is: if we don't use a laser or a distant star but a single-photon-source on the distant star (books tell me there are ways to emit single photons), then from your explanation I derive the following. Please let me know if this is correct:

    a) When the photon reaches us, the Gaussian is spread out so far that in our vicinity a plane wave is a good approximation, because (i) the wave-front is on the surface of a very large sphere (nearly plane) and (ii) the Gaussian drops off veeeery slowly, i.e. is nearly constant.

    b) Despite this approximation, it is still a Gaussian drop-off around the "main" axis of propagation.

    c) If this axis leads right through my eye, I have the highest chance to interact with the photon, i.e. I could see it if humans could see a single photon.

    d) If you are standing next to me and the wavefront hits your eye at the very same time than mine, you have a good chance to snatch the photon from me, i.e. you would see it and I would not.

    e) Our chances to see it are proportionally to the value of the Gaussian (its square?)

    f) Whether the wavefront/photon interacts with your eye or with mine, the whole spread out energy of the photon is available at the point of interaction --- collapse of the wave function (afaik, the 'collapse' is usually attributed to the Schrödinger wave function, not the electomagnetic field, but the field 'disappears' nevertheless.)

    Should these points be mostly true, I made a big step forward in understanding all that.

    Thanks a lot,
    Harald.
     
  7. May 16, 2007 #6

    Claude Bile

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    A spherical wave does not have a Gaussian profile. I believe Cepheid's intention was to compare an infinite plane wave to another type of wave that has a finite spatial extent - a Gaussian beam. The conclusions reached pertaining to Gaussian beams however do not apply to plane waves.

    For a spherical wave, the chance of observing a photon is equal at all points equidistant from the source. There is no preferred direction as is the case for the Gaussian beam.

    Claude.
     
  8. May 17, 2007 #7
    Assuming you emitted your single photon using a device which emits stuff with Gaussian profiles, I'd say you've got it!
     
  9. May 17, 2007 #8
    I was not talking about a spherical wave and he was not either, of course. But cepheid said that even the Gaussian beam spreads out, albeit slowly. I understood that the spread is not plane but rather "covers" a section on a sphere. If we wait long enough, this section may as well have an area big enough to cover earth or even the solar system. At that time the radius of the underlying sphere (i.e. the distance travelled by the wave front, if it came from a point source) is so large, that it is hardly distinguishable from a plane due to its low curvature.

    Harald.
     
  10. May 17, 2007 #9

    Claude Bile

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    Ah, I think I understand where you are coming from now.

    A wave need not exhibit a preferred direction though, this type of behaviour applies to spherical waves as well as directional beams such as Gaussian beams.

    Anyway, it sounds to me that you have grasped the concept fairly well.

    Claude.
     
  11. May 18, 2007 #10
    :bugeye: Ooooh. Are we still talking about electromagnetic fields making up a single photon?

    Harald.
     
  12. May 20, 2007 #11

    Claude Bile

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    The fields can be low enough in intensity to only "emit" (bad word choice I know :rolleyes: ) one photon at a time. Is this what you mean? Or are you referring to the "internal" fields of the photon itself?

    Claude.
     
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