View Full Version : size of photon particle
billy_boy_999
Jun21-04, 11:15 PM
how big is a photon? what are its dimensions?
if this is a meaningless question, please tell me why...
Imparcticle
Jun22-04, 12:52 AM
http://ethel.as.arizona.edu/~collins/astro/subjects/electromag6.html
I'm not sure if this is any good....http://www.fervor.demon.co.uk/photons.htm
Interestingly, I was asked this exact same question in my defense of my undergrad experimental project. At the time, I said it was a point particle.
It actually depends a bit on what you mean by "how big". An excitation of the electromagnetic field can extend arbitrarily far across space. However, we only ever detect them at points.
billy_boy_999
Jun22-04, 01:46 PM
hmm...(thanks guys)...the article says the size of a photon is inversely proportional to it's wavelength...or is it a point?
is there any way that we can say all photons - no matter their wavelength - have the same dimensions? of course, if we take them as points we can...though is a point a dimension, per se?
i'm having trouble digesting the idea that the size of a photon depends on its wavelength...if photons really are the quanta, the bare constituency of light, then surely they should have a fixed size...isn't part of the wave-particle duality the idea that light always arrives in lumps - as feynman says - and that they always have the same amount of energy? since they are always traveling at the same speed and are massless, doesn't all their energy come from their speed/momentum/velocity/whatever? so shouldn't their size also be a constant?
Silverlancer
Jun22-04, 06:45 PM
According to Quantum Mechanics, a photon, like any other point particle, has no size at all. It is a point, if treated as a particle. If it is treated as a wave, it has wavelength and amplitude instead, amplitude being the "size".
billy_boy_999
Jun22-04, 08:45 PM
The characteristic size of a photon is inversely related to its frequency: (c/f)
that is what the people at arizona.edu have to say...is this not a statement consistent with QM? is the point treatment reckoned to be a description of a physical particle? hardly, how can a point be a physical anything? is quantum mechanics the right theory to address the question of a photon's size?
arivero
Jun23-04, 02:58 AM
Now, what about a massive boson? It has a momentum p, thus a characteristic length h/p, but it also has an energy E=\sqrt{m^2 c^4 + p^2 c^2}, thus a characteristic frequency E/h, thus another length hc/E. For the photon, m=0 and both sizes coincide.
tavi_boada
Jun23-04, 05:04 AM
Aren't monochromatic waves infinite? I dont understand the syntagm "characteristic size of a photon". Can you explain??
arivero
Jun23-04, 06:20 AM
Tavi, he means, for sure, its wavelength.
russ_watters
Jun23-04, 11:36 AM
that is what the people at arizona.edu have to say...is this not a statement consistent with QM? It seems sloppily worded. Do you have a link so I can see the context?
It should be clear enough though, that since a photon can be a wave or particle depending on how you observe it, "how big is a photon?" is, if not meaningless, an incomplete question.
But, when viewing it strictly as a particle, it would have to be a point particle as stated.
techwonder
Jun23-04, 03:29 PM
Point particles have a nasty habit of having infinite energy densities :cry: , but it is true that QM assumes particles to be point like. Super String Theory assumes all particles to be small strings with a length of just about the planck length and no width, i.e. one dimensional :frown: . In that case a photon is 10e-44 meters :smile:
billy_boy_999
Jun23-04, 06:34 PM
and i was waiting for someone to mention planck length somewhere...the boson length is interesting, i'm going to have to read up on that...the sloppily worded quote comes from the first link that Imparcticle had posted...
to me it feels like there's something missing from QM's point particle treatment, it feels a deferment to the abstract...maybe that's just my bias...
Turns out that issues of photon size are very difficult -- in no small measure because photons, unlike massive particles, do not have a position operator - this was proved by Newton and Wigner in 1949. Nonetheless, it is possible to develop a probability density that tells where a photon is within a volume that is big compared to the photon's wavelength -- that is the probability of the photon triggering a photoelectric detector. Among the peculiarities of photons and QM, is the fact that the distribution of the photon's energy is spread out to regions where the probability of finding the photon is zero.
The bible on photon physics is Optical Coherence and Quantum Optics by Mandel and Wolf. It discusses, in great detail what I mentioned above. It assumes a sophisticated grasp of QM and statistics -- but it starts from ground zero, and does the basics -- state vectors, coherent fields, correlations,....-- albeit quickly. it is a great book, and it is worth the fight to read it.
Regards,
Reilly Atkinson
jtolliver
Jun25-04, 06:40 PM
If the concept of photon size is meaningful and the size is the same for all photons then it is zero or infinite because (in the absence of charged particles at least) the field equations for the electromagnetic field are scale invariant(this is a consequence of the masslessness of the photon)
It is essentialy an empty space. Current understanding is that in any case it is a complex multidimentional combination of EM field of various intensity spread within an unprecisely defined space, a worm that has very "blurry" head and tail and consists of EM field variations..... Again orientation of this worm will depend on the relative movement...HUH.
Revelationz
Jul18-04, 04:23 AM
OK, the QM world is admittedly unintuitive to macro-particles like me :confused: So can I try a re-phrase of the question?
I think, but am not sure, this statement is true: The more you try and 'pin down' the location of a QM particle, the more it tends to 'spread out'. So as the resolving power of your locating instrument increases, you find that you can detect the particle less often within a given bounding box.
That means, I think, that as resolving power of the instrument increases, a point will be reached at which the particle is detected exactly 99.00% of the time within the limits of resolution of the instrument.
So if I am looking at photons streaming in from a 'perfectly' collimated laser beam source, and using the Palomar telescope to do so, I will observe the photons a very high percentage of the time. As I shrink the aperture of the telescope, at some point I will only observe the photons 99.0% of the time, not because the beam isn't perfect, but because Heisenberg says I can't know the position of the photons that accurately.
If the above is true, then the question about size becomes something like:
A) For light of a specific frequency in the range of, say, visible red, what is the 2-D size of the aperture (3-D bounding box?) in meters, that will enclose the position of the photons 99.00% of the time?
If this question makes any more sense than the original one, then:
B)Is this aperture (bounding box?) different in size than that of photons of a different frequency, say visible blue light?
And for a somewhat related question about quantities in photons:
A photon is a wave of EM energy, oscillating at a given frequency. How many oscillations does a photon consist of?
I think you are very close to finding your own answer. Optical devices cannot be used for observation of photons because they use photons as a "bounce off" partice for imaging purposes. It we had means of control over a free moving EM frield interacting particle much smaller than any estimated size of photon then such observations would become possible. It is practically impossible to determine the length of a photon in meters, because you have to set the minimum EM field magnitude where your measurements will start and stop. Because it is believed that EM magnitude gradually increases from zero to the photon's energy range and then decreases back to zero over the length of the EM "snail" that represents photon, it is impossible to set its true length. All calculations will give only approximate results. It happens becase human brain uses different principles in processing of any information by setting definite borders to share one from another, it works in a big world like our every day world, but fails to work in the world of particles. We cannot truly realise that on a particle level we live in a world that is illusive because it is essentially nothing, i.e. something which singular example we would not normally see, but a mega multuple expample of which comrises all that we see and touch every day.
It all depends on how you think of a photon. In quantum mechanics a particle can't be thought of in the sense of a little piece of something. It's one unit of energy and other quantum values, but other than that it's more a wave. Upon measurement you could define its position within the wavelength of whatever you used to measure it. It will condense to that space when it interacts with that particle. But afterward it will again expand as a wave of possible quantum states. The same goes for whatever you used to measure the photon.
True and not. As much as anything. We are based on the knowledge of approximate nature that is few decades old. It doesn't mean it is wrong - it is developing. We won't ever know if we will never know. As it was just said: "within" -- is all it is. More alternative approaches give us a better chance.
jnorman
Aug14-04, 01:29 PM
NO NO NO - where did you guys get the idea that a photon is a point particle? photons are NOT particles in the sense that they have some specific size, location, or physical attributes. they do not have a size. the bottom line here is:
you cannot know ANYTHING about a photon between the time it is emitted and the time it is absorbed.
No-one said it's a point particle. It cannot be a point particle because inbetween its emission and absorbtion points it is in motion at C. It can have a location, because any ray of light can be located. Ultimate conclusions that something "cannot be done" are limits of the accepted theory. Humans have been only progressing for about 300 years as a culture.
danitaber
Aug19-04, 09:16 AM
I don't know if Billy boy 999 is even reading this anymore, but just in case, I will say that what Billy Boy 999 just witnessed is an example of a physicist's aversion to the phrase "I don't know" (ha!)
Personally, I think it's a meaningless question for the following reason:
We can't measure its "size". No one knows what "size" even is in regard to fundamental or quasi-fundamental particles. We can talk about wavelength, we can talk about points, we can talk about distributions and wave-packets, we can talk about strings, and we can talk about the loudness of a "blip" on a computer. We can't talk about "size". We have a hard time talking about "reality", even! (see many worlds thread, LOL)
We are less concerned with what it is than what it does. I'm sure I'm going to get jumped on (I'm slowly getting used to it) but there it is. OK, who's first for the danitaber-kabob? Refutations are welcome.
ZapperZ
Aug19-04, 09:32 AM
I don't know if Billy boy 999 is even reading this anymore, but just in case, I will say that what Billy Boy 999 just witnessed is an example of a physicist's aversion to the phrase "I don't know" (ha!)
Personally, I think it's a meaningless question for the following reason:
We can't measure its "size". No one knows what "size" even is in regard to fundamental or quasi-fundamental particles. We can talk about wavelength, we can talk about points, we can talk about distributions and wave-packets, we can talk about strings, and we can talk about the loudness of a "blip" on a computer. We can't talk about "size". We have a hard time talking about "reality", even! (see many worlds thread, LOL)
We are less concerned with what it is than what it does. I'm sure I'm going to get jumped on (I'm slowly getting used to it) but there it is. OK, who's first for the danitaber-kabob? Refutations are welcome.
Danitaber kabob.... YUM! I'll have seconds if it comes with couscous.
Actually, I don't find anything wrong with what you just said (or maybe I'm not in such a bad mood this morning). There is one thing I will pick on, though. "What it is" is DEFINED by "what it does". If you consider carefully, everything that we define is based on its physical properties. An electron has a set of properties that defines what it is. The same with a photon. You recognize a friend or family member based on physical characteristics that you observe, the sound of his/her voice, his/her behavior, etc. These are all physical properties and characteristics that DEFINE what it is.
A photon was NEVER defined as a "particle" with definite spatial boundaries, like a ping-pong ball. It was defined as a particle in the sense that each one of them carry a definite quantity of energy. The implication from this is very subtle but important, especially when we delve into things like "single-photon" sources.
Zz.
kebab! kebab! kebab! or kebap to absoultely correct, or even kabab, but never kabob!!!!!
And while I'm on the subject how can you have vegtable kebabs? kebap literally means 'broiled meat'.
danitaber
Aug19-04, 09:59 AM
There is one thing I will pick on, though. "What it is" is DEFINED by "what it does". If you consider carefully, everything that we define is based on its physical properties. An electron has a set of properties that defines what it is. The same with a photon. You recognize a friend or family member based on physical characteristics that you observe, the sound of his/her voice, his/her behavior, etc. These are all physical properties and characteristics that DEFINE what it is.
Thanks for not slaughtering and broiling me, Zapper.
OK, I'll agree with you there on a technical level; I'll stand by my original phrasing for metaphysical purposes only. (i.e., I am not the sum of red hair and stunning good looks :uhh: , etc. I am essentially something that I don't believe can be quantized.)
And I'll amend "kabob" to "kebab", but that's all the ground I'm giving. :biggrin:
Tau_Muon_PlanetEater
Aug19-04, 04:49 PM
Reillly said'
Among the peculiarities of photons and QM, is the fact that the distribution of the photon's energy is spread out to regions where the probability of finding the photon is zero.
- Question, How can the probability of finding the photon be zero at the extremes of the distribution? Wouldn't they approach zero but never hit it?
Fredrik
Aug22-04, 08:57 PM
i'm having trouble digesting the idea that the size of a photon depends on its wavelength...if photons really are the quanta, the bare constituency of light, then surely they should have a fixed size...
A photon has a perfectly well-defined wavelength only when it's in a momentum eigenstate, i.e. when it has a perfectly well-defined momentum (and energy). This never happens. A photon is always in a superposition of momentum eigenstates:
\lvert\alpha\rangle=\int{d^3p}\ f(\mathbf{p}) \lvert \mathbf{p}\rangle
The only quantity that we might want to call the "size" of the photon is the width of the Fourier transform of the momentum-space wave function, f, i.e. the uncertainty in the photon's position. This uncertainty could be anything between zero and infinity. (I'm ignoring Planck-scale effects here). Since it can be arbitrarily close to zero, it makes sense to call the photon a "point particle".
However, if we assume that the uncertainty in momentum is proportional to the magnitude of the momentum (which is the only thing we can assume if we know nothing about the state), the uncertainty in position is proportional to Planck's constant divided by p (the magnitude of the momentum). Since p is inversely proportional to the wavelength, the uncertainty in position is proportional to the wavelength.
So it makes sense to think of the wavelength as the "size" of the photon (or at least as something proportional to it). This may seem strange, but it is at least consistent with e.g. the fact that microwaves (with wavelengths of order 1 cm) won't go through a metal net with millimeter-sized holes (like the net that covers the window of your microwave oven), but they will go through a net with much larger holes.
You might wonder what the effect of the Planck scale is, since I said I ignored it. A Planck-energy photon will have a wavelength (i.e. "size") near the Planck length. This implies that the energy density of the region where the photon is located will be just large enough to form a black hole. :eek:
This is a pretty good reason to expect QED (and all other quantum field theories) to break down at the Planck scale.
Fredrik
Aug22-04, 09:10 PM
Reillly said'
Among the peculiarities of photons and QM, is the fact that the distribution of the photon's energy is spread out to regions where the probability of finding the photon is zero.
- Question, How can the probability of finding the photon be zero at the extremes of the distribution? Wouldn't they approach zero but never hit it
You are absolutely right, but the numbers we are talking about are so ridicilously small that for all practical purposes they are zero. One meter away from the expected position, the probability is less than
\frac{1}{10^{10^{30}}}
teal4two
Dec23-06, 01:19 AM
I have been wondering about the size of photons as well, or even what sense that question makes.
1) My microwave oven has metal mesh in the door. Can visible light photons from a bulb in the interior pass through the holes in the mesh? Can microwave photons pass through the holes in the mesh?
2) X-rays are scattered by crystals, but they are only scattered by part of the crystal. How big is this part?
2) X-rays are scattered by crystals, but they are only scattered by part of the crystal. How big is this part?
uh.. where'd you get this idea from?
No-one said it's a point particle.
Well, a photon IS a point particle...BUT in energy space. A photon is defined as a chunk of energy, nothing more. Ofcourse it is NOT defined as a point particle that has finite spatial boundaries, so asking about the "magnitude" of a photons is useless because this concept is defined (in most cases :wink:) based upon spatial coordinates.
Finally, the only photon quantity we know that is defined using spatial coordinates is the photon's wavelength. BUT, we are now talking about the wavelike properties of a photons while in paragraph one, we were using the particle like properties. Ofcourse, no problem with that thanks to the particle wave duality. We should however not forget that because we use a wavelike concept to describe a physical object (ie the photon) that we, intuitively, like to imagine as being a particle (ie a little ball :wink:).
So, one can use this concept to talk about "a photon's magnitude" as long as we keep in our minds what the special nature of the "photon versus magnitude" discussion is.
regards
marlon
teal4two
Dec23-06, 04:43 PM
uh.. where'd you get this idea from?Do you know anything about x-ray crystallography?
Crystals are regular geometric arrays of atoms (molecules, ions). The electrons of the atoms scatter x-rays and by measuring the position and intensity of the scattered rays, information about the structure can be obtained.
Often the structure is not homogeneous. If there is local disruption, the scattering pattern shows the average local structure. If the differences are being larger domains, each containing the same local structure, a superposition of the two patterns is observed. The difference between local disorder and larger scale differences seems like it has to be based on photon size;i.e., how much of the crystal scatters the x-ray.
1) If the entire crystal scattered (large photons), all we would see is the scattering pattern of the averaged structure.
2) If the photon was tiny compared to atoms, we would not see the interference pattern of spots.
Well, a photon IS a point particle...BUT in energy space. A photon is defined as a chunk of energy, nothing more. Ofcourse it is NOT defined as a point particle that has finite spatial boundaries, so asking about the "magnitude" of a photons is useless because this concept is defined (in most cases :wink:) based upon spatial coordinates.
Back in grad school, I was taught the same, but recently, there has been work on this by Margaret Hawton which says otherwise. The original was published in Phys Rev A, 59, 954 (1999):
http://prola.aps.org/abstract/PRA/v59/i2/p954_1
The latest arXiv preprint from the same author extends this:
In the standard formulation of quantum mechanics the real space wave function is the projection of the state vector onto an orthonormal basis of eigenvectors of a Hermitian position operator. However, it has been claimed since the early days of quantum mechanics that there is no position operator that defines such a basis for the photon. Here we will briefly review our recent work on the construction of a photon position operator and obtain a photon wave function by projecting onto its eigenvectors.
http://www.arxiv.org/abs/quant-ph/0609086
My explorations of QM have been in applications of geometric / Clifford algebra to density operator states, and it is very important to me that a position representation exist for the photon. Another Hawton article, that reads more closely to what I am doing because it discusses the gauge problem is:
Angular momentum and the geometrical gauge of localized photon states
Margaret Hawton, William E. Baylis, 2004
Localized photon states have non-zero angular momentum that varies with the non-unique choice of a transverse basis and is changed by gauge transformations of the geometric vector potential "a". The position operator must depend on the choice of gauge, but a complete gauge transformation of a physically distinct state has no observable effects. The potential "a" has a Dirac string singularity that is related to an optical vortex of the electric field.
http://www.arxiv.org/abs/quant-ph/0408017
Another article is:
Photon Position Operators and Localized Bases
Phys. Rev. A64, 012101 (2001)
Margaret Hawton, William E. Baylis
http://www.arxiv.org/abs/quant-ph/0101011
These articles have received very little attention. As an aside, Baylis seems to have his fingers in a remarkable number of interesting pies that intersect what I am working on:
Comments on "Dirac theory in spacetime algebra"
J. Phys. A: Math. Gen.35, 4791-4796
William E. Baylis, 2002
In contrast to formulations of the Dirac theory by Hestenes and by the current author, the formulation recently presented by W. P. Joyce [J. Phys. A: Math. Gen. 34 (2001) 1991--2005] is equivalent to the usual Dirac equation only in the case of vanishing mass. For nonzero mass, solutions to Joyce's equation can be solutions either of the Dirac equation in the Hestenes form or of the same equation with the sign of the mass reversed, and in general they are mixtures of the two possibilities. Because of this relationship, Joyce obtains twice as many linearly independent plane-wave solutions for a given momentum eigenvalue as exist in the conventional theory. A misconception about the symmetry of the Hestenes equation and the geometric significance of the algebraic spinors is also briefly discussed.
http://www.arxiv.org/abs/quant-ph/0202060
Hestenes geometrized spinors, I've been applying his techniques to the density matrix formalism. Here's a somewhat similar idea couauthored by Baylis, sort of half spinor and half "idempotent" theory:
A Geometric Basis for the Standard-Model Gauge Group
J. Phys. A. 34, 3309-3324 (2001).
Greg Trayling, William E. Baylis, 2001
http://www.arxiv.org/abs/hep-th/0103137
An introduction to the Clifford algebra / geometric algebra ideas we are using is here:
Relativity in Introductory Physics
William E. Baylis
http://www.arxiv.org/abs/physics/0406158
Carl
ZapperZ
Dec24-06, 04:56 AM
But to what extent is the "position" of a photon has anything to do with the "size" of a photon?
Note that anytime we detect a photon, be it on a CCD or via a PMT, we have narrowed down its position. So a photon's location in space is not meaningless concept simply based on experimental observation. However, this doesn't mean that its "size" is a meaningful concept.
Zz.
Do you know anything about x-ray crystallography?
Crystals are regular geometric arrays of atoms (molecules, ions). The electrons of the atoms scatter x-rays and by measuring the position and intensity of the scattered rays, information about the structure can be obtained.
Often the structure is not homogeneous. If there is local disruption, the scattering pattern shows the average local structure. If the differences are being larger domains, each containing the same local structure, a superposition of the two patterns is observed. The difference between local disorder and larger scale differences seems like it has to be based on photon size;i.e., how much of the crystal scatters the x-ray.
1) If the entire crystal scattered (large photons), all we would see is the scattering pattern of the averaged structure.
2) If the photon was tiny compared to atoms, we would not see the interference pattern of spots.
I've performed numerous XRD and SAXS experiments while working in an x-ray laboratory and taken many courses in solid state and x-ray physics so yes, I do know something about crystallography. When one derives the elastic scattering cross section, Bragg's law etc. the photons "size" does not come up in the calculation. Wavelength does but it doesn't equal size. As many others have pointed out it's meaningless to talk about a photons size. And photon-solid interactions are many body processes in the end so it's also pointless to talk about only a part of the solid being involved in the process. Unless of course one has a beam of a certain size and a sample of a certain size, but that's not what you were after.
teal4two
Dec24-06, 05:27 PM
However, this doesn't mean that its "size" is a meaningful concept.Let me rephrase an earlier question:
Light photons can pass through the metal mesh in a microwave oven; microwaves can't.
1) If this isn't a measure of something similar to "size" for a macroscopic object, then what is it a measure of?
2) If there are different techniques to measure this quantity, do they give similar (or at least relatively similar) values?
[If no explanation is possible in simple terms, please don't waste your time responding. The physicist Joyce referenced earlier seems as impenetrable as the literary one. I'm spinoring about eigenvectors on a string trying to gauge what any of that means.]
teal4two
Dec24-06, 05:33 PM
Thanks for the response Inha. It sounds like you have the background to answer my question. [The x-ray question I asked is not as intuitive to me as the microwave oven one.] If size doesn't matter (in the x-ray experiment), then what is this author (found via Google) referring to?
Dauter, Z. (2003). Acta Cryst. D59, 2004-2016.
"Introduction. ...
The size of the twin domains is large in comparison with the crystal cell dimensions, so that X-rays diffracted by the separate parts do not interfere and the total diffraction corresponds to the sum of intensities originating from individual domains. This is in contrast to disorder in a crystal, where there are differences in the location of the content of the neighbouring unit cells. In the case of disorder (either static or dynamic), diffracted X-rays represent the averaged (spatially or temporally) content of all unit cells."
Let me rephrase an earlier question:
Light photons can pass through the metal mesh in a microwave oven; microwaves can't. 1) If this isn't a measure of something similar to "size" for a macroscopic object, then what is it a measure of?
Wavelength. Electrons act similarly, but are still considered point particles with no size.
ZapperZ
Dec24-06, 09:24 PM
Let me rephrase an earlier question:
Light photons can pass through the metal mesh in a microwave oven; microwaves can't.
1) If this isn't a measure of something similar to "size" for a macroscopic object, then what is it a measure of?
It is a measure of the wavelength of the microwave. We have already had a discussion on the vagueness of associating this with "size". If you think that this is the "size" of a microwave photon, try making a slit of this size or smaller and tell me if you see any microwave passing through that slit.
2) If there are different techniques to measure this quantity, do they give similar (or at least relatively similar) values?
Different techniques? I'd settle for ONE technique that would claim to measure the size of a photon AND the physics that defines such a size.
Zz.
jnorman
Dec24-06, 11:10 PM
inre:
"In the standard formulation of quantum mechanics the real space wave function is the projection of the state vector onto an orthonormal basis of eigenvectors of a Hermitian position operator. "
about all i can say to that is HA HA HA - is that actually supposed to mean something?? oh, i guess i am just not smart enough to understand it....
Amr Morsi
Dec25-06, 01:02 AM
Hi,
Just to comment, it is the rest mass of photon which is zero, not its mass. Its mass is h*nu / c^2, where nu is the frequency of the "wave".
As believed today, according to the standard model, photons are point particles, i.e. they don't have size. I just also want to comment as previously mentioned in the thread, that there is a difference between real size and width of distribution of the probability function. No one has ever dealt with photon through Shrodinger Eq. (or even Dirac's.). QFT, especially QED, are concerning the quantized field due to the probability of the source (charges). Epsi Involved in the QED's equation are still for the particles (charges), not photons, which are the mediators.
Amr Morsi.
jnorman
Dec25-06, 03:30 AM
again, photons are NOT point particles. the concept of point particle implies localization. photons do not exhibit localization. yes, they have no size, but not because they are similar in some way to, for example, an electron. we simply cannot say anything meaningful about a photon in between the time it is emitted and the time it is absorbed.
Amr Morsi
Dec25-06, 10:29 AM
O.K. Jnorman,
Then, we agree that it has no size.
But, this (photon is not a particle as a consequence) will imply that Shrodinger's (or, Dirac's or, Klien-Gordon's) can't be applied upon the photon itself. But, they, however can be applied on particles, and can account for quantization (spreading) of photons (fields) only through QFT's.
As far as I can understand the word "localization" from your words, photon is an effect. Or, do you mean by localization that it cannot be said to be at a certain position (even if "probably")? If so, then this is the case of elementary particles.
Am I true?
Fredrik
Dec25-06, 11:06 AM
It is a measure of the wavelength of the microwave. We have already had a discussion on the vagueness of associating this with "size". If you think that this is the "size" of a microwave photon, try making a slit of this size or smaller and tell me if you see any microwave passing through that slit.
I don't think it makes any more sense to say that this is a measure of the wavelength of the photon. It's just as acceptable, or just as unacceptable if you prefer, to say that it's a measure of the size of the photon.
A photon state is in general a superposition of many wavelengths. The same state can also be described as a superposition of many different positions. If the position space wavefunction has a peak, it makes some sense to say that the width of this peak is "the size of the photon". It also makes some sense to say that the location of the peak of the momentum space wave function is "the momentum of the photon". Then we can also use the standard formulas to calculate a wavelength that corresponds to this "momentum", and call it "the wavelength of the photon".
None of the concepts "size", "position" and "wavelength" are well defined here, but with a little effort we can find things in the quantum mechanical formalism that correspond to all of these classical concepts. This is of course not the same as saying that the things we find are the size, position and momentum of a photon. Those concepts aren't well defined in quantum mechanics.
Fredrik
Dec25-06, 11:16 AM
again, photons are NOT point particles.
I would say that they are. I would also say that "point particle" means something different in quantum mechanics than in classical mechanics.
again, photons are NOT point particles. the concept of point particle implies localization. photons do not exhibit localization.
Photons are used to provide images on photographic plates. The images consist of a collection of spots that have absorbed photons.
yes, they have no size, but not because they are similar in some way to, for example, an electron. we simply cannot say anything meaningful about a photon in between the time it is emitted and the time it is absorbed.
In the quantum theory, one cannot say anything meaningful about an electron in between the time it is emitted and the time it is absorbed. This is the basis for the wave particle duality interpretation of the electron slit and interference experiments. The behavior of electrons and photons in QM is not easily distinguished.
On the other hand, the phrase "point" and "size" do not need to have any meaning in the quantum theory. They're just collections of words, the quantum theory is a bunch of rules on how to make calculations. Arguing about words may not be the best way to learn anything new.
I looked up "point particle" in wikipedia, and they said it means that the particle is assumed to be elementary, with no constituent parts. This definition would apply to photons as well as electrons. I think we've exhausted this particular discussion.
teal4two
Dec26-06, 08:19 PM
Wavelength. Electrons act similarly, but are still considered point particles with no size.Thanks. Do you have some resource(s) (website, book) to recommend for a more detailed, yet simple, explanation? [The websites listed at the start of this thread are dead.] To give you an idea of my current level of comprehension, I think of wavelength as being a property in the direction of propagation and thus it is hard to see how it is related to hole size - e.g., a pencil can fit through a hole much smaller than a pencil length.
ZapperZ
Dec26-06, 08:27 PM
A photon state is in general a superposition of many wavelengths. The same state can also be described as a superposition of many different positions. If the position space wavefunction has a peak, it makes some sense to say that the width of this peak is "the size of the photon". It also makes some sense to say that the location of the peak of the momentum space wave function is "the momentum of the photon". Then we can also use the standard formulas to calculate a wavelength that corresponds to this "momentum", and call it "the wavelength of the photon".
I disagree.
The "peak" in the position tells you the most probable location that the photon will be found. The "width" of that peak is in no way correspond to the "width" of that particle. It is simply the statistical width of where one would find that particle.
If what you say is true, then all those "resonance peaks" that we observe in high energy physics experiments would somehow imply the "size" of that particle. No such inference has ever been made in all those experiments. So your connection between the two is neither obvious, nor standard.
Zz.
Photons are used to provide images on photographic plates. The images consist of a collection of spots that have absorbed photons.
Hey CarlB,
I know you wrote the above post as a response to someone who wrote "point particle implies locality". Ofcourse this person is wrong because he mixes locality in coordinate space with locality in an energy basis. I looked through your references but i still do not get how my original post contains something incorrect. A photon is a point particle in energy space because it is a chunk of energy. Again, a point particle is defined as a particle with finite boundaries in some chosen base (spatial, energy, etc etc).
Concerning, the photons you "see" in a detector. What we see there is the result of an interaction between a photon and the detector. The spots we see do not, however, show us actual photons. So, IMO, saying that photons are seen through spots on a detector is NOT a reason to say that photons are not point particles (in a spatial base).
I looked up "point particle" in wikipedia, and they said it means that the particle is assumed to be elementary, with no constituent parts. This definition would apply to photons as well as electrons.
Well Wikipedia, i would not use this website (though it is a good one) for the definition of basic physical concepts.
I am sure you agree with me that if 99.9 % of the people talk about "point particle", they are referring to a particle with finite spatial boundaries. If we reread most of the posts in this very thread, we can only conclude that the same is happening here. I believe that we clearly need to make the distcintion between the spatial base and the energy base : THAT IS THE MAIN ASPECT HERE.
greets
marlon
Marlon,
Part of my problem here is that so many different arguments are very convincing. I don't see what's incorrect about your first post. But the belief that photons can only be point particles in the energy basis is so common that I wanted to bring up the Hawton result, which is not well known.
If one takes to heart the concept that the intermediate states cannot be described by the theory, (i.e. so as to avoid issues with the two slit interference problem), then the situation of a photon being emitted by a radioactive atom, and absorbed by a grain of photographic emulsion, seems to me to be about as close an example as one can have of a point emission and a point absorption.
In particular, the object that emits the photon can have dimensions far smaller than the wavelength of the photon, and it can be localized in space (before the emission) to dimensions far smaller than the emitted photon wavelength. (I.e. a heavy nucleus emitting a low energy photon.)
So it's really not surprising that there is a position representation for the photon. What is surprising is that it was denied for so long. It really makes me wonder how many other odd facts about QM are wrong.
Carl
jnorman
Dec27-06, 01:20 PM
carl - if i read cohen, et al, correctly, it sounds to me like they are implying that localization effect of photons in energy space may be an artifact of RMT strategy. http://physics.bgu.ac.il/~dcohen/ARCHIVE/wbr_PRL.pdf
still, energy space is hardly a concrete or measureable entity, and to indicate that photons demonstrate "point particle" characteristics similar to that of an electron (which can actually be identified as having a scpecific location, HUP notwithstanding), surely seems a misconception to me. i do not doubt that my understanding of this issue is incomplete, but i am just not following your logic here. i stand by my comment that you cannot say anything meaningful about a photon during its "travels" - it simply cannot be defined as having any location. i do agree, of course, that photons are always detected as particles, but this results from collapse of the probablilty function at the time of absorption, and not from any macro-perspective measureable characteristic of photons. feel free to elucidate further for my edification. danke.
teal4two
Dec27-06, 02:36 PM
Light scattering.
A visible light photon comes from the sun and enters the atmosphere. An atom in the atmosphere has electrons, whose positions can be described as probabilities (orbitals). The electric field of the photon can cause an electron in the atom to vibrate and in turn the electron radiates another photon.
What determines whether a given photon and a given electron will interact? Is this an all-or-nothing interaction (photon is either scattered or it isn't)? If the photon is on one side of the earth and the atom on the other, presumably there will be no interaction. How close do they have to be?
lightarrow
Dec28-06, 01:49 PM
Light scattering.
A visible light photon comes from the sun and enters the atmosphere. An atom in the atmosphere has electrons, whose positions can be described as probabilities (orbitals). The electric field of the photon can cause an electron in the atom to vibrate and in turn the electron radiates another photon.
What determines whether a given photon and a given electron will interact? Is this an all-or-nothing interaction (photon is either scattered or it isn't)? If the photon is on one side of the earth and the atom on the other, presumably there will be no interaction. How close do they have to be?The photon is the quantized interaction between light and matter. Just my opinion, of course.
No Virginia, there ain't no point particles, except on the pages of physics books and papers. Any particle with charge necessarily has an internal structure due to a cloud of (virtual) photons, e-p pairs, quarks, ....... it then follows that any particle, massless or massive cannot be a point particle, but necessarily has an internal structure.
The problem is: we don't know how to compute, even approximately, such structures, form factors, if you will. But the structure of QFT interactions says those structures must exist. Further, we can measure nucleon form factors with electron scattering, for example. But whatever structure electrons or photons have is beyond our capability to measure or compute -- other than basic QED phenomena like the perturbative results for the Lamb Shift or the anomalous magnetic moment of the electron.
Nonetheless, QFT almost always deals with point particles, for mathematical convenience. For the most part, the physics community accepts this approximation as appropriate for comon practice.
If one, for the moment, accepts point particles, then position operators for massive spinning particles, at least, are a bit peculiar. As is noted in the work referenced by CarlB, the Lorentz transformation of a standard position operator, +i d/dp, in 1-D with the momentum=p, gets tangled up with spin, and, more generally, with angular momentum. this happens because, in the Poincare Group, the generators of Lorentz transformations do not commute with the generators of rotation, which in turn causes the product of two Lorentz transformations to result in a rotation followed by a Lorentz xform, or vica versa.
So, the net effect is that under a Lorentz transformation, the transformed position of a particle depends on the spin state -- that is the transformed position of an spin-up electron at (T,X) will not be the same as the position of a spin-down similarly at (T,X). While I realize the dangers of ascribing classical ideas to the arcane phenomena of the quantum world, I think of this as suggesting an extended object -- and there is some classical justification for this, as discussed by Moller in his Relativity text -- a classical object with an internal angular momentum, the earth or a baseball thrown as a curveball,..., has a radius greater than |L|/(c M0), where L is the angular momentum, M0 is the rest mass, and c is the speed of light.(See section 64 in Moller's Theory of Relativity.)
It seems to be the case that these fine points of position, spin, and Lorentz transformations are, at present, more curiousities that anything else. but certainly, these ideas make the notions of positions of point particles with spin rather a dicey proposition.
Regards,
Reilly Atkinson
Careful
Dec30-06, 04:29 AM
If one, for the moment, accepts point particles, then position operators for massive spinning particles, at least, are a bit peculiar. As is noted in the work referenced by CarlB, the Lorentz transformation of a standard position operator, +i d/dp, in 1-D with the momentum=p, gets tangled up with spin, and, more generally, with angular momentum. this happens because, in the Poincare Group, the generators of Lorentz transformations do not commute with the generators of rotation, which in turn causes the product of two Lorentz transformations to result in a rotation followed by a Lorentz xform, or vica versa.
So, the net effect is that under a Lorentz transformation, the transformed position of a particle depends on the spin state -- that is the transformed position of an spin-up electron at (T,X) will not be the same as the position of a spin-down similarly at (T,X). While I realize the dangers of ascribing classical ideas to the arcane phenomena of the quantum world, I think of this as suggesting an extended object -- and there is some classical justification for this, as discussed by Moller in his Relativity text -- a classical object with an internal angular momentum, the earth or a baseball thrown as a curveball,..., has a radius greater than |L|/(c M0), where L is the angular momentum, M0 is the rest mass, and c is the speed of light.(See section 64 in Moller's Theory of Relativity.)
It seems to be the case that these fine points of position, spin, and Lorentz transformations are, at present, more curiousities that anything else. but certainly, these ideas make the notions of positions of point particles with spin rather a dicey proposition.
Regards,
Reilly Atkinson
I agree; let me first add that spin has to enter the position operator for massless particles if you want it to commute with the helicity operator.
On the other hand it wouldn't be unreasonable at all that the spin operator is included even for massive spinorial particles (apart from the obvious reason that you want the limit for m -> 0 to be well defined) : Zitterbewegung (due to spin) makes it hard to ``localize'' -say- an electron, so it is not unreasonable to think that position and spin operators are noncommuting. Indeed, in the book ``Relativistic quantum mechanics and introduction to field theory'', F.J. Ynduarain gives at page 191 a ``Newton - Wigner'' operator for spin 1/2 particles containing the spin operators.
http://www.pbs.org/wgbh/nova/programs/ht/tm/3501.html?site=10&pl=wmp&rate=hi&ch=10
Talks about slowing down light and length of photon.
ZapperZ
Jan27-08, 12:53 PM
http://www.pbs.org/wgbh/nova/programs/ht/tm/3501.html?site=10&pl=wmp&rate=hi&ch=10
Talks about slowing down light and length of photon.
I saw this when it first aired. Where exactly did it talk about "length of photon"? Maybe length of a light pulse, but not length of a photon.
You do know that this thread is more than a year old, don't you?
Zz.
I'd forgotten about this thread. Margaret Hawton's position operator was recently discussed briefly on sci.physics.research in the topic of the title
"EM Field of Photon":
http://groups.google.com/group/sci.physics.research/topics?hl=en
Hawton herself chimed in. I put together a blog post showing the relationship between her photon operator and the consistent histories interpretation of quantum mechanics and she linked it into the discussion. The blog post is here:
http://carlbrannen.wordpress.com/2008/01/14/consistent-histories-and-density-operator-formalism/
As you can see from the sci.physics.research discussion, the local expert is not convinced that a photon position operator makes sense, but he also makes it quite clear that he doesn't bother to read the articles that are linked in, LOL. This was from a few days ago, and I suspect that Hawton is busily working on another paper extending these ideas.
jambaugh
Jan27-08, 03:25 PM
According to Quantum Mechanics, a photon, like any other point particle, has no size at all. It is a point, if treated as a particle. If it is treated as a wave, it has wavelength and amplitude instead, amplitude being the "size".
Not quite. True it "has no size at all" but that means it nether has zero size nor some other value. It isn't a classical object with a size (even zero) while a "point particle" is such an animal. A photon is a process of quantized electro-magnetic energy transmission. This can happen in the very small volume of the nucleus of an atom or equally spread over a planet sized antenna array. "It has no size" in that size of any numerical value isn't a property of the photon.
agkyriak
Mar12-08, 10:14 AM
Well, a photon IS a point particle...BUT in energy space. A photon is defined as a chunk of energy, nothing more. Ofcourse it is NOT defined as a point particle that has finite spatial boundaries, so asking about the "magnitude" of a photons is useless because this concept is defined (in most cases :wink:) based upon spatial coordinates.
Finally, the only photon quantity we know that is defined using spatial coordinates is the photon's wavelength.
It is absolutely right. From a theoretical point of view the proof was given long years ago in the work of Landau and Peierls and confirmed recently in the works of other scientists Cook, Inagaki and others. Let us consider briefly this proof, using the book of (Akhiezer and Berestetskii, 1969):
The wave function of photon is here introduced as follows. The vectors of the EM field \vec {{\rm E}} and \vec {{\rm H}}, as the solutions of the wave equation of the second order, which follow from the Maxwell equations, are considered as the classical wave functions \vec {\varepsilon }\left({\vec {r},t} \right) and \vec {H}\left( {\vec {r},t} \right).
Representing the wave equation as multiplication of two equations for the advanced and retarded waves, we obtain two linear equations, which correspond to the wave vector \vec {f}_k and is a certain generalization of vectors of EM field. The equation for this function is equivalent to the system of the Maxwell equations. For this reason it is possible to consider the Maxwell's equation as the equation of one photon (Gersten, 2001). The quantization of classical wave function is produced by means of the quantization of energy of this wave by the introduction of the relationship \varepsilon =\hbar \omega . It turned out that in this case the function \vec {f}_k could be interpreted as the quantum wave function of photon in the momentum space.
But with the attempt to introduce the function of photon in the coordinate representation was revealed the insurmountable difficulty According to the analysis of Landau and Peierls (Landau and Peierls, 1930), and later of Cook (Cook, 1982a; 1982b) and Inagaki (Inagaki, 1994), the wave function of photon by its nature is nonlocal (see also the review (Bialynicki-Birula, 1994)). .
Actually, after completing the inverse Fourier transformation of above function \vec {f}_k we obtain:
\frac{1}{\left( {2\pi } \right)^3}\int {\vec {f}_k e^{i\vec {k}\vec {r}}d^3k=\vec {f}\left( {\vec {r},t} \right)} .
It seems that it is possible to determine \vec {f}\left( {\vec {r},t} \right) as the wave function of photon in the coordinate representation. Actually, because of normalization condition for \vec {f}_k the function \vec {f}\left( {\vec {r},t} \right) will be also normalized by the usual method: \int {\left|{\vec {f}\left( {\vec {r},t} \right)} \right|} ^2d^3x=1
However, the value \left| {f(\vec {r},t)} \right|^2 will not have the sense of the probability density distribution to find the photon at the given point of space. Actually, the presence of photon can be established only by its interaction with the charges.
This interaction is determined by the values of the EM field vectors \vec {{\rm E}} and \vec {{\rm H}} at the given point, but these fields are not determined by the value of the wave function \vec {f}\left( {\vec {r},t} \right) at the same point, and they are defined by its values in entire space.
In fact, the component of the Fourier field vectors, expressed byf_k , contain the factor \sqrt k . Formally this can be written down in the form
\[\vec {\varepsilon }\left( {\vec {r},t} \right)=\sqrt[4]{-\Delta }\vec {f}\left( {\vec {r},t} \right)\]
where \Delta is the Laplace operator. But \sqrt[4]{-\Delta } is integral operator, and therefore the relationship between \vec {\varepsilon }\left( {\vec {r},t} \right) and \vec {f}\left( {\vec {r},t} \right) is not local, but integral. In other words, \vec {f}(\vec {r},t) is not determined by field value \vec {{\rm E}}(\vec {r},t) at the same point, but it depends on field distribution IN A CERTAIN REGION, WHOSE SIZE IS THE ORDER OF WAVELENGTH.
This means that localization of photon in the smaller region is impossible and, therefore, the concept of the probability density distribution to find the photon at the fixed point of space does not have a sense.
This conclusion of theory is confirmed by experiment, since all measurements with the use of EM waves or photons (interference, diffraction and so forth) can be carried out to the region, not smaller as wavelength.
Akhiezer, A.I. and Berestetskiy, V.B. (1969). Quantum electrodynamics.
Bialynicki-Birula, Iwo (1994) On the wave function of the photon. Acta physica polonica, 86, 97-116),
Cook, R.J. (1982a). Photon dynamics. A25, 2164
Cook, R.J. (1982b). Lorentz covariance of photon dynamics. A26, 2754
Gersten, A. (2001) Maxwell of equation - the of one-photon of quantum of equation. Found. of Phys., Vol.31, No. 8, August).
Inagaki, T. (1994). Quantum-mechanical of approach to a of free of photon. Phys. Rev. A49, 2839.
Landau, L.D and Peierls, R. (1930). Quantenelekrtodynamik in konfigurationsraum. Zs. F. Phys., 62, 188.
peter0302
Mar12-08, 11:57 AM
The "amplitudes" of the EM field of a photon at any given point refer not to physical space in which there is a field, but rather to the strength of the EM field at that point, right?
Mentz114
Mar12-08, 03:31 PM
agkyriak :
Nice post. It's always good to see one's prejudices (only mass can be localised) supported by good maths. I've long argued against ascribing 'particle' properties to photons.
You do know that this thread is more than a year old, don't you?
lol - It's brand new to me. I've only just read just now.
In a year from my 'now' this thread will be a year old.
time must be some kind of illusion. The 'now' is very much like a point particle.
but interesting none the less :)
lightarrow
Mar29-08, 01:42 PM
The 'now' is very much like a point particle. ...or as a non-localized field.
Yes, it's a completely different point of view... :smile:
heterotic_
Jul5-10, 07:30 PM
I'm bringing this conversation back from the dead, because I feel like it has some sort of meaning.
If we talk about size in a non-conventional sense of the word, but as the average minimum amount of space it occupies, doesn't everything have to have a size? If two variables "try to" occupy the same amount of space, at points defined by x, y, z, in our universe, don't they interact (either by repulsion [in which they do not occupy the same space], attraction [bind to form something "new"] or annihilation), instead of occupying the same point in space?
Eynstone
Jul6-10, 11:47 PM
I wonder what we'd get if we tried to calculate the 'radius' of photon on the lines of the classical radius of electron.
On the other hand, don't we treat particles as points right at the outset of quantum mechanics?
heterotic_
Jul7-10, 12:06 AM
The actual radius of an electron is calculable, it is just dependent on its environment, so shouldn't this be also true of a photon?
Everything has a center, even if it is so small, its entire volume is its center. It is how far it is spread out that is changeable (or rather, how concentrated the energy is in relation to its center, or a "point") I believe that point must be the minimum space it can occupy, and therefore must have a size, even if we have no way to fathom it.
The wave function of photon is here introduced as follows. The vectors of the EM field \vec {{\rm E}} and \vec {{\rm H}}, as the solutions of the wave equation of the second order, which follow from the Maxwell equations, are considered as the classical wave functions \vec {\varepsilon }\left({\vec {r},t} \right) and \vec {H}\left( {\vec {r},t} \right).
That's a wonderful analysis! But isn't it dependent on the simple assumption that the photon IS the wave rather than the alternative that the photon is the structure that generates the wave?
P.S. Sorry, I didn't realize at first that that post was several years old.
I was also perplexed as to what possible answer could be there when we talk about the size of the photon. Even the concept of measuring the radius of the electron is not justified in quantum theory, because if we consider the electron to be a sphere with the charge smeared on its surface or throughout the volume of the sphere, then we will have to explain the spin of the electron in terms of the an actual spin, as we do for the case of, say, the earth's spin. But that is completely wrong description of the electron, as it results in absurd values of speed of the surface of the electron.
I think that quantum mechanics starts with the assumption that the elementary particles are point like objects, which do not have any dimensions. Furthermore, if we talk about quantum mechanics, then visualising the electrons, photons, etc. as classical particles would be a mistake. The only thing that we have to guide us in QM is the wavefunction of these supposed "particles", and their evolution according to the Schrodinger's equation.
heterotic_
Jul7-10, 01:12 PM
That's a wonderful analysis! But isn't it dependent on the simple assumption that the photon IS the wave rather than the alternative that the photon is the structure that generates the wave?
P.S. Sorry, I didn't realize at first that that post was several years old.
Redundancy would be silly, so I brought it back.
I would say that we cannot calculate the "classical" radius of a photon like an electron, because a photon always moves at the speed of light.
But what about calculating the mass of a photon depending on its energy? I know people say the photon has no mass, based on the definition of mass, but it does. It doesn't have a "rest mass", because it is never at rest, but the best definition of mass is that it is "the measure of inertia". Inertia is "the amount of resistance to change in velocity". Change in velocity is "an increase or decrease in Kinetic Energy (and Potential Energy, since Total energy is conserved)".
So, inertia is also "resistance to change in Energy"; the inertia of a body is dependent on its energy content. So mass then, is a measure of resistance to change in energy.
So instead of asking, can we calculate the mass of a photon, we are asking, can we calculate the measure of a photon's resistance to change in energy? YES.
Next, if we talk about the fact that a photon exhibits the characteristics of both a particle and wave, we cannot assume that a photon is only a wave. How it behaves is what it is. So how does it behave?
Now, bear with me, I'm only an undergrad, so if I'm way off base, feel free to correct me. If the photon has a center, and a wavelength, if we look at an electron, which has similar properties of a photon in that its radius is variable and it exhibits the wave-particle duality, we can get an idea of what a photon probably is. In a Hydrogen atom, the wavelength of its electron is equal to the Bohr circumference of the electron's orbit. Can we use this information to describe the duality of a photon?
We know that a photon is polarized and has a spin. So what if, it is a sphere (energy concentrated at a center, and pointing outward in all directions to create a force field) that swells and contracts? Wouldn't that explain the particle-wave duality? If this were true, the radius of a photon would then be variable, but at its swell would be its greatest radius, which we could calculate using its wavelength.
My thoughts:
r(photon-E) \leq \frac{ch}{2E}
and
m(photon-E) = \frac{h^2}{Ec}
I was also perplexed as to what possible answer could be there when we talk about the size of the photon. Even the concept of measuring the radius of the electron is not justified in quantum theory, because if we consider the electron to be a sphere with the charge smeared on its surface or throughout the volume of the sphere, then we will have to explain the spin of the electron in terms of the an actual spin, as we do for the case of, say, the earth's spin. But that is completely wrong description of the electron, as it results in absurd values of speed of the surface of the electron.
I think that quantum mechanics starts with the assumption that the elementary particles are point like objects, which do not have any dimensions. Furthermore, if we talk about quantum mechanics, then visualising the electrons, photons, etc. as classical particles would be a mistake. The only thing that we have to guide us in QM is the wavefunction of these supposed "particles", and their evolution according to the Schrodinger's equation.
My problem with QM and GR for that matter, is that they both work, but not really together. That just is not a good answer for me.
We talk about the spin of an electron and a photon, but what if it's not a spin, like you said in terms of the Earth? What if it is more like... the electricity shifting in a uniform motion throughout its volume in one direction or the other, as a result of the polarization?
From this book: Lecture Notes in Physics: The Nature of the Elementary Particle, 1978. Malcolm H. MacGregor, it is stated (in quotes):
"The particle properties of the photon emerge most clearly from Compton scattering, in which a high-energy photon makes a billiard-ball type of collision with an electron, and hence delivers its energy and momentum into a volume that has dimensions on the order of 10^-11 cm (the Compton wavelength of the electron)."
The "particle" aspect of the photon from Compton scattering experiments represents a "MeV photon" compared to an "optical-frequency wave photon" that is about 10^6 times larger in dimension.
"In dealing with the photon, we must take into account both its wave aspects and its particle aspects. But since these two aspects differ dimensionally by many orders of magnitude, they are in a practical sense separate entities, so that we can discuss the wave properties of the photon without having to consider its particle properties, and vice versa."
heterotic_
Jul7-10, 02:36 PM
From this book: Lecture Notes in Physics: The Nature of the Elementary Particle, 1978. Malcolm H. MacGregor, it is stated (in quotes):
"The particle properties of the photon emerge most clearly from Compton scattering, in which a high-energy photon makes a billiard-ball type of collision with an electron, and hence delivers its energy and momentum into a volume that has dimensions on the order of 10^-11 cm (the Compton wavelength of the electron)."
The "particle" aspect of the photon from Compton scattering experiments represents a "MeV photon" compared to an "optical-frequency wave photon" that is about 10^6 times larger in dimension.
"In dealing with the photon, we must take into account both its wave aspects and its particle aspects. But since these two aspects differ dimensionally by many orders of magnitude, they are in a practical sense separate entities, so that we can discuss the wave properties of the photon without having to consider its particle properties, and vice versa."
Realistically, that is only a good conclusion when you discuss the outcomes the two produce independent of each other.
However, we are discussing bridging the gap, so saying not to, is just illogical. If we think about what a wave really is, we envision that oscillating line we produce with technology, or a ripple from a linear perspective so we see the amplitude of a physical wave. The problem with that is that it is a perspective, it isn't the entirety of the actual wave itself. It is only a description from a reference point, and requires ignoring the rest of the "picture", as it were. It is also representing a point · moving across an oscillating line from the "side", but a straight line from the "top". You really cannot picture the wave that way, it is merely a representation to represent wavelength and amplitude, based on what we observe as physical waves from a sideview, like the water oscillating in a ripple.
I think a "real view" of the wave would produce a different picture more like:
O>·<O>·<O>·<O>·<O>·<O
With that (granted, think of the swells with more of the same wave length of a particular frequency), we could even find the radii of all the "sizes" that the photon takes on spherical calculations of each moment of the swell.
Pinewater
Sep22-10, 01:57 AM
Just an fyi, in scientific literature the 'size' of a photon generally refers to its coherence length. (which is definitely NOT to say that this is the only correct way to think about a photon's size!)
You can look up coherence length on the web, but I understand it best in the context of an interferometer. In this setting, the coherence length is the maximum distance that a single photon can travel and still interfere with itself (which means that it produces an interference pattern at the screen).
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