Are Photons Actually Infinitely Small Particles?

[A. Neumaier]

Sure if choosing between those two it could be detailed and difficult, but a third option not mentioned is partculate photon.

What is a particulate photon?

 

[Q-reeus]

Absolutely not. I was making the point that you would have to put enough energy into the metal to get it to white heat (thermionic, if you like) to do the job, for one electron that just one optical photon can do.

OK' agreed - I hadn't understood your point properly earlier. So, going back to your concluding remarks in #26, should one conclude we have to get by with a mathematical model having no conceptually clear physical structure? Appears so from most participants remarks.

 

[Q-reeus]

What is a particulate photon?

I had misspelt it, but I'm sure you know what was meant - a highly localized entity assumed to contain the entire energy/mommentum of said field quantum. No chance that was a loaded question?

 

[A. Neumaier]

I had misspelt it, but I'm sure you know what was meant - a highly localized entity assumed to contain the entire energy/mommentum of said field quantum. No chance that was a loaded question?

If it is a single photon, spread out over a large sphere, then there is essentially no chance to detect it
experimentally except by putting very sensitive detectors on a large fraction of the sphere. And if you get somewhere a recording event, how do you know that it came from your source and not from somewhere else, unless you ensure that the whole huge sphere you were entertaining in your thought experiment is completely dark - an impossibility on a large scale.

Thus your ''particular photon'' assumption is quite infeasible to test.

 

[sophiecentaur]

OK' agreed - I hadn't understood your point properly earlier. So, going back to your concluding remarks in #26, should one conclude we have to get by with a mathematical model having no conceptually clear physical structure? Appears so from most participants remarks.

I think so. It's too much of a luxury to expect everything in Science to relate back to what we already know in a comfortable way. There was the same problem in appreciating the consequences of SR, surely, and we have (mostly) got over that by now.

 

[Naty1]

A photon is the size of the appropriately vibrating (bosonic) string.

So in this view it's size depends on which dimension you measure.

 

[Q-reeus]

If it is a single photon, spread out over a large sphere, then there is essentially no chance to detect it
experimentally except by putting very sensitive detectors on a large fraction of the sphere. And if you get somewhere a recording event, how do you know that it came from your source and not from somewhere else, unless you ensure that the whole huge sphere you were entertaining in your thought experiment is completely dark - an impossibility on a large scale.
Thus your ''particular photon'' assumption is quite infeasible to test.

Suspect there has been a mistranslation: 'particulate photon' simply meant 'photon(s) as particle(s)', not as you seem to have interpreted it 'a particular (i.e. single) photon'. Without computing what realistic heat loss rates from screen to environmet might be for a [STRIKE]particular[/STRIKE] specific setup, it nonetheless seems likely one could have say hundreds or thousands of field quanta impinging per second, yet that input power is lost to the environment at a rate too high for any significant energy accumulation in the screen between photoejection events. Not a problem for particle picture, since all the needed photoemission energy is there at point of any given impact. Anyway we are arguing around in circles on this - if I can find some cryogenic photoelectric effect studies, will post on it.
BTW, interested to know how you would explain gamma ray scintillation detection on a spreading wave basis - assuming a weak/distant gamma ray source?

 

[MikeGomez]

Good thread. Missed it by a couple of days, but if posters are still interested I have a couple of questions which I think are closely related to what the OP was asking.

We have a wall of material which is opaque (as can be) to visible light as well as radio frequencies. First we pass some visible light through a 1 inch diameter hole such that we can see the spot where the light hits a screen behind the wall. Note that I would like to know the nature of individual photons, but I have said that we can see a spot on the screen which of course implies many photons. So actually no, we don’t see the spot with our eyes, we detect photons individually using some apparatus.

The main point about using visible light in this example is that wavelength is much smaller than the diameter of the hole. So then, what about when we attempt to pass photons of radio frequency through the hole of the same size?

In this case of course, we have some different kind of apparatus which can detect these longer wavelength photons (perhaps a receiving antenna instead of a screen). Now the wavelength of the photon is much larger than is the diameter of the hole. But does that matter? Will the antenna still detect our radio photon?

 

[jtbell]

The main point about using visible light in this example is that wavelength is much smaller than the diameter of the hole. So then, what about when we attempt to pass photons of radio frequency through the hole of the same size?

With a large number of photons, we get a diffraction pattern just like the one we get with light when we use a much smaller hole:

http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/cirapp.html

Single photons arrive at the screen or detector randomly according to a probability distribution which is just the classical intensity distribution:

http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/cirapp2.html#c2

As the wavelength increases, for the same diameter hole, the width of the central maximum increases. With radio waves (e.g. UHF with a wavelength of around 0.2 m) going through a hole 0.02 m in diameter, the central peak of the diffraction pattern more than fills the entire forward hemisphere beyond the aperture, so the aperture behaves almost as a "point" source of radio waves.

 

[sophiecentaur]

As you say, light has a very short wavelength and we instinctively picture a 'hole' as being bigger than one wavelength. The 'bullets idea' then very easily follows. But, if you follow the wave model, any gap will have a diffraction pattern - so some energy must get through. The actual amount of energy would be given by the energy flux density times the area of the hole. That (from hf) tells you how many photons must be getting through every second. This is right for 1500m wavelength em and a keyhole. The pattern of such a narrow aperture will be more or less hemispherical - that implies that photons are equally likely to be detected (the Archers of Radio 4 Long wave, even) at any angle.
I really do love waves. Such uncomplicated things.

 

[Q-reeus]

In this case of course, we have some different kind of apparatus which can detect these longer wavelength photons (perhaps a receiving antenna instead of a screen). Now the wavelength of the photon is much larger than is the diameter of the hole. But does that matter? Will the antenna still detect our radio photon?

A lot will depend on the details of the opaque plate - is it a good conductor, is it thick or thin wrt hole diameter. Assuming a thin metallic plate, and hole diameter << wavelength, a wave incident normal to the plate will induce currents making the hole act as an effective magnetic dipole oscillator that leaks a small amplitude wave through. Microwave theory shows the amplitude is proportional to the cube of aperture diameter. In terms of photons, one would have to translate that into a probability of transmission. If the plate is relatively thick wrt aperture size, then transmission is greatly reduced, the aperture then acting as a small length of below-cutoff waveguide. I think, but not completely sure, the angular intensity profile for a small aperture in a screen of large extent is just that of a small dipole oscillator.
[I see you already have some answers in #39 & 40]

 

[edguy99]

As you say, light has a very short wavelength and we instinctively picture a 'hole' as being bigger than one wavelength. The 'bullets idea' then very easily follows. But, if you follow the wave model, any gap will have a diffraction pattern - so some energy must get through. The actual amount of energy would be given by the energy flux density times the area of the hole. That (from hf) tells you how many photons must be getting through every second. This is right for 1500m wavelength em and a keyhole. The pattern of such a narrow aperture will be more or less hemispherical - that implies that photons are equally likely to be detected (the Archers of Radio 4 Long wave, even) at any angle.
I really do love waves. Such uncomplicated things.

Very clearly stated. The bullet idea is a good start but clearly needs to be expanded to properly explain things since much more is known about photons in todays world. Somehow that bullet, in addition to direction, speed and momentum/energy, has to have a number of additional properites:

1. spin that it appears to impart to electrons
2. some internal energy
3. a polarization component
4. finally, a phase component that varies with time and space.

Consider a photon with a phase component (it knows where it is in the wave equation). It is very tiny (less then 3 fm), flies through the air at the speed of light, but on a periodic basis expands in its direction of travel, up to 1/2 its wave length in size (in this case 315nm), then down to tiny again. This picture shows a 630 nm photon (normal helium/neon laser) trapped in a cavity that is 1260 nm big bouncing back and forth between 2 mirrors.

This type of photon should get through the tiny keyhole and would undergo defraction like a wave since it basically tracks a wave.

 

[Q-reeus]

...Consider a photon with a phase component (it knows where it is in the wave equation). It is very tiny (less then 3 fm), flies through the air at the speed of light, but on a periodic basis expands in its direction of travel, up to 1/2 its wave length in size (in this case 315nm), then down to tiny again...

Don't want to sound too sceptical but that concept is certainly new to me! Can you cite a published article expounding this model in some detail?

 

[edguy99]

Don't want to sound too sceptical but that concept is certainly new to me! Can you cite a published article expounding this model in some detail?

One that comes to mind is a series of lectures http://video.google.com/videoplay?docid=1501838765715417418#.

I don't remember the location, but he talks extensively about the "probability" of an event happening, varies with time as the photon moves through space. Specifically when talking about the probability of reflection from a surface or a small layer of glass. If the layer is 1/2 wavelength wide, then the photon has a low probability of passing. If the layer of glass is a full wavelength wide (or multiples of that wavelength), the photon has a higher probability of passing through. The picture represents this probability as a "length" that varies over time: long line = high probability of reflection, short line = low probability of reflection

 

[sophiecentaur]

It strikes me that people still seem to want to hang on to a particle view of the photon which is 'familiar' and classical, rather than just to treat it as a point (no extent) that just interacts with other particles or systems according to the state of a wave at any point. How does it help in any way to give it an extent that would vary, according to circumstances? It seems to be required to 'bend to fit' each different circumstance.

Polarisation seems a red herring to me - taken care of by the wave model, entirely (afaics). A classical wave with circular polarisation carries angular momentum so the statistics of the photons 'in' that wave allows the photons to have spin. Linear polarisation can be regarded as consisting of suitably paired spinning photons.
The phase of a traveling or standing wave can predict how the photon is likely to react (statistics again). Imagine a 50Hz wave and then it's obvious that you will get more interactions around the peaks of voltage rather than at the zero crossings. Again, using wavelengths that are not optical, can open the view of what goes on; does a particle that would have to extend to most of a country make sense in a capacitor that is 2cm long?

I started this thread in an attempt to put to bed the very classical picture of a photon that most people seem to want (going so far as to construct animations, even). It seems to be a very diehard concept; almost like angels of a pinhead.

 

[Cthugha]

Consider a photon with a phase component (it knows where it is in the wave equation). It is very tiny (less then 3 fm), flies through the air at the speed of light, but on a periodic basis expands in its direction of travel, up to 1/2 its wave length in size (in this case 315nm), then down to tiny again.

The phase of a traveling or standing wave can predict how the photon is likely to react (statistics again).

Here one really has to take care. You get something similar to an uncertainty relation when discussing phase and that is an uncertainty in the field quadratures which roughly translates into an uncertainty between phase and photon number. So if you really want to talk about a single photon (n=1), then the phase of the corresponding light field is ill defined. If you talk about single detection events of a coherent field, its ok.

 

[YummyFur]

Just how 'big' is a photon?

That sounds suspiciously like 'how big is a photon, really'

 

[sophiecentaur]

Don't you recognise irony when you see it, dear boy?

 

[YummyFur]

Of course. The above post is ironic. I think you mean 'facetious'. [wink]

 

[sophiecentaur]

Here one really has to take care. You get something similar to an uncertainty relation when discussing phase and that is an uncertainty in the field quadratures which roughly translates into an uncertainty between phase and photon number. So if you really want to talk about a single photon (n=1), then the phase of the corresponding light field is ill defined. If you talk about single detection events of a coherent field, its ok.

Well yes, of course. But a 'single' photon would be part of some experiment which would involve a system which was defined in some way (a notional transmitter and a detector and some geometry). Without that, you couldn't decide on an energy for the photon or anything about the photon that could suggest some likely statistics for it (whether it was traveling through free space or whether it was in a cavity etc.). Phase always refers to some time origin so, unless you could be more specific then the concept of the phase of a single photon would have no meaning - unless you could say something about when it was 'created'.
This all confirms my opinion that the photon concept is not clear in peoples' minds. The only thing we can say fairly definitely is that a photon is a defined amount of energy that can be transferred when em power interacts with a system. The geometry (other than that which is used in the context of the wave) seems almost irrelevant to me.

 

[sophiecentaur]

Of course. The above post is ironic. I think you mean 'facetious'. [wink]

I have really enjoyed this thread. I'd been meaning to start one like this for ages.

That's a really really, really.
It's almost a 'model' thread.

 

[Cthugha]

Phase always refers to some time origin so, unless you could be more specific then the concept of the phase of a single photon would have no meaning - unless you could say something about when it was 'created'.

Yes, of course. What I was trying to say was that even for an ensemble of measurements performed on equally prepared single photons, the phase is ill defined. The phase of a single photon indeed has no meaning.

The only thing we can say fairly definitely is that a photon is a defined amount of energy that can be transferred when em power interacts with a system.

In fact, not even the exact amount of energy is well defined as most experimentally realized single photon sources are pretty broad spectrally. However, we can say that the energy transfer is quantized. While this is more or less indeed the only thing we can say fairly definitely, I think it is already quite a remarkable and non-trivial finding.

Nevertheless the "classical concept" of a photon, as you call it, should indeed be dead and buried.

 

[sophiecentaur]

Nevertheless the "classical concept" of a photon, as you call it, should indeed be dead and buried.

That's ma boy!
If only the unthinking public could say as much.

 

[Q-reeus]

Polarisation seems a red herring to me - taken care of by the wave model, entirely (afaics). A classical wave with circular polarisation carries angular momentum so the statistics of the photons 'in' that wave allows the photons to have spin. Linear polarisation can be regarded as consisting of suitably paired spinning photons.

Last bit is fine, but of course for a classical EM plane wave that works both ways - circular polarization can be decomposed into orthogonal linearly polarized waves. The matter of associating spin = intrinsic angular momentum, with CP (circular polarization) is a bit tricky. While it's easy to show that say crossed dipole antennas as a source of CP waves react on each other to give a net time-averaged torque, there is a seemingly paradoxical lack of any EM reaction torque when a normal incident CP wave is absorbed by a resistive sheet say. Which makes it very hard to reconcile photon spin with field CP. One is forced to find the field angular momentum as due to a net non-radial component in the Poynting vector of the combined radiation field of the CP source emitter.

Consequently just how a 'point' particle photon can carry an intrinsic angular momentum, apart from mathematical postulate, is hard if not impossible to visualize. In the case of an electron say, there is this concept of the 'dressed' charge and spin angular momentum might be considered as residing in some finite effective volume of virtual particles surrounding the 'bare' charge. But for a photon - is there a feasible 'point particle corkscrew motion' model applicable, or must one accept sheer mathematical postulate only? But then I suppose we are meant to accept the lesson is 'stop trying to visualize - there are no physical models that work, period!'

...Again, using wavelengths that are not optical, can open the view of what goes on; does a particle that would have to extend to most of a country make sense in a capacitor that is 2cm long?

Not sure of the point here. At 50Hz the capacitor field is local = near-field = virtual photons right out to such distances it becomes negligible. There is some finite radiation, but incredibly weak.

 

[sophiecentaur]

Last bit is fine, but of course for a classical EM plane wave that works both ways - circular polarization can be decomposed into orthogonal linearly polarized waves. The matter of associating spin = intrinsic angular momentum, with CP (circular polarization) is a bit tricky. While it's easy to show that say crossed dipole antennas as a source of CP waves react on each other to give a net time-averaged torque, there is a seemingly paradoxical lack of any EM reaction torque when a normal incident CP wave is absorbed by a resistive sheet say. Which makes it very hard to reconcile photon spin with field CP. One is forced to find the field angular momentum as due to a net non-radial component in the Poynting vector of the combined radiation field of the CP source emitter.

Consequently just how a 'point' particle photon can carry an intrinsic angular momentum, apart from mathematical postulate, is hard if not impossible to visualize. In the case of an electron say, there is this concept of the 'dressed' charge and spin angular momentum might be considered as residing in some finite effective volume of virtual particles surrounding the 'bare' charge. But for a photon - is there a feasible 'point particle corkscrew motion' model applicable, or must one accept sheer mathematical postulate only? But then I suppose we are meant to accept the lesson is 'stop trying to visualize - there are no physical models that work, period!'

So - no classical torque resulting from CP absorption? Awkward. Would there be no circular induced currents, to account for it?

Not sure of the point here. At 50Hz the capacitor field is local = near-field = virtual photons right out to such distances it becomes negligible. There is some finite radiation, but incredibly weak.

My point was that one has to 'bend' the geometrical characteristic of the photon in order to 'fit' the practical situation. You have introduced the notion of virtual photons to take care of what you call near field and that could be ok, I suppose. If you look at the fields due to a transmitting dipole, the fields change from E & H in quadrature in the near field and E & H in phase in the far field. The virtual photons presumably relate to this local quadrature fields?

There still seems, to me, to be a wierdness with wanting photons to, somehow, be different from individual to individual. A photon that is sourced in a distant star must, surely, be identical to one sourced locally if the two of them can interact in the same way with the same receiver.

 

[Q-reeus]

So - no classical torque resulting from CP absorption? Awkward. Would there be no circular induced currents, to account for it?

Well there is a net circular acting sheet current, but the magnetic interaction, making the usual assumption purely transverse motion of charges applies, yields no net torque. This is most easily seen by considering the CP wave as two spatially and temporally orthogonal linearly polarized plane waves. For each such component, E and B are mutually orthogonal and transverse to the propagation vector k. Consequently the B component of one wave is parallel to the current (which is in the direction of E) induced by the other wave. And as we know from Lorentz force law, when J and B are parallel, there is no magnetic force. Hence no mutual interaction, regardless of relative phase.

When field strengths get very high and frequency is not too great, as for EM propagation through a weakly ionized plasma, there is an induced longitudinal motion, but it tends to be very small wrt to transverse motion.
It may be of comfort to know some researchers can show a CP wave will induce angular momentum in media, in some circumstances at least: http://www.opticsinfobase.org/abstract.cfm?URI=oe-13-14-5315 (unlocked file, so can be downloaded)

My point was that one has to 'bend' the geometrical characteristic of the photon in order to 'fit' the practical situation. You have introduced the notion of virtual photons to take care of what you call near field and that could be ok, I suppose.

Well vp's are not my original idea - and I'm aware the notion is hotly disputed here at PF/QM. Was merely equating 'accepted' terminology. Personally I'm agnostic as to reality of vp's - out of sheer ignorance of all the subtle arguments if nothing else.

If you look at the fields due to a transmitting dipole, the fields change from E & H in quadrature in the near field and E & H in phase in the far field. The virtual photons presumably relate to this local quadrature fields?

That more or less fits many peoples view - quadrature = zero time-averaged Poynting vector = reactive field(s) = 'virtual photons'. We left out static fields but let's not go there!

There still seems, to me, to be a wierdness with wanting photons to, somehow, be different from individual to individual. A photon that is sourced in a distant star must, surely, be identical to one sourced locally if the two of them can interact in the same way with the same receiver.

I agree. Yet while the search for a physical model that is universally applicable may be futile, surely along the way we can gain insights, if nothing else by eliminating models that just do *not* work other than on an ad hoc basis.

 

[Dickfore]

Photons do not even have a position, because, unlike non-relativistic QM, in relativistic QFT there is no probabilistic meaning attached to the wave function considered as a function of space-time coordinates.

Instead, its meaning is revealed in Second Quantization. Namely, the wave function gets promoted to a field operator that creates (annihilates) at a particular point in space, at a particular instant in time.

In QM, as well as QFT, elementary particles (those with which we associate a wave function or a field) are truly point particles. So, even if you had asked what is the size of the electron, you would have gotten an answer that it is a point particle.

 

[MikeGomez]

Re my question in post #38.

With a large number of photons, we get a diffraction pattern just like the one we get with light when we use a much smaller hole:

http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/cirapp.html

Single photons arrive at the screen or detector randomly according to a probability distribution which is just the classical intensity distribution:

http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/cirapp2.html#c2

As the wavelength increases, for the same diameter hole, the width of the central maximum increases. With radio waves (e.g. UHF with a wavelength of around 0.2 m) going through a hole 0.02 m in diameter, the central peak of the diffraction pattern more than fills the entire forward hemisphere beyond the aperture, so the aperture behaves almost as a "point" source of radio waves.

O.k. so with the radio waves I have simply described circular aperture diffraction at a different scale (than visible light). But what I was really wondering about is what you called the classical intensity distribution. The number of photons received, not the diffraction pattern.

If the wavelength is 100 times the hole diameter, and this causes a severe reduction in the probability of receiving a given photon, then that does seem to imply that a photon is somehow extended perpendicular to its direction of motion. The photon was directed at the center of the hole, let’s say in a direction we call the Z-axis. Now the photon’s ability to pass through in the XY plane depends on its wavelength!

I can only think of two ways this can happen.

1: The photon possesses a virtual field which extends perpendicular to its direction of motion, which travels faster than the speed of light in order to detect the edge of the hole and react with it. No, I don’t really believe this, so please don’t think I’m a crackpot.

2: The space that the hole occupies possesses virtual EM fields (ZPE). The virtual fields already know how big the hole is long before the real photon arrives. By this I mean that the virtual fields have natural preferred resonance frequencies which correspond with the size of the hole, due to interactions within the inside edges.

The real photon is some manifestation of the EM field, and it therefore seems to me quite reasonable for it to react with the virtual EM fields in such a way as to create diffraction patterns, probability distributions, etc.

Or are there other explanations?

 

[Q-reeus]

Photons do not even have a position, because, unlike non-relativistic QM, in relativistic QFT there is no probabilistic meaning attached to the wave function considered as a function of space-time coordinates.

Instead, its meaning is revealed in Second Quantization. Namely, the wave function gets promoted to a field operator that creates (annihilates) at a particular point in space, at a particular instant in time.

Which in order to square with Lorentz invariance, would seem to demand truly point particles, yes?
I'm wondering if QFT creation/annihilation is somehow the answer to the problem I posed here https://www.physicsforums.com/showthread.php?t=457544 but still without takers second time around (shameless plug :shy:)?

In QM, as well as QFT, elementary particles (those with which we associate a wave function or a field) are truly point particles. So, even if you had asked what is the size of the electron, you would have gotten an answer that it is a point particle.

Implying that for electron, virtual particle dressing cannot be invoked as seat of intrinsic angular momentum? If so this is a little disturbing because it signifies a total departure from any classical notion of what angular momentum entails at minimum - a finite moment arm! Showing my ignorance of QM/QFT here.

 

[Dickfore]

angular momentum in quantum mechanics is the generator of rotations and is not defined as in classical mechanics. But, you are derailing this thread.