Comparing Photons & EM Waves in Physics

[mysearch]

I am trying to review some basic physics, purely for my own interest, and have been looking at particle physics in the form of the Bohr model and EM theory in the form of Maxwell’s equation. While I still have a way to go on these topics, I was wondering if anybody is in a position to help me with some fairly fundamental issues concerning the wave-particle duality of light in vacuum?

I realize that the debate about the wave-particle duality has been rumbling on since the time of Newton and Huygens. However, initially when I started out, I got the impression that light could be considered in terms of either a wave or particle model, but having now looked into the some of the details, it appears that `duality` only applies under certain circumstances, e.g. refraction doesn’t appear to work very well for the particle model, but is explained by the wave model. The following questions are only intended to convey the nature of some of the issues I am trying to resolve in my own mind:

1. Are EM waves are only generated by `free` charged particles?
2. Are photons are only generated by electrons bound to a nucleus?​

It is difficult to summarise all the issues in 1 post and I guess it will depend on whether anybody is interested in this specific topic as to whether it is worth listing further details in subsequent posts. However, I will try to initially rationalise some of the background issues I looking to confirm:

o An EM wave has both electric and magnetic fields, which oscillate in phase, but perpendicular to each other and to the direction of energy propagation?

o An electric field is a quantity that exists between two-charge particles rather than a quantity associated with one particle?

o The only way the electric field strength would oscillate in value, at a given point is if the charge particle itself was oscillating around a central point?

o A magnetic field results in the presence of a changing or moving of an electric charge?​

I am assuming that within the classical model, the charge of the ‘orbiting` electrons might be neutralised by the charge of the central proton(s) and, as such, an atom would have no net charge that would radiate an EM wave? If so, do only ‘free` charged structures in motion, e.g. electrons or ions, create a radiating EM wave?

o A photon is emitted when an electron transitions to a lower energy orbit and absorbed when an electron transitions to a higher energy orbit within an atom?

o The energy of a photon is described as a discrete quantum of energy [E=hf]?

o A photon has no rest mass, but a kinetic mass m=E/c^2 and no externalised charge?​

There doesn’t seem to be accepted or definitive structure of a photon, but it is said to have no charge, but somehow maintains both an electric and magnetic fields. However, a discrete photon seems to be very different to an EM wave that is radiating energy, in all directions, from a moving charged source?

o Photons are discrete entities that propagate in one direction?

o Do the electric and magnetic fields within a photon oscillate in phase?

o If so, there must be a point in the cycle where there is both zero E-M field values and therefore no forward propagation?

o How is the photon energy E=hf reconciled to the energy associated with the oscillating E-M fields, i.e. is it aggregated over 1 cycle?​

This is probably more than enough to see if anybody is interested in discussing these issues any further. Thanks

 

[pallidin]

In my opinion, you have done extremely well with posting your concerns. Much better than I have seen from some others. Good job!

Perhaps those more knowledgeable than I can jump-in and selectively comment.

 

[ZapperZ]

=

1. Are EM waves are only generated by `free` charged particles?
2. Are photons are only generated by electrons bound to a nucleus?​

There's simply way too many things to tackle here, so I'm going to address just this two issues, which appear to be central to your question.

1 and 2 are self-contradictory. EM waves ARE photons.

If you go to a synchrotron center all over the world, they generate EM waves by using free electrons. Yet, if you look at what they do with these EM waves, you'll see that they often use it in the form of "photons", i.e. they can use it to do photoemission experiment, etc.

Light is light, no matter HOW they are generated. However, depending on what you use it for, and depending on its nature (for example, it's wavelength), then it may be easier to treat it as "photons" or as classical waves. However, it doesn't mean that there isn't ONE single consistent formulation for ALL EM waves. This has already been explained in the FAQ in the General Physics forum.

Furthermore, many other so-called wave-like behavior such as diffraction and multiple-slit interference can be derived and described via QM description (do a search on the Marcella paper that I've referenced many times). The reason why we don't use that often when we deal with the typical interference effect is because it is tedious, and the classical wave picture is more convenient and easier for students just learning physics. However, just because we switch back and forth doesn't mean that there isn't a means to describe all light phenomena using just ONE single consistent description.

Zz.

 

[cmos]

I'll try to answer some of your questions in a very brief and concise manner. I won't go into any great detail and I hope that you are able to reference any doubts you might have.

o An EM wave has both electric and magnetic fields, which oscillate in phase, but perpendicular to each other and to the direction of energy propagation?

True under most elementary cases; however, there are exceptions (e.g. the magnetic field lags behind the electric field for wave propagation through a conductor).

o An electric field is a quantity that exists between two-charge particles rather than a quantity associated with one particle?

False. The electric field of a single point charge 'q' goes as q/r²; I'm sure you have seen this. The initial concept for a field is to uncouple the necessity for two particles in a theory of force. A force requires two particles, but the field only requires one. The force between two particles is then the effect of placing one particle into the field of the other and vice versa.

o The only way the electric field strength would oscillate in value, at a given point is if the charge particle itself was oscillating around a central point?

For an isolated charge, yes.

o A magnetic field results in the presence of a changing or moving of an electric charge?

True. But it can also result from a changing electric field (Maxwell's displacement current).

o A photon is emitted when an electron transitions to a lower energy orbit and absorbed when an electron transitions to a higher energy orbit within an atom?

True.

o The energy of a photon is described as a discrete quantum of energy [E=hf]?

True.

o A photon has no rest mass, but a kinetic mass m=E/c^2 and no externalised charge?

False. A photon is a massless particle. It carries energy 'hv' and momentum 'E/c' but no mass. Correct about the charge.

o Photons are discrete entities that propagate in one direction?

This can get a bit muddy. Quantum arguments do come into play. It also what you mean by "one direction." The phenomena of reflection and refraction introduce additional directions.

o Do the electric and magnetic fields within a photon oscillate in phase?

As ZapperZ stated, the quantization of the electromagnetic field is the photon.

o If so, there must be a point in the cycle where there is both zero E-M field values and therefore no forward propagation?

The propagation is the dynamical nature of your quantities. True, at some instance there may be zero instantaneous power, but at the very next instance this will not be the case. So therefore, by definition, something is propagating.

o How is the photon energy E=hf reconciled to the energy associated with the oscillating E-M fields, i.e. is it aggregated over 1 cycle?

By your equation; for a monochromatic electromagnetic wave, the energy of a single photon is given by the frequency of oscillation. The total energy of a wave packet (you can figure this out from Poynting's theorem) is then the sum of the energies of all the photons that make up that wave packet.

 

[Meir Achuz]

I haven't gone through your whole post, but thils might clarify things for you a bit:
A classical EM wave is just a wave that has a huge number of photons.

 

[Naty1]

For the viewpoint from classical (continuous wave) physics look for the concepts of special and general relativity and Maxwells equations; for the quantum (discrete) view of physics consider quantum theory, quantum mechanics, quantum gravity and string theory. All have some overlap and some distinct irreconciable differences.

Wikipedia is one online source but it can get complex and very mathematical with inadequate explanation; one good introduction is THE BOHR MODEL at

http://en.wikipedia.org/wiki/Bohr_Model

You might also find some Q& A at
http://math.ucr.edu/home/baez/physics/

interesting.

And another site with lots of references: http://www.physics.ucdavis.edu/Text/Carlip.html#Hawkrad

 

[jtbell]

o A photon is emitted when an electron transitions to a lower energy orbit and absorbed when an electron transitions to a higher energy orbit within an atom?

That statement is true as far as it goes. However, those processes are not the only ones in which photons are emitted or absorbed.

o A photon has no rest mass, but a kinetic mass m=E/c^2 and no externalised charge?

With "kinetic mass," you're probably thinking of the "relativistic mass" defined via [itex]E = mc^2[/itex]. Most physicists don't use "relativistic mass" in their work nowadays. When most physicists say "mass," they mean "invariant mass" (often called "rest mass") which is zero for a photon.

Nevertheless, many popular and introductory treatments of relativity use "relativistic mass," and some people here like to use it, which leads to confusion and debates about which kind of "mass" is "better."

o Photons are discrete entities that propagate in one direction?

Don't think of photons as tiny classical pellets that travel in well-defined paths. All we "really know" about a photon is that we produce it at one location, and detect it at another location, and the elapsed time between the two is consistent with a speed of c along a straight line (in vacuum). But we cannot actually observe the photon in between without destroying it (and preventing it from arriving at its original "destination"). The mathematics of quantum mechanics (more precisely, quantum field theory) does not address the question, what is the photon "really doing" in between? That is the subject of interpretations of quantum mechanics, about which there is no general agreement. In this forum you find frequent debates about such interpretation questions.

o If so, there must be a point in the cycle where there is both zero E-M field values and therefore no forward propagation?

Consider an ordinary transverse wave traveling along a stretched string. At any instant, there are points where the displacement of the string is zero, but nevertheless the wave travels. The points of zero displacement travel along with the wave. So it is also with electromagnetic waves.

o How is the photon energy E=hf reconciled to the energy associated with the oscillating E-M fields, i.e. is it aggregated over 1 cycle?

If you have a classical EM wave with a given amplitude falling on a surface with a given area, you can calculate the amount of energy falling on that surface per second, and you can translate that into an equivalent number of photons per second via E=hf. But that is about as far as you can go. You cannot associate an individual photon with a discrete spatial "chunk" of the EM wave.

It's best to think of photons versus classical EM waves as being alternative descriptions of light etc., which directly correspond to each other only in some ways. The detailed relationship (going between classical electrodynamics and quantum field theory) is complex and subtle.

 

[mikelepore]

As far as I know, the idea of photons is only useful for describing the instants when the energy is absorbed or emitted (photoelectric effect, Compton effect, etc.), and the wave picture applies to all of the traveling (including diffraction). Someone please tell me if there are some exceptions to my statement. Several writers above seem to have expressed the opposite view ("EM waves ARE photons"; "A classical EM wave is just a wave that has a huge number of photons.")

 

[marmot]

From what I gathered from my modern physics class, it seems that the photon was introduced as a way to explain how light would transfer momentum to electrons, kicking them out. So it seems that the wave aspects of EM radiation have to do with how it can have destructive and constructive interference and other wave properties, and the particle model as a way to deal with momentum transfer. I suppose its kindof like when we deal with quantum effects of matter particles, where the wave aspects have to do with things seen in diffraction gratings, and while the particle properties have to do with momentum.

 

[mysearch]

Thank you for all the helpful responses to my questions in post #1. Given the scope of responses I would like to address some comments, but rather than confuse the situation by putting all comments in 1 post, I will raise a separate posts for each.

Again, thanks for all the help, I appreciate that many of you must repeatedly have to answer questions like mine, i.e. from people who are coming to these topics for the first time.

 

[mysearch]

Response to #3

1 and 2 are self-contradictory. EM waves ARE photons.

I accept that this may well be the case, but I guess my questions were more focused on understanding why. Starting out, as I did in classical physics, i.e. Bohr model and Maxwell’s equations, the wave-particle duality appeared to raising many unresolved issues in my mind. I accept that the answers may well lie in quantum mechanics. I have acquired a copy of Feynman’s lecture on QED, but I am not totally sure whether his ideas are still in `vogue`?

You made several external references without directly providing links, which after a bit of searching I found and list below with some general comments:

FAQ in the General Physics Forum.
https://www.physicsforums.com/showthread.php?t=104715
I think post #3 is the most relevant to the present discussion.
This might be a good place to put the others links below if appropriate and useful.
As a general point, while I searched the classical forum for the relevant threads, I missed the FAQ thread as it was in another forum. Maybe the classic forum should replicated the FAQ.

http://www.src.wisc.edu/
While I took a quick look at the synchrotron center website, I struggled to find any direct relevance to my basic questions. Maybe I simply didn't find the right section.

http://arxiv.org/abs/quant-ph/0703126
I did a site search on the forum for Marcella paper but finding a link wasn’t easy, e.g.
“site:https://www.physicsforums.com Marcella”. Eventually, I found a paper, which I am assuming is the right one. While I haven’t had time to read properly, I am not sure it is intended for beginners.

However, just because we switch back and forth doesn't mean that there isn't a means to describe all light phenomena using just ONE single consistent description.

A good point to highlight as I suspect that it is one that may not be obvious to those initially trying to read deeper into this subject area for the first time.

 

[mysearch]

Response to #4

Thank you for taking the trouble to try to respond to all my generalised questions.

True under most elementary cases; however, there are exceptions (e.g. the magnetic field lags behind the electric field for wave propagation through a conductor).

I had seen some information to this effect, which is why I restricted my questions to propagation in vacuum.

False. The electric field of a single point charge 'q' goes as q/r²; I'm sure you have seen this.

While I accept your statement, can you actually measure an electric field without introducing a second unit test charge probe?

False. A photon is a massless particle. It carries energy 'hv' and momentum 'E/c' but no mass. Correct about the charge.

Again, I accept your statement, but the issue I was generally raising related to relativistic mass vs invariant mass. Jtbell covers this issue in post#7

This can get a bit muddy. Quantum arguments do come into play. It also what you mean by "one direction." The phenomena of reflection and refraction introduce additional directions.

I need to better understand the quantum aspects.

As ZapperZ stated, the quantization of the electromagnetic field is the photon.

Sorry, I not sure whether I fully understand your statement, but I think you are saying that the E-M fields associated with a photon must be in phase, because as Zapper has indicated “EM waves ARE photons”

The propagation is the dynamical nature of your quantities. True, at some instance there may be zero instantaneous power, but at the very next instance this will not be the case. So therefore, by definition, something is propagating.

The reason why the phase issue arose in my mind was linked to a perception of a mechanical wave, where the energy associated with a wave is conserved in the form of potential and kinetic energy. At any point on this type of wave, when the amplitude of the wave is zero, the propagation media, at that point, has acquired maximum kinetic energy. Initially, I assumed that a point of minimum electric field might correspond to a maximum magnetic field as an analogous process in EM waves.

The total energy of a wave packet (you can figure this out from Poynting's theorem) is then the sum of the energies of all the photons that make up that wave packet.

While I am not familiar with Poynting's theorem, I will look it up. Many thanks again.

 

[mysearch]

Response to #5

A classical EM wave is just a wave that has a huge number of photons.

While this may be a useful visualisation, I not sure that it doesn’t lead you back to the dichotomy of wave-particle duality, e.g. refraction, which I believe Zapper and others were trying to highlight can only be resolved by quantum mechanics. This is an assumption on my part not a statement of fact. Thanks

 

[mysearch]

Response to #6

For the viewpoint from classical (continuous wave) physics look for the concepts of special and general relativity and Maxwells equations; for the quantum (discrete) view of physics consider quantum theory, quantum mechanics, quantum gravity and string theory. All have some overlap and some distinct irreconciable differences.

Clearly, some of my issues overlap into other areas of modern physics, so your point is well taken. One issue that I did raise (yes, I had more) relates to the fact that a magnetic field normally corresponds to a moving charge, but within relativity, another frame of reference might consider the charge to be stationary and therefore would have no magnetic field. However, this is possibly not the time to raise more questions, but if anybody could give me an insight into this issue, it would be appreciated. Thanks for all the links.

 

[mysearch]

Response to #7

That statement is true as far as it goes. However, those processes are not the only ones in which photons are emitted or absorbed.

Could you provide some pointers as to where to look for more information?

With "kinetic mass," you're probably thinking of the "relativistic mass" defined via [itex]E = mc^2[/itex]. Most physicists don't use "relativistic mass" in their work nowadays. When most physicists say "mass," they mean "invariant mass" (often called "rest mass") which is zero for a photon. Nevertheless, many popular and introductory treatments of relativity use "relativistic mass," and some people here like to use it, which leads to confusion and debates about which kind of "mass" is "better."

You are absolutely right. As you may have gathered, I am not a professional physicist and was only highlighting an assumption that photons can undergo a particle-like collision.

Don't think of photons as tiny classical pellets that travel in well-defined paths. All we "really know" about a photon is that we produce it at one location, and detect it at another location, and the elapsed time between the two is consistent with a speed of c along a straight line (in vacuum). But we cannot actually observe the photon in between without destroying it (and preventing it from arriving at its original "destination"). The mathematics of quantum mechanics (more precisely, quantum field theory) does not address the question, what is the photon "really doing" in between? That is the subject of interpretations of quantum mechanics, about which there is no general agreement. In this forum you find frequent debates about such interpretation questions.

I think this is a very good point to highlight. While I won’t bring speculative theories into this discussion, I have seen a number of ideas that define the nature of photons in quite different terms. I am not saying they are right, but I am not sure that anybody fully understands all the implications of this topic, not even a quantum physicist.

Consider an ordinary transverse wave traveling along a stretched string. At any instant, there are points where the displacement of the string is zero, but nevertheless the wave travels. The points of zero displacement travel along with the wave. So it is also with electromagnetic waves.

I have a slightly different take on this analogy, but accept that I might be completely wrong. If you are discussing a mechanical wave, a point of zero displacement of a traverse wave, i.e. zero potential energy, corresponds to the propagating media having maximum kinetic energy. I mentioned this point in response to cmos when highlighting why I raised the issue of the E-M field phase, i.e. initially I thought they might be out of phase as per potential and kinetic energy in a SHM pendulum, which might then explain how they self propagate, but clearly I was wrong on this point, so will have to look harder for the accepted explanation.

If you have a classical EM wave with a given amplitude falling on a surface with a given area, you can calculate the amount of energy falling on that surface per second, and you can translate that into an equivalent number of photons per second via E=hf. But that is about as far as you can go. You cannot associate an individual photon with a discrete spatial "chunk" of the EM wave.

I guess one of the things that confused me regarding the perception of EM waves vs photons related to the nature of the source. For example, if I concentrated a large number of charged particles in a given region of space and then could oscillate them around a point would it be correct to say, classically, that they would radiate an E-M wave in all directions? If so, would these waves be continuous while the charged source continued to oscillate? In 3-D space, the energy at any given radius would be distributed over the increasing size of the surface area of a sphere, i.e. the inverse square law? I think what you and Zapper are trying to highlight is that duality implies that this energy has to also correspond to the discrete distribution of photon energy, i.e. E=hf, over this same surface area. Sorry, if I have got the wrong idea.

It's best to think of photons versus classical EM waves as being alternative descriptions of light etc., which directly correspond to each other only in some ways. The detailed relationship (going between classical electrodynamics and quantum field theory) is complex and subtle.

Now that I am beginning to understand. Many thanks.

 

[mysearch]

Response to #8

As far as I know, the idea of photons is only useful for describing the instants when the energy is absorbed or emitted (photoelectric effect, Compton effect, etc.), and the wave picture applies to all of the traveling (including diffraction")

The following table seems to summarise some of the duality problems as perceived from a classical perspective. In part, this was my starting point:

Effect……….Wave…...Particle
Reflection……Yes…...Yes
Refraction……Yes…...Partially
Interference..Yes…...No
Diffraction…….Yes…...No
Polarization…..Yes…...No
Photoelectric….No…...Yes

 

[mysearch]

Response to #9

From what I gathered from my modern physics class, it seems that the photon was introduced as a way to explain how light would transfer momentum to electrons, kicking them out.

My understanding is that photons were an idea linked to Planck’s idea of a quantum (1900) and Einstein’s work (1905) on the photoelectric effect. Subsequently, these ideas became `entangled`, in more ways that one, in the ideas of quantum mechanics, However, I am sure that there are others more qualified to respond.

 

[ZapperZ]

My understanding is that photons were an idea linked to Planck’s idea of a quantum (1900) and Einstein’s work (1905) on the photoelectric effect. Subsequently, these ideas became `entangled`, in more ways that one, in the ideas of quantum mechanics, However, I am sure that there are others more qualified to respond.

You are correct. The idea of the "corpuscular" theory of light came clearly out of Einstein's photoelectric effect model, although historically, I think Newton did dabble in such an idea as well, but not fully baked. It is with Einstein's photoelectric model (which, to some people is http://physicsworld.com/cws/article/print/21818" than his SR theory) that such an idea became very well-defined.

Zz.

 

[cmos]

I'll just reply to what I think are still the unresolved issues. Let me know if I missed something.

While I accept your statement, can you actually measure an electric field without introducing a second unit test charge probe?

Without getting into specifics of such an apparatus, yes it is true that you need to introduce something additional into your experiment to make a measurement. This doesn't mean the electric field isn't there before you measure it. What isn't there is the force on the second particle.

Sorry, I not sure whether I fully understand your statement, but I think you are saying that the E-M fields associated with a photon must be in phase, because as Zapper has indicated “EM waves ARE photons”

My statement was to reaffirm what ZapperZ said.

While I am not familiar with Poynting's theorem, I will look it up. Many thanks again.

It's just a way to determine the flow of electromagnetic power. I mentioned it to give an explicit example of how to reconcile the energies associated with the classical field and the quantum photons.

 

[cmos]

Response to #8

The following table seems to summarise some of the duality problems as perceived from a classical perspective. In part, this was my starting point:

Effect……….Wave…...Particle
Reflection……Yes…...Yes
Refraction……Yes…...Partially
Interference..Yes…...No
Diffraction…….Yes…...No
Polarization…..Yes…...No
Photoelectric….No…...Yes

Actually, INDIVIDUAL photons have been shown the interfere with themselves. An example is to perform the double slit experiment but to send only one photon at a time. Obviously you would need to have a photosensitive film in place of a screen. Over time, the usual diffraction pattern would form. I believe Taylor was the first to perform such an experiment; his experiment, I believe, lasted for several months.

 

[mikelepore]

Double slit experiment, one photon at a time ... the way I would interpret the results, it was a particle property when it was emitted, it was a wave property when it traveled through space and when it passed through the slits, and it was a particle property when it affected the screen. I don't see any gain in explanatory or predictive power by saying that the photon traveled through space or that the photon "interfered with itself."

 

[mikelepore]

Effect……….Wave…...Particle
Refraction……Yes…...Partially

I recall reading that a stream of particles that increases speed when it crosses a boundary would bend toward the normal, and a wave that decreases speed when it crosses a boundary would bend toward the normal, so a particle model fails to explain refraction.

 

[cmos]

Double slit experiment, one photon at a time ... the way I would interpret the results, it was a particle property when it was emitted, it was a wave property when it traveled through space and when it passed through the slits, and it was a particle property when it affected the screen. I don't see any gain in explanatory or predictive power by saying that the photon traveled through space or that the photon "interfered with itself."

Your argument is flawed in that, by your example, it is necessary to say precisely that at point A the disturbance acts as a wave, at point B it acts as a particle, and so forth. The uncertainty principle, however, forbids you to do this. The resolution is to allow for the particle-wave duality; this leads directly to single-photon interference as well as photon motion.

 

[meopemuk]

Response to #8

The following table seems to summarise some of the duality problems as perceived from a classical perspective. In part, this was my starting point:

Effect……….Wave…...Particle
Reflection……Yes…...Yes
Refraction……Yes…...Partially
Interference..Yes…...No
Diffraction…….Yes…...No
Polarization…..Yes…...No
Photoelectric….No…...Yes

I think you should put "Yes" everywhere in your rightmost ("Particle") column. All these effects are properly explained by treating photons as particles. Of course, these explanations require you to abandon the naive classical picture of a particle as a material point moving along well-defined trajectory. The propagation of photons-particles must be described in terms of quantum mechanics.

The entire particle-wave duality conundrum can be described like this: "Each time we observe or measure the photon (e.g., by a photographic plate) it looks like a particle; each time we do not observe the photon (e.g., when it passes through the double-slit) it behaves as a wave." If you assume (as I do) that physics is a science about observations and measurements, then the particle picture of the photon is the physical one. The wave picture of the photon refers to situations when the photon is not observed. So, this picture belongs to philosophy rather than physics. Something along the lines of the ethernal question: "Is the moon there when nobody is looking?"

 

[cmos]

The entire particle-wave duality conundrum can be described like this: "Each time we observe or measure the photon (e.g., by a photographic plate) it looks like a particle; each time we do not observe the photon (e.g., when it passes through the double-slit) it behaves as a wave." If you assume (as I do) that physics is a science about observations and measurements, then the particle picture of the photon is the physical one. The wave picture of the photon refers to situations when the photon is not observed. So, this picture belongs to philosophy rather than physics. Something along the lines of the ethernal question: "Is the moon there when nobody is looking?"

This is just absurd.There is a mathematical description of the wave nature of light hence a physical reality. How to observe it? Simply illuminate a diffraction grating and let diffracted light hit the wall. You are now observing the wave nature of light as was first realized in the 1800's.

 

[meopemuk]

This is just absurd.There is a mathematical description of the wave nature of light hence a physical reality. How to observe it? Simply illuminate a diffraction grating and let diffracted light hit the wall. You are now observing the wave nature of light as was first realized in the 1800's.

I was talking not about wave nature of light (=billions of photons), but about wave nature of a single photon. Single photon cannot produce diffraction/interference picture. When hitting the wall or the photographic plate, it makes a single dot.

 

[lightarrow]

I think you should put "Yes" everywhere in your rightmost ("Particle") column. All these effects are properly explained by treating photons as particles. Of course, these explanations require you to abandon the naive classical picture of a particle as a material point moving along well-defined trajectory. The propagation of photons-particles must be described in terms of quantum mechanics.

The entire particle-wave duality conundrum can be described like this: "Each time we observe or measure the photon (e.g., by a photographic plate) it looks like a particle; each time we do not observe the photon (e.g., when it passes through the double-slit) it behaves as a wave." If you assume (as I do) that physics is a science about observations and measurements, then the particle picture of the photon is the physical one. The wave picture of the photon refers to situations when the photon is not observed. So, this picture belongs to philosophy rather than physics. Something along the lines of the ethernal question: "Is the moon there when nobody is looking?"

Ok, quite correct; the problem is that if you only take the "particle" view, you can't say anything about why you have detected it in a specific point of the screen; instead, with the "wave" view, at the opposite, you will know the "global pattern" without knowing which point it is; you can't escape complementarity. Saying that the particle view is somewhat better because it refers to a measure while the wave doesn't it's not correct: you should say that the particle view refers to a "single" detection, while the wave one to many at once.

The fact is: "particle" = detection event; wave = global behaviour of detection events.

However I know that you already know very well all of this...

 

[meopemuk]

Ok, quite correct; the problem is that if you only take the "particle" view, you can't say anything about why you have detected it in a specific point of the screen;

Why not? When I say "particle" I mean "particle propagating according to the laws of quantum mechanics". These laws do not allow us to predict which specific point on the screen will be hit by the given particle. However, these laws do allow us to calculate probabilities and, therefore, the "global behavior of detection events". The entire wave-like behavior of particles is fully described by the laws of quantum mechanics.

Yes, these laws involve something called "wave function". Does this mean that photon (or electron) is a wave? No, I don't think so. The wave function is a mathematical abstraction which is useful for calculating probabilities. That's all. Quite similarly, vectors in the Hilbert space are mathematical abstractions, which help us to calculate quantum probabilities. Does this mean that we should now say that "particle is a vector in the Hilbert space"? No.

 

[lightarrow]

Why not? When I say "particle" I mean "particle propagating according to the laws of quantum mechanics". These laws do not allow us to predict which specific point on the screen will be hit by the given particle. However, these laws do allow us to calculate probabilities and, therefore, the "global behavior of detection events". The entire wave-like behavior of particles is fully described by the laws of quantum mechanics.

Well; then I could say, in the same way, that when we talk of "waves" we intend "waves propagating and detected according to the laws of quantum mechanics". How are detected? Simple: at random, a detector in a point "clicks" with a probability proportional to <psi|psi>.

 

[meopemuk]

Well; then I could say, in the same way, that when we talk of "waves" we intend "waves propagating and detected according to the laws of quantum mechanics". How are detected? Simple: at random, a detector in a point "clicks" with a probability proportional to <psi|psi>.

If the "wave" you are talking about is an abstract mathematical thing that is found only in QM equations, but not in nature, then I agree. However, I wouldn't like to think that real electron is such a wave, that electron's charge is really spread out over large volume, and that at the instant of the "click" this "cloud" collapses to a point. This weird behavior is OK for the wave treated as a mathematical abstraction, but it doesn't look OK if the wave is a real physical object.

 

[lightarrow]

If the "wave" you are talking about is an abstract mathematical thing that is found only in QM equations, but not in nature, then I agree. However, I wouldn't like to think that real electron is such a wave, that electron's charge is really spread out over large volume, and that at the instant of the "click" this "cloud" collapses to a point. This weird behavior is OK for the wave treated as a mathematical abstraction, but it doesn't look OK if the wave is a real physical object.

Ok, infact I don't say the wave is a real physical object, but, in the same way, you can't say that [what you mean with "particle"] is a real physical object, in my hopinion, since the only physical object here is the "click" of the detector. Don't you agree?

 

[ZapperZ]

Ok, infact I don't say the wave is a real physical object, but, in the same way, you can't say that [what you mean with "particle"] is a real physical object, in my hopinion, since the only physical object here is the "click" of the detector. Don't you agree?

But the "particle" in question is defined as having a localized energy, and a "click" in a detector at a particular location is consistent with such a definition. The "wave" of light that is commonly used is a realwave that exists in real space, i.e. it is similar to seeing water waves. This is NOT the wavefunction of light as described in QM, which exists in configuration space. That is the distinction that is being confused here.

Zz.

 

[meopemuk]

Ok, infact I don't say the wave is a real physical object, but, in the same way, you can't say that [what you mean with "particle"] is a real physical object, in my hopinion, since the only physical object here is the "click" of the detector. Don't you agree?

I think we can agree here. "Clicks" is the only objective reality that we observe. Both "particles" and "waves" are simply mental constructs that we invent to visualize our mathematical description of "clicks". In my opinion the notion of "particles" makes this visualization more clear and less contradictory.