The Debate on Light: Particle or Wave?

[dodo]

May I add a layman's question to the thread?

19th century postulated an 'ether', as the medium on which light waves would propagate. The current story they tell you at school is that EM waves "carry their own media", being composed of, I seem to recall, two orthogonally-polarized components, one being the "E" and the other the "M". If this goes back to Maxwell, I suppose it might not be quite the "current" story. Then, which is?

And... in which way can this be related to gravity? In a sense, the "media" of gravity is the space itself, which is distorted by the presence of masses. Can (or cannot) be the same for light? Imagine light as a ripple in space - in which way could this account for the "alternate" particle explanation for light?

I apologize in advance for the complete sci-fi tone of the question - in the hope that the act of explaining to a layman can in itself be a source of enlightened amusement.

 

[russ_watters]

Dodo, quite a number of experiments have been done in an effort to find a medium on which light propagates. None have been successful and the laws of physics work just fine without one. It appears that there isn't one (the Electric and Magnetic fields are not the same as the classical ether).

 

[terrabyte]

lol why delete my link.

i just pulled the first one i found that explained photon-atom interactions. i did not "base" my entire physics knowledge on "crackpot's website"

you people are way too touchy.

not going to even bother taking this any further. <wave>

 

[Integral]

lol why delete my link.

i just pulled the first one i found that explained photon-atom interactions. i did not "base" my entire physics knowledge on "crackpot's website"

you people are way too touchy.

not going to even bother taking this any further. <wave>

Why delete your link? Because it did not contain valid physics. This thread has been answered quite throughly by someone who knows what they are talking about. The fact that you were unable to identify the link as nonsense says that you either did not have sufficient knowledge of Physics to tell, or are a crackpot pushing bad information. I do not know, and I do not care, which. I just do not want crackpot links in the middle of a good thread. Please be more careful in the future. If you do not know, ask questions, do not post random links from the web.

 

[terrabyte]

Are you saying photon-atom interaction is wrong?

i could have swore that was common physics. Energy kicks electrons into higher but unstable orbits, atom randomly kicks back down to lower orbit and emits a photon in a uncertain direction.

the rest of the page was pure gibberish of course, but that's not why i linked it.

anyways, i said i was gone <wave again>

 

[JohnDubYa]

light has both the nature of wave and particle, as according to de broglie he showed his work as folllowing

And what physical property does this "matter wave" represent? If you draw a picture of this wave, what physical properties do the horizontal and vertical axes correspond?

 

[dodo]

Found this in Google's cache - guess it points to what I was looking for. Thank you all!

http://216.239.59.104/search?q=cache:SWDu2I0gBTMJ:arxiv.org/abs/gr-qc/9701034

 

[Tsunami]

I think I know what you are getting at here. Unfortunately, as you can see from the previous post, it can easily be misinterpreted or misrepresented. The biggest obstacle in this issue is trying to convey the meaning of the mathematics. QFT (and QED) makes NO provision for classical fields. Therefore, classical fields (and thus, "waves") essentially do not exist and are meaningless in QFT.

Then what are we left with? I hate to say we have a "particle field", because that again can be misinterpreted (refer to my discussion what classical particle is and why this is NOT what a photon is). Can I get away with saying a "quantum field" without grossly misrepresenting the mathematics? Maybe. If we stick by that, then I would not want to leave the impression that photons can flip back and forth between two apparently different perpective.

Zz.

(Know that this comment comes from a student who only has introductory knowledge of QM: )

Maybe a nice way to explain the wave/particle behaviour of light, is that light behaves more like a particle when its energy increases? For instance, it's pretty useful to talk about microwaves as if they were particles. However, X-rays almost always behave like particle streams.
One could say that it's less probable that light with low energy behaves as particles than that light with high energy behaves like particles.. I think.

 

[dextercioby]

(Know that this comment comes from a student who only has introductory knowledge of QM: )

Maybe a nice way to explain the wave/particle behaviour of light, is that light behaves more like a particle when its energy increases? For instance, it's pretty useful to talk about microwaves as if they were particles. However, X-rays almost always behave like particle streams.
One could say that it's less probable that light with low energy behaves as particles than that light with high energy behaves like particles.. I think.

Nope,it's incorrect.Light is either all wave (macroscopical/classical description,or all particle (microscopical/quantum description).There's no overlapping...And quantum/photonic description can account for macroscopical phenomena as well,it's just that the photons need to form a statistical ensemble (plus coherence,...).

Daniel.

 

[Tom Something]

could someone describe how diffraction can be explained by the photons rather than waves?

 

[misogynisticfeminist]

Particle. No experiment has ever been performed on a photon that demonstrated wave properties to my knowledge.

no no no. What about thomas young's experiment?

Photons are interpreted in terms of its wavefunction, the photon field in which photons are the quanta.

 

[ZapperZ]

could someone describe how diffraction can be explained by the photons rather than waves?

You may want to read this thread:

https://www.physicsforums.com/showthread.php?t=68917

In addition, my journal entry on the misconception of the Heisenberg Uncertainty Principle illustrates this using the diffraction from a single slit due to photons. So you may wish to read that also.

Zz.

 

[Tsunami]

Nope,it's incorrect.Light is either all wave (macroscopical/classical description,or all particle (microscopical/quantum description).There's no overlapping...And quantum/photonic description can account for macroscopical phenomena as well,it's just that the photons need to form a statistical ensemble (plus coherence,...).

Daniel.

Bah... now you got me confused again...

so both the wave and particle description are part of the same universal description of light (dixit ZapperZ) yet at the same time light is either all wave, or all particle...?

Something tells me that the case is :

either
1) I'm not going to understand any of this until I see a more expansive course about QFT.
or
2) Virtually nobody completely understands this and this is why the interpretations of you and ZapperZ (at least seem to) logically contradict.

 

[marlon]

so both the wave and particle description are part of the same universal description of light (dixit ZapperZ) yet at the same time light is either all wave, or all particle...?

Zapper is correct. The duality exists only because of our 'classical minds' ; we want to think in terms of either particles or waves. There is no problem with that but we do need to keep the correct perspective on things here. First of all 'particles' in this case does not mean little objects with finite boundaries. It means little finite pieces of energy (this is the actual quantization , right ?)

Secondly, and this is what dextercioby meant, in QM we have experiments that are better explained with the wave-like notion (eg the double slit experiment) and we have those experiments that are better described with the particle-like notion (eg photo-electric effect). However in the end both descriptions are just ONE SINGLE way of describing the physical properties of light...that is all.

Another common misconception is the fact that the photo-electric effect proved the existence of photons. That is not true because this photo-electric effect can be described in terms of the wavelike-notion of the incident EM-radiation too. It is only the atoms of the target electrode that are treated with QM. However, the particle-like notion of light is suggested by this experiment. If you want to read more, check out my journal and find the article on creating an entangeled photon-state in an undergrad lab

marlon

EDIT : well, here's the article : Create entangled photons yourself in an undergrad laboratory :
http://marcus.whitman.edu/~beckmk/QM/grangier/Thorn_ajp.pdf

 

[ZapperZ]

Bah... now you got me confused again...

so both the wave and particle description are part of the same universal description of light (dixit ZapperZ) yet at the same time light is either all wave, or all particle...?

Something tells me that the case is :

either
1) I'm not going to understand any of this until I see a more expansive course about QFT.
or
2) Virtually nobody completely understands this and this is why the interpretations of you and ZapperZ (at least seem to) logically contradict.

Without even going into QFT, let's make sure we make something very clear here:

All of the properties of light can be described by QM, and even so-called wavelike properties can be obtained using the photon description.

Now, contrary to popular beliefs, especially among students, physics instructors are not heartless masochists who will force the students to use the photon description when the classical wave picture is easier and more direct to be used. That is why the classical wave theory are still used when we describe diffraction and interference effects, especially in classical optics classes. It doesn't mean, however, a unified QM description doesn't exist for such things. It is just more involved and requires a bit more of a sophistication in knowledge to do it. The classical wave picture is simply a "short cut" to getting what we want to get.

Zz.

 

[marlon]

Just to add what Zapper has said, QFT does not 'explain' the duality more or better as QM. You know, everytime somebody drops the term QFT, my heart goes...You know why ? Well, because lot's of people like the use the posh-sounding epitheton QFT but i always feel the urge to ask : do you know what it means. I mean, if i were a professor teaching QFT, my first chapter in my course would explain why we need the quantum-part, why we need the fields part and why we need the relativistic part. Every principle of QM is copied, to some extent, by QFT since QFT really is the unification of both QM and special relativity. Beware, i said SPECIAL relativity and not GENERAL relativity...

marlon

 

[Thomas2]

You read wrong. What I said was...

"Having said that, the most common explanation for the "wave-particle duality" is that light behaves as waves in experiments such as the double slit, and behaves as particles when we do things like the photoelectric effect."

The photoelectric effect can well be explained in terms of the wave picture if the wave-atom interaction is properly being considered. In contrast, the particle theory of light is completely inconsistent and would in fact not enable photoionization at all because (due to its small mass) the photon could not transfer enough energy to the photoelectron.
A further proof for the incorrectness of the particle model for light is the experimental fact that photoelectrons are not primarily released in the direction of propagation of light (as one should expect it for particles) but perpendicular to this in the direction of the electric field vector .

See my webpage http://www.physicsmyths.org.uk/photons.htm for more details (I can't copy the page here because it contains a number of formulae).

 

[marlon]

In contrast, the particle theory of light is completely inconsistent and would in fact not enable photoionization at all because (due to its small mass) the photon could not transfer enough energy to the photoelectron.

Well, i don't want to be impolite but do you really mean this ? Anyhow, you are totally wrong...I suggest you actually check the history of the photo-electric effect-explanation and just to be sure, also double check what is meant by the words : "particle theory of light"...Why do you even bring the 'small mass' of photons...That small mass is ZERO

marlon

 

[Thomas2]

...Why do you even bring the 'small mass' of photons...That small mass is ZERO

Well ,that would be even worse, because then it could not transfer any energy at all. But I think you mix up mass with 'rest mass' here. Only the latter is zero in the relativistic theory of the photon, but since the photon does not rest in any reference frame it has a finite mass (or at least it is supposed to have).

 

[ZapperZ]

The photoelectric effect can well be explained in terms of the wave picture if the wave-atom interaction is properly being considered. In contrast, the particle theory of light is completely inconsistent and would in fact not enable photoionization at all because (due to its small mass) the photon could not transfer enough energy to the photoelectron.
See my webpage http://www.physicsmyths.org.uk/photons.htm for more details (I can't copy the page here because it contains a number of formulae).

I disagree. I worked in photoemission spectroscopy for my postdoc, and if you look at the work by Spicer, who almost single-handedly pioneered this technique, you will see why what you said is incorrect.

Furthermore, since when does a photon have a "small mass"? The photon mass is irrelevant in a photoemission process. In fact, if a photon HAS a mass, a bunch of things from photoemission spectroscopy would be wrong, and the band structure we used in the semiconductors in your modern electronics should not work. This is because photoemission spectroscopy is the FIRST such technique that could independently verify the theoretical calculations of band structure of solids. And they all use the photon picture!

And I have said this so many times, I am beginning to bore myself. Look up "multiphoton photoemission". Now explain how the amount of photoelectrons detected as a function of the photon intensity have discrete dependence that just HAPPENS to coincide with the discrete number of photons being absorbed to cause that particular transition. Don't tell me the energy level of the material is discrete since this is a transition from a continuous conduction band into a continuous vacuum band. While the photoelectric alone cannot rule out the wave picture, I haven't seen even a single attempt at reconcilling the multiphoton process with the wave picture.

Zz.

 

[marlon]

Well ,that would be even worse, because then it could not transfer any energy at all. But I think you mix up mass with 'rest mass' here. Only the latter is zero in the relativistic theory of the photon, but since the photon does not rest in any reference frame it has a finite mass (or at least it is supposed to have).

Then let's call it the photon's energy, ok ? Just to be clear. Now, again i ask you to double check the actual proof of the photo-electric effect. This will prove you wrong.

regards
marlon

 

[dextercioby]

Nice debate here,is it still in General Physics ?? Oh,i know Thomas2...Huge misconceptions...The bad part is that he don't seem to be willing to learn anything...

Daniel.

 

[ZapperZ]

Well ,that would be even worse, because then it could not transfer any energy at all. But I think you mix up mass with 'rest mass' here. Only the latter is zero in the relativistic theory of the photon, but since the photon does not rest in any reference frame it has a finite mass (or at least it is supposed to have).

Typically, when someone says something like this, it shows that this person never went through any formal study of SR and QM. They are always surprised that light can have energy and momentum, but NO MASS! Horrors! It shows they haven't seen the complete derivation of the relativistic effects. If that is the case, most of the time, we always have to keep taking several steps backwards each time we introduce an explanation. This can be quite exasperating (at least for me), so I'll let others with more patience on this take over.

Zz [with 2 extra cups of coffee this morning]

Zz.

 

[marlon]

This can be quite exasperating (at least for me), so I'll let others with more patience on this take over.

Thanks but no thanks
I believe the guy with the promotion should do the 'most difficult' tasks

Zz [with 2 extra cups of coffee this morning]

Zz.

Same here...

marlon

 

[Thomas2]

Furthermore, since when does a photon have a "small mass"? The photon mass is irrelevant in a photoemission process. In fact, if a photon HAS a mass, a bunch of things from photoemission spectroscopy would be wrong

Well, this would prove then that light has to be considered as a wave (see also my reply to Marlon above)

And I have said this so many times, I am beginning to bore myself. Look up "multiphoton photoemission". Now explain how the amount of photoelectrons detected as a function of the photon intensity have discrete dependence that just HAPPENS to coincide with the discrete number of photons being absorbed to cause that particular transition. Don't tell me the energy level of the material is discrete since this is a transition from a continuous conduction band into a continuous vacuum band. While the photoelectric alone cannot rule out the wave picture, I haven't seen even a single attempt at reconcilling the multiphoton process with the wave picture.

How should a multiphoton transition work in the particle picture if you can't even make an individual one work? On the other hand, it is no problem with the wave-atom interaction model if the first transition is to a state below the ionization threshold and the lifetime of the level is long enough so that the electron can absorb another wave frequency that ionizes it finally.

Also, as I mentioned above already, the experimental fact that photoelectrons are primarily emitted into the direction of the electric field vector (i.e. perpendicular to the direction of propagation of light) clearly invalidates the particle picture.

 

[Thomas2]

Typically, when someone says something like this, it shows that this person never went through any formal study of SR and QM. They are always surprised that light can have energy and momentum, but NO MASS! Horrors! It shows they haven't seen the complete derivation of the relativistic effects.

It seems it is you who should brush up on his Relativity. Through E=m*c^2 each photon with energy E can be assigned a relativistic mass m=E/c^2 .

 

[Doc Al]

It seems it is you who should brush up on his Relativity. Through E=m*c^2 each photon with energy E can be assigned a relativistic mass m=E/c^2 .

Given your track record on this site regarding relativity, you are in no position to lecture anyone. Unless specifically stated otherwise, "mass" generally means the invariant or rest mass.

 

[Thomas2]

Given your track record on this site regarding relativity, you are in no position to lecture anyone. Unless specifically stated otherwise, "mass" generally means the invariant or rest mass.

On the contrary, ZapperZ is in no position to make the suggestion 'this person never went through any formal study of SR and QM'.

 

[dextercioby]

If u did (go),why don't u prove it?


Daniel.

 

[Doc Al]

On the contrary, ZapperZ is in no position to make the suggestion 'this person never went through any formal study of SR and QM'.

Formal study or not, anyone curious about your grasp of special relativity is encouraged to review the posts you've made in the relativity forum. They will remove any doubt as to your mastery of the subject.

 

[ZapperZ]

How should a multiphoton transition work in the particle picture if you can't even make an individual one work? On the other hand, it is no problem with the wave-atom interaction model if the first transition is to a state below the ionization threshold and the lifetime of the level is long enough so that the electron can absorb another wave frequency that ionizes it finally.

Correction. It is YOU who can't make a single-photon photoelectric effect to work. The rest of us can.

Also, as I mentioned above already, the experimental fact that photoelectrons are primarily emitted into the direction of the electric field vector (i.e. perpendicular to the direction of propagation of light) clearly invalidates the particle picture.

Excuse me, but can you point out to me how you could tell that from a simple photoelectric effect experiment? The light source is UNPOLARIZED. Don't believe me? Try it.

I worked with polarized photoemission spectroscopy from a synchrotron. My AVATAR that you see with my profile is the data taken with a light source polarized along one of the high symmetry direction of crystal structure. There is ZERO problem in dealing with such a thing with the photon picture.

Again, refer to ALL of Bill Spicer's work in photoemission and show me where there is a "problem" in dealing with the photon picture. So far, all you have shown is your inability to understand what has been formulated. You should never criticize something based on ignorance.

Zz.

 

[Thomas2]

If u did (go),why don't u prove it?

I don't have to prove anything to somebody whose major contribution in this forum (or in this thread anyway) is to personally attack other people anonymously under the cover of his username. Put your words where your mouth is and try to argue scientifically against what I said on my webpage http://www.physicsmyths.org.uk/photons.htm .

 

[Thomas2]

Formal study or not, anyone curious about your grasp of special relativity is encouraged to review the posts you've made in the relativity forum. They will remove any doubt as to your mastery of the subject.

Maybe I have studied special relativity just better and deeper than you have and have in fact a better insight and grasp of it (remember, everything is relative). But the 'mass' issue raised above is anyway only a semantic problem here. The important point is that a consequent application of the energy- and momentum conservation laws should not enable photoionization at all with the particle picture, whereas on the other hand the wave model is in fact consistent with the short times required for photoionization if one considers the wave-atom interaction properly (as shown on my page http://www.physicsmyths.org.uk/photons.htm ).

 

[Thomas2]

Correction. It is YOU who can't make a single-photon photoelectric effect to work. The rest of us can.

Only if you close your eyes to what I said on my page http://www.physicsmyths.org.uk/photons.htm . But anyway, it is more important that the wave model can, contray to common opinion, indeed explain the photoeffect if a proper wave-atom interaction model is used.

Excuse me, but can you point out to me how you could tell that from a simple photoelectric effect experiment? The light source is UNPOLARIZED. Don't believe me? Try it.

I am not quite sure what your point is here. If you use polarized light, the photoelectrons will be emitted along the corresponding electric field vector; if you use unpolarized light, they will be emitted in the plane that contains all field vectors. In no case will they be emitted in the forward direction (and the the latter is what you would expect with particle collisions).

 

[ZapperZ]

Only if you close your eyes to what I said on my page http://www.physicsmyths.org.uk/photons.htm . But anyway, it is more important that the wave model can, contray to common opinion, indeed explain the photoeffect if a proper wave-atom interaction model is used.

But this is saying nothing new. The photoelectric effect is KNOWN to be a STRONG evidence for the photon picture, but it also cannot rule out completely the wave picture. But you, on the other hand, somehow thinks it rules out the photon picture because of your inability to understand what "photons" are and confusing it with the need to have mass (i.e. your lack of knowledge of Special Relativity). This is bogus.

I am not quite sure what your point is here. If you use polarized light, the photoelectrons will be emitted along the corresponding electric field vector; if you use unpolarized light, they will be emitted in the plane that contains all field vectors. In no case will they be emitted in the forward direction (and the the latter is what you would expect with particle collisions).

Emitted in the FORWARD direction? What the...?

I could shine a plane polarized light, having the polarization vector PARALLEL to the surface of the photocathode. Guess what? If I scan for photoelectrons, I get then in ALL DIRECTIONS! Not only that, this is in the BACKWARDS direction with respect to the direction of the incoming light. We see this ALL THE TIME when we do ANGLE-RESOLVED photoemission spectroscopy (ARPES).

Question: how many times have you done the simple photoelectric effect that we ask students to perform in undergraduate labs, and how many times have you done the more sophisticated photoemission spectroscopy of any kind?

Zz.