The Debate on Light: Particle or Wave?

[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.

 

[ZapperZ]

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 ).

Correction: a bastardizaton of the energy-momentum conservation laws will not allow for the photon picture to work. A CORRECT application of it will. You have made ZERO reference to spicer's work, and all the subsequent advancement in photoemission spectroscopy in the study of materials. If all of these were based on the WRONG principles, there's zero reason why they work so accurately in determining the electronics properties of materials (open any photoemission text). This fact seems to be glaringly omitted in your criticism.

And please stop using your webpage as a reference. It's all "relative", remember? Having a webpage requires ZERO knowledge of physics. Just look at Crank Dot Net. If you think having webpages is how physics is done, then you've revealed an additional level of ignorance.

Zz.

 

[Thomas2]

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.

The photoelectric effect is supposed to be evidence for the particle picture and against the wave picture (see for instance http://hyperphysics.phy-astr.gsu.edu/hbase/mod1.html#c4 (Wave Particle Duality)).

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 do not somehow think, but have shown on my page http://www.physicsmyths.org.uk/photons.htm that the photoelectric effect is in fact evidence for the wave picture rather than the particle picture. If you think it is bogus, then you should give detailed reasons why.

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).

Your photoemission work is hardly relevant in this context: is obvious that from a solid surface photoelectrons will be emitted in all directions since they suffer scattering in the material. You need individual atoms i.e. a gas as a target to show the original angular distribution of photoelectrons, and this shows that they are primarily ejected along the electric field vector (see for instance http://prola.aps.org/abstract/PR/v37/i10/p1233_1 ). You can find this also theoretically derived in some Quantum Mechanics textbooks that deal with photoionization (e.g. Blochinzev), but most textbooks actually don't mention this.

a bastardizaton of the energy-momentum conservation laws will not allow for the photon picture to work. A CORRECT application of it will.

Exactly, that's why most people consider the photoeffect erroneously to be evidence of the particle picture.

You have made ZERO reference to spicer's work, and all the subsequent advancement in photoemission spectroscopy in the study of materials

Spicer's work is hardly relevant for having established the photoeffect as evidence for the particle nature of light, so I don't know why you keep on mentioning it . Try to argue with basic and generally known experiments.

And please stop using your webpage as a reference...Having a webpage requires ZERO knowledge of physics

So does posting in Physicsforums. But if you have done some logic in the past, then you should know that either of them is not synonymous with zero knowledge. It is suggesting that it is which shows a level of igorance or maybe even of bad intentions.

 

[Tsunami]

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.

Ok...

so, we have light as quanta of energy... little blocks of energy. In certain situations, these little blocks of energy make up something that behaves like waves... this is especially through for blocks of energy with low frequency, or low-energy quanta. Is this correct?

Then, with all what has been said before, my puny mind only sees these options:

1) Wave-like behaviour is only true for low energy photons, while particle-like behaviour can describe all phenomena - thus, the wave-like behaviour is something like an approximation of the particle-like behaviour.
2) Wave-like behaviour can be applied to all phenomena, and particle-like behaviour can be applied to all phenomena - thus, there is a dual description, but both descriptions are always completely reconcilable (this doesn't contradict what ZapperZ said only if the unified description allows for both descriptions in every situation... the descriptions would simply be different approaches).
3) Some third option because I'm too tired to think straight at this moment.

 

[ZapperZ]

The photoelectric effect is supposed to be evidence for the particle picture and against the wave picture (see for instance http://hyperphysics.phy-astr.gsu.edu/hbase/mod1.html#c4 (Wave Particle Duality)).

I do not somehow think, but have shown on my page http://www.physicsmyths.org.uk/photons.htm that the photoelectric effect is in fact evidence for the wave picture rather than the particle picture. If you think it is bogus, then you should give detailed reasons why.

I did already! You seem to be confusing "energy" with evidence of "mass" in the photoelectric effect! I can play your game too. If you think you have any validity, send it into a peer-reviewed journal. Now THAT would determine if what you think is right is bogus or not.

Your photoemission work is hardly relevant in this context: is obvious that from a solid surface photoelectrons will be emitted in all directions since they suffer scattering in the material. You need individual atoms i.e. a gas as a target to show the original angular distribution of photoelectrons, and this shows that they are primarily ejected along the electric field vector (see for instance http://prola.aps.org/abstract/PR/v37/i10/p1233_1 ). You can find this also theoretically derived in some Quantum Mechanics textbooks that deal with photoionization (e.g. Blochinzev), but most textbooks actually don't mention this.

And this is where you clearly show your ignorance.

The photoelectric effect as described in the Einstein model is the effect done from a SOLID surface, and in particular, a METALLIC solid surface. This is where such a model is applied to, and NOT from a gas! The excitation comes out of a continuous conduction BAND, and not from discrete energy states exhibit by gas atoms and molecules!

So in effect you are applying a description in which it wasn't MEANT to be applied to. This is consistent with what you were trying to do when you implied that light has a "small mass". This is what I call the bastardization of physics principles - you simply use something without clearly understand when and where it should apply. So OF COURSE you get nonsensical answers! But don't blame the physics or our understanding of it for your mistakes.

Spicer's work is hardly relevant for having established the photoeffect as evidence for the particle nature of light, so I don't know why you keep on mentioning it . Try to argue with basic and generally known experiments.

Then you haven't a clue what spicer has already accomplished.

So does posting in Physicsforums. But if you have done some logic in the past, then you should know that either of them is not synonymous with zero knowledge. It is suggesting that it is which shows a level of igorance or maybe even of bad intentions.

I have published several papers on the photoemission spectroscopy, including in the Physical Review journals, the same journal that you cited above. So what have YOU done to show that what you have concluded is valid?

Zz.

 

[ZapperZ]

Ok...

so, we have light as quanta of energy... little blocks of energy. In certain situations, these little blocks of energy make up something that behaves like waves... this is especially through for blocks of energy with low frequency, or low-energy quanta. Is this correct?

Then, with all what has been said before, my puny mind only sees these options:

1) Wave-like behaviour is only true for low energy photons, while particle-like behaviour can describe all phenomena - thus, the wave-like behaviour is something like an approximation of the particle-like behaviour.
2) Wave-like behaviour can be applied to all phenomena, and particle-like behaviour can be applied to all phenomena - thus, there is a dual description, but both descriptions are always completely reconcilable (this doesn't contradict what ZapperZ said only if the unified description allows for both descriptions in every situation... the descriptions would simply be different approaches).
3) Some third option because I'm too tired to think straight at this moment.

The problem here is that it isn't just a matter of the "source" here, which in this case, is light. It is also a matter of our DETECTION method, and that has nothing to do with light itself.

Again, the double slit, for example, isn't really a "test" of the light - it really is how light is reacting to the setup and the fact that it is given the ability to go through two different path equally well. Thus, if you look at the Feynman description of it, it really is the superposition of path, not the "interference of single photons", that is the origin of the interference pattern. Once we realize this, then we will notice that this isn't restricted to just light, but ANY object in which our ability to exactly track which "slit" that particle goes through is unknown.

With this in mind, look at how we handle and detect EM radiation of different wavelenths. You will notice that the longer the wavelength, our detection of it differs and goes more into the classical regime. Our instruments are now larger, something that we can start to comprehend, and begin to incorporate a lot more decoherence. So this is a function of not only the light property itself, but how we handle such properties.

Zz.

 

[Tsunami]

Ok, thanks ZapperZ, I don't think I understand all of that, but it's enough to protect me from major misconceptions.

(And to ask more explanation than you've already given me would not only be asking more than is polite, but also plain lazy.)

 

[quantumdude]

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).

Please. It's obvious to anyone who knows SR that your understanding of it contains rudimentary errors that stem from severe misconceptions that you are either unwilling or unable to let go of.

If you think it is bogus, then you should give detailed reasons why.

People do that on a daily basis here.

Thomas2, your userid has been banned. I am now going to ban your new userid, Thomas3. Do not start another one. Your contributions are no longer welcome here.