Is the Wave Aspect of Light More Like a Snake or a Wave in Water?

[Arc_Central]

Thanks hypnagogue

But isn't the Schrodinger wave in relation to electron orbits? In other words - This is not a depiction of a photon in the dead of space?

I'm wondering if the pic is a 3D depiction after it is laid out on 2D paper. Do you have a link to the site you got this pic from?

If it is what I think it is. I am quite happy with what I am seeing, and I can still commence with my own depiction of a photon in 3D terms.

 

[arcnets]

Originally posted by Ivan Seeking
In another context, we might mean the electric and magnetic field strengths associated with a photon for example.

Here, I can't resist re-stating something that I have stated many times in these forums before:
I think there exists such a thing as a 'single electron wavefunction'. But such a thing as a 'single photon electromagnetic field' does not exist.

 

[hypnagogue]

Originally posted by Arc_Central
Thanks hypnagogue

But isn't the Schrodinger wave in relation to electron orbits? In other words - This is not a depiction of a photon in the dead of space?

I'm wondering if the pic is a 3D depiction after it is laid out on 2D paper. Do you have a link to the site you got this pic from?

If it is what I think it is. I am quite happy with what I am seeing, and I can still commence with my own depiction of a photon in 3D terms.

As I understand it, the Schrodinger wave is a mathematical description of a subatomic particle. It isn't constrained to describing only electron orbits, although electron orbits may be subsumed under the general description of the wave.

The picture I gave a link to depicted a 2-dimensional drum top in a state of vibration in 3 dimensions. Remember, though, that there is not necessarily any physical meaning behind the Schrodinger wave, at least in the sense that we are familiar with.

The site that this picture comes from is pretty irrelevant-- I think it's something about programming and/or making 3-dimensional images. Actually, come to think of it, that might be relevant for what you are doing. To get to the site delete the "images/..." part of the URL.

 

[Ivan Seeking]

You can't draw a photon. You can't draw a wave equation. Any respresentation of such would be wrong and misleading. This whole notion results from a misunderstanding of these ideas. You can only draw a representation for the probability of finding a photon.

 

[Ivan Seeking]

Originally posted by arcnets
Here, I can't resist re-stating something that I have stated many times in these forums before:
I think there exists such a thing as a 'single electron wavefunction'. But such a thing as a 'single photon electromagnetic field' does not exist.

Interesting. Can you explain? We can produce single photons that are described by the EM wave equation. Actually, the more I think about it, I don't see where you get this.

 

[Ivan Seeking]

Originally posted by Arc_Central
Let us say that we have one photon with a travel distance of say 100 light years, and two detectors are at that distance, and they are separated by 25 light years. Either one could detect the photon - correct?

Yes.

In regards to the wave collapsing to one location - Does the whole of the wave collapse instantaneously?

Yes.

EDIT: Whoops. I think this is wrong. I am thinking that a very short time interval is required eg 10^-31 sec. If may be 20 to 45 orders of magnitude less than this,; I would have to check. This also sets the speed limit for Quantum Computers.

Also in regards to detection - Are you saying that detection requires the ejection of an electron? Is there any other way to detect the wave? Can a wave excite an electron and still not eject it, and can we detect that?

I believe that the only means for detecting an incident photon is the release of an electron. If we only excite an electron, then when it falls back into its ground state, another photon is released.

 

[Arc_Central]

I hate to ask all these questions of you Ivan, but I gots one more.

Lets say we have the Earth and it's all there is - plus two detectors. One detector is one light year away from Earth in one direction, and the other dectector is on the opposite side of Earth one light year away. I fire a laser toward one of the detectors. Is there any chance that the detector on the opposite side of Earth can detect any of the photons at the one year time after firing the laser?

 

[Ivan Seeking]

Originally posted by Arc_Central
I hate to ask all these questions of you Ivan, but I gots one more.

Lets say we have the Earth and it's all there is - plus two detectors. One detector is one light year away from Earth in one direction, and the other dectector is on the opposite side of Earth one light year away. I fire a laser toward one of the detectors. Is there any chance that the detector on the opposite side of Earth can detect any of the photons at the one year time after firing the laser?

Yes.

 

[Ace-of-Spades]

Thats QM tunneling right?

 

[Ivan Seeking]

Originally posted by Ace-of-Spades
Thats QM tunneling right?

No...it need not be. The photon could follow any possible path and arrive at the detector. I think the odds of a photon tunneling through the Earth would be absolutely astronomical...like 1:e^500!
It could just go around the Earth though. A very small chance for this exists. Of course, at any time now I could get stomped...but I think this is correct.

 

[Arc_Central]

Something tells me this may not be possible - That the best possible detection angle would be 180 degrees. Does anyone know of any experiments that detected a 360 degree possibility for detection?

 

[Ivan Seeking]

Originally posted by Arc_Central
Something tells me this may not be possible - That the best possible detection angle would be 180 degrees. Does anyone know of any experiments that detected a 360 degree possibility for detection?

I am taking this directly from QED.

 

[wimms]

I guess Arc_Central is right here, as you Ivan also said, "The photon could follow any possible path", which excludes obstructions, like in doubleslit experiment.

Is it okay to picture single photon as expanding probability sphere with radius in lightyears, at the 'surface' of which (excluding obstructed paths) there is equal probability of detecting that same photon, anywhere on the surface? Ie. if we detect photon that's been inflight for 15B lightyears, we can assume that its 'sphere' has 30B diameter, and after probability collapses in detection, all of that probability 'sphere' disappears, including 30B lyrs away, instantly?

 

[Ivan Seeking]

Originally posted by wimms
I guess Arc_Central is right here, as you Ivan also said, "The photon could follow any possible path", which excludes obstructions, like in doubleslit experiment.

Is it okay to picture single photon as expanding probability sphere with radius in lightyears, at the 'surface' of which (excluding obstructed paths) there is equal probability of detecting that same photon, anywhere on the surface? Ie. if we detect photon that's been inflight for 15B lightyears, we can assume that its 'sphere' has 30B diameter, and after probability collapses in detection, all of that probability 'sphere' disappears, including 30B lyrs away, instantly?

I didn't mean to go through the earth; I meant to go around it. I think this qualifies as a possible path. I am hunting for my QED book as we speak...or type...

Edit: One reason I think this is correct is the explanation that we use for attraction with electric fields. The argument that a probability exists for the photon from electron A, to come in from behind positron B [meaning that the electron and the positron face each other ], is less than zero; and this fact ultimately accounts for the attraction between the charges. In this way I think that even nonsensical paths must be considered. I must admit though that I could be in trouble here…but I don’t think so yet.

 

[Arc_Central]

Ivan

With your 360 degree wave - The chances are pretty high that a photon will collapse, reflect, diffract, or whatever on the Earth. Somewhere toward 50% chance that the laser light pointed away from Earth will end up doing something on Earth in some way. This aint jiving with me at all. Somehow I don't think the universe would look the same either. Would it mean that some sunshine from the other side of the sun is warming good ole Earth?

 

[Ivan Seeking]

Originally posted by Arc_Central
Ivan

With your 360 degree wave - The chances are pretty high that a photon will collapse, reflect, diffract, or whatever on the Earth. Somewhere toward 50% chance that the laser light pointed away from Earth will end up doing something on Earth in some way. This aint jiving with me at all. Somehow I don't think the universe would look the same either. Would it mean that some sunshine from the other side of the sun is warming good ole Earth?

There is a chance. There is a chance for any possible path...whether it makes sense or not. This is one example of where the particle model really flops. But, until I find my book in this mess, I must admit to less than 100% certainty on the correct interpretation here.

 

[Loren Booda]

On land, snakes can locomote 4 distinct ways.

 

[Arc_Central]

There is a chance.

From your point of view - There is a high probability. If I understand your interpretation correctly. Any point in the sphere of influence would have an equal chance for detection. The Earth would influence close to 50% of the photons from that laser. Unless of course the photons go through the Earth.

 

[arcnets]

(Repeat: There is no such thing as a 'single photon e.m. field.)

Originally posted by Ivan Seeking
Interesting. Can you explain? We can produce single photons that are described by the EM wave equation. Actually, the more I think about it, I don't see where you get this.

OK, I'll try to explain. A photon, by definition, has sharp energy. Because it is massless, it thus has sharp momentum. This means [del]p = 0. Now from the uncertainty relation [del]p [del]x >= h, we find that such a state has [del]x = [oo] and thus is totally unlocalized.

Now if we want to find any probability distribution that makes any sense, then we must have the photons be absorbed somewhere, meaning [del]x is finite. This means that the wave describing the process has [del]p > 0, and thus corresponds to a mixture of photons.

This does not apply to particles which have mass. As you know, the bound electron states in an atom have [del]E = 0, but [del]p and [del]x both finite.

 

[Ivan Seeking]

Originally posted by Arc_Central
From your point of view - There is a high probability. If I understand your interpretation correctly. Any point in the sphere of influence would have an equal chance for detection. The Earth would influence close to 50% of the photons from that laser. Unless of course the photons go through the Earth.

Oh no I never said anything about equal probabilities. The greatest chance is that a photon will appear to travel in a straight line like a bullet. As we deviate from this path, the chance for the path gets smaller and smaller. For this example, I speak of probabilities that are very small. But you asked if ANY chance existed for such paths.

 

[Ivan Seeking]

Originally posted by arcnets
(Repeat: There is no such thing as a 'single photon e.m. field.)
Now if we want to find any probability distribution that makes any sense, then we must have the photons be absorbed somewhere, meaning [del]x is finite. This means that the wave describing the process has [del]p > 0, and thus corresponds to a mixture of photons.

But you shift gears from a propogating wave function, to one of a collapsed wave function. Your statement only appears to be correct once the photon "strikes" a target. We have a completely different animal at this point.

 

[Ivan Seeking]

Originally posted by arcnets
This does not apply to particles which have mass. As you know, the bound electron states in an atom have [del]E = 0, but [del]p and [del]x both finite. [/B]

But they need not be. I can force a [del]p or [del]x equal to zero for an electron also. You consider only the bound state; then you compare this to an unbound wave-function.

 

[arcnets]

Originally posted by Ivan Seeking
Your statement only appears to be correct once the photon "strikes" a target.

That is correct. In fact, I'm trying to point out that probabiltiy statements about photons only make sense if these are detected (and not 'in dead space').
Imagine you do a photon experiment, like double-slit or anything. In doing so, you detect the photons with some detector device, say a photographic film. For high intensities, you get a distribution of the energy input per unit time on the screen which can be calculated from the e.m. field (IIRC, E(x)² + B(x)² or so).
If you dim down the source, you observe the screen being hit randomly by single photons. If you let this accumulate for a long time, you see that the probability distribution on the film converges to the one calculated from the e.m. field.
Now you could of course scale down the e.m. field so it represents one photon per unit time. But it's still a wavefunction describing a flow of photons, not a single photon.

I also have my doubts whether an e.m. field ever 'collapses'. I guess an e.m. field is ruled by Maxwell's equations, and I'm afraid these don't allow a 'collapse'.

But they need not be. I can force a p or x equal to zero for an electron also.

'Course you can. What I'm trying to say is, you can have [del]E = 0 for an electron and still get an interesting spatial distribution. But a photon must have [del]E = 0, and is then spread out all over space. Unless you detect it, see above.

You consider only the bound state; then you compare this to an unbound wave-function.

OK, bad wording by me. I should have said:
"This does not apply to particles which have mass. As you know, there are (bound or free) electron states which have [del]E = 0, but [del]p and [del]x both finite."