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Shape of the wave of a single photon


DParlevliet

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then the conclusion is that a photon is/has a wave which on large distances (in extreme: in the whole universe) is a wave with constant amplitude.

And that is yet another problem with the "the photon goes through both slits" interpretation. Like most interpretations/analogies it doesn't make too much sense when you examine it too closely.

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If the wave is there exactly at the moment the photon is emitted, and is gone (collapsed) when the photon is absorbed, then the wave has non-locality. The wave is everywhere at once, also at large distances, so faster then light.

Of course after emission the wave propagates with the speed of light.

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If the wave is there exactly at the moment the photon is emitted, and is gone (collapsed) when the photon is absorbed, then the wave has non-locality. The wave is everywhere at once, also at large distances, so faster then light.

 

Probably more accurate to say the photon has non-locality. (But yes.)

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Probably more accurate to say the photon has non-locality. (But yes.)

 

Could be, but I am not fully sure. It can have non-locality properties (like in EPR). But the particle postion can be (average) calculated between source and absorption with the speed of light. So that property has locality.

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Could be, but I am not fully sure. It can have non-locality properties (like in EPR). But the particle postion can be (average) calculated between source and absorption with the speed of light. So that property has locality.

 

You can consider that the photon went in a straight line (or a series of straight lines, in the case of diffraction) from the source to where you detect it. But ... for QED to calculate the probability of that path, it has to take into account everything the photon could have encountered (i.e. non-local effects). Also, if you try and test whether the photon really did take the path you assume, then you change the outcome of the experiment (by forcing it to be localized).

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It is about the question if the waves are flat (A2) or spherical (A1). blue = emittor, red = photon

A2 will give in the double slit a different interference pattern then A1.

forum4.GIF

According Feynman it is A1. QED is right, so that must be the answer.

However the implication is that if the wave travels, new waves must come out of the ploint where the photon was emitted. How far the photon travels, always somewhere in spare there will be this point.

I know QM is weird, but this does not feel right

 

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You are right. I think because what has been mentioned in the topic diffraction. The interfering waves are secondary spherical waves, so it does not matter at what angle the primary wave hits the slits, as long as they are in phase.

 

A1 means also that the amplitide (intensity) is 1/r2. But Feynman arrows are propability over an area (flux, sum of amplitudes). So the amplitudes A, a1, a2 in the formula are flux. This shows also in the double slit:

 

Narrow one slit to the half. Then the interference pattern will also be half (B). So propability of a single photon depends on the "amount" of wave passing through the slits.

 

forum5.GIF

Edited by DParlevliet
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Then the most important question: what is the function of the wave. The double-slit shows clearly that it determines the probability of absorption of the photon at the detector. See an example of only a detector:

forum6.GIF

When the photon is absorbed as above, we know what average path it has followed (not exactly). When calculating with speed of light we also know where is was (most of the time) between emittor and detector. But according the wave amplitude, which is equal in space, the propability elsewhere, like a or b, is the same as at the detector. That cannot be true of course. So the conclusion should be that the wave is not everywhere a propability wave, only at the place of absorption.

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Remarkable, I have never heard of this wave. I was told that a real probability wave must be able to normalise: integrated over the whole universe should be 100% probability. That is not possible with this wave, I think.

Another result of the above situation B: if one slit is narrowed to half, then the absorption probability changes, but not the energy of the photon. So the wave does not contain energy in its area/volume. This is obvious for a probability wave, but that arises the question: where is the energy? That must be in the particle. So my conclusion would be that wave and particle must be both present.

And what is the path of this particle? Some propose that the particle goes through both slits. Because the particle cannot be split, the explanation would be that the slits has non-locality. In the classic world you see two separate slits, but in the quantum world it would it be one slit. But how should this non-locality work, only for the slits. And in situation B, how does the non-locality know that there is a full and half slit, so the photon must be divided in 2/3 and 1/3? That looks far searched.

So my guess would be that the particle really goes through one slit, independent if we detect/know or not.

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Remarkable, ...........................................etc

 

Unless i misunderstand you. you have come to the conclusion I mentioned much earlier in this thead. That there are in fact two waves. One the probaliity wave , which is concerned with the location of the photon. and two the sinusoidal wave of constant peak amplitude carrying the light energy. Whether these two waves are somehow one of the same but two aspects of the same wave [ maybe they are ].

i do not know. Fenyman spoke of the probability wave as well as the actual photon [ which is a sine wave of constant peak amplitude ] as an entity.

 

what i have just said is maybe more of a muddle. please say so, as you seem to be getting your head around it , in the context of the double slit experiment.

 

there is one of Fenyman lectures , where he reviews simple reflection. here the photon takes all possible paths to the object. but the ones that went along paths not following angle of incidence equals angle of reflection. somehow the other paths were further out the probability wave, and somehow canceled , leaving mainly i = r .MOSTLY but maybe a little elsewhere !

 

When you get a clear insight of what a photon actually is, PLEASE rattle my cage and tell me what you SEE .

 

mike

Edited by Mike Smith Cosmos
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you have come to the conclusion I mentioned much earlier in this thead. That there are in fact two waves.

 

Two waves.... ?

 

I have 2 usd cheap laser at hand that has 5 mW max output, and wavelength 650 nm.

 

1 W = 1 J/s

so 5 mW = 5 mJ/s = 0.005 J/s

 

Photon with wavelength 650 nm has:

E=h*c/650 nm = 6.62607*10^-34 * 299792458 / 650 *10^-9 = ~3.056*10^-19 J energy...

 

Assuming (unreal) that there is no lost and all energy is converted to photons with such wavelength, this mine tiny laser will produce

0.005 J / 3.056*10^-19 J = 16,360,879,214,888,979 photons emitted per 1 second of work..

That's 16,360,879 per nano second.

 

When such tremendous quantity of photons will collide with atoms in wall we see dot. From the all directions we look at it. Photons are reflected in the all directions

Edited by Sensei
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Unless i misunderstand you. you have come to the conclusion I mentioned much earlier in this thead. That there are in fact two waves. One the probaliity wave , which is concerned with the location of the photon. and two the sinusoidal wave of constant peak amplitude carrying the light energy.

 

I don't think so. The double slit shows one wave which is partly probability (only at the place of absorption) without energy.

 

How would you test this conjecture? Your guess is contrary to what we know of quantum mechanics.

 

That cannot be tested, it is a guess. Based of logic and simplicity as explained above, but still a guess.

 

As far as I know QM does not talk about what is physically happening. It just provides formulae. With the double slit it uses only the wave property, because that is sufficient. I have seen that some people talk about particles going through both slits, but as a guess too, without proof or test.

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That cannot be tested, it is a guess. Based of logic and simplicity as explained above, but still a guess.

 

As far as I know QM does not talk about what is physically happening. It just provides formulae. With the double slit it uses only the wave property, because that is sufficient. I have seen that some people talk about particles going through both slits, but as a guess too, without proof or test.

 

It's not science if it can't be tested. The fact that single photons interfere indicates that both slits are involved.

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Then discussing about particles splitting up is no science either. But there are branches of science, like history, where testing is not possible at all, and theories are based on logic arguing. Nobody forbids to think about possible explanations, even if you cannot test it (yet or never).

You talk about a photon (= particle + wave). The photon interferes with its wave. I talk about the particle.

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Then discussing about particles splitting up is no science either. But there are branches of science, like history, where testing is not possible at all, and theories are based on logic arguing. Nobody forbids to think about possible explanations, even if you cannot test it (yet or never).

 

You talk about a photon (= particle + wave). The photon interferes with its wave. I talk about the particle.

 

 

Having listened to all that has been discussed in this thread and elsewhere, i am beginning to pick up an image of roughly what a photon looks like and its nature.

 

Due to its micro size , the de Broglie wave is significant. so whatever the photon is it can be anywhere across the de brodlie probability wave pattern previously illustrated going from a very small probability right though to a max and back down to a minimum. Let us say the wave packet represents the extent to which the photon " heart " is thought of as being.

 

Provided this width is greater than the double slit distance , then it is possible for the photon to go through both slits. whether it does is another matter.

The wave packet nonetheless can be 'seen' as being bigger than the two slits and thus going through both as a sort of wave front, thus causing interference patterns the other side of the slits.

 

Now this wave packet looks a bit like a vibrating bubble with this dense "heart" in its middle , being the place where the photon heart mostly is. Like a bee buzzing around in a bottle.

 

This whole thing is buzzing with the frequency given by the De Broglie equation as well as the Energy equation.. as f

 

This is in fact the frequency of the Light photon with its constant peak amplitude.

 

But due to the Quantum nature and Plank , this photon only lasts for its requisit time. start bzzzz stop.

 

This whole bullet of buzzing at f ( frequency) is the wave packet or particle.

 

Now . Try and get a hold of this photon by detection and the whole thing collapses to a point, like smashing the bottle and swatting the bee [ the heart of the photon ] Got you you B#### . It it grounded at one location only.

 

This does quite beg the question , what quite is this bee heart that is buzzing about ?

 

 

How does that sound like a visualisation ?

 

post-33514-0-00600400-1383156700_thumb.jpg

Edited by Mike Smith Cosmos
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Then discussing about particles splitting up is no science either. But there are branches of science, like history, where testing is not possible at all, and theories are based on logic arguing. Nobody forbids to think about possible explanations, even if you cannot test it (yet or never).

 

You talk about a photon (= particle + wave). The photon interferes with its wave. I talk about the particle.

 

 

 

History is not a branch of science. Even so, a scientific approach can be used, and it must rely on evidence, otherwise it is conjecture.

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History is not a branch of science. Even so, a scientific approach can be used, and it must rely on evidence, otherwise it is conjecture.

 

Yes, my conclusion was just good argued guess = conjecture.

Not so important, still interesting to think about.

 

 

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Having listened to all that has been discussed in this thread and elsewhere, i am beginning to pick up an image

 

How does that sound like a visualisation ?

attachicon.gifimage.jpg

This has an echo of the behaviour of the electron in orbit. Electrons bound by the orbital constraints.Except now out of the atom the space occupied by the photon is more , much more than the atom size. Seems like the photon is the electron escaped ( like the genie out of the bottle ). Now that is a neat picture.

 

post-33514-0-35817500-1383157890_thumb.jpg

Edited by Mike Smith Cosmos
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This has an echo of the behaviour of the electron in orbit. Electrons bound by the orbital constraints.Except now out of the atom the space occupied by the photon is more , much more than the atom size. Seems like the photon is the electron escaped ( like the genie out of the bottle ). Now that is a neat picture.

 

If we limit in this topic to what the double-slit shows, then there is no wave packet, but a wave in the whole universe with amplitude 1/r2. It is not likely that there are more waves.

Another interesting example. Place in front of the double slit a mirror. Now nothing will change, there is still the same interference pattern

forum7.GIF

What happens in the mirror?

- The wave: That is easy. First I thought that the wave would be reflected by the atoms in the mirror, but according the topic diffraction it is easier. Reflection is caused by the space in front of the mirror where every point radiates a spherical wave: so according the Huygens principle

- The particle is more interesting. It is elastically scattered by one atom, but spherical. So how does the particle know that it must go down?

 

Before going further I want to ask what is suppose to be happening between photon and atom during reflecting:

1 the photon is absorbed by the atom and a new equal photon is transmitted by the atom

2 or the photon is only scattered by the atom, so remains the same.

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Before going further I want to ask what is suppose to be happening between photon and atom during reflecting:

1 the photon is absorbed by the atom and a new equal photon is transmitted by the atom

2 or the photon is only scattered by the atom, so remains the same.

 

Better leave that to swansont , he is the expert on all this!

 

Mike

Edited by Mike Smith Cosmos
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[...] The particle is more interesting. It is elastically scattered by one atom, but spherical. [...]

Your conception of wave and particle is wrong, and this misleads you.

 

A wave is a particle (or several particles) because it can interact very locally if needed, for instance at one pixel of a camera, and concentrate there everything it has. But nothing beyond.

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My experience is that everybody has a different explanation about the particle-wave and then tell me that my conception is wrong.

In several places I have read that in a mirror a photon is absorbed by an atom and re-emitted. But then the mirror result cannot be explained. If the photon is absorbed, then at the same time its wave is gone. If the photon is re-emitted, how does it know that it must be emitted at the same angle as the original photon? Is has no remembrance of the previous photon. So the conclusion is that the photon is not absorbed but scattered.

And a second conclusion would be that the scattered particle is driven by the wave, and not the other way around. Because a single atom can scatted a particle in all directions. It is apparently the wave-property of the photon which is causing its particle-property to go in the mirrored directlion.

Edited by DParlevliet
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