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a reply to: ImaFungi

There is no dispute about numerous photons going in all directions from a source as the diagram I previously posted shows:

originally posted by: Arbitrageur

The concept we disagreed with was your idea that a single photon would expand in all directions. It doesn't. The diagram shows many photons going in different directions.

Let's say you were able to lower the photon production rate to 1 per second. One photon would go in one direction, the next photon would go in a different direction, and so on, such that you'd eventually get the same pattern as shown in the diagram where adding up the pattern of many photons would result in a somewhat uniform spherical radiation pattern.

You seem to think this spherical pattern would represent what a single photon would do, but it doesn't. I hope you would have figured this out by now.

The difference between a single photon versus the statistical distribution of numerous photons is also demonstrated in some double-slit videos on youtube. You can't see any pattern at all after the first few photons strike the target. The distribution pattern only becomes apparent after a statistically significant number of photons has struck the target.

Will a photon change property/wavelength as soon as it hits an object and reflects of it and moves on until the photon hits an other abject?

Light is one reason to why we can observe motion compared to stationary Objects.
Light is also the main Source we use to observe changes in matter.
Light is also Our main Source of time.

originally posted by: spy66
Will a photon change property/wavelength as soon as it hits an object and reflects of it and moves on until the photon hits an other abject?

Light is one reason to why we can observe motion compared to stationary Objects.
Light is also the main Source we use to observe changes in matter.
Light is also Our main Source of time.

Would it surprise you if i told you a photon doesnt reflect off an object? It is a copy of the EM field that we see as reflected light. Well on a mirror anyway thats why they dont scatter light and instead re radiate it.

originally posted by: dragonridr

originally posted by: spy66
Will a photon change property/wavelength as soon as it hits an object and reflects of it and moves on until the photon hits an other abject?

Light is one reason to why we can observe motion compared to stationary Objects.
Light is also the main Source we use to observe changes in matter.
Light is also Our main Source of time.

Would it surprise you if i told you a photon doesnt reflect off an object? It is a copy of the EM field that we see as reflected light.

So it actually reflects and copy the EM Field of the Objects it interacts with?

EDIT.

If that is the case. A photon would still change right?

a reply to: spy66

A photon would still change right?

Do you mean change in direction? If that's not what you mean, what kind of change are you talking about? I'm not seeing any wavelength (color) changes in this photograph in the reflection, are you?



There are only a handful of videos of Richard Feynman's lectures. Four are from his university of Auckland lectures and one of them is on reflection of light. He describes the problems with classical wave theory, and classical particle theory and how they both fell short of explaining specific reflection behavior for centuries since Newton, so it's not the simplest topic. Then he describes how we can model what happens in reflection, though he admits it's not an intuitive model, like the rest of quantum mechanics.

QED: Fits of Reflection and Transmission -- Quantum Behaviour -- Richard Feynman (2/4)

originally posted by: Arbitrageur
a reply to: spy66

A photon would still change right?

Do you mean change in direction? If that's not what you mean, what kind of change are you talking about? I'm not seeing any wavelength (color) changes in this photograph in the reflection, are you?

There are only a handful of videos of Richard Feynman's lectures. Four are from his university of Auckland lectures and one of them is on reflection of light. He describes the problems with classical wave theory, and classical particle theory and how they both fell short of explaining specific reflection behavior for centuries since Newton, so it's not the simplest topic. Then he describes how we can model what happens in reflection, though he admits it's not an intuitive model, like the rest of quantum mechanics.

QED: Fits of Reflection and Transmission -- Quantum Behaviour -- Richard Feynman (2/4)

Thank you, this was actually very good.

What i actually ment was this: The light Source that lights up "lets say the landscape in the image you displayed" is from the sun initially and than the photons hit the landscape and reflect onto Your view. And we get a image With differnt colours, and you get a perspective of the landscape, its distance, shape and dimention compare to its background.

Since the inititial light is from the sun. The reflected light from the landscape would have changed so that we can view the landscape?
Isent this true?

a reply to: spy66

Ok i think i see what your asking the photons dont change however light is absorbed making us perceive color. Scientifically speaking, color is the amount of light absorbed by an object. When light hits an object, the eye sees a certain amount of light and also certain types of light. but there isnt a frequency change the photons still remain the same just basically less or more of a certain type.

a reply to: spy66
a reply to: dragonridr

Light from the sun has all the colors, so there are red, blue, green, yellow, etc photons hitting the landscape.

When we see green trees, the green photons are reflected from the sun hitting the leaves, to our eyes, without changing color. Many of the photons of colors other than green are absorbed by the leaves and that is the energy the tree uses for photosynthesis. The changes in color with distance are related to atmospheric absorption, so when the sun turns orange at sunset, the photons aren't shifting wavelength, rather reddish photons with longer wavelength don't interact with as many atmosphere molecules so we see more of them. The blue photons interact with more air molecules and get scattered more so that's why the sky is blue.

That's not to say that photons can't change their frequency when interacting with matter, as happens in Compton scattering, but as the name suggests, this is a type of "scattering" and we don't refer to it as "reflection".

Compton scattering is an inelastic scattering of a photon by a quasi-free charged particle, usually an electron. It results in a decrease in energy (increase in wavelength) of the photon (which may be an X-ray or gamma ray photon), called the Compton effect. Part of the energy of the photon is transferred to the recoiling electron. Inverse Compton scattering also exists, in which a charged particle transfers part of its energy to a photon.

a reply to: Arbitrageur

Thanks to both of you. Got the idea now

a reply to: Arbitrageur

The concept we disagreed with was your idea that a single photon would expand in all directions. It doesn't.

That's not strictly true.

The photon wavefunction gives nonzero probability values for position at all locations; it isn't until the photon is detected that we can say which direction it went.

originally posted by: Astyanax
a reply to: Arbitrageur

The concept we disagreed with was your idea that a single photon would expand in all directions. It doesn't.

That's not strictly true.

The photon wavefunction gives nonzero probability values for position at all locations; it isn't until the photon is detected that we can say which direction it went.

We need to draw a distinction between the photon itself, and the wavefunction of the photon. To say the wavefunction spreads out is correct, but the photon itself does not spread out.

Photons do not spread out as they propagate

experiments confirm that the photon is not a short pulse of electromagnetic radiation; it does not spread out as it propagates

Let me repeat for emphasis:

it does not spread out as it propagates
it does not spread out as it propagates
it does not spread out as it propagates

I don't think ImaFungi was referring to the wavefunction, unless I misinterpreted what he said.

a reply to: Arbitrageur

Sure, but that photon is only shown to be present wherever it was present after it was detected. Until then, it could be anyplace. Because it could be any photon emitted by the source in any direction. What gives it location and direction is the act of identifying the particular photon — aka detection.

There's no photon till it's detected. Until then, there's only a wavefunction.

originally posted by: Astyanax
There's no photon till it's detected. Until then, there's only a wavefunction.

Do we teach this in graduate school? Yes. As dragonridr said we teach graduate students the Copenhagen interpretation as if it's true without much emphasis on alternative interpretations (That I'm aware of).

Is this correct? We don't know, and I don't assume that it is. I suspect probably not. As discussed in the OP, and OP video, where physicist Sean Carroll prefers an alternate interpretation of quantum mechanics (Everett), he believes it's not true. In the DeBroglie-Bohm interpretation that is also not true. As Dr Carroll explains, when one investigates QM interpretation critically, one is not necessarily drawn toward his preferred Everett interpretation specifically, but one is driven away from the traditional Copenhagen interpretation (according to him). Here is the Debroglie-Bohm view which experimentally cannot yet be distinguished from Copenhagen, so this could in fact be true:

de Broglie–Bohm theory of quantum mechanics

The de Broglie–Bohm theory of quantum mechanics is a theory by Louis de Broglie and extended later by David Bohm to include measurements. Particles, which always have positions, are guided by the wavefunction. The wavefunction evolves according to the Schrödinger wave equation, and the wavefunction never collapses. The theory takes place in a single space-time, is non-local, and is deterministic. The simultaneous determination of a particle's position and velocity is subject to the usual uncertainty principle constraint. The theory is considered to be a hidden variable theory, and by embracing non-locality it satisfies Bell's inequality. The measurement problem is resolved, since the particles have definite positions at all times.

So if this interpretation is correct, there is a photon before it's detected, and the wavefunction is guiding it.

Think of a galaxy cluster 40 billion light years away, which 12 billion years ago emitted a photon, like COSMOS-AzTEC3 for example. If the wavefunction for that photon is now 80 billion light years across and gives the photon a chance of being observed 80 billion light years away, or here on Earth, isn't the DeBroglie Bohm interpretation a little easier to swallow with the photon having a position before it's observed, than Copenhagen where it doesn't? I'm not saying DeBroglie Bohm is right, as nobody knows, but I do entertain the possibility that it could be correct, and in that example at least, it seems to be less problematic than the Copenhagen interpretation.

a reply to: Arbitrageur

I'm not arguing interpretations. What I'm saying applies to any of the interpretations. It doesn't matter whether you believe in a real or purely stochastic existence for the particle before detection.

Consider a many-worlds scenario in which that photon is in different locations along different worldlines. Unless it emerges from a coherent emitter, 'that' photon is one of many emerging from the source in all directions. This is true in all worldlines. So how do you identify 'your' photon as the same in all worldlines? How do you tell it apart from all the other photons being emitted by the source? You can't. You put a detector in the way of a photon; but wherever you put the detector relative to the source you'd still detect a photon. By what means could you identify one specific photon in multiple worldlines?

As far as I know, you can't.

The same problem arises in the de Broglie-Bohm interpretation. 'The 'photon is always somewhere, but where is it? How do you tell it apart from all the others emitted by the source?

Even a coherent emitter doesn't really solve the problem. Which of the stream of photons being emitted by the source did your detector detect? How did it differ from all the other photons being emitted by the source until you detect it? Is there even a difference, strictly speaking, at all?

You can devise a source that produces one photon at a time, of course, but even in that case the wavefunction would take nonzero values at all locations but one. Possibly not in the de Broglie case; I can't work through the mathematics, so there may be some wiggle room there for establishing a unique position for the particle before detection. Is there?

By the way, I should mention, for the general information of everyone, that this a very esoteric consideration and doesn't really apply to the discussion you were having earlier with spy66. Still, for the sake of being pedantic...

a reply to: Arbitrageur

Well me personally i dont believe the copenhagen interpretation is correct. Ive seen things that just cant be explained with it and this brings up hidden variables to cover it. In the 1960s, John Bell proved an important theorem about hidden variables theories. He showed that any deterministic hiddenn variables theory capable of giving all the statistical results of standard quantum mechanics must allow for superluminal connections, in violation of Einstein's assertion that no signals can move faster than light. We really didnt discuss this part of Bells theorem but it tells us among other things that we cant explain observations without including something that goes faster than light.

a reply to: dragonridr


I have another question:

A photon can not have mass, if it did it would not travel With the exact speed of light c in a vacuum.

A photon has no electrical charge and is therefor stable. That means it is a constant from E.

Can you explain to me how, or at what time in the light cone time frame a photon gets its spin?

Would the photon have a spin already in the past light cone, or would the photon recieve the spin and vector from the Object it interacts With before it becomes the New light cone?

Or does the photon spin just change vector constantly at the speed of light "as it interects with With the motion and electrical charge of a particle"?

- What i mean is: From the past light cone a Photon brings With it a vector and spin, and when it interact With a particle it changes vector and momentom and becomes the New light cone.

Is this true?

If it true, the photons does change both its momentom and vector when it interacts With a particle.

The light cone i am refering to: "The past light cone E is the future light cone inverted at the constant speed of light".

originally posted by: spy66
a reply to: dragonridr


I have another question:

A photon can not have mass, if it did it would not travel With the exact speed of light c in a vacuum.

A photon has no electrical charge and is therefor stable. That means it is a constant from E.

Can you explain to me how, or at what time in the light cone time frame a photon gets its spin?

Would the photon have a spin already in the past light cone, or would the photon recieve the spin and vector from the Object it interacts With before it becomes the New light cone?

Or does the photon spin just change vector constantly at the speed of light "as it interects with With the motion and electrical charge of a particle"?

- What i mean is: From the past light cone a Photon brings With it a vector and spin, and when it interact With a particle it changes vector and momentom and becomes the New light cone.

Is this true?

If it true, the photons does change both its momentom and vector when it interacts With a particle.

The light cone i am refering to: "The past light cone E is the future light cone inverted at the constant speed of light".

Ok first a photon may have mass its just so small that we cant detect it. never really liked it when people say they dont have mass. One mass and energy are interchangeable remember e=mc2 but that aside. If i can push against an object with a laser than there is mass involved. Now a Neutrino on the other hand can pass through an entire planet without slowing down or even hitting another particle. So maybe they have no mass at all or so little as to have almost no interactions.

As for spin your making it very complicated its not. Spin is created when the photon is created. The part i think your confusing is there are two types of angular momentum. There is light spin angular momentum abbreviated SAM and light orbital angular momentum also called OAM . First SAM an electromagnetic wave is said to have circular polarization when its electric and magnetic fields rotate continuously around the beam axis during the propagation. so our electric and magnetic waves among others travel in a cork screw around a central axis. We can think of this as our photon spinning. Now OAM is different and i think this is what your asking about.

OAM has to do with the shape of our wave front. This has to do with relativity and how we observe our wave front. If i have a light beam we can say it has linear momentum and spin can be caused by the shape of our wave front. For example when we pass light through a lens we can change its wave front actually causing our light beam to spin. Think of this as focusing our beam the more spin we put on the wave front the tighter the beam of our laser. Since we can plot the spin of our light beam we can say its time dependent and also dependent on where our observer is located. We call this paraxial approximation and heres the math.

en.wikipedia.org...

originally posted by: Astyanax
The same problem arises in the de Broglie-Bohm interpretation. 'The 'photon is always somewhere, but where is it? How do you tell it apart from all the others emitted by the source?

In deBroglie-Bohm, the photon you detected on earth from 12 billion years ago was traveling in the direction of where the Earth would eventually be now, for the 12 billion years. You can't say there was no photon before the observation, because in the DeBroglie-Bohm interpretation there was, and there is no wave-function collapse upon observation as there is in the Copenhagen interpretation.

So let's say .01 light seconds before you detect the photon, it was .01 light seconds away from your detector according to deBroglie-Bohm. Maybe you didn't know where it was then, but you know now where it was then and it wasn't equally likely to have been 80 billion light years away, as the Copenhagen interpretation of wave function collapse might suggest.

a reply to: dragonridr
You managed to answer that question without even mentioning light cones. I didn't even attempt an answer because I didn't see what light cones had to do with it, and apparently you didn't either since you didn't mention them.

@spy66
The past and future light cones are conceptual, and I didn't understand the relevance to what happens to a single photon as in your question. The past light cone is just a way of saying that if a star is 11,000 light years away, the past light cone shows a circle 10,000 years ago with a radius if 10,000 light years, that will not encompass that star 11,000 light years away (assuming the relative distance to the star hasn't changed much in the last 10,000 years which for a star in the milky way may be a valid assumption).

a reply to: Arbitrageur

So let's say .01 light seconds before you detect the photon, it was .01 light seconds away from your detector according to deBroglie-Bohm. Maybe you didn't know where it was then, but you know now where it was then and it wasn't equally likely to have been 80 billion light years away, as the Copenhagen interpretation of wave function collapse might suggest.

Thanks. That's a very lucid explanation.

a reply to: Astyanax
I suppose, except for my wrong units (only one should be light seconds, not both. The first reference should be just seconds, but I didn't see it before the edit window expired). I'm glad you understood my point in spite of that.