Shape of the wave of a single photon

[swansont]

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.

 

Photons are identical particles so far as their basic properties go. You could not identify a photon as being different if it was absorbed and re-emitted.

 

Asking how does the photon know anthropomorphizes the photon. It doesn't "know" anything nor does it have to "remember" anything, and you do not get to tell it how it behave. The photon reflects at the same angle because that's the result that follows the laws of nature — conserves energy and momentum, etc.

[DParlevliet]

Photons are identical particles so far as their basic properties go. You could not identify a photon as being different if it was absorbed and re-emitted.

 

Indeed, but is has a certain direction of travel in space

 

Asking how does the photon know anthropomorphizes the photon. It doesn't "know" anything nor does it have to "remember" anything, and you do not get to tell it how it behave. The photon reflects at the same angle because that's the result that follows the laws of nature — conserves energy and momentum, etc.

 

Let me phrase better: why does the reemiited photon travel at that specific angle?

Suppose in a mirror a photon is absorbed-reemitted. During absorption the wave collapses. When a new photon is emitted it has a new wave, which is spherical like all emitted photons, so it can travel in all directions. I see now way to explain why it will always travel at the reflected angle.

[swansont]

 

Indeed, but is has a certain direction of travel in space

 

 

Let me phrase better: why does the reemiited photon travel at that specific angle?

Suppose in a mirror a photon is absorbed-reemitted. During absorption the wave collapses. When a new photon is emitted it has a new wave, which is spherical like all emitted photons, so it can travel in all directions. I see now way to explain why it will always travel at the reflected angle.

 

Because of the boundary conditions that have to be met. The electric and magnetic field components have to behave properly. Another method is to apply Fermat's principle.

[DParlevliet]

Because of the boundary conditions that have to be met. The electric and magnetic field components have to behave properly. Another method is to apply Fermat's principle.

 

Fermat and Feynman describe what happens, but does not explain why. If photons are scattered/bounced they behaves according Fermat and Feynman and you can explain why on the level of one photon. With absorpted-reemitted you cannot.

[Strange]

 

Fermat and Feynman describe what happens, but does not explain why.

 

Physics doesn't really do "why".

 

Here is Feynman talking about just that issue: http://www.youtube.com/watch?v=qjmtJpzoW0o

[swansont]

 

Fermat and Feynman describe what happens, but does not explain why. If photons are scattered/bounced they behaves according Fermat and Feynman and you can explain why on the level of one photon. With absorpted-reemitted you cannot.

 

I don't see how Fermat fails to work. The path of least time is the path of least time regardless of whether it's a wave or photon.

[DParlevliet]

Do you mean with absorpted-reemitted?

[swansont]

Fermat doesn't rely on either model, AFAIK

[DParlevliet]

A photon is emitted by a single atom in the center of an empty spherical detector. Is the probability of getting detected everywhere on the inner surface the same?

[swansont]

A photon is emitted by a single atom in the center of an empty spherical detector. Is the probability of getting detected everywhere on the inner surface the same?

 

Absent other restrictions, it should be.

[DParlevliet]

The same with a reemitted photon?

Edited November 1, 2013 by DParlevliet

[swansont]

The same with a reemitted photon?

 

From real state or a virtual state?

[DParlevliet]

From an atom that just before absorpted a photon

[Mike Smith Cosmos]

Surely a photon can in its entirety ,

 

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move off in one unique direction( all be it that the wave front on the wave packet is as wide as the probability wave /de broglie ). If this were not so the energy density of the photon would diminish by the inverse square law greater by the surface of the expanding spherical surface as r increases. The photon does not do this. It goes it's own unique single way on its way to infinity without loosing energy.

 

A normal wave front ( non photon) would decrease in amplitude by the inverse square law. As the energy is dissipated over a larger and larger surface area of the sphere.

 

Many photons could be present at a source . Then the density of photons at the increasing surface area would be diluted. But an individual photon from this source would continue its journey to infinity, provided it did not collide with a planet or foot or something else.

Edited November 1, 2013 by Mike Smith Cosmos

[swansont]

From an atom that just before absorpted a photon

 

Involving real state or a virtual state?

 

I'm not asking this to be frivolous. The reply is different depending on the answer.

Surely a photon can in its entirety ,

 

O

N

L

Y

 

move off in one unique direction( all be it that the wave front on the wave packet is as wide as the probability wave /de broglie ). If this were not so the energy density of the photon would diminish by the inverse square law greater by the surface of the expanding spherical surface as r increases. The photon does not do this. It goes it's own unique single way on its way to infinity without loosing energy.

If that is "surely" how a photon behaves, you should be able to experimentally confirm it. How do you check the path of a photon in free space without interacting with it?

 

A normal wave front ( non photon) would decrease in amplitude by the inverse square law. As the energy is dissipated over a larger and larger surface area of the sphere.

But we are discussing quantum mechanics. There are probabilities for finding particles in space, but we only detect the particle in one place when we look. All you've shown with the above example is that a photon does not behave in a way that nobody expects it to behave.

[Mike Smith Cosmos]

 

 

If that is "surely" how a photon behaves, you should be able to experimentally confirm it. How do you check the path of a photon in free space without interacting with it?

 

But we are discussing quantum mechanics. There are probabilities for finding particles in space, but we only detect the particle in one place when we look. All you've shown with the above example is that a photon does not behave in a way that nobody expects it to behave.

 

Yes, I can see the probability wave allows for the photon to not necessarily, take a completely straight path. Perhaps dodging about a bit sideways as the probability wave allows. But I doubt this would be little if any away from the straight line it started off with or it would never get to its aimed for destination.

 

Edited November 1, 2013 by Mike Smith Cosmos

[Strange]

Surely a photon can in its entirety ,

 

O

N

L

Y

 

move off in one unique direction(

 

You can't think of photons like little bullets.

[DParlevliet]

Involving real state or a virtual state?

 

I'm not asking this to be frivolous. The reply is different depending on the answer.

 

Then I don't know the difference. A photon is elasticly scattered by an atom by aborption/re-emmission. Does that deliver a real of virtual state? I suppose a real state?

Edited November 1, 2013 by DParlevliet

[swansont]

 

Then I don't know the difference. A photon is elasticly scattered by an atom by aborption/re-emmission. Does that deliver a real of virtual state? I suppose a real state?

 

If it's a real state then the emission pattern is symmetric. If it's a virtual state, then it's not — the atom can't absorb any energy or momentum. In reflection or transmission, absorption would be by a virtual state.

 

Yes, I can see the probability wave allows for the photon to not necessarily, take a completely straight path. Perhaps dodging about a bit sideways as the probability wave allows. But I doubt this would be little if any away from the straight line it started off with or it would never get to its aimed for destination.

 

PHOTON TO MOON.jpg

 

How would you measure this, though? If you can't, then you actually don't know it.

[DParlevliet]

If it's a real state then the emission pattern is symmetric. If it's a virtual state, then it's not — the atom can't absorb any energy or momentum. In reflection or transmission, absorption would be by a virtual state.

 

At the moment a photon is reflected in a mirror, does its wave collapse?

[Mike Smith Cosmos]

 

How would you measure this, though? If you can't, then you actually don't know it.

One can't because as we know the wave function collapses. However, we have measured by detection a number of photons being aimed at the moon reflector and returning to earth. It is almost beyond comprehension , that an individual photon would have bounced around the universe to get to the moon reflector.

 

What an experimenter could do is fire an INDIVIDUAL photon , a number of times at the moon reflector. Once having got sight line up. Then check they all come back. I think that would be evidence that individual photons travel in straight lines despite theirs probability flexibility .

 

If some do not return , it would indicate a deviation from straight line say by probability deviation. Or some other reason , like a seagull flying by. What say you ?

Edited November 2, 2013 by Mike Smith Cosmos

[swansont]

One can't because as we know the wave function collapses. However, we have measured by detection a number of photons being aimed at the moon reflector and returning to earth. It is almost beyond comprehension , that an individual photon would have bounced around the universe to get to the moon reflector.

 

What an experimenter could do is fire an INDIVIDUAL photon , a number of times at the moon reflector. Once having got sight line up. Then check they all come back. I think that would be evidence that individual photons travel in straight lines despite theirs probability flexibility .

 

If some do not return , it would indicate a deviation from straight line say by probability deviation. Or some other reason , like a seagull flying by. What say you ?

 

This is from a laser, and not a situation where the photon could go in any direction. Still, the photons don't all hit at the same point, and only a small fraction of the emitted photons come back.

 

At the moment a photon is reflected in a mirror, does its wave collapse?

 

I don't know.

[Enthalpy]

 

At the moment a photon is reflected in a mirror, does its wave collapse?

To my limited understanding, we say "collapse" when the receiving object, for instance an atom or a camera pixel, can interact with the incoming particle (here the photon) only if this particle is in a special state, more restrictive than the set of states the particle can have.

 

So for instance if the photon can have any polarization but the mirror reflects only the horizontal one, we'd say "collapse". Possibly even if it leaves many states allowed to the reflected photon. Or if the photon can have varied energies but the mirror is selective.

 

If the mirror is big enough and reflects all energies, polarizations... I would not say "collapse" because the reflected photon can be as diverse as when arriving.

 

Special mention to the size. A photon arriving from a distant star is extremely wide and gets detected a one camera pixel. In this case, where again the detector has constrained the wide diversity of behaviours available to the particle, we don't say "collapse". I suppose this wording is because the photon concentrated in one pixel after light-years travel is not an eigenstate of light's equation, so the event is not the choice of one eigenstates among several available.

 

But below the question of wording, an interesting part is that the particle determines its behaviour only when needed (at the detector), not before - this was an open question long ago, and the experimental answer is "no hidden parameter". The particle IS undetermined until it must "choose".

 

An other interesting part is that the particle determines itself at the interaction, but only as much as is needed. Say, if the detector allows for some tolerance on the position, then the momentum will keep some precision. Or imagine a screen with two slits: a photon must first "choose" if it hits the screen or not, but if passing through, it does not "choose" through which slit, and this allows interferences downstream the slits.

 

As long as you treat photons like waves, they propagate as expected, most often... Until you detect them, and then they're granular, hence particles. It's most often the same for any particle. If you consider propagation like a wave and interactions like a particle, with choice ("collapse") at the interaction, you're often right - the true worries come with entanglement.

 

From such considerations, that particles restrict they behaviour only when needed, and only as much as needed, I take my integrist wording about particles, for instance:

- "A valence electron in an atom has the size of the atom". The electron can be smaller, but only if the experiment can force it so. It can be as small as needed for all our existing experiments, hence "point-like". But only if the measure needs it. When the electron is an orbital, we have no reason to allege it's smaller.

- "The size of a photon is the extension of the wave packet", because we have no other means to determine the size. Once detected, the photon can be 3µm*3µm. Prior to that, the wave packet is the only thing that defines a size; if emitted at 21cm by neutral hydrogen, the wave packet is several million light-years long, and I say this is the size of the photon.

 

Take with caution.

[DParlevliet]

I limit myself to the simple results of a simple measurement.
I understood that according QM the wave of a photon collapses when the photon is absorbed. \

It is my opinion that the mirror shows that the wave during reflection does not collapse, because there is a (angular) relation between the wave before and after the mirror. If the wave would collapse during reflection, there will a new outgoing wave which cannot have information from the collapsed wave (I have never heard of a wave un-collapsing ).

No let us approach it from another way with the question: does the (probability) wave of a photon interferes with the wave of another photon?
Imagine two electronically triggered emitters in front of a detector.

 

Both emit at about the same time a photon with the same energy. Does the wave from photon b interferes with the wave of photon a? In other words, is the probability of detecting photon somewhere on the detector influenced by the wave of photon b?

I guess not.

[Toffo]

No let us approach it from another way with the question: does the (probability) wave of a photon interferes with the wave of another photon?

Imagine two electronically triggered emitters in front of a detector.

 

Yes they do interfere. You need two lasers, you aim those at the same spot, where you will see an interference pattern. This experiment has been done.

 

You can turn the other laser off too. That just makes that laser's photon wave really weak and non-coherent. The interference pattern is the interference pattern of one strong beam and very many really weak beams in very many different phases. (The turned off laser emits thermal radiation, you see)

 

Here's a nice link:

http://www.researchgate.net/post/The_interference_between_two_different_lasers_at_the_same_wavelength

 

At the moment a photon is reflected in a mirror, does its wave collapse?

 

Of course not.

 

Does the mirror think it has been hit by a small bullet? No.

Does some quantum wave disappear? No.

Is some part of a quantum wave teleported to another place? No.

 

 

Let's say a quantum wave has been trapped in a small cavity for a long time, and it has been leaking out slowly.

Let's say the cavity is inspected and found to be empty. This operation causes the quantum wave in the cavity to become teleported away from the cavity to everywhere where the leaked out quantum wave exists. This is a collapse of a quantum wave.

Edited November 3, 2013 by Toffo