Photons Detector not yielding which-path info.

[peter0302]

Oh it's not a pain, it's just bringing back painful memories of going mad! :)

Right I know what you're saying. I'll make it even more simple. Take away the prism and instead use a converging lens and put a single detector at the focal point. What gives?

What gives is that it's still only a small minority of the photons that are being detected. The rest float off into space and help destroy the interference pattern. The coincidence circuitry, among other things, isolates the photons that actually _are_ detected so that the analysis can performed on just those. Without that, if you just looked at D0, there'd be far too much noise.

You've got to get the overwhelming majority of idler photons from the BBO crystal to hit the detector. I don't know how you can do that without disturbing the interference pattern because if you place anything too close to the crystal you learn which-path (in principle) but if it's too far away there's too much uncertainty in the location of the photons to capture enough of them.

That's why I thought of using lenses behind the slits to focus the beams but I don't think the HUP will let us control both the position and direction of the photons with such precision.

 

[nrqed]

Oh it's not a pain, it's just bringing back painful memories of going mad! :)

Right I know what you're saying. I'll make it even more simple. Take away the prism and instead use a converging lens and put a single detector at the focal point. What gives?

Right

What gives is that it's still only a small minority of the photons that are being detected.
.

Why? We are taking about a gedanken experiment so I am not sure if your point is about a practical limitation or a fundamental one. Why can't we assume that they are all detected?
If th elimitation is only a question of experimental precision, then it makes the rest of the argument shaky

The rest float off into space and help destroy the interference pattern. .

how?

.

The coincidence circuitry, among other things, isolates the photons that actually _are_ detected so that the analysis can performed on just those. Without that, if you just looked at D0, there'd be far too much noise.
.

Why? It seems to me that every single pair now has gone through a set up in which the which-way information is not available. Hence an interference pattern. I don't see how losing some of the photons could change the result. The ones that are not counted, it seems to me, would still be part of the interefrence pattern since no which way information si available for those as well. So we simply take an interference pattern and removes some of the photons, we simply get a weaker (in the sense of less intense) interference pattern, no?


.

You've got to get the overwhelming majority of idler photons from the BBO crystal to hit the detector. I don't know how you can do that without disturbing the interference pattern because if you place anything too close to the crystal you learn which-path (in principle) but if it's too far away there's too much uncertainty in the location of the photons to capture enough of them.

Again it seems to me as if you are talking about practical limitations, not fundamental ones. And again, if there is no which path information for any of the photons, it seems to me that it does not matter if we don't catch all of them anyway. I don't see how to take away photons from an interference pattern and leaves something that will look like the combination of two interference pattern shifted by pi! (and *even* if that was possible, it would be incredible to think that the photons missed are just the right ones to accomplish this!). So there is something I am clearly missing!

That's why I thought of using lenses behind the slits to focus the beams but I don't think the HUP will let us control both the position and direction of the photons with such precision.

But we could instead simply use some mirrors to redirect the idler photons to a single detector.


Thanks for your input.

 

[ueit]

Ok, photons aren't charged. The only relevant charged objects are the electrons in the detectors and mirrors.

The electron shells from which the photons originate are charged. The photon's properties are therefore correlated to whatever fields reach the source.

Yes there is. You're guessing because of what you believe in your gut should be right and you're ignoring 75 years of science that says you're wrong.

Can you be more precise what experiment/theoretical result has shown that?

The reason I am so sure of this is that 75 years of experiments have not proven otherwise.

Same comment as above.

You *think* I am wrong that the mirror's state is unchanged??

That's a fair assessment.

Yes indeed, which becomes embodied in a new photon reflected off the surface.

The incoming photon has a momentum m. The reflected one has -m (on the axis that makes a 90 degree angle with the mirror). Now, -m does not equal m, does it?

The negative you wish me to prove is that the microscopic workings of the experimental set up work together in a grand conspiracy to yield the results we see and that they're not truly random. I can't prove that.

So, if you can't prove that, why do you still claim that "75 years of science" have proven me wrong? Please decide which of the statements:
1. I cannot be proven wrong.
2. I have already been proven wrong,

better reflects your position because both of them cannot be true at the same time.

A "conspiracy" is usually imposed top -to- down. My suggestion is nothing of this sort, on the contrary.

Indeed. But no mainstream theory believes that the microscopic movements of the molecules in the experimental components are the answer.

A theory has no "beliefs". It is your belief that by applying the theories we have to the whole experimental setup (making, of course, simplifications, but being careful to base them on the microscopic behavior and not on a macroscopic general behavior) nothing interesting could emerge. May be you could add some substance to this claim.

I wonder why the double slit works with uncharged neutrons then...

The neutrons are made of charged quarks and they also have a non-zero magnetic moment.

 

[peter0302]

The electron shells from which the photons originate are charged. The photon's properties are therefore correlated to whatever fields reach the source.


Can you be more precise what experiment/theoretical result has shown that?

Feynman's online QED lectures talk about this in the context of reflecting light off of water. He addresses your argument: "Isn't it just a matter of precision? Couldn't we aim a photon at a precise spot on the water - say directly at an electron - and guarantee reflection or refraction?" and he goes on to state that "no, we've tried with very precise instruments and we cannot do that."

Now, I don't know if he's being literal here but I understand his point. The uncertainty principle won't allow you to aim a photon directly at a particular electron and guarantee a hit. That uncertainty carries over into the reflection probability and therefore you cannot in principle accomplish what you're talking about doing. In other words you can build a classical model of photons and electrons as billiard balls and try to analyze a beam-splitter by saying "what are the odds of the photon hitting an electron and being reflected versus the odds of it being passed through?" And you can't do it because no matter how close you look the odds will be what the odds will be, because of the HUP.

I can't point you to specific experiments. Maybe someone else here can?

The incoming photon has a momentum m. The reflected one has -m (on the axis that makes a 90 degree angle with the mirror). Now, -m does not equal m, does it?

You just answered your question. m + -m = 0, which is the change in momentum of the mirror after the reflection. Hence, the state of the mirror is unchanged.

So, if you can't prove that, why do you still claim that "75 years of science" have proven me wrong? Please decide which of the statements:
1. I cannot be proven wrong.
2. I have already been proven wrong,

If the HUP is right you are wrong, and since in 75 years the HUP has not been proven wrong, so far the odds aren't looking good for you. That doesn't mean you are wrong necessarily but I think the burden is on you at this point to prove otherwise.

A "conspiracy" is usually imposed top -to- down. My suggestion is nothing of this sort, on the contrary.

Well, the conspiracy could have begun at the big bang with the breath of god. Science can never disprove that which is why we don't like to talk about it.

A theory has no "beliefs". It is your belief that by applying the theories we have to the whole experimental setup (making, of course, simplifications, but being careful to base them on the microscopic behavior and not on a macroscopic general behavior) nothing interesting could emerge. May be you could add some substance to this claim.

I think any delayed choice experiment is the substance to my claim.

 

[peter0302]

Oh, nrqed, I am going to respond to your post when I have more time. Got super busy at work today. :) But by the way, I just want you to know that I hate you. You've awoken the beast in me like I knew you would!

 

[ueit]

Feynman's online QED lectures talk about this in the context of reflecting light off of water. He addresses your argument: "Isn't it just a matter of precision? Couldn't we aim a photon at a precise spot on the water - say directly at an electron - and guarantee reflection or refraction?" and he goes on to state that "no, we've tried with very precise instruments and we cannot do that."

Now, I don't know if he's being literal here but I understand his point. The uncertainty principle won't allow you to aim a photon directly at a particular electron and guarantee a hit. That uncertainty carries over into the reflection probability and therefore you cannot in principle accomplish what you're talking about doing. In other words you can build a classical model of photons and electrons as billiard balls and try to analyze a beam-splitter by saying "what are the odds of the photon hitting an electron and being reflected versus the odds of it being passed through?" And you can't do it because no matter how close you look the odds will be what the odds will be, because of the HUP.

What you are saying is perfectly true. HUP does not allow us to use experiment to decide on the probabilistic nature of QM. This is what I was saying in one of my previous posts. Anyway, I don't see the relevance of this to my point that the emission of a photon is influenced by an electric field. This is true, regardless of any interpretation. On the other hand a realistic interpretation of QM is probably required to understand the causal chain, but again, HUP is not a problem as the existence of such an interpretation (Bohm's) has shown us.

I can't point you to specific experiments. Maybe someone else here can?

I can wait.

You just answered your question. m + -m = 0, which is the change in momentum of the mirror after the reflection. Hence, the state of the mirror is unchanged.

No, in this example momentum is not conserved. "Conserved" means it remains the same. If you apply the above calculation to a free, non-interacting particle at two time points you get m(initial momentum) + m (final momentum, unchanged because nothing happened) = 2m. So your particle will accelerate without any force acting on it which is clearly wrong.

If the HUP is right you are wrong, and since in 75 years the HUP has not been proven wrong, so far the odds aren't looking good for you. That doesn't mean you are wrong necessarily but I think the burden is on you at this point to prove otherwise.

The HUP is right, we know that but I don't see why I should be wrong. A charged particle still interacts with other charged particles.

Well, the conspiracy could have begun at the big bang with the breath of god. Science can never disprove that which is why we don't like to talk about it.

I don't see any analogy between your poetic words and the fact that charged particles interact with other charged particles.

I think any delayed choice experiment is the substance to my claim.

Well, you didn't show me any argument, just a restatement of your original position

 

[peter0302]

No, in this example momentum is not conserved. "Conserved" means it remains the same. If you apply the above calculation to a free, non-interacting particle at two time points you get m(initial momentum) + m (final momentum, unchanged because nothing happened) = 2m. So your particle will accelerate without any force acting on it which is clearly wrong.

That's just nonsense. Go back and study mechanics.

As for the rest, I don't know what else to say to convince you. Sorry.

 

[ueit]

That's just nonsense. Go back and study mechanics.

I think you should go back to introductory arithmetics.

From wiki:

"The law of conservation of linear momentum is a fundamental law of nature, and it states that the total momentum of a closed system of objects (which has no interactions with external agents) is constant."

Do you agree with the above statement?

If yes, let's calculate the total momentum before reflection:

Mi (initial total momentum) = m (initial photon's momentum) + 0 (assume the mirror doesn't move in that direction) = m
Mf (final total momentum) = -m (photon's momentum after reflection) + 0 (you claim that the mirror remains unchanged) = -m

So, does m equal -m?

 

[peter0302]

So, does m equal -m?

For purposes of what we've been talking about, yes.

Do you understand vectors?

 

[ueit]

Do you understand vectors?

I think I do.

 

[dst]

The magnitude of momentum is conserved in all cases.

Let's apply vectors to it in momentum space, mirror is unchanged -

Before - m(p) = sqrt(x^2 + y^2)
After - m(p) = sqrt((-x)^2 + (-y)^2) = sqrt(x^2 + y^2).

So it is conserved.

 

[hyperon]

I am not exactly convinced that finding which path information necessarily collapses the photon's wavefunction. If a single detector is placed in front of only one slit, this does collapse the wavefunction if the photon passes through that slit, since the photon would have interacted with the detector.

However, if the photon passed through the other slit, there would be nothing for it to interact with before reaching the screen. This situation is then equivalent to one with no detector in front of the first slit. Which path information is still preserved because the nondetection of the photon at the detector implies that it can ideally only have passed through the second slit. In this case, we would know which path information without collapsing the photon's wavefunction.

Perhaps I am treating this too classically. Please correct me if I am wrong.

On further thought I tried considering the possibility that the detector somehow disturbs the entire wavefunction of the photon from one end of the setup where it is placed. But this would not follow from peter0302's interpretation of collapse of wavefunction by an irreversible interaction with macroscopic objects in the experiment.

 

[ueit]

The magnitude of momentum is conserved in all cases.

Let's apply vectors to it in momentum space, mirror is unchanged -

Before - m(p) = sqrt(x^2 + y^2)
After - m(p) = sqrt((-x)^2 + (-y)^2) = sqrt(x^2 + y^2).

So it is conserved.

Do you claim that only the magnitude and not also the direction of the momentum vector is conserved?

 

[Count Iblis]

You'll never get a paradox if you try to use a mirror to extract the "which path information" by measuring the mirror's recoil. To get an interference pattern the wavefunction of the mirror must be such that the mirror is confined within some fraction of the photon's wavelength. This means that the mirror's wavefunction in momentum space will have a width that is larger than the photon's momentum (this is the uncertainty principle).

So, by measuring the momentum of the mirror, you cannot be sure whter or not the mirror has recoiled. Now, you could still make statistical predictons, because when the mirror recoils its's wavefunction does change. The sharper peaked the mirrors wavefunction in momentum space is the better you see if the mirror has recoiled. But then the wavefunction in ordinary space will become wider and the interference pattern will become less visible anyway.

According to quantum mechanics, the interference term is proportional to the overlap of the wavefunctions of the mirror when it recoils and when it doesn't. The interference pattern can only vanish completely if this overlap is zero, i.e. if the two states of the mirror are orthogonal. In that case they can be considered to be eigenstates of an operator which upon measurement would yield the which path information. But if the two states are not orthogonal, then you cannot have an operator such that the two states are the eigenstates because two eigenstates corresponding to different eigenvalues are always orthogonal.

 

[peter0302]

Do you claim that only the magnitude and not also the direction of the momentum vector is conserved?

Ueit, you're overthinking this. When you bounce a billiard ball off the side of the table, the ball bounces off with almost all of the momentum it had before. If this were in a vacuum and there was no such thing as sound or thermodynamics, the momentum would be exactly what it was before. Do you not think momentum is conserved in that case?

Thinking classically, the momentum from the photon is SO SMALL compared to the force of the other atoms in the mirror that 99.99999999999% of the momentum is imparted back to the new photon. So there might be a miniscule change in the mirror but it's still indistinguishable from random molecular movement.

Thinking quantum mechanically, what Count Iblis said.