Photons Detector not yielding which-path info.

[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.