Quantum erase explained with waves

[Cthugha]

Now your spin explanation:
- Before Q1/Q2, is the spin of the photon + or -
- Suppose after Q1 the spin is +, what is the spin after Q2?

Sorry. I do not know which scenario you are talking about. Q1 and Q2 are at 90 degrees to each other? At what orientation are they relative to the incoming photon and what is its polarization?

 

[DParlevliet]

The scenario we are discussing about: axis Q1 and Q2 at 90 degrees, and 45 degree to the incoming photon with lineair polarization.

 

[Cthugha]

Ok, 45 degrees. So let us choose the initial polarizatiion such that we get left circular polarization (-) after Q1 and right circular polarization (+) after Q2.

At a position at the deetctor these overlap. As they overlap at some angle, you do not really get a linear polarization, not even classically, but let us ignore that for a moment. In qm you now add the probability amplitudes and then square them to get the probabilities for the possible detection events. The detection events of course rely on absorption. In absorption events, conservation of energy, momentum and angular momentum must be satisfied, so the electron or whatever particle absorbs the single photon needs to acquire all of them. Photons have an energy of \hbar \omega, spin angular momentum of \pm \hbar and linear momentum of \hbar k. Electrons decohere quickly, so after the absorption, you will find it in an eigenstate in which it gained these amounts of energy and momentum. This includes the \pm \hbar of spin angular momentum. As only these values are possible, you can easily check which amount was given to the electron and this automatically tells you, whether the photon was - or + and also which path it took. Both processes just take place with equal probability.

 

[DParlevliet]

We would no talk about waves anymore (that was my area) but about your explanations with spin.
- I suppose in your explanation after Q1 the spin is -, and after Q2 + ?
- Now what is the spin before Q1/Q2 ?

 

[Cthugha]

- Now what is the spin before Q1/Q2 ?

You wanted to have the quantum equivalent of linear polarization. That means that in the ensemble you have a coherent, fixed-phase superposition of + and -. For the single photon in the experimental run it means that it is essentially undefined at this moment which of the two it is. It is in a superposition, if you want to think about it that way.

It is of course not trivial to create such a state. The problem about conservation rules of course also concerns the emission process. The energy, spin and momentum need to be transferred from the emitter to the photon, so in principle checking the state of the emitter can tell you the state of the photon. Therefore it is important to have some emission process which is more or less not too sensitive to decoherence. For example if the emitter is very light, the recoil from the emission process will already destroy the superposition and you never have it in the first place. If it is heavy, the momentum transfer drowns in uncertainty and the emitter does not end up in a state that is orthogonal to the initial one. That is one of the reasons why non-linear optical processes like parametric downconversion using BBOs are heavily used in quantum optics.

 

[DParlevliet]

Then no superposition. But then, what spin has the incoming photon?
So back to the original measurement in a situation where the BBO emits a photon which goes through Q1 or Q2 after which one can measure through which. Now again the question: what spin does the particle have before and after Q1 or Q2 (no superposition)? You may choose yourself the incoming photon properties.

 

[Cthugha]

Then no superposition. But then, what spin has the incoming photon?

I do not get it...in the experiment, the photon is in a superposition initially. Why do you want change that situation?

So back to the original measurement in a situation where the BBO emits a photon which goes through Q1 or Q2 after which one can measure through which. Now again the question: what spin does the particle have before and after Q1 or Q2 (no superposition)? You may choose yourself the incoming photon properties.

In the original experiment, the photon pair is in a Bell state. This is a superposition of the state of the photon going to the double slit being in a horizontal superposition and the other one being in a vertical one and the other way round. However, the exact properties are undefined before going through the slit. As Bell tests tell us, they are not well defined until some measurement is performed. Only the state of the whole photon pair (the sum of the polarizations of both) is well defined.

It is not clear that the photon has predefined properties before being detected. If you assume that, you can only explain the results by taking non-local correlations between the two photons of the photon pair into account. You can also describe the experiment in a local manner, but then you need photons without predefined polarization properties. You may have heard discussions about Bell tests, non-locality and hidden variables. That is exactly what is important in the experiment at hand.

 

[DParlevliet]

In the original experiment, the photon pair is in a Bell state. This is a superposition of the state of the photon going to the double slit being in a horizontal superposition and the other one being in a vertical one and the other way round. However, the exact properties are undefined before going through the slit.

So the superposition is between the photon going to Ds and the photon going to Dp? The photon to Ds is horizontal or vertical, depending if it goes through Q1 or Q2?
If so, can you transfer this to spins? For instance if the photon goes through Q1 what is the spin before and after Q1?

 

[Cthugha]

So the superposition is between the photon going to Ds and the photon going to Dp? The photon to Ds is horizontal or vertical, depending if it goes through Q1 or Q2?
If so, can you transfer this to spins? For instance if the photon goes through Q1 what is the spin before and after Q1?

If the photon going to DS is in a horizontal superposition, the one going to Dp is in a vertical one and vice versa. This also translates to spins. If the one going to DP is vertical, you get - after Q1 and + after Q2. If it is horizontal, you get + after Q1 and - after Q2. None of this is predefined, though. The system just goes into one of these states immediately as soon as the superposition decoheres. However, the system can decohere anywhere. If the photon going to Dp is measured, this leads to decoherence and also decides what state the photon going to Ds will be found in.

 

[DParlevliet]

If the photon going to DS is in a horizontal superposition, the one going to Dp is in a vertical one and vice versa. This also translates to spins. If the one going to DP is vertical, you get - after Q1 and + after Q2...

Is it - after Q1 and + after Q2 at the same time or - after Q1 or + after Q2 depending if the photon goes through Q1 or Q2.

 

[Cthugha]

If you are in a superposition state, you can consider it an "and", BUT: as soon as some measurement collapses the collective state of the photons towards some state, the kind of measurement that has been performed matters. As soon as a measurement gets done which allows you to identify that the system is in one of the possible states, the whole system is in that well known state.

This has some important consequences. For example you can get which-way information on the photon going to Ds by performing a measurement on the photon going to Dp.
Also, the whole "truth" about the system becomes apparent only after both photons have been detected and you know both results. Therefore, it is quite complicated to say whether you have the "and" or the "or". But you will know that afterwards.

 

[DParlevliet]

You can put a small, light, frictionless disk there. If it absorbs the photon, the angular momentum gets transferred to the disk and it will start to rotate, either clockwise or counterclockwise, depending on the spin.

Now use the above measurement at Ds. You remove Dp, so this photon disappears in space without being absorbed. In what direction will the disk rotate?

 

[Cthugha]

You cannot really remove Dp, but if we assume that nothing happens to it for a while, you get a 50/50 chance of the disk rotating in either direction. The direction will now also tell you the state of the photon going to Dp.

 

[DParlevliet]

I asked "Is it - after Q1 and + after Q2 at the same time or - after Q1 or + after Q2 depending if the photon goes through Q1 or Q2" and you answered "If you are in a superposition state, you can consider it an "and"
At the moment of absorption the photon is still in superposition state, so it is "and", so you have to add the + and - coming from Q1 and Q2.

 

[DParlevliet]

I just found an article http://arxiv.org/abs/1112.4522 which I think goes the same direction as my arguments (but with more knowledge).

 

[Cthugha]

I asked "Is it - after Q1 and + after Q2 at the same time or - after Q1 or + after Q2 depending if the photon goes through Q1 or Q2" and you answered "If you are in a superposition state, you can consider it an "and"
At the moment of absorption the photon is still in superposition state, so it is "and", so you have to add the + and - coming from Q1 and Q2.

Ehm, whether you consider it the "and" or the "or" does not change anything. This is a rather philosophical question. In the end you have which way info because the single photon cannot be linearly polarized and the final state after decoherence is consistent either with the photon having been + or -.

As to your reference. Ellerman does not follow your position. He is actually taking a contrary position. You repeatedly asked what the state of a photon before the QWP or somewhere else is. Ellerman's point is that this is not a meaningful question. He especially argues against people like Wheeler who claim that you can change the past using the delayed choice quantum eraser. That is of course nonsense. His summary "We must conclude, therefore, that the loss of distinguishability is but a side effect, and that the essential feature of quantum erasure is the post-selection of subensembles with maximal fringe visibility." is similar to several explanations of quantum eraser experiments I already posted on these forums.

To summarizer his point: he says that you do not retroactively change what happened, but you just use a filtering process that leaves us with the consistent set of results. That is indeed the way "which-way" information needs to be interpreted here. Depending on your detector setting, you just keep a subset which has taken a certain way and it is the possibility to perform a measurement which allows you to pick such a subset or not which you can erase. You can backtrack, so to speak.

 

[DrChinese]

I just found an article http://arxiv.org/abs/1112.4522 which I think goes the same direction as my arguments (but with more knowledge).

I consider the arguments made in the referenced paper completely unconvincing. The Polarization Analyzer/Loop example is the most obvious one to me. The full loop is an eraser, and I don't see any kind of argument that explains this other than a bit of verbal legerdemain. I would say this paper is not a reflection of consensus.

 

[DParlevliet]

Ehm, whether you consider it the "and" or the "or" does not change anything. This is a rather philosophical question.

It is not, otherwise your superposition is philosophical too. You say that the spin is - after Q1 and + after Q2 if the photon only goes through Q1. That means it is in superposition and as you told when it is absorbed first both states of superposition must be added. If the result is wrong, then the input state or description of the QWP/slit is wrong. Then you have to change that so the output becomes right.

 

[Cthugha]

I consider the arguments made in the referenced paper completely unconvincing. The Polarization Analyzer/Loop example is the most obvious one to me. The full loop is an eraser, and I don't see any kind of argument that explains this other than a bit of verbal legerdemain. I would say this paper is not a reflection of consensus.

I think one has two consider the two different points he raises. He disagrees with Cramer and other people who claim that the quantum eraser is proof for retrocausality. He is right with that, but it should be added that Cramer's opinion is not mainstream either.

Then he goes on to say that it is not correct to say that the photon went through one slit only because the superposition was collapsed afterwards. Here he implicitly claims that the wavefunction is a realistic rntity and corresponds to some physical reality. This is philosophy and opinion.

It is not, otherwise your superposition is philosophical too.

Well, yes, it is. It is one way to interpret the meaning of the wave function. Personally I follow a different approach, but for people not familiar with the depths of quantum optics this Copenhagen-like approach is usually the easiest one to understand. We do not know, what really happens microscopically between emission and detection. You can find models which consider superposition and the wave as real and you find interpretations like Bohmian mechanics where the wave is just a guiding wave and the particle takes some well defined path and has some well defined values each time. All of these models are consistent with experiment and so far there is no experiment able to distnguish between them.

You say that the spin is - after Q1 and + after Q2 if the photon only goes through Q1. That means it is in superposition and as you told when it is absorbed first both states of superposition must be added. If the result is wrong, then the input state or description of the QWP/slit is wrong. Then you have to change that so the output becomes right.

Most of all I said that it is not a good idea to attach values to the photon before detection. My opinion was that if you do the right measurement, you know that the spin WAS - after Q1. You must add both states of superposition. This is math, not physics. You cannot conclude that this means that both states are simultaneously realized and real. That may or may not be the case. We do not know. You can interpret the wave formalism as some real property of the system or as some probabilistic mechanism without deeper physical meaning which is used to calculate what events will occur. We have no evidence that either is more justified than the other.

 

[DParlevliet]

Here he implicitly claims that the wavefunction is a realistic rntity and corresponds to some physical reality.

No, that is an often used misconception about the wave-particle duality. Copenhagen does not state that the wave is a physical reality but that a photon partly acts like a classical wave. Practical measurements shows that if you use wave equipment and -detectors, that with simple measurements one can use the classic wave formula, but without believing that it is a real wave. That is not philosophy but practice
So if simple measurement cannot be explained with waves, it smells bad.

 

[Cthugha]

No, that is an often used misconception about the wave-particle duality. Copenhagen does not state that the wave is a physical reality but that a photon partly acts like a classical wave.

First of all, that guy is not a supporter of Copenhagen and that guys considers the wave as real.
As has been pointed out several times, this way of understanding the wave-particle duality was all-out wrong. It is not tenable as there are situations in which the wave is non-local and as such non-classical. That problem is solved using quantum electrodynamics. The more modern versions of Copenhagen also take that into account.

Practical measurements shows that if you use wave equipment and -detectors, that with simple measurements one can use the classic wave formula, but without believing that it is a real wave. That is not philosophy but practice

Well, the classical wave model has very narrow limits. But, anyway, if you do not believe that it is a real wave: where is your problem and why do you always ask what the photon actually is when it goes through the slits? I am puzzled.

 

[DrChinese]

Here he implicitly claims that the wavefunction is a realistic entity and corresponds to some physical reality. This is philosophy and opinion.

Actually I too agree that the wave function corresponds to some physical "something". But yes, his argument is not really explicit (as you say) and there is still plenty of interpretation involved (also as you say). So much so that this paper, which really gives no specific new predictions or similar, doesn't really settle anything on the matter.

 

[Cthugha]

Actually I too agree that the wave function corresponds to some physical "something".

Maybe I should have emphasized that it is a perfectly valid opinion. Rereading my post it looks like a suggested otherwise. That was not my intention.

 

[DParlevliet]

If it is real or not is not important. The only thing what matters is that in simple measurements you can use the classical wave rules, but not together with particle rules. My point in the above measurement is that the waves are marked, not the particle. So after the slit you can measure through which slit a wave went, but not the particle. And because in absorption of the photon the two waves must be added (if you want interference) the marking is lost again. You did try to explain it with spin, which is a particle property, but I have not seen a consistent full explanation of the experiment.

And: every article claims that with marked waves the path of the particle in can be measured. But I have yet seen doing that.
.

 

[Cthugha]

If it is real or not is not important. The only thing what matters is that in simple measurements you can use the classical wave rules, but not together with particle rules.

What is wrong about using the correct theory, QED, instead of a simplification that is known to be wrong in many cases?

My point in the above measurement is that the waves are marked, not the particle. So after the slit you can measure through which slit a wave went, but not the particle. And because in absorption of the photon the two waves must be added (if you want interference) the marking is lost again. You did try to explain it with spin, which is a particle property, but I have not seen a consistent full explanation of the experiment.

What do you mean by "if you want interference"? When you have perfect distinguishability of the paths, the superposition does not give rise to any interference terms. You add them of course, but that does not mean there is interference. This is why you actively need to do the erasing which projects the system into a state which does contain interference terms. For the same reason the marking is not lost. If you just perform the right measurement (spin-sensitive absorption, placing a QWP and a linear polarizer or whatever) just one of the field components survives. If you can get interference, it is not possible to perform a measurement where only one field component survives.

And: every article claims that with marked waves the path of the particle in can be measured. But I have yet seen doing that.
.

Well, you deny the obvious. See for example "Observation of the Fresnel and Arago laws using the Mach-Zehnder interferometer" (Am. J. Phys. 76, 39 (2008)) for a demonstration that the addition of two light beams of orthogonal circular polarization does not give rise to interference.

 

[Dadface]

Well, you deny the obvious. See for example "Observation of the Fresnel and Arago laws using the Mach-Zehnder interferometer" (Am. J. Phys. 76, 39 (2008)) for a demonstration that the addition of two light beams of orthogonal circular polarization does not give rise to interference.

I have been trying to follow the discussion here but I'm confused by the point above. According to Pescetti and Piano:

Two light waves in orthogonal states of elliptical polarisations coming from the same unplolarised wave can interfere if brought into the same plane. Isn't this equivalent to the situation observed in the experiment under discussion where interference is observed when the "which way" information is erased?

 

[Cthugha]

Two light waves in orthogonal states of elliptical polarisations coming from the same unplolarised wave can interfere if brought into the same plane. Isn't this equivalent to the situation observed in the experiment under discussion where interference is observed when the "which way" information is erased?

Yes, that is quite right. The point I tried to make was that just adding the waves at some detector position without erasing anything will not give you any interference, so contrary to what DParlevliet claimed above the which-way info is not lost when the photon is absorbed (without erasing the which-way info).

 

[Dadface]

Yes, that is quite right. The point I tried to make was that just adding the waves at some detector position without erasing anything will not give you any interference, so contrary to what DParlevliet claimed above the which-way info is not lost when the photon is absorbed (without erasing the which-way info).

Thanks, that's clarified it.

 

[DParlevliet]

Well, you deny the obvious. See for example "Observation of the Fresnel and Arago laws using the Mach-Zehnder interferometer" (Am. J. Phys. 76, 39 (2008)) for a demonstration that the addition of two light beams of orthogonal circular polarization does not give rise to interference.

In this article the path of the photon is not measured.

 

[Cthugha]

In this article the path of the photon is not measured.

Eh? That was never the point of that paper. The point is that your claim:

And because in absorption of the photon the two waves must be added (if you want interference) the marking is lost again.

is wrong. You do not get interference if you add orthogonal states. That includes counter-circular polarizations as demonstrated. From this point on it is trivial that you have which-way info if you do not have interference. The interplay between both is given by the Englert-Greenberger duality relation.