Is Polarisation Entanglement Possible in Photon Detection?

[Zafa Pi]

Well I won't derive the inequality for you, but I will give you an argument that justifies my comments.

Consider the normal Bell experiment set up. Something like $A \leftarrow S \rightarrow B$ where some source fires off particles to Alice and Bob measuring at 0, 60 and 120. You agree we'll see a violation of the Bell inequality if the source is generating entangled states. We'll assume a source of perfectly entangled particles.

Now although it's usual to assume Alice and Bob are (roughly) equidistant from the source and measure (roughly) simultaneously, it's not necessary to do so to see the violation. Things are set up in the 'usual' fashion so that we can draw some conclusions about local hidden variables. Alice and Bob will still see a violation if Bob stores the particles, makes several cups of tea, and then does his measurements. As long as we can associate Alice's measurement and Bob's measurement on the partner particles we'll still see the violation.

The distance of the source to the 2 parties is also irrelevant (for inequality violation) so we'll still get the inequality violation with the following set up
$$A \leftarrow S \rightarrow \rightarrow \rightarrow \rightarrow \rightarrow B$$

So we now imagine the source to be In Alice's lab. Now what happens when Alice makes a measurement? If she makes a measurement then she'll know, by virtue of the entanglement, what state is on its merry way to Bob. So all Alice has to do is to make a measurement, stop the particle going off to Bob, and now prepare a new single particle (unentangled with anything) in the state that would have gone on to Bob - which she knows, because she's made a measurement.

You should now be able to see that we can drop the entangled source and measurement part altogether. Alice simply prepares particles uniformly at random in one the six possible measurement eigenstates and sends them off to Bob. There'll be a Bell inequality violation between the preparation data of Alice and the measurement data of Bob.

Of course, we can't draw any of the important conclusions about local variables from doing things this way, but it does show us that we don't actually need entanglement to see a violation of the mathematical inequality - and also, if you think about it, does show us the equivalence of the proper/improper mixed states.

BTW, it seems that you could accomplish what you want in an easier fashion. Alice randomly selects an observable the flips a coin to decide either +1 or -1 which determines a state, then calls up Bob and tells him the state, then Bob makes a measurement by his randomly selected observable and they violate Bell's inequality.
Of course this nothing to with the usual Bell set up. But it's possible I am not following you.

 

[Simon Phoenix]

BTW, it seems that you could accomplish what you want in an easier fashion. Alice randomly selects an observable the flips a coin to decide either +1 or -1 which determines a state, then calls up Bob and tells him the state, then Bob makes a measurement by his randomly selected observable and they violate Bell's inequality

Yes, but Alice hasn't sent him a state to work with. So in this case Bob would first have to prepare a particle in the state indicated by Alice (in each timeslot) and then make a randomly selected measurement on it. In this case Bob would see a violation of a Bell inequality between his preparation data (supplied by Alice) and his measurement data.

 

[zonde]

You think that the Copenhagen interpretation considers $|\psi\rangle$ and $e^{i \theta} |\psi\rangle$ to be different states? I don't think that's true.

Yes, I think that $|\psi\rangle$ and $e^{i \theta} |\psi\rangle$ are different vectors so they are different states (if we defined "state" as state vector).
Look, we can add different vectors and get third vector. We can't do that with rays. I have quote from Neumaier to back up what I say (as I am not so confident about my math):

One can add state vectors and gets another state vector, but adding two distinct rays produces a 2-diemnsional subspace and not a ray.

I'm not so sure about my knowledge of Copenhagen but thinking in terms of probability amplitudes we need phase factor to add probability amplitudes correctly. If we throw away phase factor we can't get interference effect. So I would say that phase factor is important whenever we talk about interference.

 

[zonde]

The 'single particle' violation shows us that the actual violation of the inequality (that is the violation of the mathematical inequality on its own) has little to do with locality or non-locality.

I would say that violation of the inequality tells us something about independence of two measurements. Locality is just a way how we try to implement independence of two measurements.
Your example certainly does not suggest independence. So it does not give any insights if one thinks in terms of dependence/independence of two measurements.

 

[Simon Phoenix]

zonde : "an entangled particle can't be in a mixed state"

This is incorrect

Stevendaryl : "it can always be in an improper mixed state"

This is correct, in fact if the global state is pure then for an entangled state of 2 particles, each particle on its own IS in a mixed state, by definition (an entangled state is one that cannot be written as a product of pure states). In this case we call it an 'improper' mixture because we know it has been derived from entanglement - but as been said many times the actual mathematical description of this 'improper' state is identical to the mixed state we get by simply preparing single particles in a statistical mixture to which it corresponds.

So it's probably best to think of the terms 'proper' and 'improper' as just shorthand terms to indicate where the (same) mixed state has come from. In terms of measurements on that particle alone both are entirely equivalent. As I said earlier we can also go the other way so that every proper mixed state can be thought of as an improper mixed state of a component part of a larger entangled pure system (this is the process of 'purification').

This 'purification' procedure is quite useful - for example if we consider an EM field mode in a thermal (mixed) state - so a 'proper' mixture - then it's the same as considering it to be one mode of an entangled pure state of 2 field modes (in this case it is the 2-mode squeezed state). So we can do our calculations for our mode of interest but using a larger pure state, which can make calculations more straightforward.

 

[Simon Phoenix]

Your example certainly does not suggest independence. So it does not give any insights if one thinks in terms of dependence/independence of two measurements.

The preparation by Alice is independent of the measurement by Bob.

Like I said this example has a practical benefit in that the Ekert protocol for establishing a quantum key distribution, which relies on entangled particles and checking to see whether or not the Bell inequality is violated as the eavesdropper detection step, can be achieved with single particles. It's much easier from a technology standpoint to work with single particles (in this case photons).

 

[zonde]

The preparation by Alice is independent of the measurement by Bob.

Yes indeed, that much your example shows. So if we make a distinction between mutual independence and one-way independence then we can see that we can violate inequality with one-way independence. But this is sort of clear from explanation of entanglement measurements using collapse. Non-local collapse is enough to get consistent explanation.

 

[Simon Phoenix]

But this is sort of clear from explanation of entanglement measurements using collapse

Yes - it is very obvious when you see it. But hopefully you now see that by viewing the measurement procedure as a preparation procedure as in this example it is also obvious that proper and improper mixtures are equivalent.

 

[rubi]

nubi

I'm sorry you're so butthurt for being wrong, but this is physics and not politics, so I'm not going to make a compromise. Wrong is wrong. If you don't want to be wrong in the future, then just don't claim anything if you don't know for sure that it's correct.

Given the notation from my post #67, is it correct that ½|0⟩⟨0| + ½|1⟩⟨1| is the mixed state of one of the photons from |J⟩ = √½(|00⟩ + |11⟩)?

Yes, that's the mixed state of Alice's particle, as we have been discussing since 7 pages. But you can also represent it differently, that would make no difference.

I would like to see how someone proposes to replicate the correlations that can be exhibited by a pair from |J⟩ with mixed states as suggested in post #72. E.g. derive a Bell inequality and violate it with with mixed states that are separated in the usual fashion and locality is assumed. I am confident this cannot be done, otherwise why bother with entanglement.

Trivially, you cannot compute correlations between Alice and Bob if you only have a description of Alice's system. Just like you cannot compute correlations between Alice and the Andromeda galaxy if you only have a description of the composite Alice/Bob system. Nevertheless, Alice's particle is in a mixed state, the composite Alice/Bob system is in the EPRB state and the composite Alice/Bob/Andromeda system is in some complicated state we don't know. Here's a completely classical analogy: You have an urn with 10 cards that have a color (green/blue) on the front and on the back (we can tell the difference between the front side and the back side). They are distributed as follows (front/back): 3x blue/blue, 3x blue/green, 1x green/blue and 3x green/green. There will be some non-trivial correlations between the front and the back sides. However, I can just forget about the back side and calculate the distribution of the front sides: 6x blue, 4x green. Obviously, from only knowing the colors of the front sides, I cannot calculate the correlations between the colors on the front and the back sides, because there are several possible distributions of the complete system that match the distribution of the front sides and I don't know which one is the correct one. Nevertheless, the latter distribution describes the state of affair of the front sides completely. The situation is exactly identical for the partial trace of the EPRB state.

Yes, I think that $|\psi\rangle$ and $e^{i \theta} |\psi\rangle$ are different vectors so they are different states (if we defined "state" as state vector).

Copenhagenists will disagree. You can check Weinberg, who certainly isn't (like Ballentine) guilty of advocating the ensemble interpretation:

In quantum mechanics state vectors that differ by a constant factor are regarded as representing the same physical state.

Look, we can add different vectors and get third vector. We can't do that with rays. I have quote from Neumaier to back up what I say

So what? You can use any vector to define a state, even those that you got by adding individual vectors. Nevertheless, the vectors aren't themselves states. It is very important to not call them states. Quantum mechanics would be broken if vectors were states, because the observed symmetries of nature would be violated. (By the way, @A. Neumaier will certainly agree that the only valid definition of states is by density matrices.)

I'm not so sure about my knowledge of Copenhagen but thinking in terms of probability amplitudes we need phase factor to add probability amplitudes correctly. If we throw away phase factor we can't get interference effect. So I would say that phase factor is important whenever we talk about interference.

These are relative phase factors and not global phase factors. They are conserved in the density matrix. Only the ambiguity of the global phase factor goes away.

 

[morrobay]

BTW, it seems that you could accomplish what you want in an easier fashion. Alice randomly selects an observable the flips a coin to decide either +1 or -1 which determines a state, then calls up Bob and tells him the state, then Bob makes a measurement by his randomly selected observable and they violate Bell's inequality.
Of course this nothing to with the usual Bell set up. But it's possible I am not following you.

Can you show how there is single particle inequality violation ?

 

[A. Neumaier]

I'm not so sure about my knowledge of Copenhagen but thinking in terms of probability amplitudes we need phase factor to add probability amplitudes correctly. If we throw away phase factor we can't get interference effect. So I would say that phase factor is important whenever we talk about interference.

This has nothing to do with interpretations!

There is a confusion between pure states and state vectors. Interference is about observing a pure state given by the ray associated with a state vector obtained as a linear combination of state vectors in a preferred basis. The pure states themselves are always rays, forming a projective space, not a vector space, while the state vectors form a Hilbert space. Only the latter can be added.

Of course, informally, one often talks about a state vector as a state, but this is just short hand for the correct mathematical view. Therefore one needs to be careful. For example, $\psi$ and $-\psi$ are the same state in this loose sense but their 50-50% superposition with another state is quite different!

 

[stevendaryl]

Yes, I think that $|\psi\rangle$ and $e^{i \theta} |\psi\rangle$ are different vectors so they are different states (if we defined "state" as state vector).

Yes, I know. I'm saying that we shouldn't define "state" that way. They aren't different states from the point of view of the Pauli exclusion principle, for example. They aren't different states from the point of view of predictions for the results of experiments.

Look, we can add different vectors and get third vector. We can't do that with rays.

That's true. As a matter of fact, I think I said that in a previous post. The mathematics of Hilbert space is about vectors, not rays. But two different vectors do not correspond to different states.

I guess before continuing this argument, I'd like to back up to why we are arguing in the first place. I've forgotten what is supposed to follow from whether or not the phase is part of the state.

 

[stevendaryl]

It shows us in a rather striking way another example of the equivalence of proper and improper mixtures, and from an applications point of view one can use this to provide a quantum key distribution scheme based on violation of a Bell inequality with single particles (in this scheme any eavesdropper would be equivalent to introducing a hidden variable thus ensuring no violation, which can be detected).

Well, your modified EPR experiment is basically the "collapse" interpretation, where the collapse is actually performed explicitly by Alice. In the collapse interpretation, after Alice measures spin-up for her particle, the state of Bob's particle "collapses" to spin-down. In your alternative scenario, Alice explicitly creates a spin-down particle and sends it to Bob.

So I don't think it should be too surprising that your altered scenario can violate Bell's inequalities--it's always been known that instantaneous wave function collapse was a way to explain EPR, but that was rejected by people who dislike the notion of an objective instantaneous collapse (since that would be an FTL effect).

 

[stevendaryl]

So it's probably best to think of the terms 'proper' and 'improper' as just shorthand terms to indicate where the (same) mixed state has come from. In terms of measurements on that particle alone both are entirely equivalent.

Yes, and that's a weird fact about quantum mechanics, that Bob can't distinguish between a proper and improper mixed state. In the case of a proper mixed state, Bob's particle is really either spin-up or spin-down, he just doesn't know which. In the case of an improper mixed state, his particle is neither spin-up nor spin-down until after he measures it. So it seems that these are different situations. But QM absolutely rules out an experiment that could distinguish them.

 

[vanhees71]

You are simply giving particular definition of "state". There is not much to understand except that this is different definition from Copenhagen's one.

No, the choice of interpretation has nothing to do with the formalism. In any formulation of QT this is the definition of a pure state, and nothing else. As you demonstrate very well, there's a lot to understand concerning the concept of state, even if you leave aside any interpretation issue (which I highly recommend to do; you have to understand the formalism first, and without wanting to be rude, from what you say I have the strong impression that you don't understand it yet).

 

[zonde]

Yes, I know. I'm saying that we shouldn't define "state" that way.

I understood that. I will try and look how it goes for me.
So I will have to "translate" all the statements like these:
Quantum superposition on wikipedia: "Quantum superposition is a fundamental principle of quantum mechanics. It states that, much like waves in classical physics, any two (or more) quantum states can be added together ("superposed") and the result will be another valid quantum state; and conversely, that every quantum state can be represented as a sum of two or more other distinct states."
Measurement problem on wikipedia: "Prior to observation, according to the Schrödinger equation, the cat is apparently evolving into a linear combination of states that can be characterized as an "alive cat" and states that can be characterized as a "dead cat". Each of these possibilities is associated with a specific nonzero probability amplitude; the cat seems to be in some kind of "combination" state called a "quantum superposition"."
and so on.
Well, I suppose that at least "wave function" includes this arbitrary phase factor.

I guess before continuing this argument, I'd like to back up to why we are arguing in the first place. I've forgotten what is supposed to follow from whether or not the phase is part of the state.

Good idea.

 

[stevendaryl]

I understood that. I will try and look how it goes for me.
So I will have to "translate" all the statements like these:
Quantum superposition on wikipedia: "Quantum superposition is a fundamental principle of quantum mechanics. It states that, much like waves in classical physics, any two (or more) quantum states can be added together ("superposed") and the result will be another valid quantum state; and conversely, that every quantum state can be represented as a sum of two or more other distinct states."

The distinction between a Hilbert space vector and a state is kind of nitpicky one, and many people don't bother making the distinction. Working with something concrete like a vector rather than something abstract like an equivalence class is a bother, which only makes a difference in very special cases. It's sort of like with rational numbers. If you have a fraction, people often talk about the numerator and the denominator of that fraction, but actually, there is no such thing as the numerator of a fraction, because a fraction is an equivalence class of objects with different numerators and denominators: $1/2 = 3/6 = 4/8 = ...$. If you find a reference that talks about the numerator of a rational number, that doesn't prove that rationals aren't equivalence classes, it's just an example of being loose with language.

It's very often in mathematics easier to deal with a specific representative of an equivalence class, rather than the class itself.

 

[vanhees71]

I disagree strongly. That pure states are represented by rays rather than vectors is vital for quantum theory. There wouldn't be a working non-relativistic quantum mechanics a la Heisenberg, Schrödinger, and Dirac if vectors would represent pure states rather than rays, no half-integer spin particles etc. etc. Instead of working with the cumbersome rays, you can work right away with the statistical operator. The only specialty about pure state is that they are projection operators of the form $|\psi \rangle \langle \psi|$.

 

[stevendaryl]

Well, I suppose that at least "wave function" includes this arbitrary phase factor.

I would not say that. Let's take the case of a spin-1/2 particle, where we only consider the spin degrees of freedom. Then an arbitrary normalized state vector can be written in the form $|\psi\rangle = e^{i \chi} (cos(\frac{\theta}{2}) e^{-i \frac{\phi}{2}} |U\rangle + sin(\frac{\theta}{2}) e^{+i\frac{\phi}{2}} |D\rangle)$. The relative phase $e^{i \phi}$ is certainly important, but the overall phase $e^{i \chi}$ is not.

 

[zonde]

I guess before continuing this argument, I'd like to back up to why we are arguing in the first place. I've forgotten what is supposed to follow from whether or not the phase is part of the state.

If we build a larger model that includes Bob modeling his mixed state as statistical mixture of certain orthogonal pure states and Alice modeling her mixed state as statistical mixture of the same orthogonal pure states and we include in this larger model means of pairing up Alice's detections with Bob's detections there will be measurement angles for Bob and Alice for which we would not be able to combine Alice's model with Bob's model in such a way that it reproduces predictions for correlations of entangled particles.

 

[vanhees71]

That I've already answered above: To address correlations, of course you cannot use the reduced single-photon states, because these neglect the correlations. Again, I can only suggest to first learn the fundamental facts about quantum theory before you go to the more complicated aspects!

 

[zonde]

I would not say that. Let's take the case of a spin-1/2 particle, where we only consider the spin degrees of freedom. Then an arbitrary normalized state vector can be written in the form $|\psi\rangle = e^{i \chi} (cos(\frac{\theta}{2}) e^{-i \frac{\phi}{2}} |U\rangle + sin(\frac{\theta}{2}) e^{+i\frac{\phi}{2}} |D\rangle)$. The relative phase $e^{i \phi}$ is certainly important, but the overall phase $e^{i \chi}$ is not.

And if you take particle in a box. There overall phase factor is function of time, right?

 

[vanhees71]

An energy eigenstate is time-independent. That's why the time dependence of the corresponding wave function is a phase factor $\exp(-\mathrm{i} E t/\hbar)$. This is underlining again the importance of describing the state as a ray or the corresponding statistical operator rather than the vector. It's only clear that an energy eigenstate is time-indpendent when considering the correct representants of the states (rays or statistical operator).

 

[zonde]

That I've already answered above: To address correlations, of course you cannot use the reduced single-photon states, because these neglect the correlations. Again, I can only suggest to first learn the fundamental facts about quantum theory before you go to the more complicated aspects!

But I am not using reduced single-photon states. I rather model Bob (doing something) and Alice (doing something).

 

[vanhees71]

Then I'm no knowing what you are talking about.

 

[A. Neumaier]

Yes, and that's a weird fact about quantum mechanics, that Bob can't distinguish between a proper and improper mixed state.

The state encodes by definition everything that can be said about a system once it is prepared,

Thus the distinction between proper and improper (i.e., how it was prepared) is operationally irrelevant.

We have a similar situation classically: We cannot determine from a glass of water at room temperature whether it was prepared by letting ice melt or by letting hot water cool down. A very ordinary fact! Why should it be thought of as weird in the quantum case?

 

[zonde]

Then I'm no knowing what you are talking about.

Can Bob use reduced single-photon states (he does not care about correlations)? Can Alice use reduced single-photon states (she does not care about correlations either)?

 

[stevendaryl]

If we build a larger model that includes Bob modeling his mixed state as statistical mixture of certain orthogonal pure states and Alice modeling her mixed state as statistical mixture of the same orthogonal pure states and we include in this larger model means of pairing up Alice's detections with Bob's detections there will be measurement angles for Bob and Alice for which we would not be able to combine Alice's model with Bob's model in such a way that it reproduces predictions for correlations of entangled particles.

Well, yes. This has been said many times before: when you perform a trace to get a single-particle mixed state from a two-particle pure state, you throw away information about correlations. The pure state for the two-particle system contains more information than the sum of the mixed states for the single-particles. The whole is more than the sum of the parts.

I don't think there is any disagreement about that. That's one of the weird features of quantum mechanics that has no analog in classical mechanics. Classical mechanics is reductionistic, in the sense that the most complete description of the parts of a composite system give you the most complete description of the composite. Quantum mechanics is not like that, because there is nonlocal correlation information.

 

[A. Neumaier]

at least "wave function" includes this arbitrary phase factor.

Yes. A wave function is a function, and these can be added and superimposed, while states cannot.

Hence calling a state vector a state is (sometimes harmful) sloppiness, even though wikipedia does it. (Remember that wikipedia also endorses virtual particles popping in and out existence and similar nonsense, because it takes literally what is meant sloppily.)

 

[vanhees71]

That's not true in the case of entangled states. The here discussed example of polarization entangled two-photon states clearly demonstrates this. If you take the partial trace over one photon, all you get are unpolarized single-photon states, and that's indeed what can be found by only looking at one of the entangled photons. However, if you do correlation measurements on the polarization of both photons (e.g., measuring the polarization ins different (non-orthogonal) directions, you can demonstrate the violation of Bell's inequality).

Your glass-of-water analogy is different. It just tells you that equibrium states do not contain any information about the history of how this state was reached. That's almost a definition of the equilibrium (maximum-entropy) state, no more no less.

 

[stevendaryl]

The state encodes by definition everything that can be said about a system once it is prepared,

Thus the distinction between proper and improper (i.e., how it was prepared) is operationally irrelevant.

It's not though. If Bob has a proper mixed state due to ignorance of the true state, then even though he can't tell the difference, someone else who knows the true state, can. If Alice flips a coin, and with probability 1/2 sends a spin-up particle to Bob, and with probability 1/2 sends a spin-down particle to Bob, then Alice knows ahead of time what Bob's spin measurement result will be. So for Alice, that's different from the case of an improper mixed state, where nobody knows ahead of time what Bob's result will be.

 

[vanhees71]

Can Bob use reduced single-photon states (he does not care about correlations)? Can Alice use reduced single-photon states (she does not care about correlations either)?

Sure, the reduced single-photon states describe precisely what either Alice of Bob will find when looking only at one of the photons. In our case just unpolarized ones.

 

[zonde]

Well, yes. This has been said many times before: when you perform a trace to get a single-particle mixed state from a two-particle pure state, you throw away information about correlations. The pure state for the two-particle system contains more information than the sum of the mixed states for the single-particles. The whole is more than the sum of the parts.

I don't think there is any disagreement about that. That's one of the weird features of quantum mechanics that has no analog in classical mechanics. Classical mechanics is reductionistic, in the sense that the most complete description of the parts of a composite system give you the most complete description of the composite. Quantum mechanics is not like that, because there is nonlocal correlation information.

What I meant to say with my model that even if you are allowed to include back whatever information you want you can't make the model consistent.

 

[rubi]

What I meant to say with my model that even if you are allowed to include back whatever information you want you can't make the model consistent.

This claim is false. (See post #38 for the proof.)

 

[A. Neumaier]

However, if you do correlation measurements on the polarization of both photons

This requires having a proper pure state of the big system! But if all you have and measure is the state of the small system, you cannot distinguish it. That's why it is called a state!

Similarly with a glass of water. If you consider the bigger system that includes a camera that had observed the process of warming or cooling, you can recover from its state additional information about the history of the glass of water.