Undergrad Is Polarisation Entanglement Possible in Photon Detection?

[rubi]

rubi, you are contradicting yourself. In one sentence you write that state vectors and state operators are equivalent but in next sentence you write that vectors contain "ambiguous phase factor" and therefore state operator is correct expression for state and vector is not. So please make up your mind.

I'm not contradicting myself. A vector defines a state, but it is not a state itself. The state is the vector with the phase factor stripped off. It's a point in the projective Hilbert space, not the Hilbert space. The mathematical description of such a state is given by a density matrix. A state can be pure or mixed, but even if it is mixed, one can find a vector that defines the state, as I have demonstrated in post #115. Whether you use vectors or density matrices to define a state is completely irrelevant. The difference between a state and a vector that defines a state is also very important. In general, a vector is not invariant under Galilei transformations, but the state it defines (i.e. the density matrix) is invariant.

Ballentine also explains this in his book by the way:

Since τ2τ1 and τ3 are the same space–time transformations, we require that
U(τ2)U(τ1)|Ψ and U(τ3)|Ψ describe the same state. This does not mean
that they must be the same vector, since two vectors differing only in their
complex phases are physically equivalent, but they may differ at most by a
phase factor.

 

[vanhees71]

One last time: A pure state is represented by a RAY IN HILBERT SPACE, i.e., the set $[\psi]=\{\exp(\mathrm{i} \phi) |\psi \rangle|\phi \in \mathbb{R} \rangle \}$, where $\langle \psi|\psi \rangle=1$. It is very important to understand this to understand quantum theory as a whole.

Equivalently you can introduce states as a being represented by statistical operators $\hat{\rho}$. Pure states are those statistical operators that are projection operators, $\hat{\rho}_{\psi} = |\psi \rangle \langle \psi|$.

 

[zonde]

I'm not contradicting myself. A vector defines a state, but it is not a state itself. The state is the vector with the phase factor stripped off. It's a point in the projective Hilbert space, not the Hilbert space. The mathematical description of such a state is given by a density matrix. A state can be pure or mixed, but even if it is mixed, one can find a vector that defines the state, as I have demonstrated in post #115. Whether you use vectors or density matrices to define a state is completely irrelevant. The difference between a state and a vector that defines a state is also very important. In general, a vector is not invariant under Galilei transformations, but the state it defines (i.e. the density matrix) is invariant.

Yes, you explained how you see the difference between state vector and state operator. The point is that there is a difference even if it's only matter of phase factor.
Returning to Ballentine. He emphasize that Copenhagen state vector is complete description of every system in ensemble. That is important difference between Copenhagen and Ensemble interpretation. In Ensemble interpretation you allow differences between systems in ensemble (described as "state"). So "mixed state" makes sense in Ensemble interpretation but it does not really make sense in Copenhagen (if you stick to single interpretation of "state").
You might say it's just semantics but people (I at least) try to form mental pictures about important concepts in the topic like "state". If one has formed mental picture of state based on Copenhagen he will say that mixed state gives complete description of every system in ensemble associated with that state. If you need example look at the post #48 by Simon Phoenix:

Let's suppose we have a spin-1/2 particle in the mixed state described by |0><0| + |1><1| (the omitted normalization constant is 1/2, but I really must learn LaTex one of these days) where the |0> and |1> are eigenstates of the spin-z operator. Now transform basis to the eigenstates of the spin-x operator which we label as |0*> and |1*>. In this new basis the density operator is |0*><0*| + |1*><1*|. So do we have a statistical mixture of the pure states |0> and |1>, or do we have a statistical mixture of the states |0*> and |1*>?

The answer to this questions is 'yes'

Both are entirely equivalent descriptions of the same mixed state.

 

[zonde]

One last time: A pure state is represented by a RAY IN HILBERT SPACE, i.e., the set $[\psi]=\{\exp(\mathrm{i} \phi) |\psi \rangle|\phi \in \mathbb{R} \rangle \}$, where $\langle \psi|\psi \rangle=1$. It is very important to understand this to understand quantum theory as a whole.

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

 

[Simon Phoenix]

If one has formed mental picture of state based on Copenhagen he will say that mixed state gives complete description of every system in ensemble associated with that state.

I'm not sure I would say 'complete' here because there's an element of subjectivity to the density matrix we assign based on our knowledge. Let's suppose Alice prepares a (pure) state of a spin-1/2 particle - so let's suppose spin up in the spin-z direction. Now she sends this particle to Bob, but all Bob knows is that it could have been prepared in a spin-z eigenstate, or it could have been prepared in a spin-x eigenstate. The density operator Alice assigns is just a projection, but Bob has to assign a mixed state to this. Both descriptions are consistent with the results of a single measurement.

This is important in quantum key distribution in which measurements are made on single quantum particles (usually photons). In this scheme Alice sends single particles in a given timeslot. In each timeslot a random choice is made whether to prepare the particle in one of the 4 eigenstates of the spin-z or spin-x operators.

It is Bob's lack of knowledge together with his inability to discover that knowledge with a single measurement, which is reflected in him having to assign a mixed state to each transmitted particle, that is responsible for the security. To Alice, each transmitted particle is in a pure state which she knows. To Bob, each particle is in a mixed state (if he could actually assign a pure state to each timeslot then he would be able to recover the key).

 

[rubi]

Yes, you explained how you see the difference between state vector and state operator. The point is that there is a difference even if it's only matter of phase factor.

There is no difference between a state defined by a state vector and a state directly defined by a density matrix. For every state that is given by a density matrix, a vector formulation can be found. If you are (for whatever reason) committed to only using vectors, then you can just take any density matrix state and convert it into a vector description. Hence, there is no difference between the density matrix formulation and the vector formulation, except that the vector formulation contains an additional ambiguous phase factor, which you will also get if you convert a density matrix into a vector. It is completely irrelevant whether you arrange the information of the state in a 2-dimensional array ("matrix") or a list ("vector"). The information content is identical.

Returning to Ballentine. He emphasize that Copenhagen state vector is complete description of every system in ensemble. That is important difference between Copenhagen and Ensemble interpretation. In Ensemble interpretation you allow differences between systems in ensemble (described as "state"). So "mixed state" makes sense in Ensemble interpretation but it does not really make sense in Copenhagen (if you stick to single interpretation of "state").

No, Ballentine never makes any difference in the description between the Copenhagen interpretation and the ensemble interpretation. Both use the exact same mathematics and the exact same axioms. The only difference is the interpretation of the states (which are defined in exactly the same way in both interpretations). In the Copenhagen interpretation the state refers to individual systems and in the ensemble interpretation, it just refers to an ensemble of identically prepared systems. However, this is just a difference in the interpretation of the state, which is described by the exact same mathematics in both interpretations. It is definitely not the case that only the ensemble interpretation uses mixed states. The Copenhagen interpretation uses them in exactly the same way.

Moreover, you can convert any density matrix into a vector, so what is your problem with density matrices? If you don't like the reduced density matrix we calculated for Alice's particle, just take the vector I defined in post #115 and work with that. It defines the exact same state as the density matrix, but is given by a vector. It behaves exactly like a vector in the Copenhagen interpretation should behave. You can evolve it, collapse it, apply Born's rule and do everything a Copenhagenist would like to do. It is the state of Alice's particle also in the Copenhagen interpretation and it leads to the correct, experimentally confirmed predictions.

You might say it's just semantics but people (I at least) try to form mental pictures about important concepts in the topic like "state". If one has formed mental picture of state based on Copenhagen he will say that mixed state gives complete description of every system in ensemble associated with that state. If you need example look at the post #48 by Simon Phoenix:

Your mental picture is plain wrong and not shared by any physicists. Copenhagenists work with density matrices on a regular basis and have no problem calling them states.

By the way, nobody in this thread has been considering the ensemble interpretation or interpretations in general. What has been said about states applies to all existing interpretations.

 

[stevendaryl]

If you describe state using density matrix you are not broadening concept of state, you are changing it. And it can lead to quite a mess if you do not announce that clearly enough.

I suppose that's fair, although it is possible to do all of quantum mechanics using density matrices as the only notion of "state".

 

[stevendaryl]

Yes, you explained how you see the difference between state vector and state operator. The point is that there is a difference even if it's only matter of phase factor.

But because of the phase factor, a vector can't be a state, since vectors with different phases represent the same state. (Actually, it's not just phase---as I understand it, the vectors of Hilbert space can have different normalizations. To get a state, you have to normalize.) So it's never the case that a vector is a state. As vanhees points out, in the usual way of doing quantum mechanics (without density matrices), a state is not a vector, but an equivalence class of vectors. Mathematically, you can paper over this as being nit-picky, because you can just pick one arbitrary element of the equivalence class to represent the entire class, but technically, a vector is not a state in QM.

Returning to Ballentine. He emphasize that Copenhagen state vector is complete description of every system in ensemble. That is important difference between Copenhagen and Ensemble interpretation. In Ensemble interpretation you allow differences between systems in ensemble (described as "state"). So "mixed state" makes sense in Ensemble interpretation but it does not really make sense in Copenhagen (if you stick to single interpretation of "state").

There is an analogy with two different ways of doing classical mechanics:

1. You can view the state of a system to be a point in phase space (that is, you give a position and momentum to each particle, and that specifies the state of the system). Newton's laws tell how this point in phase space changes with time.
   
2. You can view the state of a system to be a probability distribution on phase space. Newton's laws can be used to say how this distribution changes with time.

I don't think that there is any commitment to a particular interpretation when you choose one way or another. It's a matter of which approach is easier to work with.

I think the same thing is true of the state vector versus density matrix formulations of quantum mechanics. There is no commitment to a particular interpretation of quantum mechanics implied. Even if you believe that particles have a "real" wave function that is a vector in Hilbert space, you can be in the situation (doing physics in the presence of statistical uncertainty) where you don't know that wave function exactly. Then using density matrices is the best approach.

 

[stevendaryl]

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

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.

 

[rubi]

I suppose that's fair, although it is possible to do all of quantum mechanics using density matrices as the only notion of "state".

Density matrices are not a broadening of the concept of state. They are completely equivalent to descriptions in terms of vectors. It's physically impossible to distinguish them. That's a mathematical theorem called the GNS construction. The use of density matrices instead of vectors nothing but a convention. Instead of computing partial traces, we could instead apply the GNS construction to the observables of the subsystem and arrive at a vector state directly. Of course, partial traces are much more convenient and hence everybody is using density matrices instead of vectors.

 

[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.

Your classical example allows communication between Alice and Bob and thus becomes a simple math exercise with no physics. It does clearly show the necessity for the usual set up where Alice and Bob are separated and perform their experiments at nearly the sametime in order to show how entanglement provides the spooky correlations that can't be replicated classically.

 

[Simon Phoenix]

Your classical example allows communication between Alice and Bob and thus becomes a simple math exercise with no physics

Well I beg to differ (of course). First off the example is far from being 'classical'. It shows us that the violation of the Bell inequality in QM does not strictly require entangled states. It requires entangled states (and spacelike measurements) if one wishes to draw some conclusions about local hidden variables, but violation of the inequality itself is a feature of non-commuting observables in QM and not strictly entanglement.

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).

 

[Zafa Pi]

I have a feeling that I know what’s going on with the latter part of this thread mostly involving nubi, Simon, steven, vanhees, zondi, and me. Feedback might revise that feeling. I’ll start with a minor epiphany that happened last night.

I was looking at van Gogh’s irises and listening to Ravel when the vision of a chemist came along and scraped off all the paint and decomposed it into little piles of all the different elements, cadmium chromium, etc. He did the same with the canvas, little piles of carbon, sulfur, etc.. Yet I could still see the painting between the two piles. He says here is a list of the weights of the elements in the paint, “the mixture of the paint” which describes the paint, and I have the same for the canvas. I said, no way, those lists do not describe the paint or canvas because together they miss the very essence of the painting. He says each by themselves are perfectly adequate descriptions. I say, nope they are not adequate. He says, you’re missing the point ...

Each of you will likely find something inadequate about this as an analogy for the issue in the thread. My main complaint about the analogy is that the paint piles and canvas piles cannot exhibit spooky action at a distance except in my mind’s eye.

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⟩)? If not what is?

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.

 

[Simon Phoenix]

I am confident this cannot be done, otherwise why bother with entanglement

Your confidence is misplaced

But, and it's a big 'but' (so to speak), the violation exists between the preparation data (of Alice) and the measurement data (of Bob). So we don't need entanglement to see a violation of the mathematical inequality we call Bell's inequality. If you actually did the experiment as I suggest with single particles the data set would be (statistically) indistinguishable from a data set obtained by an experiment with entangled particles.

Think of it this way - suppose Bob in his lab was only told that he was working with particles from entangled pairs, but in reality he'd been duped and was working with single particles supplied to him by Alice as indicated. After he's done all his measurements he compares his data with that of Alice (who presents her data as if it were measurement data). Can he tell he's been duped? Is there any statistical test he can do to uncover the deception? There is not.

The 'Bell' inequality was effectively known about a century before Bell. It was derived by George Boole who showed that if we had three random binary variables $A$, $B$ and $C$ then the joint distributions $P(A,B)$, $P(A,C)$ and $P(B,C)$ could only be constructed as marginals of the distribution $P(A,B,C)$ provided a 'Bell' inequality is satisfied. Of course we should note that in writing down $P(A,B,C)$ as it might apply to spin variables we're already making a 'classical' assumption by effectively assuming such a quantity would be meaningful in QM.

We bother with entanglement (a) because we need it to say something about local hidden variable theories and (b) it has lots of other nice features other than violation of a Bell inequality

 

[Simon Phoenix]

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.

I should add that of course we can't do a 'usual' Bell experiment (i.e., with spacelike measurements) in which Alice and Bob have (unentangled) mixed states and see a violation - but I thought I had made that clear (sorry if it wasn't). 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. We force it to say something about locality vs non-locality by requiring spacelike measurements in the context of a hidden variable model.

 

[Zafa Pi]

Well I beg to differ (of course). First off the example is far from being 'classical'. It shows us that the violation of the Bell inequality in QM does not strictly require entangled states. It requires entangled states (and spacelike measurements) if one wishes to draw some conclusions about local hidden variables, but violation of the inequality itself is a feature of non-commuting observables in QM and not strictly entanglement.

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).

I don't follow you. With the usual separation of Alice and Bob can you give an explicit way that they produce values of + and -1 that violate the CHSH inequality without employing entangled photons?

 

[Simon Phoenix]

With the usual separation of Alice and Bob can you give an explicit way that they produce values of + and -1 that violate the CHSH inequality without employing entangled photons?

If by 'usual separation' you actually mean Alice and Bob perform spacelike measurements on a state given by $$ \rho = \frac 1 {2} \left[ | 0 \rangle \langle 0 | + | 1 \rangle \langle 1 | \right] \otimes \frac 1 {2} \left[ | 0 \rangle \langle 0 | + | 1 \rangle \langle 1 | \right]$$ then no I cannot - but then this isn't what I was saying anyway.

If you want to see a violation with spacelike measurements then, sure, you need entangled states. Then you can draw those rather profound conclusions about local hidden variables.

I was (attempting) to show that one can still violate the mathematical inequality with single (and therefore unentangled) particles. In this latter case one cannot draw any conclusions about locality or non-locality (obviously).

 

[Zafa Pi]

I should add that of course we can't do a 'usual' Bell experiment (i.e., with spacelike measurements) in which Alice and Bob have (unentangled) mixed states and see a violation - but I thought I had made that clear (sorry if it wasn't).

It wasn't clear to me. This the first I've seen you say this. All I've ever considered was the 'usual' Bell experiment.

 

[Simon Phoenix]

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.

Of course, we can't draw any of the important conclusions about local variables from doing things this way

It requires entangled states (and spacelike measurements) if one wishes to draw some conclusions about local hidden variables

I should add that of course we can't do a 'usual' Bell experiment (i.e., with spacelike measurements) in which Alice and Bob have (unentangled) mixed states and see a violation

I had hoped that these various previous comments would have made it clear - apologies if it wasn't. I first realized this when working on some QKD stuff and it came as a bit of a shock to me too, initially. After all, I'd always gone with the assumption 'violation of inequality' = 'entanglement'. But it's kind of a nice example (I think) in the context of this discussion because it shows that we can get these non-classical correlations (i.e. violation of a Bell inequality) with single particles by utilizing the equivalence of improper and proper mixtures.

 

[Zafa Pi]

I had hoped that these various previous comments would have made it clear - apologies if it wasn't. I first realized this when working on some QKD stuff and it came as a bit of a shock to me too, initially. After all, I'd always gone with the assumption 'violation of inequality' = 'entanglement'. But it's kind of a nice example (I think) in the context of this discussion because it shows that we can get these non-classical correlations (i.e. violation of a Bell inequality) with single particles by utilizing the equivalence of improper and proper mixtures.

It was the last of your quotes that made it clear to me and I commented on it.
As for the equivalence of improper and proper mixtures, why in post #19 would steven respond to zonde's saying "an entangled particle can't be in a mixed state", with: "it can always be in an improper mixed state" if steven thought they were the same? I find that confusing.

 

[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 ?