Is Polarisation Entanglement Possible in Photon Detection?

[zonde]

Something has gone seriously wrong with this discussion. That tracing out one component of a two-component system produces a density matrix that looks like a mixed state is just simply a fact. Is it really a mixed state? Yes--a mixed state is a density matrix that cannot be written in the form $|\psi\rangle \langle \psi|$

There is no disputing these facts. They are simply facts. You can argue about what this implies about interpretations of quantum mechanics, but you can't dispute what's plainly true. I don't understand what's going on here, at all.

I agree that there something wrong with this discussion.
Phrases "state $\rho$" and "mixed state is a density matrix" make a category error. A state is a vector. Density matrix is not a vector or anything similar.

 

[Simon Phoenix]

I don't understand what's going on here, at all.

I'm having the same difficulty. I think what's happening is that there's this feeling that because a component part has come from an entangled pair then somehow this entanglement must be 'reflected' in the 'state' of the component part - so it has to be more than just a classical-like mixture. I think this is the sticking point.

Or not, I'm not totally sure lol.

 

[zonde]

I think what's happening is that there's this feeling that because a component part has come from an entangled pair then somehow this entanglement must be 'reflected' in the 'state' of the component part - so it has to be more than just a classical-like mixture. I think this is the sticking point.

I think one problem is with "component part" of entangled state. As much us I understand what a product space is there can be no independent descriptions for each side ("component parts") within that space. So the closest thing to "component part" that we can have is a trace that gives us density matrix. But such density matrix is not a component of entangled state (a vector).

 

[vanhees71]

Look at post #19 or improper mixed states on line. Do I need to prove a Bell inequality for you when Alice and Bob have mixed states, they are classical you realize.
Indeed vanhees71is mistaken. The physical distinction lies in the entangled correlations that cannot be replicated with "proper" mixed states.

That's of course true, but for this you need to entire (in this case the two-photon state). It's not described by the reduced states of either photon.

 

[rubi]

I agree that there something wrong with this discussion.
Phrases "state $\rho$" and "mixed state is a density matrix" make a category error. A state is a vector. Density matrix is not a vector or anything similar.

No that's not correct. If anything deserves the name "state" in quantum mechanics, it's the density matrix. Vectors are ambiguous due to the arbitrary phase factor. If we didn't acknowledge this, we couldn't even explain spin or the Galilei invariance of the hydrogen atom, so it has practical importance to use the term correctly, i.e. not for vectors. In practice, we are often lazy and refer to vectors as states, but it is strictly speaking not correct. Moreover, we could also represent density matrices as vectors if we cared to purify them.

I think one problem is with "component part" of entangled state. As much us I understand what a product space is there can be no independent descriptions for each side ("component parts") within that space. So the closest thing to "component part" that we can have is a trace that gives us density matrix. But such density matrix is not a component of entangled state (a vector).

Well, you understand incorrectly. Everything that can be known about a component of a system is contained in the mixed state computed by the partial trace operation. The situation is completely analogous to the composite system, which is also just a component of an even larger system. Again, I urge you to confront a linear algebra textbook to clear up this misunderstanding. Your reasoning is mathematically erroneous.

 

[vanhees71]

The point is that taking the partial trace over a part of a composite system means that you forget about all of the traced out part. What you get is a statistical operator that describes the statistics for the outcome of measurements on this partial system. By definition you don't care about how it's entangled with the rest of the world, and that's the key issue to resolve all the apparent "problems" with entangled states brought up by EPR and much better a bit later by Einstein himself.

Take the standard example of the polarization entangled two-photon states. Only looking at the polarizations and keeping in mind that it refers to photons measured far distant places A (Alice) and B (Bob), you have the pure entangled state vector
$$|\Psi \rangle=\frac{1}{\sqrt{2}}(|HV \rangle + |VH \rangle),$$
and the state is described by the statistical operator
$$\hat{\rho}_{AB}=|\Psi \rangle \langle \Psi|.$$
The state of the single photons is given by tracing out the other photon, and as shown way up in the thread this leads to
$$\hat{\rho}_{A}=\hat{\rho}_B=\frac{1}{2} \hat{1}.$$
Both photons are perfectly unpolarized. In practice this means A and B measure on an ensemble of such prepaared photons with 50% probability H and 50% probability V polarized photons, and no matter to which direction of the polarization filter this measurement refers. There's this and only this information you can get measuring only the single-photon polarizations, and it doesn't matter, how in detail you prepare this state. They could as well come from a thermal source with a filter concerning the energy (frequency) of the photons.

Only if you make measurements on the two-photon system you can figure out the entanglement. For that you must measure both photons at A and B and make sure to relate only photon pairs which were prepared in the entangled pair. The latter is usually done by doing a coincidence measurement, i.e., A and B keep precise enough time stamps of their registration events and then can figure out the correlations described by the entanglement.

This is important, because with all the wild speculations about differences between states that are described by the same statistical operator QT wouldn't be a consistent description, and you'd have to find a better definition of states. To my knowledge there's not the slightest hint that this is necessary.

 

[kith]

If two systems are entangled, we cannot assign a state vector to the subsystems individually. So we have two choices:
1) Simply accept that the individual subsystem doesn't have a quantum state on its own.
2) Trace out the other subsystem to obtain the reduced density matrix and broaden the concept of what a quantum state is to include it.

@zonde and @Zafa Pi, you don't have to agree with option 2) but you have to acknowledge that this is the standard terminology used in quantum physics. You have both been here for a while and there have been quite a few fruitless discussions which have been at least complicated by terminology issues. Why don't you learn and use the standard terminology in order to discuss the physical content you have in mind?

 

[kith]

Here is also an argument, why the distinction between proper mixed states (obtained by mixing ensembles) and improper mixed states (obtained by tracing out a part of an entangled system) isn't as clear cut as one might think.

If we want to mix two ensembles physically, we need to interact with them somehow to bring them together. How can we be sure that we have obtained a proper mixed state? We can't because it would involve checking correlations with the environment for which it is almost always impossible to measure all the relevant degrees of freedom. And because interaction typically leads to entanglement, we have every reason to expect that we have obtained an improper mixed state.

So if one wants to make this distinction, he needs to outline how to obtain a proper mixed state in the first place.

 

[forcefield]

I think the root cause of confusion here is the simple fact that getting first the measurement result "vertical" and then the measurement result "horizontal" is not, physically, the same as getting first "horizontal" and then "vertical". However, QM (of course) makes the same probabilistic prediction for both sequences. And discussing about (order of) individual measurement results goes beyond QM.

 

[rubi]

Trace out the other subsystem to obtain the reduced density matrix and broaden the concept of what a quantum state is to include it.

I wouldn't say that it is a broadening of the concept of state. If zonde isn't happy with a matrix $\rho=\frac{1}{2}\left(\begin{array}{cc}1 & 0 \\ 0 & 1 \\\end{array}\right)$ on $\mathcal H=\mathbb C^2$, I will just give him $\mathcal H = \mathbb C^4$ and $\left|\Psi\right>=\frac{1}{\sqrt{2}}\left(\begin{array}{c} 1 \\ 0 \\ 0 \\1 \end{array}\right)$. Instead of calculating expectation values with the trace formula $\mathrm{Tr}(\rho A)$, he can calculate them with the standard formula $\left<\Psi\right|\tilde A\left|\Psi\right>$, where $\tilde A = \left(\begin{array}{cc}A & 0 \\ 0 & A \\\end{array}\right)$ is just the block matrix with $A$ on the diagonal. Of course all of this is completely unnecessary, but if zonde wants a state vector, he gets a state vector.

I think the root cause of confusion here is the simple fact that getting first the measurement result "vertical" and then the measurement result "horizontal" is not, physically, the same as getting first "horizontal" and then "vertical".

We're not talking about sequential measurements here, so that's not an issue.

 

[forcefield]

We're not talking about sequential measurements here, so that's not an issue.

Well, I meant the usual QM measurement, which requires an ensemble.

 

[stevendaryl]

I think one problem is with "component part" of entangled state. As much us I understand what a product space is there can be no independent descriptions for each side ("component parts") within that space. So the closest thing to "component part" that we can have is a trace that gives us density matrix. But such density matrix is not a component of entangled state (a vector).

Well, in quantum mechanics, both vectors and density matrices are referred to as "states". But since a vector is a special case of a density matrix (or corresponds to a special case of a density matrix), we may as well view density matrices as the general notion of a quantum-mechanical state, and pure states (ones that arise from vectors) are a special case. A "state" for quantum mechanics is just a mathematical object that summarizes the statistics for possible measurement results on the system. That broad definition covers density matrices and vectors.

The weird thing about quantum mechanics is that the same density matrix can arise from two very different processes: (1) tracing out the degrees of freedom of another component of a composite system, and (2) doing a classical "mixture" of two pure states to reflect ignorance about which pure state is applicable.

 

[zonde]

If two systems are entangled, we cannot assign a state vector to the subsystems individually. So we have two choices:
1) Simply accept that the individual subsystem doesn't have a quantum state on its own.
2) Trace out the other subsystem to obtain the reduced density matrix and broaden the concept of what a quantum state is to include it.

@zonde and @Zafa Pi, you don't have to agree with option 2) but you have to acknowledge that this is the standard terminology used in quantum physics. You have both been here for a while and there have been quite a few fruitless discussions which have been at least complicated by terminology issues. Why don't you learn and use the standard terminology in order to discuss the physical content you have in mind?

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 tried to look up Ballentine and run across this:
"It allows us to discriminate between the two principal classes of interpretations.
A. A pure state $|\Psi\rangle$ provides a complete and exhaustive description of an individual system. A dynamical variable represented by the operator Q has a value (q, say) if and only if $Q|\Psi\rangle = q|\Psi\rangle$.
B. A pure state describes the statistical properties of an ensemble of similarly prepared systems."

Pay attention that in case A "state" is a state vector but in case B it's a state operator (at the start of the book Ballentine defines "state" as possible statistics of preparation procedure and describes it with a state operator).
So it seems that distinction between "state" as a vector or as an operator comes bundled with different interpretations.

 

[rubi]

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.

As I have explained in post #115, density matrices and vectors are exactly equivalent. Nobody is changing the concept of state except yourself.

I tried to look up Ballentine and run across this:
"It allows us to discriminate between the two principal classes of interpretations.
A. A pure state $|\Psi\rangle$ provides a complete and exhaustive description of an individual system. A dynamical variable represented by the operator Q has a value (q, say) if and only if $Q|\Psi\rangle = q|\Psi\rangle$.
B. A pure state describes the statistical properties of an ensemble of similarly prepared systems."

Note, that he is talking about pure states in both cases. That's because he is not at all trying to saying what you are trying to put in his mouth. He is concerned with the difference between the Copenhagen interpretation and the ensemble interpretation, which he advocates in his book. Both interpretations work with pure states and mixed states.

Pay attention that in case A "state" is a state vector but in case B it's a state operator

No it's not. It's a "pure state" in both cases and you are free to write it as a matrix or a vector, since both notions are completely equivalent.

(at the start of the book Ballentine defines "state" as possible statistics of preparation procedure and describes it with a state operator).

Ballentine correctly defines a state as a density matrix, just like everybody else, because vectors contain an ambiguous phase factor. That has nothing to do with the quote you cited. According to Ballentine, a pure state is a density matrix (or a ray if you want) in both cases.

So it seems that distinction between "state" as a vector or as an operator comes bundled with different interpretations.

Maybe it seems like that to you, but quantum physicists universally don't agree with your personal opinion. And it's quite despicable that you're trying to put words in the mouth of a respected physicists just in order to defend your lost position.

 

[zonde]

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.

It's a "pure state" in both cases and you are free to write it as a matrix or a vector, since both notions are completely equivalent.

Ballentine correctly defines a state as a density matrix, just like everybody else, because vectors contain an ambiguous phase factor.

 

[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.