I Is Polarisation Entanglement Possible in Photon Detection?

[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.

 

[stevendaryl]

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.

I'm sorry, I don't understand what you mean. What's not consistent?

 

[A. Neumaier]

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.

The underlying assumption here is that the single system has a true state, and that this state is pure. Both are unprovable assumptions.

In general, if someone has a wrong state due to ignorance he will make wrong predictions of the full observable statistics. Ignorance therefore has no place in physics - Nature behaves independent of what we choose to know or ignore.

 

[vanhees71]

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.

I agree, of course, with that. Maybe, I've misunderstood your previous posting.

 

[stevendaryl]

The underlying assumption here is that the single system has a true state, and that this state is pure. Both are unprovable assumptions.

Nothing in science is provable, but the assumption that it is possible for Alice to produce a pure spin-up state for Bob is empirically testable, in the sense that Alice can repeat the same experiment over and over and note how often it is that when she prepares a state that is spin-up in the z-direction, Bob measures spin-up in the z-direction. The hypothesis that it is a pure spin-up state can be tested, and all tests are consistent with that assumption.

 

[stevendaryl]

In general, if someone has a wrong state due to ignorance he will make wrong predictions of the full observable statistics. Ignorance therefore has no place in physics - Nature behaves independent of what we choose to know or ignore.

You're contradicting yourself here. By definition, an improper mixture does reflect ignorance about the true state.

And you're wrong that ignorance has no place in physics. The use of probability gives us a way to reason in the presence of uncertainty/ignorance.

Anyway, you're venturing into philosophy here, and I don't find your philosophy of science very compelling. Let's stick to the physics. The situation in which Alice flips a coin and sends a spin-up state to Bob if her coin is "heads" and sends a spin-down state if her coin is "tails" is certainly a different situation than the case where Bob measures the spin of one member of an entangled two-particle system. In the first case, Alice knows ahead of time what result Bob will get, and in the second case, she doesn't. Those are clearly different situations. But from Bob's point of view, they are both described by the mixed state with equal weights of spin-up and spin-down. In the first case, the mixture reflects Bob's ignorance and in the second it does not. You say "ignorance has no place in physics", but I think that's silly.

 

[stevendaryl]

My pet peeve is that often when people say "Philosophy has no place in physics", what they really mean is "No philosophy that is in disagreement with mine has a place in physics".

 

[zonde]

I'm sorry, I don't understand what you mean. What's not consistent?

Let's say Alice receives photon, she argues that it is either H polarized or V polarized. Now she measures it at some angle $\alpha$ and gets click in either channel#1 behind PBS or the channel#2. Let's say it was channel#1. Now she says, well it was either H photon that with probability $p_1$ went into channel#1 or it was V photon that with probability $p_2$ went into channel#1. Bob does the same with angle $\beta$. Now we (who observe both Alice and Bob doing their reasoning) argue that either Alice and Bob both got H photon (say we created entangled state with correlated polarizations) or both got V photons. We try out every possibility with that condition in place and there is none that is consistent with QM predictions for pure state probabilities (expected $p_1$ and $p_2$ for Alice and Bob). We argue that: well, maybe we picked wrong pure states. They were actually not H and V photons but say +45deg./-45deg. photons. But it turns out that if Alice and Bob would use different angles $\alpha_2$ and $\beta_2$ we end up with the same conclusion that we can't get pure state probabilities right whatever we try.

 

[Simon Phoenix]

So I don't think it should be too surprising that your altered scenario can violate Bell's inequalities

Indeed it isn't - it surprised me way back when when we first realized it and applied it to generate a QKD protocol, but that's because I had been unthinkingly applying the mantra "violation = entanglement".

I certainly think the 'easiest' way to understand all of the QKD work is in terms of collapse - and it's the way I approach the problem being confident that I'm not going to get the wrong predictions. Of course we can interpret all of the QKD stuff in whatever 'interpretation' we like and get the same answers. Speaking purely for myself I find the 'collapse' way of thinking about it to be cleaner and to leave aside all of the legitimate concerns about FTL 'changes of state' that it implies to a lovely Summer sunset with a glass or two of beer

 

[stevendaryl]

Let's say Alice receives photon, she argues that it is either H polarized or V polarized. Now she measures it at some angle $\alpha$ and gets click in either channel#1 behind PBS or the channel#2. Let's say it was channel#1. Now she says, well it was either H photon that with probability $p_1$ went into channel#1 or it was V photon that with probability $p_2$ went into channel#1. Bob does the same with angle $\beta$. Now we (who observe both Alice and Bob doing their reasoning) argue that either Alice and Bob both got H photon (say we created entangled state with correlated polarizations) or both got V photons. We try out every possibility with that condition in place and there is none that is consistent with QM predictions for pure state probabilities (expected $p_1$ and $p_2$ for Alice and Bob). We argue that: well, maybe we picked wrong pure states. They were actually not H and V photons but say +45deg./-45deg. photons. But it turns out that if Alice and Bob would use different angles $\alpha_2$ and $\beta_2$ we end up with the same conclusion that we can't get pure state probabilities right whatever we try.

Sorry, I still don't understand what you are talking about. What do you mean when you say "none that is consistent with QM predictions for pure state probabilities"?

In the entangled two-photon state, you have the state \frac{1}{\sqrt{2}} (|H\rangle |H\rangle + |V\rangle |V\rangle). We do the trace business and Alice uses the mixed state \frac{1}{2} |H\rangle \langle H| + \frac{1}{2} |V\rangle \langle V|. Bob uses the same mixed state. The meaning is that if Alice measures the polarization, she'll get H with probability 50% and V with probability 50%. Actually, that density matrix says that if Alice measures the polarization at ANY angle, she will get H with probability 50% and V with probability 50%. Same with Bob.

Now, if you ask Alice what the probability is that she and Bob get the same polarization, she can't answer that question using her's and Bob's density matrices, because those have thrown away correlation information.

So what inconsistency are you talking about? Do you mean an inconsistency of the type: "The theory says we will find X, but in actually X is not the case."

 

[stevendaryl]

Indeed it isn't - it surprised me way back when when we first realized it and applied it to generate a QKD protocol, but that's because I had been unthinkingly applying the mantra "violation = entanglement".

I certainly think the 'easiest' way to understand all of the QKD work is in terms of collapse - and it's the way I approach the problem being confident that I'm not going to get the wrong predictions. Of course we can interpret all of the QKD stuff in whatever 'interpretation' we like and get the same answers. Speaking purely for myself I find the 'collapse' way of thinking about it to be cleaner and to leave aside all of the legitimate concerns about FTL 'changes of state' that it implies to a lovely Summer sunset with a glass or two of beer

Yes, this reminds me of philosophical discussions about the nature of mathematical objects. Platonism is the position that mathematical objects (such as numbers or sets or functions) exist independently of humans, and when we do mathematics, we are just discovering pre-existing truths. When people are seriously talking about the philosophy of mathematics, they tend to reject platonism as silly: What does it mean that these things exist? Where do they exist? But when you're actually doing mathematics, Platonism seems as good a philosophy as any, and it allows you to get on with your work without worrying too much about the deeper meaning of what it is that you are doing.

The collapse interpretation is almost universally rejected by people who think deeply about physics. But if you're just doing physics, and wanting to get answers that you can compare with experiment, then the collapse interpretation works as well as any.

 

[zonde]

Sorry, I still don't understand what you are talking about. What do you mean when you say "none that is consistent with QM predictions for pure state probabilities"?

In the entangled two-photon state, you have the state \frac{1}{\sqrt{2}} (|H\rangle |H\rangle + |V\rangle |V\rangle). We do the trace business and Alice uses the mixed state \frac{1}{2} |H\rangle \langle H| + \frac{1}{2} |V\rangle \langle V|. Bob uses the same mixed state. The meaning is that if Alice measures the polarization, she'll get H with probability 50% and V with probability 50%. Actually, that density matrix says that if Alice measures the polarization at ANY angle, she will get H with probability 50% and V with probability 50%. Same with Bob.

Now, if you ask Alice what the probability is that she and Bob get the same polarization, she can't answer that question using her's and Bob's density matrices, because those have thrown away correlation information.

So what inconsistency are you talking about? Do you mean an inconsistency of the type: "The theory says we will find X, but in actually X is not the case."

Can we speak about mixed state before Alice actually makes her measurement? I suppose so. Let's say we prepare for Alice beam of light that for first 5 seconds is H polarized and for next 5 seconds we switch polarization to V. If she measures her beam with polarizer at an angle $\alpha$ then for first 5 seconds her "click" rate will be say $p_1$ and for next 5 seconds it will be $p_2$. And $p_1+p_2$ gives probability 0.5 of the whole expected photon count.

 

[stevendaryl]

Can we speak about mixed state before Alice actually makes her measurement? I suppose so. Let's say we prepare for Alice beam of light that for first 5 seconds is H polarized and for next 5 seconds we switch polarization to V. If she measures her beam with polarizer at an angle $\alpha$ then for first 5 seconds her "click" rate will be say $p_1$ and for next 5 seconds it will be $p_2$. And $p_1+p_2$ gives probability 0.5 of the whole expected photon count.

Okay, but I don't understand what this scenario is supposed to be illustrating.

 

[zonde]

Okay, but I don't understand what this scenario is supposed to be illustrating.

These $p_1$ and $p_2$ agree (within some limits) with predicted probabilities of QM for pure states. This does not work within in my extended model of Alice and Bob i.e. you can't get similar $p_1$ and $p_2$ right.

 

[stevendaryl]

These $p_1$ and $p_2$ agree (within some limits) with predicted probabilities of QM for pure states. This does not work within in my extended model of Alice and Bob i.e. you can't get similar $p_1$ and $p_2$ right.

Something is not clicking for me. I don't know what you mean by "you can't get similar p_1 and p_2".

Maybe a little elaboration on the meaning of density matrices. Although there is something a little subjective about density matrices (they include uncertainty about the true state), if you have a way to reproduce the same situation over and over again, then it is possible by statistics to zero in on a precise density matrix. If you get statistics for measurement results at a variety of filter orientations, there will only be one density matrix consistent with those statistics (I'm pretty sure). In the scenario you're talking about, the procedure for producing photons is changing with time, so I don't think that there will be a unique density matrix that can be discovered empirically (unless the time dependence is periodic, and the same pattern repeats over and over). So in terms of repeated measurements, you can distinguish between a pure state and a mixed state. But what you can't distinguish, empirically, is proper versus improper mixed states. Statistics for measuring at a variety of filter orientations is not going to tell you whether you have a proper mixed state (a pure state that is chosen randomly, with the same probabilities, over and over) and an improper mixed state (due to looking at one component of an entangled pure state).

There definitely is a distinction between the pure state \frac{1}{\sqrt{2}} |H\rangle + \frac{1}{\sqrt{2}} |V\rangle and the mixed state \frac{1}{2} |H\rangle \langle H| + \frac{1}{2} |V\rangle\langle V|, and you can distinguish them through statistics. But you have to perform measurements at a variety of orientations to see the difference.

 

[zonde]

But what you can't distinguish, empirically, is proper versus improper mixed states. Statistics for measuring at a variety of filter orientations is not going to tell you whether you have a proper mixed state (a pure state that is chosen randomly, with the same probabilities, over and over) and an improper mixed state (due to looking at one component of an entangled pure state).

Yes certainly.

Something is not clicking for me.

Maybe you are trying to see in my argument more than I'm actually trying to claim. I'm trying to say that improper mixed state can't be modeled as statistical mixture of pure states (proper mixed state) even so observable statistics are the same for both cases (because model goes beyond observed statistics).

I'm thinking about your statement that tracing out one side just strips away information. Maybe you have a point.

 

[rubi]

I'm trying to say that improper mixed state can't be modeled as statistical mixture of pure states (proper mixed state) even so observable statistics are the same for both cases (because model goes beyond observed statistics).

And you're wrong, because it can be modeled that way.

However, it makes no sense to argue as long as you haven't learned the formalism. It can't be understood without knowledge about the quantum formalism.

 

[forcefield]

Maybe you are trying to see in my argument more than I'm actually trying to claim. I'm trying to say that improper mixed state can't be modeled as statistical mixture of pure states (proper mixed state) even so observable statistics are the same for both cases (because model goes beyond observed statistics).

I suggest that you add a physical collapse and you should see that it can be modeled.

 

[zonde]

I suggest that you add a physical collapse and you should see that it can be modeled.

Idea is that you have to specify particular pure states for statistical mixture before Alice and Bob has performed measurements.
Do you still think it can be modeled?

 

[forcefield]

Idea is that you have to specify particular pure states for statistical mixture before Alice and Bob has performed measurements.
Do you still think it can be modeled?

Yes. Alice always measures before Bob or vice versa.

 

[Zafa Pi]

You're contradicting yourself here. By definition, an improper mixture does reflect ignorance about the true state.

And you're wrong that ignorance has no place in physics. The use of probability gives us a way to reason in the presence of uncertainty/ignorance.

Anyway, you're venturing into philosophy here, and I don't find your philosophy of science very compelling. Let's stick to the physics. The situation in which Alice flips a coin and sends a spin-up state to Bob if her coin is "heads" and sends a spin-down state if her coin is "tails" is certainly a different situation than the case where Bob measures the spin of one member of an entangled two-particle system. In the first case, Alice knows ahead of time what result Bob will get, and in the second case, she doesn't. Those are clearly different situations. But from Bob's point of view, they are both described by the mixed state with equal weights of spin-up and spin-down. In the first case, the mixture reflects Bob's ignorance and in the second it does not. You say "ignorance has no place in physics", but I think that's silly.

Hi - I find your comment quite interesting. What if in the 1st case Alice flips the coin but doesn't see it land and instead a machine sends off the appropriate state to Bob. Are cases 1 and 2 still different situations, i.e. the first case reflects Bob's ignorance and in the second it doesn't?

 

[morrobay]

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).

So this single particle inequality violation is just constructed by Alice to demonstrate FTL effect.
I don't understand why more attention is not given to CFD to account for inequality violations : https://arxiv.org/pdf/1605.04889.pdf
Instead of relying on a bizarre feature that has no known physical mechanism.

 

[Simon Phoenix]

Instead of relying on a bizarre feature that has no known physical mechanism.

If we assume that a 'state' is some objective property of a system, and we accept the 'axiom' of quantum mechanics that says if we make a von Neumann measurement of type I (basically what has been called a 'filter' type measurement) then the result of the measurement will be an eigenvalue of the measurement operator and the measured system will be in an eigenstate of the measurement operator after the measurement, then there is also no known physical mechanism that will achieve this.

The only way out of this (that I can see) is to assume that there is no meaning to a system being 'in' a state and the word 'state' means a mathematical quantity that is merely descriptive of our knowledge and not descriptive of some objective physical property of an entity. Measurement is then simply an 'updating' of our knowledge, a bit like (but not exactly like) when we update a probability distribution based on new data (measurements). In the quantum case we're not updating a probability but something from which we can derive probabilities. This doesn't really tell us what a measurement 'is' in physical terms but just describes its effect on our knowledge (whatever that rather vague term actually means). It also doesn't really explain (to my mind, at least) why our 'knowledge' has to be encoded in a mathematical object that evolves according the Schrodinger equation (involving physical things like energy and interactions), lives in an abstract complex space, has such close connections at a deeper level to classical mechanics, and yet is not supposed to model 'reality' in any objective way. I would (grudgingly) agree that this 'knowledge' viewpoint makes more coherent logical sense, but as a physicist it leaves me very unsatisfied because I no longer have any real physical 'picture' of what's happening but must deal with things in a very operational way using vague terms like 'knowledge' or 'what can be known' in order to interpret things.

Decoherence will give us the correct density operator if we remain ignorant of the actual result, but it does not explain why we get a particular pure state after measurement (if we know the result decoherence predicts an incorrect density operator). Of course in order for these comments to make sense we must believe that the formalism of QM applies to single measurements on single systems - and not just to ensembles.

 

[vanhees71]

If you prepare the value of only one observable (e.g., by filtering) then in general you don't know in which eigenstate the system is. Suppose you determine observable $A$ to have the value $a$ and if $|a,b \rangle$ is a complete orthonormal set of eigenvectors, and if we assume for simplicity that $b \in \{1,2,\ldots,n\}$ ("$n$-fold degeneracy"), then you'd rather associate the state
$$\hat{\rho}=\frac{1}{n} \sum_{b=1}^n |a,b \rangle \langle a,b|.$$
That choice is due to the maximum-entropy principle (or the principle of least prejudice) in the sense of information theory. You must not assume something you don't know. You have a pure state if and only if $n=1$.

 

[A. Neumaier]

in the sense that Alice can repeat the same experiment over and over and note how often it is that when she prepares a state that is spin-up in the z-direction, Bob measures spin-up in the z-direction.

This is a test of the preparation of an ensemble, not of a single state.

By definition, an improper mixture does reflect ignorance about the true state.

This assumes that ignorance affects the outcome of physical experiments, and hence the state of a system.

Now I am ignorant about most experiments done in the word, but my ignorance obviously doesn't affect their outcome.

In fact, nobody's ignorance can make a difference since ignorance is a property of the state of a brain while measured is some information about the state of a tiny quantum object usually completely unclupled to any brain. Most measurements are done automatically without anyone observing the details, so there is no distinguished knower whose ignorance might be relevant.

Ignorance about a (pure or mixed) state simply means that when one assigns an arbitrary state for it it is likely to give wrong predictions - unless this arbitrary state happens to be the physical state.

 

[stevendaryl]

This is a test of the preparation of an ensemble, not of a single state.

You're making a philosophical point that I disagree with. To me, if someone flips a coin and hides the result, then I use probabilities to reflect my ignorance about the fine details of the coin-flipping process. In my opinion, bringing up ensembles is unnecessary and unhelpful.