Undergrad 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.