I Violation of Bell Inequality with unentangled photons

[pines-demon]

$$(\alpha|x\rangle+\beta|p\rangle)(\alpha^*\langle x|+\beta^*\langle p|)=|\alpha|^2|x\rangle\langle x|+|\beta|^2|p\rangle\langle p|+\text{interference terms}$$
$$\text{interference terms} = \alpha\beta^*|x\rangle\langle p| +\alpha^*\beta|p\rangle\langle x|$$

Is this related to the state? to the observables or to what? I am lost here. There are clearly interference terms in whatever you are doing there.

 

[PeterDonis]

pines-demon tried to get rid of the continuous measurement outcomes, to be closer to typical Bell measurements

I agree that you can't just bin things into binary measurements. I'm not sure exact continuity is required. For example, with a momentum entangled state, I think you could have a spherical array of momentum detectors with finite angular resolution; you wouldn't need exact, infinite angular resolution. You'd just need a lot more than only two such detectors (one per hemisphere) to spot Bell inequality violations, since you'd have to make measurements at various relative angles, not just 180 degrees.

 

[gentzen]

Is this related to the state? to the observables or to what? I am lost here. There are clearly interference terms in whatever you are doing there.

This is an observable. Your $S_x$ and $S_p$ were observables too. Maybe to clarify, we have $\alpha=\cos(\theta)$ and $\beta=\sin(\theta)$ in your case.

 

[pines-demon]

This is an observable. Your $S_x$ and $S_p$ were observables too. Maybe to clarify, we have $\alpha=\cos(\theta)$ and $\beta=\sin(\theta)$ in your case.

Isn’t $\alpha\beta=\cos\theta\sin\theta$ different than 0 for most angles?

 

[gentzen]

Isn’t $\alpha\beta=\cos\theta\sin\theta$ different than 0 for most angles?

Yes, of course.

I guess what you didn't understand is that your $|\alpha|^2S_x+|\beta|^2S_p$ is missing those interference terms. That was my point, why I showed you an observable where those terms are not missing.

 

[pines-demon]

Yes, of course.

I guess what you didn't understand is that your $|\alpha|^2S_x+|\beta|^2S_p$ is missing those interference terms. That was my point, why I showed you an observable where those terms are not missing.

This is getting very cryptic. I take $S_\theta=S_x\cos\theta+S_p\sin\theta=\alpha S_x+\beta S_p$, I do what then? $S_\theta^2$ or $S_\theta S_\theta^\dagger$? Either way I get interference term right? Is not that different to your calculation. What am I missing?

 

[gentzen]

What am I missing?

You are missing

If you define it like this, then $S_x$ and $S_p$ are already matrices (or rather operators), basically ...

Your sign measurement operators $S_x$ and $S_p$ are no longer representable by a single state, hence you have to represent them as matrices/operators. But forming the superposition of matrices doesn't give you interference. You only get interference, if you form the superposition of two single states, which represented two measurements.

..., basically $S_x=\int_0^\infty dx\, |x\rangle\langle x|$ and $S_p=\int_0^\infty dp\, |p\rangle\langle p|$.

If you want to be more accurate, it would be $S_x=\int_{-\infty}^\infty dx\, sign(x)|x\rangle\langle x|$. But the crucial part is that $|x\rangle\langle x|$ got already formed.

 

[javisot]

In the EPR paper, the entangled particles cannot be used for Bell's inequality violations, and it was the fist paper about entanglement.

If this were correct then EPR is not a quantum experiment, but this is not the case.

 

[pines-demon]

You are missing

Who is being cryptic?

Your sign measurement operators $S_x$ and $S_p$ are no longer representable by a single state, hence you have to represent them as matrices/operators. But forming the superposition of matrices doesn't give you interference.

1) Please be explicit on what is the calculation to perform to check if there is interference.
2) Why is it needed?

 

[javisot]

I have a naive doubt about Bell's inequalities.

Let us suppose an experiment (X) that can be explained with the same precision with both classical and quantum mechanics. Let us also assume that in the quantum version there is a violation of Bell's inequalities.

Should Bell's inequalities be violated in the classical version?

(I understand that by definition it is not possible, right?)

 

[PeterDonis]

Let us suppose an experiment (X) that can be explained with the same precision with both classical and quantum mechanics. Let us also assume that in the quantum version there is a violation of Bell's inequalities.

Your two assumptions are inconsistent. If a classical model matches the experimental results, the experimental results cannot violate the Bell inequalities.

 

[pines-demon]

I have a naive doubt about Bell's inequalities.

Let us suppose an experiment (X) that can be explained with the same precision with both classical and quantum mechanics. Let us also assume that in the quantum version there is a violation of Bell's inequalities.

Should Bell's inequalities be violated in the classical version?

(I understand that by definition it is not possible, right?)

This is not possible. If the classical and quantum set ups give similar results then neither of them violates Bell inequalities. Most quantum setups do not violate Bell inequalities, it is only for specific states.

 

[gentzen]

Who is being cryptic?

Maybe you should also start to react with "of course" from time to time, otherwise I have little idea where I lost you, or where we disagree. Take for example

If you define it like this, then $S_x$ and $S_p$ are already matrices (or rather operators),

Just answer something like: "Of course, my $S_x$ and $S_p$ are operators. Observables in QM are self-adjoint operators after all." Or also: "No, I disagree. My $S_p$ is exactly analogous to $S_z$ in spin measurements."

basically $S_x=\int_0^\infty dx\, |x\rangle\langle x|$ and $S_p=\int_0^\infty dp\, |p\rangle\langle p|$.

also here "Yes, basically. But my $S_x$ has eigenvalues 1 and -1, while yours has 1 and 0."

If you want to be more accurate, it would be $S_x=\int_{-\infty}^\infty dx\, sign(x)|x\rangle\langle x|$. But the crucial part is that $|x\rangle\langle x|$ got already formed.

or here: "Yes, I agree." Or if you prefer: "Why do you bother me with this. Why is it important?"

This is getting very cryptic. I take $S_\theta=S_x\cos\theta+S_p\sin\theta=\alpha S_x+\beta S_p$, I do what then?

Looks to me like you think I am being cryptic. Correct?

If you have an observable $S_\theta$ and want to check whether it is analogous to the spin observables in Bell measurements, I guess you should check whether its eigenvalues are +1 and -1.

$S_\theta^2$ or $S_\theta S_\theta^\dagger$? Either way I get interference term right? Is not that different to your calculation. What am I missing?

No, you can compute $S_\theta^2$ or $S_\theta S_\theta^\dagger$ if you want, but it doesn't seem relevant to anything we are discussing here.

1) Please be explicit on what is the calculation to perform to check if there is interference.
2) Why is it needed?

The "Why is it needed?" is a good question. For the Bell experiment, the more relevant question would be to check the spectrum/eigenvalues of your $S_\theta$. Even so the spectrum/eigenvalues are hard to compute, checking a specific eigenvalue is doable, "in principle".

What I mean by "interference" is terms which can have a negative sign. Say if I have light in a horizontally homogeneous layer, I will often first compute amplitudes $A_{up}$ and $A_{down}$. If I am only interested in the general intensity distribution without standing waves, I could compute $I_{up}+I_{down}$ with $I_{up}=|A_{up}|^2$ and $I_{down}=|A_{down}|^2$. If the light is perfectly monochromatic and I am interested in the standing waves, I could compute $I_{up}+I_{down}+2\operatorname{Re}(A_{up}^*A_{down})$. The first two terms are non-negative, hence it is the third term which can have a negative sign. (But of course, it goes in both directions, because there is also constructive interference.)

 

[pines-demon]

Maybe you should also start to react with "of course" from time to time, otherwise I have little idea where I lost you, or where we disagree. Take for example

or here: "Yes, I agree." Or if you prefer: "Why do you bother me with this. Why is it important?"

Let's try, but I am not going to go point by point validating your arguments.

Just answer something like: "Of course, my $S_x$ and $S_p$ are operators. Observables in QM are self-adjoint operators after all." Or also: "No, I disagree. My $S_p$ is exactly analogous to $S_z$ in spin measurements."

also here "Yes, basically. But my $S_x$ has eigenvalues 1 and -1, while yours has 1 and 0."

Looks to me like you think I am being cryptic. Correct?

If you have an observable $S_\theta$ and want to check whether it is analogous to the spin observables in Bell measurements, I guess you should check whether its eigenvalues are +1 and -1.

The "Why is it needed?" is a good question. For the Bell experiment, the more relevant question would be to check the spectrum/eigenvalues of your $S_\theta$. Even so the spectrum/eigenvalues are hard to compute, checking a specific eigenvalue is doable, "in principle".

(Of course, but) I was not sure where the 1/0 vs 1/-1 will play. Should I consider that you are putting traps in your arguments to see if I get them?

No, you can compute $S_\theta^2$ or $S_\theta S_\theta^\dagger$ if you want, but it doesn't seem relevant to anything we are discussing here.

I proposed examples to try to pin down the calculation.

What I mean by "interference" is terms which can have a negative sign. Say if I have light in a horizontally homogeneous layer, I will often first compute amplitudes $A_{up}$ and $A_{down}$. If I am only interested in the general intensity distribution without standing waves, I could compute $I_{up}+I_{down}$ with $I_{up}=|A_{up}|^2$ and $I_{down}=|A_{down}|^2$. If the light is perfectly monochromatic and I am interested in the standing waves, I could compute $I_{up}+I_{down}+2\operatorname{Re}(A_{up}^*A_{down})$. The first two terms are non-negative, hence it is the third term which can have a negative sign. (But of course, it goes in both directions, because there is also constructive interference.)

Of course, I do get what interference is about. To me that seems like the usual definition of interference, no issues here. Let's say that we are in the usual spin Bell test, can you specify what is your $A_{up}$? It is not catchy question in anyway but you keep bringing interference without specifying why it fails here, a specific calculation (you don't have to do the calculation just specify it) would be more helpful.

 

[gentzen]

(Of course, but) I was not sure where the 1/0 vs 1/-1 will play. Should I consider that you are putting traps in your arguments to see if I get them?

This again tell me more about your image from me, than about where I might have misinterpreted you, where I might have lost you, or where we might disagree.

No, I am not putting traps in my arguments. I am just lazy like everybody else, make mistakes from time to time, like (nearly) everybody, and generally simply try to communicate.

Especially if I try to compute your "proposed example" in detail, I risk to make many minor or major mistakes. For example, I did some computations for your example yesterday, many with results more complicated than I would have preferred. My simplest results was $S_x|p\rangle = 2i \int \frac{dp'}{p-p'}|p'\rangle$. Is it correct? I guess! But maybe the factor $2i$ is wrong, who knows, or I mixed-up some signs?

(I also noticed that you quoted many of my equations above, but didn't manage to write "OK" or "I agree" for even a single one. Looks like our discussion feels more like a fight to you at the moment. What can I do to change that?)

 

[Demystifier]

Let me give a simple explanation how is symmetrization (or anti-symmetrization) due to indistinguishability of particles different from entanglement in a narrow sense.

Consider the symmetrized state
$$|\psi\rangle = \frac{1}{\sqrt{2}}[|A\rangle|B\rangle+|B\rangle|A\rangle]$$
If we think of it as an entangled state of distinguishable particles, then, upon measurement, the state can collapse into a non-symmetrized state, e.g.
$$|\psi\rangle \to |A\rangle|B\rangle$$
It is such collapses upon measurement that give rise to measurable correlations that can violate Bell inequalities or produce other effects that cannot be interpreted in terms of local hidden variables. That's what is interesting about entanglement in a narrow sense.

On the other hand, if we think of $|\psi\rangle$ as a state of indistinguishable particles, then the two particles must always be in a symmetrized state, so a collapse into a non-symmetrized state is impossible. This removes a large set of possibilities for nontrivial measurable correlations.

For example, in the original EPR argument we have particles without spin with entangled momenta
$$|\psi\rangle = \frac{1}{\sqrt{2}}[|p\rangle|-p\rangle+|-p\rangle|p\rangle]$$
If we measure the momentum of the first particle and find the value $p$, then we know that the momentum of the second particle is $-p$. Then we can also measure the position of the second particle, so assuming that the act of measurement cannot change reality due to locality, EPR conclude that we can know simultaneously both momentum and position, which makes QM incomplete. But this argument does not work if the particles are indistinguishable. Namely, a measurement by which we find that the first and the second particle have momenta $p$ and $-p$, respectively, corresponds to the collapse
$$|\psi\rangle \to |p\rangle|-p\rangle$$
which is impossible because the final state is not symmetrized. (And if we symmetrize it, we are turning back to the original state $|\psi\rangle$ before the collapse.) Thus the EPR argument in its original form does not work for indistinguishable particles.

 

[pines-demon]

This again tell me more about your image from me, than about where I might have misinterpreted you, where I might have lost you, or where we might disagree.

I proposed a state and some operators. You replied by saying that there is no interference in that case. I asked what interference, you kept bringing interference and definitions of interference but I still do not know why is it important or how to check that interference is not there based on the state and the operators.

No, I am not putting traps in my arguments. I am just lazy like everybody else, make mistakes from time to time, like (nearly) everybody, and generally simply try to communicate.

Thus no need to bring a mistake if it is unnecessary to the discussion so far.

Especially if I try to compute your "proposed example" in detail, I risk to make many minor or major mistakes. For example, I did some computations for your example yesterday, many with results more complicated than I would have preferred. My simplest results was $S_x|p\rangle = 2i \int \frac{dp'}{p-p'}|p'\rangle$. Is it correct? I guess! But maybe the factor $2i$ is wrong, who knows, or I mixed-up some signs?

I agree, I get the same if I use $S_x=\int \mathrm dx\, \mathrm{sign}(x) |x\rangle \langle x|$.

(I also noticed that you quoted many of my equations above, but didn't manage to write "OK" or "I agree" for even a single one.

I agreed with your definition of interference if that is what your are asking. Did I miss any equation?

Looks like our discussion feels more like a fight to you at the moment. What can I do to change that?)

I am sorry if that looks that way, I am trying hard to pin down your counterarguments before moving on with a more specific analysis.

 

[pines-demon]

which is impossible because the final state is not symmetrized. (And if we symmetrize it, we are turning back to the original state $|\psi\rangle$ before the collapse.) Thus the EPR argument in its original form does not work for indistinguishable particles.

Are you saying that the original EPR does not work because particles are spinless (and indistinguishable)?

 

[gentzen]

I am trying hard to pin down your counterarguments before moving on with a more specific analysis.

I had misinterpreted what you wanted to do with your $S_x$ and $S_p$: Your intention was to use them like the spin measurement operators $S_z:= \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$ and $S_x:= \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$, where you form $S_\theta:= S_z \cos \theta+S_x \sin \theta$. Then $S_\theta =\begin{bmatrix} \cos \theta & \sin \theta \\ \sin \theta & -\cos \theta \end{bmatrix}$, which has trace 0 and determinant -1, hence eigenvalues +1 and -1. This means that you should be totally happy that they "are already matrices (or rather operators)". But you never corrected my misinterpretation of your intentions. You just waited until I found out myself, when you wrote "2) Why is it needed?".

So for your interpretation of what you want to do, the important question is whether your operator $S_\theta$ has only eigenvalues +1 and -1 like in the case of spin measurements. If not, and your operator $S_\theta$ has more than two eigenvalues or even a continuous spectrum, then you cannot directly reproduce what one does for spin measurements with it.

 

[pines-demon]

I had misinterpreted what you wanted to do with your $S_x$ and $S_p$: Your intention was to use them like the spin measurement operators $S_z:= \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$ and $S_x:= \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$, where you form $S_\theta:= S_z \cos \theta+S_x \sin \theta$. Then $S_\theta =\begin{bmatrix} \cos \theta & \sin \theta \\ \sin \theta & -\cos \theta \end{bmatrix}$, which has trace 0 and determinant -1, hence eigenvalues +1 and -1. This means that you should be totally happy that they "are already matrices (or rather operators)". But you never corrected my misinterpretation of your intentions. You just waited until I found out myself, when you wrote "2) Why is it needed?".

I thought that was clear, I did not wait until you found out, more like I was discussing without knowing that you got something off (I still do not know what was your previous interpretation, is this related to your point on interferences?).

So for your interpretation of what you want to do, the important question is whether your operator $S_\theta$ has only eigenvalues +1 and -1 like in the case of spin measurements. If not, and your operator $S_\theta$ has more than two eigenvalues or even a continuous spectrum, then you cannot directly reproduce what one does for spin measurements with it.

You have a point here. Let me think about it before I propose another $S_\theta$.

 

[Demystifier]

Are you saying that the original EPR does not work because particles are spinless (and indistinguishable)?

Not at all, it works for particles which are distinguishable. For example, one particle can be a hydrogen atom made of a proton and an electron, so that the total spin is zero. The other particle can be a helium atom, also with total spin zero. The full system is then a hydrogen atom and a helium atom, which are distinguishable from each other.

 

[pines-demon]

In my defense, this is not an isolated problem. The problem of Bell-testing the original EPR state has been discussed in research. Bell wrote a proof in Speakable and unspeakable in quantum mechanics, suggesting that the original EPR state has a non-negative Wigner function so it can be reproduced with a local hidden variables model. This is not necessarily true, it has been challenged, see https://https://doi.org/10.1103/PhysRevA.71.022103. Also there are many version of continuous variable Bell inequalities out there, see for example https://journals.aps.org/pra/abstract/10.1103/PhysRevA.67.012105.

 

[pines-demon]

Not at all, it works for particles which are distinguishable. For example, one particle can be a hydrogen atom made of a proton and an electron, so that the total spin is zero. The other particle can be a helium atom, also with total spin zero. The full system is then a hydrogen atom and a helium atom, which are distinguishable from each other.

What about two electrons? Then you don't have issues because spins handle the symmetrization part.

 

[Morbert]

What about two electrons? Then you don't have issues because spins handle the symmetrization part.

As an aside: When we write Bell states like $$\ket{\psi} = \frac{1}{\sqrt{2}}(\ket{\uparrow}_1\ket{\downarrow}_2 + \ket{\downarrow}_1\ket{\uparrow}_2)$$The labels 1,2 are not strictly particle labels, but instead addressable mode labels. I.e. A spin-measurement doesn't tell us something like "electron 1 had spin up and electron 2 has spin down" but rather, "The electron in the mode addressable by apparatus 1 yielded spin up and the electron in the mode addressable by apparatus 2 yielded (or will yield) spin down."

Hence, indistinguishable particles like electrons in a mode framework have an "effective distinguishability" that yields Bell-inequality-violating correlations with the appropriate experiments.