Violation of Bell Inequality with unentangled photons

[martinbn]

Thanks for stepping in. I hoped that there was something stronger that I was missing. Trying to perform Bell tests on the original EPR set up is a hard task, but there is no argument to discard it right out of the bat.

Difficult as an actual experiment? That is not interesting to me. Or dificult to do theoretically?

 

[pines-demon]

Difficult as an actual experiment? That is not interesting to me. Or dificult to do theoretically?

Experiment.

 

[gentzen]

Also why take a different sign convention to mine?

Because starting with |p>|-p> corresponds to the EPR description of the experiment (the total momentum is zero). In principle, one should be able to compute the position representation from that. I just guessed that it would be the integral over |x>|x>, because your position and moment representations had different signs too, and because it would make sense that both particles get created at the same position.

 

[gentzen]

What is the issue with that? just take a finite time after that? Why are you focusing on that point?

Because the spin in the normal "measurements showing Bell's inequality violations" is constant in time. The position is not, which on the one hand gives the possibility to measure "some mixture between momentum and position at creation time". And on the other hand, it constitutes an important difference to polarization/spin based experiments.

You could instead start with some weird potential where that is its prepared ground-state, at least to first approximation.

But there is no potential in the EPR description of the experiment. And a potential would indeed turn the continuous states of the EPR experiment into discrete states, which would already be closer to polarization/spin based experiments.

 

[pines-demon]

Because starting with |p>|-p> corresponds to the EPR description of the experiment (the total momentum is zero). In principle, one should be able to compute the position representation from that. I just guessed that it would be the integral over |x>|x>, because your position and moment representations had different signs too, and because it would make sense that both particles get created at the same position.

Fine take $|p\rangle|-p\rangle$ what is the problem of being in the same position? We can figure out the spin or use a boson.

 

[gentzen]

Fine take $|p\rangle|-p\rangle$ what is the problem of being in the same position?

It just makes it easy for me to see that the state is not time invariant.
With the signs of your state, I would have been a bit less sure, even so I guess that it won't be time invariant either.

We can figure out the spin or use a boson.

No, I had no problem with being in the same position in that sense. I just wanted to nail down the details of the experiment and its mathematical description.

 

[pines-demon]

It just makes it easy for me to see that the state is not time invariant.
With the signs of your state, I would have been a bit less sure, even so I guess that it won't be time invariant either.


No, I had no problem with being in the same position in that sense. I just wanted to nail down the details of the experiment and its mathematical description.

Ok lets try. Prepare your state in whatever version you prefer $\int \mathrm d x\, |x\rangle|\pm x\rangle$. Now you have two options either you keep continuous variables and then you need a continuous variables extension of the Bell inequalities (there are various approaches). Or you bin, by binning you just look at the sign of the observable. You have $S_x$ as the sign of $x$, $S_p$ as the sign of $p$ and you do a CHSH test where your spin observables are replaced by $S_{\theta}=S_x\cos\theta+S_p\sin\theta$ for a given abstract angle $\theta$.

Edit: missing integral

 

[gentzen]

Ok lets try. Prepare your state in whatever version you prefer $|x\rangle|\pm x\rangle$. Now you have two options either you keep continuous variables and then you need a continuous variables extension of the Bell inequalities (there are various approaches). Or you bin, by binning you just look at the sign of the observable. You have $S_x$ as the sign of $x$, $S_p$ as the sign of $p$ and you do a CHSH test where your spin observables are replaced by $S_{\theta}=S_x\cos\theta+S_p\sin\theta$ for a given abstract angle $\theta$.

If you define it like this, then $S_x$ and $S_p$ are already matrices (or rather operators), basically $S_x=\int_0^\infty dx\, |x\rangle\langle x|$ and $S_p=\int_0^\infty dp\, |p\rangle\langle p|$. And your $S_{\theta}$ then doesn't contain interferences between $|x\rangle$ and $|p\rangle$ like you have in polarization/spin based Bell inequality violation experiments.

 

[PeterDonis]

Ok lets try.

You need to prepare an entangled state, as @gentzen has pointed out. For example, a momentum entangled state could be (ignoring normalization) $\ket{p} \ket{-p} - \ket{-p} \ket{p}$. This would describe a state in which you don't know which particle went in which direction (because the particles are indistinguishable--in this case, as should be evident from what I wrote down, they're fermions), but you know the total momentum is zero.

 

[pines-demon]

You need to prepare an entangled state, as @gentzen has pointed out. For example, a momentum entangled state could be (ignoring normalization) $\ket{p} \ket{-p} - \ket{-p} \ket{p}$. This would describe a state in which you don't know which particle went in which direction (because the particles are indistinguishable--in this case, as should be evident from what I wrote down, they're fermions), but you know the total momentum is zero.

We were writing in short form, the initial state has an integral $\int \mathrm d x\, |x\rangle |\pm x\rangle$. (I will edit it)

Also it is not clear that you have fermions or not with your state because we don't know what is going on with spin. Not that it matters.

 

[pines-demon]

If you define it like this, then $S_x$ and $S_p$ are already matrices (or rather operators), basically $S_x=\int_0^\infty dx\, |x\rangle\langle x|$ and $S_p=\int_0^\infty dp\, |p\rangle\langle p|$. And your $S_{\theta}$ then doesn't contain interferences between $|x\rangle$ and $|p\rangle$ like you have in polarization/spin based Bell inequality violation experiments.

Can you be more explicit? What do you mean that it does not contain interferences?

 

[PeterDonis]

We were writing in short form

Even in "short form", your state was not entangled. It was a product state. That's obvious from what you wrote.

it is not clear that you have fermions or not with your state because we don't know what is going on with spin.

Assume the particles are spin zero. Then the state I wrote is indeed a fermion state.

What do you mean that it does not contain interferences?

He means just what I told you: the state you wrote is not entangled. It's a product state.

I think you need to take a step back and think through what you're proposing. You're right that it is possible to construct an entangled state in configuration space (for example, as I said above, using spin zero particles whose only degrees of freedom are the position/momentum ones), and then to specify measurements that would demonstrate the equivalent in configuration space of the Bell inequalities that Bell wrote down for spin. But you're not doing it correctly.

 

[pines-demon]

Even in "short form", your state was not entangled. It was a product state. That's obvious from what you wrote.

He means just what I told you: the state you wrote is not entangled. It's a product state.

I think you need to take a step back and think through what you're proposing. You're right that it is possible to construct an entangled state in configuration space (for example, as I said above, using spin zero particles whose only degrees of freedom are the position/momentum ones), and then to specify measurements that would demonstrate the equivalent in configuration space of the Bell inequalities that Bell wrote down for spin. But you're not doing it correctly.

I think you misunderstood, see the edit to post #37, are you telling me that you can write
$$|\psi\rangle=\int \mathrm d x |x\rangle |x\rangle$$
as the product of two states? How?

Assume the particles are spin zero. Then the state I wrote is indeed a fermion state.

Spin zero fermions?

 

[PeterDonis]

Spin zero fermions?

Fair enough, replace my minus sign with a plus sign and make them bosons.

 

[PeterDonis]

I think you misunderstood see the edit to post #37

I see, basically you're constructing a state in which two particles are known to be at the exact same position, but we don't know what that position is--it could be anywhere. Yes, that's an entangled state.

However, since "total position" isn't a conserved quantity, I'm not sure you can construct a Bell inequality based on it. Doing it with momentum would be better because total momentum is a conserved quantity.

 

[gentzen]

Can you be more explicit? What do you mean that it does not contain interferences?

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

 

[PeterDonis]

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

How does it even make sense to combine position and momentum eigenkets? I don't understand what you're doing here.

 

[gentzen]

How does it even make sense to combine position and momentum eigenkets? I don't understand what you're doing here.

You mean, because they are not normalizable? But they are well defined in a suitable space of distributions, and one can form their complex superposition in that space. Less sure about the $|x\rangle\langle x|$ part. Products of distributions can be nasty, and taking the dual is also dangerous.

But the point of that reply are the $\text{interference terms}$, which were unclear to pines-demon.

 

[PeterDonis]

You mean, because they are not normalizable?

No.

the point of that reply are the ,

Interference between position and momentum eigenkets? What does that mean? I don't understand.

To me the key point is whether or not the state is entangled. "Interference" is not the same thing.

 

[gentzen]

To me the key point is whether or not the state is entangled. "Interference" is not the same thing.

The state is entangled, but pines-demon tried to get rid of the continuous measurement outcomes, to be closer to typical Bell measurements. I told him before

... noticed that position and momentum give continuous results, but "measurements showing Bell's inequality violations" work with binary results. So you suggest binning. My guess it that binning will probably destroy the delicate quantum entanglement. ...

So when pines-demon came up with a sufficiently well specified experiment with binning, I tried to explain to him why that specific experiment won't show Bell inequality violations.

 

[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?)