Expected result of such an experiment?

[LsT]

The experiment I am thinking of is simple:

- A typical (ideal) photon polarization quantum entanglement setup with Alice and Bob.
- Alice and Bob detectors are constant at 90 deg. angle difference (such as they allways get perfect corelation with different result)
Plus:
- Between Alice and source we place a polarizer "Aa", in relative angle to Alice 45 deg.
- And we place a detector "Da" between Alice and Aa such as we know when a photon has passed polarizer Aa and it is heading towards Alice.
- We record the results of Alice and Bob, only if Da detects.

(if there is a technical problem of detecting a single photon after Aa without "stoping" it, please consider the experiment with something similar, such as light pulses)

Now what will happen?
Does not Da collapses wave function in QM?
Is there a possibility that the perfect correlation between Alice and Bob will continue?
Is there a similar experiment?

Thanks

 

[zonde]

The experiment I am thinking of is simple:

- A typical (ideal) photon polarization quantum entanglement setup with Alice and Bob.
- Alice and Bob detectors are constant at 90 deg. angle difference (such as they allways get perfect corelation with different result)
Plus:
- Between Alice and source we place a polarizer "Aa", in relative angle to Alice 45 deg.
- And we place a detector "Da" between Alice and Aa such as we know when a photon has passed polarizer Aa and it is heading towards Alice.
- We record the results of Alice and Bob, only if Da detects.

(if there is a technical problem of detecting a single photon after Aa without "stoping" it, please consider the experiment with something similar, such as light pulses)

Now what will happen?
Does not Da collapses wave function in QM?
Is there a possibility that the perfect correlation between Alice and Bob will continue?
Is there a similar experiment?

Polarizer "Aa" will determine correlation. So you don't get perfect anticorrelation but rather no correlation. Other polarizer (the one you call detector) just reduce amount of photons and have no effect on correlation.

And you can't detect photon without "stoping" it (as you suspected) but light pulses can not be entangled. So your setup is not realistic.

 

[LsT]

Thanks zonde,

I think they have made pulses entangled http://arxiv.org/pdf/1111.2073 but anyway that's beyond me and not my point, so I'll try to reform the example to hopefully make it realistic for photons:

• Typical ideal setup for polarization entangled photon pair experiment. Alice with detector constant at 0 deg. and Bob constant at 90 deg. (for perfect anti-correlation)
  
• We now place also Victor "next" to Alice with a similar detector at the same angle (0 deg)
  
• Between Alice and source we place a birefringent polarizer (eg ideal wollaston prism or something like that) "Aa". The one output of this polarizer is headed towards Alice at 45 deg. polarization angle, and the other output at Victor at -45 deg. That way we know the polarization of any photon after Aa, and before reaching Alice or Victor.

Now if Aa determines correlation, half of the times that Bob detects, also will detect Alice or Victor.
If the perfect anti-correlation between Bob and Alice/Victor holds (lets say), then when Bob detects, neither Alice or Victor will, and when Alice or Victor detects, Bob will not.

 

[zonde]

Yes, seems ok.

 

[LsT]

Ok then I guess there is consensus that in such a setup, polariser "Aa" determines correlation.

So the hypothetical case in which perfect anti-correlation between Alice/Victor and Bob is preserved is against QM?

 

[Khashishi]

If Bob detects, then Alice and Victor have a partial chance of detecting as well, and vice versa. This is just like putting three polarizers in series, at 0, 45, and 90 degrees. If you have two polarizers at 0 and 90 degrees, none go through. But if you add the third polarizer, you can have some photons through. In your case, you use a birefringent prism for the 45 degree polarizer, but that is a minor detail.

 

[zonde]

So the hypothetical case in which perfect anti-correlation between Alice/Victor and Bob is preserved is against QM?

Yes

 

[zonde]

So the hypothetical case in which perfect anti-correlation between Alice/Victor and Bob is preserved is against QM?

Look, detector that detects polarized photons is actually two devices - polarizer plus detector. If you replace "detector" polarizer with wollaston prism then you have two wollaston prisms in a row. And say we can insert second wollaston prism ("Aa" polarizer) between first wollaston prism and detector. You see, you can't assign specific role to polarizer before detector and polarization beam splitter that you have inserted. You have to sort it out without saying which one was there before and which one was inserted later.

 

[LsT]

If Bob detects, then Alice and Victor have a partial chance of detecting as well, and vice versa. This is just like putting three polarizers in series, at 0, 45, and 90 degrees. If you have two polarizers at 0 and 90 degrees, none go through. But if you add the third polarizer, you can have some photons through. In your case, you use a birefringent prism for the 45 degree polarizer, but that is a minor detail.

I agree with you. The thing is not that I don't get at all malus law, I just try to see how are you sure that this will happen also in the case of entangled photons.

Look, detector that detects polarized photons is actually two devices - polarizer plus detector. If you replace "detector" polarizer with wollaston prism then you have two wollaston prisms in a row. And say we can insert second wollaston prism ("Aa" polarizer) between first wollaston prism and detector. You see, you can't assign specific role to polarizer before detector and polarization beam splitter that you have inserted. You have to sort it out without saying which one was there before and which one was inserted later.

I am not sure I get your point, so I made a sketch of the setup, with separate polarisers and detectors:

My question is what if it turns out that alice/victor and bob continue to get perfect anticorrelation despite the fact that the wollaston prism polariser is there? Or in other words, how are we sure that "w" (Aa) will "destroy" entanglement?

 

[Khashishi]

Bob and Alice's measurements will be independent. I'd say this is a result of standard quantum mechanics and not an open question at all.
P(Bob)=50%
P(Alice)=25%
P(Alice ^ Bob)=12.5%
P(Alice ^ Victor)=0

 

[zonde]

Or in other words, how are we sure that "w" (Aa) will "destroy" entanglement?

There is more rigorous way how entangled state can be analyzed.
Keeping it simple after the source each beam of entangled photons contains H-polarized and V-polarized photon modes. In first beam splitter they can interfere and create so called non classical effects. But at the outputs of first beam splitter there are pure H-polarized (in one output) and V-polarized (in other output) modes so there is nothing that can interfere in subsequent polarizer.

 

[LsT]

after the source each beam of entangled photons contains H-polarized and V-polarized photon modes. In first beam splitter they can interfere and create so called non classical effects

That's not an interpretation?

Anyway I have no argument against what the expected result should be, my only remaining question is now whether this expected result is just expected, or it is also experimental fact.

Thinking about the typical quantum eraser experiment, for example, despite the fact that circular polarizers are placed after each slit, photons continue to be entangled, because "path information erasure" is done through entanglement at later time, right? So what is the difference there that prevents wave funtion "collapse"?

 

[zonde]

That's not an interpretation?

There are specific observations in real experiments that you would explain using this approach. You can take a look in this paper:
http://arxiv.org/abs/quant-ph/0205171

You can include phase $\phi$ in a bit more general description of entangled photon state $|\psi\rangle = |H\rangle|H\rangle + e^{i\phi} |V\rangle|V\rangle$ and if you change the phase you can observe (experimentally) smooth transition from entangled state to separable state (hidden entanglement).

Anyway I have no argument against what the expected result should be, my only remaining question is now whether this expected result is just expected, or it is also experimental fact.

Thinking about the typical quantum eraser experiment, for example, despite the fact that circular polarizers are placed after each slit, photons continue to be entangled, because "path information erasure" is done through entanglement at later time, right? So what is the difference there that prevents wave funtion "collapse"?

In quantum eraser it's not circular polarizers but quarter wave plates (it changes pure linear polarization into circular polarization). But there are H and V polarization modes in entangled photon beam and quarter wave plate changes this phase $\phi$ between H and V modes. So you get different interference between H and V modes when beam encounters PBS or polarizer at +/-45 deg.

 

[LsT]

You can include phase ϕ\phi in a bit more general description of entangled photon state ##|ψ⟩=|H⟩|H⟩+eiϕ|V⟩|V⟩|\psi\rangle = |H\rangle|H\rangle + e^{i\phi} |V\rangle|V\rangle## and if you change the phase you can observe (experimentally) smooth transition from entangled state to separable state (hidden entanglement).

Thats interesting, but I am not sure I get it right. It means when you go from circular to more and more elliptical polarization you start to get no more non-classical effects? How you can know this angle? From the properties of the crystal that creates the entangled pair? Can this angle be zero? Also, I suspect that we talk about the superposition of a photons polarization, because in the eigenstate the photon is allways circularly polarized?

 

[zonde]

Thats interesting, but I am not sure I get it right. It means when you go from circular to more and more elliptical polarization you start to get no more non-classical effects?

No, it's the other way around. When you go from linear to circular you get no more non-classical effects. Interference term (non-classical effects) vanishes when two terms are orthogonal in complex plane (one term has "i" in front of it).

How you can know this angle? From the properties of the crystal that creates the entangled pair? Can this angle be zero?

You can find out this angle by measuring entangled photons but you can change it by inserting quarter wave plates in photon paths. I am afraid to be wrong but I think you need it in only one arm to alter this angle.
And yes this angle can be zero. You get two different entangled states for $\phi=0$ and $\pi$ and you get classical product states for $\phi=\pm\pi/2$

Also, I suspect that we talk about the superposition of a photons polarization, because in the eigenstate the photon is allways circularly polarized?

One of the basic things in QM is that any pure state (eigenstate as you say) can be written as superposition of other pure states. So I would say that circular polarization, linear polarizations in H/V basis and linear polarization in +45/-45 deg basis are all equally "basic" in QM. But linear polarizations can be measured directly with polarizers and PBS so I rather tend to perceive them as more "basic".

 

[LsT]

Ok, so the "spooky action" of the entangled pair (the polarization correlation) "stops" when the H/V angle phase of the photons is $\pi/2$. And, I suppose that the zero and $\pi$ phase difference between Horizontal and Vertical modes of polarization cannot exist but in the sense of superposition, right? (how else can you have H and V polarization at the same time?)

 

[zonde]

And, I suppose that the zero and π\pi phase difference between Horizontal and Vertical modes of polarization cannot exist but in the sense of superposition, right? (how else can you have H and V polarization at the same time?)

I believe it's not possible to attribute superposition to single isolated photon. Does this answer your question?