Just to be clear, the part you are wrong about is this:
> If you only have a single member of the pair, then you will see the same interference pattern in a double slit experiment than with a not-entangled particle.
This bit:
> It doesn't matter if the other particle has collided with a brick, went thru a double slit experiment, went thru a bad double slit experiment, or is flying to Andromeda.
is correct.
Also, there is an interference pattern in the results of a double-slit run on entangled particles, but it is not "the same" as you get with non-entangled particles, and the procedure you have to go through to observe this interference pattern is radically different.
> Here's how it works. We send a pair of EPR photons through a pair of two-slit
apparati each of which has a polarization rotator on one of the slits. On one side
of the apparatus (side A) we install a polarization filter which filters out
interference on that side and makes it visible. We can filter out interference on
the other side (side B) of the apparatus as follows: on side A we keep a record of
which photons passed through the filter and which were reflected. On side B we
keep a record of where each photon landed on the screen. We then take these
two records and combine them: for each photon that was passed through the
filter on side A, we take the corresponding photon on side B and note where it
landed on the screen. The end result is a (visible) interference pattern. It was
there all along, but the only way we can filter it out so we can see it is to combine
information from both sides of the experiment. And that is the last nail in the
coffin of superluminal communication via entangled photons.
Let's suppose you are in lab B and measure where the photons hit the screen. There are no visible interference patters. But just before you call to lab A, it's is nuked from orbit.
Now, you are unsure if the people that was generating the photons were sending pairs of entangled photons to A and B, or they were just sending pairs of normal photons.
Can you look at the data you collected in B and discover what the people in the generator were doing?
Now imagine the same experiment with a rebuild lab A', but you remove the polarization rotator. Now you see the interference pattern. And A' gets nuked again. Can you look at the data you collected in B and discover what the people in the generator were doing?
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I understand that you can take a plane and collect all the pieces of A and A' are reconstruct them, after all Classic Mechanics and Quantum Mechanics without the measurement rule are reversible, so it's theoretically possible, but very impractical.
I think this discussed in section 5. For the measurement problem, I prefer the something-something-decoherence solution. I call it something-something-decoherence because there are still a lot of work to be done before it's clear if it's the correct solution.
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About the experiment in 4.2:
I'm 99% sure after adding the polarizer at 45° there will not be interference. You can split the polarizer into two smaller polarizers with the same angle, one for each slit. The exchange the order of the rotator+polarizer in one slit to polarizer+rotator. Note that after the exchange, the new polarizer must be rotated 90°, so it's polarizer at "-45°". Now the first thing in one slit is a polarizer at "+45°" and the other is at "-45°", so they will select orthogonal states and not get interference even after rotating one of them.
I think this can be fixed using a quarter-wave plate, but the calculation is slightly more complicated.
[Sorry for the looooong delay. I missed your reply last week and I just saw it yesterday night.]
Now I'm confused. I had to read the link carefully and try to translate the experiment with polarizer to the experiment with the double slit.
I'm still not sure, but an important point is that in a usual double slit experiment there is a single slit before that acts like a collimator and ensure the photons have no preference for each slit. Let's suppose you see isolated in lab B:
* If you add the collimator, then all effect of entanglement are destroyed and you see the usual interference pattern.
* If you don't add the collimator, you have and ensemble of particles that go to each slit, and cause no interference pattern. It's not necessary to add a polarization rotator to one of them.
* If the experiments are far away, ¿does it count as an implicit single slit collimator? I'm confused here. I'd prefer a version with only polarization or other property that is not mixed with the setup of the experiment.
Yeah, it's confusing. I should probably write a new paper just on this topic because there isn't really a good explanation anywhere.
> translate the experiment with polarizer to the experiment with the double slit
Best is not to get too hung up on the physical details. What matters is that a stream of particles can get separated along two separate paths and brought back together, and this can produce an interference pattern. The particular degree of freedom along with the separation takes place (position, polarization, spin, whatever), or the details of how they are split and brought back together (two-slit, half-silvered mirrors, Stern-Gehrlach apparatus, whatever) is mostly irrelevant. What matters is:
1. When unentangled particles are sent through one of these split-combine setups they produce an interference pattern.
2. When you "measure" the degree of freedom along which the particles are split in one of these split-combine setups, the interference pattern disappears and is replaced by a non-interference pattern. (This is just basic quantum mechanics 101.)
3. When you send entangled particles through a split-combine setup what you get is a non-interference pattern, exactly the same as the one you get when you "measure" an unentangled particle. But...
4. If you go through a rather elaborate process (see below) you can separate the entangled particles into two groups, each of which exhibits an interference pattern, and these two interference patterns will add up to make a non-interference pattern. (Even more interesting, there is more than one way that you can do this separation, each of which will produce a different pair of interference patterns, but any given pair will add up to the same non-interference pattern.)
The "elaborate process" involves making measurements on the complimentary observable for one member of each entangled pair, and classifying the other member of the pair into one of two groups based on the outcome of that measurement. This is where the physical details get really complicated for anything other than polarization, where the complimentary observable is just polarization along an axis rotated by 45 degrees to the original.
Note also that the reason that #3 above is true is that measurement and entanglement are actually the same physical phenomenon. The mathematical description of an entangled particle and a "measured" particle is exactly the same.
> 1. When unentangled particles are sent through one of these split-combine setups they produce an interference pattern.
It depends on how you prepare the particle beam. With a laser or passing first though a single (centered) slit, you get a beam of pure state particles that cause interference.
If you use other method, you can get an ensemble that doesn't produce interference.
About the use of the polarizer at 45°, I'm still not convinced. I have to write it carefully, but now I'm super busy [1]. I hope to take some time to write it next month.
Anyway, it would be nice to replace the double slit experiment with another experiment, like a beam splitter. It's difficult to remember what the two slits and the screens do to my brackets. (I think I know now, but I must write that carefully.)
Just to be clear, the part you are wrong about is this:
> If you only have a single member of the pair, then you will see the same interference pattern in a double slit experiment than with a not-entangled particle.
This bit:
> It doesn't matter if the other particle has collided with a brick, went thru a double slit experiment, went thru a bad double slit experiment, or is flying to Andromeda.
is correct.
Also, there is an interference pattern in the results of a double-slit run on entangled particles, but it is not "the same" as you get with non-entangled particles, and the procedure you have to go through to observe this interference pattern is radically different.