My understanding that In quantum eraser experiment we can interpret the past only when we got the data from the coincidence counter. My question is why we need coincidence counter? Is this becuase the BBO produce entanglement photon and normal photon together ? i.e. if it is efficent 100% ( entanglement photons only ) then we can observe directly that our choice in the future really will change the inteference pattern ? Is there anything in quantum mechanic prevents that in the BBO? So we need always the coincidence counter?
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First, there is no "entanglement photon" and "normal photon". There are simply two photons that will give correlated measurement results. Your question really doesn't need any quantum eraser setup (which doen't bring anything new to the quantum mechanics) but requires better understanding of what quantum entanglement actually is.
Consider photons in a Bell state, \begin{equation} |\psi\rangle=\frac{1}{\sqrt{2}}|\uparrow_z\rangle_A\otimes|\downarrow_z\rangle_B+\frac{1}{\sqrt{2}}|\downarrow_z\rangle_A\otimes|\uparrow_z\rangle_B \end{equation} where $|\uparrow_z\rangle,|\downarrow_z\rangle$ are spin up, spin down states in $z$ axis. Consider for a moment the measurements on just one photon, be it $A$ or $B$ doesn't matter. Then no matter what happens with the other photon you will always get half a time spin up, half a time spin down. One photon of this (maximally) entangled pair is indistinguishable from the photon being with the $50%$ classical (without quantum superposition) probability in the state $|\uparrow_z\rangle$ and with the $50%$ probability in the state $|\downarrow_z\rangle$ (and funny thing is that for other axis it's the same) Where the quantum entanglement appears is in the correlations of the measurement results for $A$ and $B$.
E.g. you may ask with what is probability $P\Big(\uparrow_{\theta,B}|\downarrow_{\phi,A}\Big)$ that if you measure spin of $B$ in axis $\theta$ you get "up" provided that the measurement of spin of $A$ in axis $\phi$ resulted in "down". This correlations are very easily described in quantum theory without introduction of any magic. However if you try to assume the hidden variables - that the nature is actually classical and probabilistic quantum description is simply a rough aproximate picture, then you get into all these troubles with superluminal signals and signals back into the past. Contrary to what for some reason most of the popular sources will tell you nothing of that exists in the quantum mechanics and the quantum eraser experiment and its variations proves nothing new except closing another hidden variable loophole. All the quantum mechanics tells you is that you get probabilistic measurements of $A$ and $B$ that don't influence each other (i.e. measurements on $A$ are the same if you don't compare them with $B$) and there are certain correlations between them.
As I said not about quantum eraser but simpler experiment where you just measure spins of $A$ and $B$ without any extra tricks. Let's first consider the idealized situation - there is no noise and photons $A$ and $B$ always hit their respective detectors. In reality the coincidence counter also helps you significantly reduce the noise so that you more surely detect the entangled pair you are interested in.
Consider now one of the photons. The measurement of its spin is usually done in the following way - you have a detector that catches the photon irrespective of its spin but in the photon's path you put a polarization filter. So when the filter is for example is oriented in the axis $z$ and the photon is in the state $|\uparrow_z\rangle$ it is detected but if it's in the state $|\downarrow_z\rangle$ it is absorbed by the filter and there's no detection signal. If it's in the superposition e.g. in the state $|\uparrow_x\rangle$ it will come through half a time.
Remembering what I said about the Bell state if you are interested only in one photon, be it $A$ or $B$ and don't put any filter no matter what happens with the second photon you will get the same detection rate $n$. If you put a filter in any direction you will get the detection rate $\simeq n/2$, again no matter what happens with the second photon. So $P\Big(\uparrow_{\theta,A}\Big)=\frac{1}{2}$.
But now let's put the coincidence counter. Then you will get signal only when BOTH photons are measured with spin up for their corresponding detector axis. So if you get in this case the detection rate $m$, \begin{equation} \frac{n}{m}\simeq P\Big(\uparrow_{\theta,A},\uparrow_{\phi,B}\Big) \end{equation} And from here you can derive the correlations you are interested in. I think from that presentation it's obvious that it's not that when you measured spin for $A$ you somehow influenced the spin of $B$ and because of that detector at $B$ gives you different detection rate, it's just that you ignore those of the measurements when $A$ had spin down.
Now how it works in case of the quantum eraser (all images are taken from the Wikipedia). You have photon going to the double slit experiment at $D_s$ and another going directly to the detector at $D_p$. At first there's no polarization filter at $D_p$,
and at $D_s$ you get no interference (the measured picture is green).
All the coincidence counter does in this case is ensuring that the photon you measure at $D_s$ indeed belongs to the entangled pair and not just a random noise photon. There will be of course lucky noise photons that will hit both detectors simultaneously but this is rare.
In the second case you put some polarization filter before $D_p$,
and at $D_s$ you get some interference picture. But what will happen if you turn the filter at $D_p$ in the opposite way? You will get another interference picture. And when you sum up those picture you will get (surprise!) the original picture without interference!
That's because in the second case you ignore half of the measurements done on $D_p$ leaving only those that were made provided that the detector at $D_s$ produced a signal.
So here is the final answer: with the coincidence counter you measure a certain correlation between measurements for the different entangled photons AND reduce the noise coming from the non-entangled photons.
OON
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Much of the complication and apparent mystery is actually due to the design of the experiment. If you look at the following diagram, I think I can probably make things clearer.
Notice at the top of the diagram, just right of center, there is a detector labeled Dsub0. There is also an arrow and an x. That means there is a narrow vertical detection screen that is hooked up to a stepper motor. It gets shifted on schedule, collects all photons that it sees, sends a signal to the coincidence counter for each photon, all that, including the time, gets sent to the counter where a record is kept. While that process goes on and on, the other detectors send in a signal to the coincidence counter where they get recorded. As far as I can tell, these other detectors are dead simple. They just record a hit and the coincidence counter record something like, "detector Dsub1. at time Tsub. If you graphed these hits from left to right according to the time on one swing, then went back and graphed hits from left to right according to the time of the next swing, you would get a crude picture of the fringes of an interference pattern from Dsub2 and Dsub3. The record that shows up from Dsub2 is different from the one that shows up from Dsub3. The two are backwards to each other because of the way that mirrors and beam splitters work on the light to make one the horizontal "flipping" of the images. With some math to re-flip one of the images you can get a regular-looking set of interference fringes.
What the experimenters would like to find is cases where a Dsub4 or Dsub1 photon (which is not one that has been divided by a beam splitter) shows up at the same time that the movable detector is in its center position and appears to be having a photon show up there at the same time that a photon shows up at Dsub4 or Dsub 1.
There is one thing that I have to question. I guess I ought to trust the designers to know how to make lab equipment. Nevertheless, just to be sure that everybody is aware of the problem: There is a light path from the laser to Dsub0, and to all the other detectors. But what goes on the time chart kept by the coincidence includes not only the light path from laser to detector, but also the time it takes for electricity to go from the detectors to the coincidence counter. Electricity move very rapidly, but it is slower than light. A little extra wire, or components that are located at different distances from the coincidence counter all need to be taken into account.
My own account of the experiment says that you start something "quantum," in this case you fire off one photon from the one-shot laser, and there is immediately a probability for the whole interaction. But parts of this cosmic pachinko game are out of our awareness. That's part of what makes the quantum world so mysterius. Immediately, in this case:
If Dsub0 shows up as a photon that hasn't been through a beam splitter then Dsub1 or Dsub4 show up with a photon that hasn't been through a beam splitter. On the other hand, if you observe that Dsub1 shows up with a photon that hasn't been through a beam splitter, then Dsub0 must show up as a photon that hasn't been through a beam splitter. The same for Dsub4. It "demands" Dsub0 to show up as a photon that hasn't been through a beam splitter.
If Dsub2 shows up as a photon that has been formed from a photon that has been through a beam splitter, then Dsub0 must show up the same way. IfDsub3 shows up as a photon that has been formed from a photon that has been through a beam splitter, then Dsub0 must show up the same way.
Maybe the above hints will be enough to get you started.
user9695085
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