DrChinese said:
2. The quality of entanglement can be measured by how close you come to perfect correlations when setting up the experiment. So you might expect that there is always a mix of ES> + PS> statistics (Entangled and Product). Ideally, ES is 100%. But clearly, that ideal is not met in this experiment and the result will be a deviation from the QM predicted rates accordingly. But not enough to cross back into the Local Realistic side of the Bell Inequality.
It is not exactly deviation from QM. You see QM covers this PS> state too. So you don't need to resort to some other idea (LHV or anything) in any case.
I have posted this formula couple of times but maybe it will make more sense now in conjunction with real experimental setup.
P_{VV}(\alpha,\beta) = \underset{product\; terms}{\underline{sin^{2}\alpha\, sin^{2}\beta + cos^{2}\alpha\, cos^{2}\beta}} + \underset{interference\; term}{\underline{\frac{1}{4}sin 2\alpha\, sin 2\beta\, \mathbf{cos\phi}}}
This is a bit reduced (without \theta_{l} factor) equation (9) from paper -
http://arxiv.org/abs/quant-ph/0205171/" that describes type-I PDC source.
The same way can be described type-II PDC. I found this out from Kwiat et al "New High-Intensity Source of Polarization-Entangled Photon Pairs" (I won't post the link to be on the safe side with forum rules about copyrights). There equation (1) is:
|\psi\rangle=(|H_{1},V_{2}\rangle+e^{i\alpha}|V_{1},H_{2}\rangle)/\sqrt{2}
that is basically the same equation but in more QM format.
As you can see from this first formula cos\phi acts as coefficient in range from -1 to 1 and accordingly this interference term can change it's weight between maximally negative, none at all and maximally positive. QM does not place any restrictions on that.
So if interference term becomes zero and photon state reduces to completely local realistic product state it's still covered by this QM description.
Physical interpretation in QM about this cos\phi coefficient is that it characterizes transverse and longitudinal (temporal) walkoffs.
As experimenter you have a goal to get this cos\phi maximally close to either 1 or -1 and if you do not succeed for some reason then interpretation says you have not compensated those walkoffs to satisfactory level.
DrChinese said:
So are you saying that the detectors somehow influence this? I don't follow that point or what you think the implications would be. It is the setup that determines things, of which the detectors are an element. But their efficiency shouldn't matter to that setup.
It's hard for me to say something about your comment that efficiency shouldn't be a factor. That's because since some time for me it's not the question of "if" but rather "how". And to be precise it's not only efficiency of detectors but rather coincidence detection efficiency of the setup as whole.
But more to the point, I interpret this interference term as correlation in samples of detected photons meaning that they are uneven. If this unevenness is similar we have positive interference term, if this similarity is inverted we have negative interference term and if we have this unevenness in independent "directions" we don't have interference term. Obviously for efficient detection any "direction" in unevenness of sample is no more detectable.
This loss of information for efficient detection can be illustrated with example like this. Let's say we have a box with different objects in it. We have hole in the box and if we shake the box some objects fall out. Afterward we can look at the objects that are outside the box and objects that are left inside. So we can find out some probabilities whether particular object is more likely to fall out of the box or stay inside. If we always shake the box until all the objects fall out of the box (efficient detection) we loose any information about that falling out probability.
But we're not so dumb as to appreciate college THAT little.