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Consider an entanglement experiment as below

A source of entangled photons sends the entangled photons to Alice and Bob. Detectors at Alice and Bob are filters and either there is a detection or there is not. A detection is stored in data as 1 and a non-detection is stored as a 0.

The detectors are selected in a random manner. In actual experiment, all 4 detectors are fixed and the photons are directed randomly to different detectors using optical fibers, instead of rotating detectors.

The states used are of the form $$ |\Psi\rangle = \frac{1}{\sqrt{1+r^2}} \left( |V_A\rangle \otimes |H_B\rangle + r |H_A\rangle \otimes |V_B\rangle \right) $$

Value of r is -2.9.

The question is about outcomes at Alice. There are ~1750 million trials sent at Alice = a1 setup (setup a1 angle = 94.4 degrees).

According to the paper, the angles used are $a1 = 94.4^\circ$, $a2 = 62.4^\circ$, $b1 = -6.5^\circ$, $b2 = 25.5^\circ$.

It is not clear what the angles are related to, so the assumption is they are relative to 0 degree (horizontal)

a1 = Alice setting 1, a2 = Alice setting2, b1 = Bob setting1, b2 = Bob setting2.

Irrespective of correlations etc., what is QM predicted percent of trials that are expected to record a "1" in ~1750 million trials sent to a1? you may just give a rough estimate of percent.

Per my little knowledge, the percent detection (when looked at a1 outcomes only) , should be ~50% irrespective of whatever the correlations are between Alice and Bob. Is this correct? If not why it could be different and how much different.

I am looking for a rough predicted percent so, 48% would not be considered very different from 50%.

For more details about experiment - it is published at https://arxiv.org/abs/1511.03190

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closed as off-topic by Norbert Schuch, Jon Custer, David Z Dec 28 '16 at 3:44

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The states used are of the form $$ |\Psi\rangle = \frac{1}{\sqrt{1+r^2}} \left( |V_A\rangle \otimes |H_B\rangle + r |H_A\rangle \otimes |V_B\rangle \right) $$ where the entanglement parameter $r$ depends on the setup and apparently was set at $r=-2.9$ for actual runs (note that the maximally entangled state requires $r=-1$). So, in principle, the chances are $1/(1+r^2)$ for detecting a $|V_A\rangle \otimes |H_B\rangle$ pair and $r^2/(1+r^2)$ for a $|V_A\rangle \otimes |H_B\rangle$ one. But the measurements are not done along the $H$ and $V$ directions. According to the paper, the angles used are instead $a1 = 94.4^\circ$, $a2 = 62.4^\circ$, $b1 = -6.5^\circ$, $b2 = 25.5^\circ$.

Even so, assuming angles are measured relative to the horizontal (please correct this if I'm wrong), a measurement along angle $\theta$, say on Alice's side, tests for a state $$ |\theta_A\rangle = \cos(\theta) |H_A\rangle + \sin(\theta) |V_A\rangle $$ and the probability of detection on $|\Psi\rangle$ reads $$ P_A(\theta) = Tr\left[ (|\theta_A \rangle \langle \theta_A|) (|\Psi\rangle \langle \Psi|) \right] = |\langle \Psi | \theta_A \otimes H_B \rangle |^2 + |\langle \Psi | \theta_A \otimes V_B \rangle |^2 = \frac{\sin^2(\theta) + r^2 \cos^2(\theta)}{1+r^2} $$ Alternatively, you can write the states $|H_A\rangle$ and $|V_A\rangle$ as superpositions of $|\theta_A\rangle$ and its orthogonal $|\left(\theta-\frac{\pi}{2}\right)_A\rangle \equiv \sin(\theta) |H_A\rangle - \cos(\theta) |V_A\rangle$, substitute in $|\Psi\rangle$ and extract the probabilities therefrom. That is, $$ |H_A\rangle = \cos(\theta)\; |\theta_A\rangle + \sin(\theta)\; |(\theta - \frac{\pi}{2})_A\rangle $$ $$ |V_A\rangle = \sin(\theta)\; |\theta_A\rangle - \cos(\theta)\; |(\theta - \frac{\pi}{2})_A\rangle $$ and $$ |\Psi\rangle = \frac{1}{\sqrt{1+r^2}} \left[ |\theta_A\rangle \otimes \left[ \sin(\theta)\;|H_B\rangle + r \cos(\theta)\; |V_B\rangle \right] - |(\theta - \frac{\pi}{2})_A\rangle \otimes \left[\cos(\theta)\; |H_B\rangle - r \sin(\theta)\; |V_B\rangle \right] \right] $$ A quick check shows that the probability $P_A(\theta)$ is $50\%$ either if $r^2=1$, in which case $\theta$ may be arbitrary, or for $\cos^2\theta = 1/2$, which gives $\theta = ± \pi/4$.

So no, the predicted probabilities of detection for setups a1 and a2 are not $50\%$.

Exercise: estimate the correct probabilities for the actual angles employed, $a1 = 94.4^\circ$ and $a2 = 62.4^\circ$ :)

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  • $\begingroup$ Thanks a lot udrv. Per the first formula, the value for 94.4 is (0.99705275222 + 2.9*2.9*.00588)/(1+8.41) = roughly 11%. Value for 62.4 is (.7853 + 8.41*.2145)/(1+8.41) = roughly 27.5%. In any case, per formula, the probability does not seem to be less than 10%, whatever the angles are. Because the denominator is 9.41 and numerator is >= 1. And 1/9.41 = 10.62%. Do these numbers look correct? $\endgroup$ – kpv Dec 27 '16 at 21:03
  • $\begingroup$ Yep, looks like what I got. $\endgroup$ – udrv Dec 27 '16 at 21:49
  • $\begingroup$ The observed detection (in actual data) is less than 1 percent. Does that mean need to re-look at the formula, or re-do the detection counts (which I have done by the way). To verify, Is it possible to give similar formula, and value of r for another experiment published at arxiv.org/pdf/1511.03189v2.pdf ? $\endgroup$ – kpv Dec 27 '16 at 22:03
  • $\begingroup$ @udrv The data in the mentioned experiment does not describe independent clicks in the detectors. It measures the probability to have (or not have) a click in the detector given that you had (not had) a click in the other detector -- which of those, depends on the setting. So even assuming perfect detectors your formula needs to depend on both $\theta_A$ and $\theta_B$ for each setting (and, moreover, for each setting in a different way). $\endgroup$ – Norbert Schuch Dec 27 '16 at 23:15
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    $\begingroup$ @kpv Wrote this before seeing your last comment: Correct, the probabilities you obtained from the answer concern only the Alice part of the experiment, since your question asked about setup a1 "irrespective of correlations". They are not the ones used in Eq.(1) of the paper, $p_{++}(a1b1)$ etc. If you'd like to extract those, you do need to refine and account for Bob's measurements along b1, b2. That is, you need the joint probabilities for the 4 pairs of events appearing in Eq.(1). The idea is the same though. $\endgroup$ – udrv Dec 28 '16 at 1:36

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