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I'd like to build a cheap and simple quantum coin flip circuit.

I have posted in the EE stack exchange about the specific circuit design, but was wondering if anyone in physics would have enough overlap in understanding to help me.

My idea is to use simple latching circuit driving two LEDs, and have the initialization be driven by some random event. My initial thought was to compare the resistance of two photoresistors in the dark as a "quantum" event. There should be some small variation in the photoresistivity due to shot noise at very low counts. So essentially the latch would be powered up and whichever photoresistor had a slightly lower/higher resistance would drive the latch to initialize in that state and stay there until power is removed.

Does this setup sound reasonably close to a quantum measurement?

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  • $\begingroup$ No two photoresistors will be exactly identical. How will you exactly null out any resistance difference between the two before starting the counting? $\endgroup$ – Samuel Weir Jun 25 at 21:02
  • $\begingroup$ That's a good point. I was thinking I would add a variable resistor in series and then adjust it to make them approximately equal. I think it would be alright to allow there to be a 5-10% bias in the "coin." $\endgroup$ – Will Stedden Jun 25 at 22:05
  • $\begingroup$ "Does this setup sound reasonably close to a quantum measurement?" Not... really, unless you can be sure that the latching actually happens because of detection of a single photon, which is probably not going to be the case. Ask yourself what experiment you could do to prove that the latch even is quantum. $\endgroup$ – DanielSank Jun 25 at 22:37
  • $\begingroup$ I think that you would have a hard time arguing to anyone that your "quantum" coin flip random number generator is superior to just about any classical pseudorandom number generator if your quantum number generator has such a large bias. In fact, looking for bias would be the very first test that would be applied to a coin flip generator that claims to be completely random. $\endgroup$ – Samuel Weir Jun 25 at 22:38
  • $\begingroup$ @DanielSank, this might come from my misunderstanding the barrier between a quantum and a classical measurement. Is it no longer considered a quantum measurement if you were to quantize (without intermediate measurement) the output of, say, 10 photon detections? $\endgroup$ – Will Stedden Jun 25 at 22:56
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I think I found my own answer. Shot noise on a photodiode is a result of quantum mechanical effects (https://www.rp-photonics.com/shot_noise.html). So I will be comparing the resulting photocurrents to create a quantum random bit.

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