creeky belly
Posts: 205 Joined: June 2006

Quote  This experiment wouldn't test for quantum entanglements within the body (e.g. eyes to brain to fingers) but would be looking for external quantum entanglements (quantum randomizer to test subject). 
No. The quantum random number generator doesn't work this way, there's no entanglement. It's most likely a polarizing beam splitter with an input polarization at 45 degrees to it (that's how most of the commercial ones work). To get entanglement you need a twoparticle system in a superposition state, ie one that can't be written as a product state. What you're really measuring is the stochastic QM process, transmitted as classical bits. You can look at the bits long before it gets to the observer without changing the result.
Here's an example of entanglement. Suppose you had two bits, of which there are four possible states: 00,01,10, and 11. Let's say the probability of the first bit being 0 was p and the probability of the second bit being 0 was q. Then, the probability of measuring each of the four states is as follows:
00: pq 01: p*(1q) 10: (1p)*q 11: (1p)*(1q)
If p=q=0.5, then the probability of measuring any of the four states is 25%. If we write this as a vector, it looks like 0.25 * [1,1,1,1]. This is the classic representation of probability. Ok so far? Now we get to the quantum states. What are the requirements for QM? Namely the inner product of the state vector must be normalized to 1. Let's say that this is a pi0 decay, in that case the probability of measuring 00 and 11 is 0 (pi0's decay into a spinup, spindown pair, but the actual choice follows a stochastic process). From our classical equations, this means that p*q = 0, so either p or q is 0. But our equations also say that p*(1q) != 0, so then p=1 and q=0. If we look at the third equation, (1p)*q != 0, we see that there's a contradiction, since the probability for measuring 10 is not 0 in this decay. This means it can't be expressed as a classical probability of bits. In fact, the state vector would be 0.707 * [ 0, 1, 1, 0], with either +/ or i's as long as the state vector had the correct inner product. What constitutes entanglement? Take the state 0.707 * [1,1,0,0]. This is the state where 00 and 01 are equally likely, but 10 or 11 will never be measured. Does the measurement of one state fix the other? No. If we measure 0 on the first bit, there is no way to figure out what the other bit will be (50% 0, 50% 1). We can rewrite this as a tensor product of two spin 1/2 states: 0.707 * [1,0] x [1,1]. Quantum? Yes, it breaks the classical probabilities. Entangled? No.
Let's look at the state 0.707 * [0,1,1,0]. We already know it's quantum, is it entangled? If we measure 0 on the first bit does it fix the second bit? Yes. So there's no way to write [0,1,1,0] as the tensor product of two spin 1/2 systems. That's entanglement, which is not present in a quantum random number generator by design. Having any entanglement compromises the security of your protocol, which is why they use single photon generators such as quantum dots.
