Comment on Test of a prototype quantum internet runs under New York City for half a month
Sekoia@lemmy.blahaj.zone 3 weeks ago… what you said is correct, but that’s superposition, not entanglement. Entanglement is when you create a product state of several qubits that cannot be decomposed into a tensor product of basic states (a single proton/photon/whatever).
bunchberry@lemmy.world 3 weeks ago
I am factually correct, I am not here to “debate,” I am telling you how the theory works. When two systems interact such that they become statistically correlated with one another and knowing the state of one tells you the state of the other, it is no longer valid to assign a state vector to the system subsystems that are part of the interaction individually, you have to assign it to the system as a whole. When you do a partial trace on the system individually to get a reduced density matrix for the two systems, if they are perfectly entangled, then you end with a density matrix without coherence terms and thus without interference effects.
This is absolutely entanglement, this is what entanglement is. I am not misunderstanding what entanglement is, if you think what I have described here is not entanglement but a superposition of states then you don’t know what a superposition of states is. Yes, an entangled state would be in a superposition of states, but it would be a superposition of states which can only be applied to both correlated systems together and not to the individual subsystems.
Let’s say R = 1/sqrt(2) and Alice sends Bob a qubit. If the qubit has a probability of 1 of being the value 1 and Alice applies the Hadamard gate, it changes to R probability of being 0 and -R probability of being 1. In this state, if Bob were to apply a second Hadamard gate, then it undoes the first Hadamard gate and so it would have a probability of 1 of being a value of 1 due to interference effects.
However, if an eavesdropper, let’s call them Eve, measures the qubit in transit, because R and -R are equal distances from the origin, it would have an equal chance of being 0 or 1. Let’s say it’s 1. From their point of view, they would then update their probability distribution to be a probability of 1 of being the value 1 and send it off to Bob. When Bob applies the second Hadamard gate, it would then have a probability of R for being 0 and a probability of -R for being 1, and thus what should’ve been deterministic is now random noise for Bob.
Yet, this description only works from Eve’s point of view. From Alice and Bob’s point of view, neither of them measured the particle in transit, so when Bob received it, it still is probabilistic with an equal chance of being 0 and 1. So why does Bob receive different outcomes?
Because when Eve interacts with the qubit, from Alice and Bob’s perspective, it is no longer valid to assign a state vector to the qubit on its own. Eve and the qubit become correlated with one another. For Eve to know the particle’s state, there has to be some correlation between something in Eve’s brain (or, more directly, her measuring device) and the state of the particle. They are thus entangled with one another and Alice and Bob would have to assign the state vector to Eve and the qubit taken together and not to the individual parts.
Eve and the qubit taken together would have a probability distribution of R for the qubit being 0 and Eve knowing the qubit is 0, and a probability of -R of the qubit being 1 and Eve knowing the qubit is 1. There is still interference effects but only of the whole system taken together. Yet, Bob does not receive Eve and the qubit taken together. He receives only the qubit, so this probability distribution is no longer applicable to the qubit.
He instead has to do a partial trace to trace out (ignore) Eve from the equation to know how his qubit alone would behave. When he does this, he finds that the probability distribution has changed to 0.5 for 0 and 0.5 for 1. In the density matrix representation, you will see that the density matrix has all zeroes for the coherences. This is a classical probability distribution, something that cannot exhibit interference effects.
Bob simply cannot explain why his qubit loses its interference effects by Eve measuring it without entanglement, at least within the framework of quantum theory. That is just how the theory works.