News for hackers. Stuff that matters.
Re:Quantum Computing Explained Simply by comicjk on Thursday July 20, 2017 @11:12AM (#14813140)
> when the qubit is measured (read queried) it will return a value of 0 or 1 with equal probability

To clarify, qubits don't necessarily have equal probabilities of |0> and |1>. If the probabilities were always equal, we wouldn't be able to use quantum computers for anything besides generating coin flips! The "answer" from a quantum computer is encoded as output probabilities of the qubits, which are found by running the calculations a bunch of times and counting the frequency of each |0> and |1> state.

Re:Quantum Computing Explained Simply by ganonm on Thursday July 20, 2017 @11:27AM (#14813300)
I think you've misread my comment as what you're saying is incorrect. When a qubit is in the state (|0> + |1>)/sqrt(2) it does have equal probabilities of being measured as either |0> or |1>. I think what you're trying to say is that (|0> + |1>)/sqrt(2) is not the only superposition state a qubit can be in. A qubit can be in any arbitrary position of the two possible states.
Re:Quantum Computing Explained Simply by comicjk on Thursday July 20, 2017 @12:54PM (#14814185)
Your original comment is totally correct, but I thought it might mislead beginners - that's why I said "to clarify." I think there is a common confusion about whether a qubit can have probabilities besides 50:50. The Hadamard gate example explains superposition, but tends to increase this misconception, so I also wanted to emphasize the many other possible states and how an answer is extracted from them.
Re:Quantum Computing Explained Simply by gaze on Thursday July 20, 2017 @02:15PM (#14814987)
This is a bit misleading I think. There are a set of quantum algorithms with deterministic outcomes. The right answer will be measured with probability 1, or something very close...