4 ms·
> 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 probabiliti
by comicjk 9y ago
> 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.
- ganonm 9y agoI 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.
- comicjk 9y agoYour 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.
- gaze 9y agoThis 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...