3 ms·
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
by ganonm 9y ago
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.
- 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.