Quantum Computing Asked by MysteriousWaffle on May 13, 2021
From what I’m reading, building the constant-1 and constant-0 operations in a quantum computer involves building something like this, where there are two qubits being used. Why do we need two?
The bottom bit in both examples is not being used at all, so has no impact on the operation. Both operations seemingly only work if the top qubit’s initial value is 0 so surely what this is just saying is that this is an operation which either flips a 0 or leaves it alone – in which case what is the second qubit needed for? Wouldn’t a set-to-0 function set the input to 0 whatever it is and wouldn’t need one of its inputs to be predetermined?
Granted, the ‘output’ qubit is for output, but it’s value still needs to be predetermined going into the operation?
Image is from this blog but I’ve seen it come up in other blogs and videos.
Indeed, the second qubit is doing nothing in these circuits, so you can ignore it.
Answered by glS on May 13, 2021
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