5 votes
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Is the CNOT in the standard three-qubit circuit for the GHZ state necessary?

If you initialize three qubits to $|0\rangle$, apply a Hadamard gate to each, then measure each in the computational basis, the result will be an independent coin flip for each bit: that is, any of ...
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  • 639
3 votes

How many quantum gates are needed to prepare an arbitrary state?

A canonical reference for gate decompositions is Barenco et al., Elementary gates for quantum computation. In particular, it also contains recipes to decompose an arbitrary $n$-qubit unitary into ...
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2 votes
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How to create superposition of states with fixed parity with a quantum circuit?

I actually found the solution to my problem - with one caveat: the amplitudes $a,b,c,d$ can not be completely arbitrary. An $n$-qubit superposition state with even parity will have $2^{n-1}-1$ free ...
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2 votes
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How many quantum gates are needed to prepare an arbitrary state?

I believe this Q&A answers your question about decomposition in detail: Minimum number of 2 qubit gates to build any unitary In short, you are correct that the lower bound for a number of 2-qubit ...
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  • 610
1 vote

How to create superposition of states with fixed parity with a quantum circuit?

To do this in general is equivalent to creating an arbitrary $n-1$-qubit state, and then using the pattern of controlled-nots to determine the bit value of the final qubit, as you did in your solution....
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1 vote
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How to create known quantum state in Qiskit (or any other platform) comprising of two or more bits?

As Martin Vesely mentioned in the comments, you can use the initialize function to perform such a task. For instance, to create the state you desire, you can do the ...
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