Let's say my algorithm starts with the qubit state $|0^N\rangle$. Is there a possibility to end up in a superposition where every component is made of a bitstring containing exactly an $1$ at on position and only $0$'s on every other position, i.e. $\frac{1}{\sqrt{2}^N} (|1,0^{N-1}\rangle + |01,0^{N-2}\rangle + ...+|0^{N-1},1\rangle)$

I would like to only use the standard gates/unitary transformations: $X,Y,Z,H, \text{CNOT}$

  • 4
    $\begingroup$ This is known as w-state, and you can find good answers for your question here: quantumcomputing.stackexchange.com/q/4350/9474. Note that the normalization factor should be $\frac{1}{\sqrt{N}}$ $\endgroup$ Mar 24, 2023 at 21:11
  • $\begingroup$ Thanks, the question is answered. $\endgroup$
    – kerf
    Mar 24, 2023 at 22:19

1 Answer 1


It is not possible to construct the above state using the specified gates. States that can be prepared using these gates are called stabilizer states. To quote this stack exchange post, "for a stabilizer state, the 1-site reduced density matrices must be maximally mixed or pure, which [W states, like the one on your question] aren't."


Not the answer you're looking for? Browse other questions tagged or ask your own question.