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So the question came up in a book I am working through. Given a circuit with $n$ qubits, construct a state with only $n$ possible measurement results, each of which has only $1$ of $n$ qubits as $1$, such as $|0001\rangle$, $|0010\rangle$, $|0100\rangle$, $|1000\rangle$, obviously normalized.

The only way I can think to do this is to take the all $|0\rangle$ input state, apply $\operatorname{H}$ to each qubit and then used multiple-controlled $\operatorname{CNOT}$ gates to affect the change on each qubit, but I feel like this won't lead to the desired end state.

To be clear, I am enquiring how to create a $W_n$ state can be arbitrarily prepared, given $n$ qubits.

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  • $\begingroup$ I'm not sure what you mean by "n possible measurement results" here. If you have $n$ qubits then there are $2^n$ possible measurement results. The rest of the question is fine and interesting though. I edited the title to better reflect what I think you are asking. Feel free to revert the edit if I misunderstood you $\endgroup$ – glS Nov 2 '18 at 13:39
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    $\begingroup$ also, just for reference, states such as the one you refer to are often referred to as W states, and a similar question was also asked on physics.SE, see physics.stackexchange.com/q/311743/58382. Finally, as pointed out in that question, you should really specify the set of gates that you want to use $\endgroup$ – glS Nov 2 '18 at 13:40
  • $\begingroup$ @glS It appears the actual term for the state I was looking for is the W state, which has only n measurements. $\endgroup$ – luminalQubit Nov 2 '18 at 14:31
  • $\begingroup$ again, I don't know what you mean by "has only n measurements". A state of $n$ qubits can have $2^n$ possible measurement results, it doesn't matter which particular state it is $\endgroup$ – glS Nov 2 '18 at 14:33
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    $\begingroup$ @gIS: for what it's worth, I find luminalQubit's terminology fairly clear. A $W$ state does only have $n$ possible measurement results, because none of the other $2^n - n$ bit strings are 'possible measurement results' of the $W$ state. $\endgroup$ – Niel de Beaudrap Nov 2 '18 at 16:01