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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.

Sanchayan Dutta
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GaussStrife
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  • 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 – glS Nov 02 '18 at 13:39
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    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 https://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 – glS Nov 02 '18 at 13:40
  • @glS It appears the actual term for the state I was looking for is the W state, which has only n measurements. – GaussStrife Nov 02 '18 at 14:31
  • 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 – glS Nov 02 '18 at 14:33
  • @glS I am unsure how the example given there scales efficiently as the number of qubits increases, nor how they actually arrive at the state :/ – GaussStrife Nov 02 '18 at 14:33
  • depending on what you mean by "there" , you might edit your post accordingly to point out why the other question(s) do not answer yours. This will automatically put the question in the reopen queue – glS Nov 02 '18 at 14:35
  • @glS Yes, a state has 2^n possible measurements, but only if it has been prepared in a fully parallel state. The W state has only n measurements, where n is the number of qubits. – GaussStrife Nov 02 '18 at 14:35
  • no it doesn't, and again, you need to define what you mean by "has only $n$ measurements" – glS Nov 02 '18 at 14:36
  • @glS The W state, as described in the links provided here, in the linear combination of n states, each with equal probability of measurement, each with only a single qubit shifted to 1, while the rest are 0. How does this state not only have n possible measurements? – GaussStrife Nov 02 '18 at 14:39
  • @glS Also, what do you mean no it doesn't? If each qubit is not set up as an equal probabilistic distribution between 0 and 1, then how can the full range of 2^n states be achieved by measurement? – GaussStrife Nov 02 '18 at 14:51
  • as I said, you need to define what you mean by "has only $n$ measurements", as it is not standard terminology and is not clear what it means. If what you mean is "there are only $n$ measurement outcomes corresponding to a positive probability", then sure, this is true – glS Nov 02 '18 at 14:59
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    @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. – Niel de Beaudrap Nov 02 '18 at 16:01
  • @luminalQubit Hi. Welcome to Quantum Computing SE! You say: "I am unsure how the example given there scales efficiently as the number of qubits increases, nor how they actually arrive at the state". I suspect that you're confused about some particular steps/claims/assumptions in the answers to the duplicate question. If so, could you please elaborate which specific parts of those answers you do not understand, and edit it into the original question? Then we will be able to reopen your question. – Sanchayan Dutta Nov 02 '18 at 18:04

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