Can someone specify a quantum circuit that will deterministically output the $3$-qubit $|W \rangle$ state, if the input to the circuit is $|0,0,0 \rangle$? Or, is there a quantum circuit with a different $3$-qubit input state that will output the $3$-qubit $|W \rangle$ state deterministically? I am looking for an abstract description, devoid of any physical system. If this is contained in a reference, one can just list the reference.

For those not familiar with the representation of the $3$-qubit $|W \rangle$ state in the computational basis, it is:

$$|W \rangle = \frac{1}{\sqrt{3}} \bigg( |0,0,1\rangle + |0,1,0\rangle + |1,0,0\rangle \bigg) .$$

Generalizations to the $n$-qubit case would also be greatly appreciated. For clarity on the allowable gates in the circuit, please use either $1$-qubit or $2$-qubit gates. Also, it'd be fine if the answer used a $3$-qubit gate such as Toffoli or Fredkin (and these $3$-qubit gates would not need to be further decomposed).

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    $\begingroup$ You should specify what gates you're prepared to accept in your quantum circuit for your question to be well-posed. Otherwise, I could just say "let $U$ be the three-qubit gate that takes ..." and it would be a valid answer. Presumably you want to rule that out, but unless you provide actual hard lines, you're just making people guess at what you'll find acceptable or not. $\endgroup$ Feb 13, 2017 at 1:44
  • $\begingroup$ OP probably wants it using single and double qubit gates. $\endgroup$
    – biryani
    Feb 13, 2017 at 4:23
  • $\begingroup$ Emilio this seems pedantic, but I see your point. @biryani thanks, you got what I was after. $\endgroup$
    – sunspots
    Feb 13, 2017 at 23:39
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    $\begingroup$ What does W stand for in W-state? $\endgroup$
    – W. Voltera
    Oct 21, 2018 at 16:07
  • $\begingroup$ @W.Voltera perhaps "Wolfgang," since the state emanates from arxiv.org/pdf/quant-ph/0005115.pdf $\endgroup$
    – sunspots
    Jul 9, 2019 at 1:37

1 Answer 1


Given a gate schema $G(p) = \begin{bmatrix} \sqrt{1-p} & -\sqrt{p} \\ \sqrt{p} & \sqrt{1-p} \end{bmatrix}$, you can use $G(1/3)$ to make a circuit that does the trick.

First you make a balanced superposition of three states. Any three states. That's where the $G(1/3)$ comes in. We'd also need a $G(1/2)$ but $H$ will do the trick. Second, you permute the states so the non-zero amplitudes are on the states you want.

For example, I created a W-state circuit in my quantum simulator Quirk. The inline state displays make it a bit easier to see what's going on:

W-state circuit

$G(1/3)$ is not really a "standard" gate. Since I didn't want to use some crazy mix of allowed gates to approximate it, I just defined it as a custom gate:

Defining the gate

But as the comments on the question point out, you can also just directly define a gate that maps straight from $|000\rangle$ to the W-state. The core of whether this problem is easy or hard, whether it requires approximations or can be done exactly, is the question what gates do you allow yourself?

0-to-W gate definition


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