Can someone specify a quantum circuit that will deterministically output the $3$-qubit $W$ 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 $W$ 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 $W$ state in the computational basis, it is:

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

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). Please just avoid the trivial answer as written by Emilio below in the comments.

  • $\begingroup$ 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. $\endgroup$ – Emilio Pisanty Feb 13 '17 at 1:44
  • $\begingroup$ OP probably wants it using single and double qubit gates. $\endgroup$ – biryani Feb 13 '17 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 '17 at 23:39
  • 2
    $\begingroup$ What does W stand for in W-state? $\endgroup$ – W. Voltera Oct 21 '18 at 16:07

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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