The $ n $ qubit $ W $ state is defined here https://en.wikipedia.org/wiki/W_state

The stabilizer rank of a quantum state $|\psi\rangle$ is the minimal $r$ such that \begin{equation} |{\psi}\rangle = \sum_{j=1}^{r} c_j |φ_{j}\rangle. \end{equation} for $c_j \in \mathbb{C}$ and stabilizer states $|φ_j\rangle$.

What is the stabilizer rank of the $ n $ qubit $ W $ state?

The $ 1 $ qubit $ W $ state is just the ket $| 1 \rangle$, a stabilizer state with stabilizer generator $ -Z $. And the $ 2 $ qubit $ W $ state is the Bell state $ |{\Psi^+}\rangle= \frac{1}{\sqrt{2}} \Big(|{01}\rangle+|{10}\rangle \Big) $, a stabilize state with stabilizer generators $ -ZZ, XX $. Thus for $ n=1,2 $ the stabilizer rank is $ 1 $.

It is known, see What is the stabilizer group of a $|W\rangle$ state? , that for all $ n \geq 3 $ the $ W $ state is not a stabilizer state, thus the stabilizer rank is strictly greater than $ 1 $ for all $ n \geq 3 $.

And the stabilizer rank of the $ n $ qubit $ W $ state is at most $ n $ since we can just use all the weight $ 1 $ computational basis kets as our $ n $ stabilizer states. Thus the stabilizer rank must be some number at most $ n $ but strictly greater than $ 1 $.


1 Answer 1


I think it's doable to get at least an upper bound of $\lceil n/2 \rceil$ for the stabiliser rank of $W_n$.

Note that, as you wrote, $W_2 = |01\rangle + |10\rangle $ is a stabiliser state, so has stabiliser rank 1.

But since we have $W_4 = W_2 \otimes |00\rangle + |00\rangle \otimes W_2$, this shows that $\chi(W_4) \leq 2$.

More generally, we can in this way we see that from $$W_{2^k} = W_{2^{k-1}} \otimes |0\dots0\rangle + |0\dots 0\rangle \otimes W_{2^{k-1}}$$

we obtain $\chi(W_{2^k})\leq 2^{k-1}$.

This means for all $n = 2^k$ that $\chi (W_n) \leq n/2$.

Now, say $\chi (W_n) = f(n)$. Then,

$$W_{n+1} = |10\dots 0\rangle + |0\rangle \otimes W_n \Rightarrow \chi(W_{n+1}) \leq \chi(W_n) + 1$$

$$W_{n+2} = (|10\rangle + |01\rangle ) \otimes |0\dots 0\rangle + |0\rangle \otimes W_n \Rightarrow \chi(W_{n+1}) \leq \chi(W_{n+2}) + 1$$

So $f(n)$ satisfies $f(n+2) \leq f(n) + 1$. So in particular, $f(n) \leq \lceil n/2 \rceil$ by induction.

Of course finding lower bounds on stabiliser rank is an extremely subtle topic, so these kind of observations are the low hanging fruits. For example, if the stabiliser rank of $W_n$ would be really $\lceil n/2 \rceil $, it would most likely be quite tricky to find a proof for the lower bound...

  • $\begingroup$ Nice induction! So beyond the observation in the question that $ \chi(W_1)=1=\chi(W_2) $ your upper bound proves that $ \chi(W_3)=2=\chi(W_4) $. Is there any way to rule out that, for example, the possibility that $ \chi(W_n)=2 $ for all $ n \geq 3 $? $\endgroup$ Commented Jun 5 at 23:52
  • $\begingroup$ My approach would be the following: The support of a stabiliser state in is always an affine subspace of $\mathbb{F}_2^n$, see, e.g., eq (1) in arxiv.org/pdf/2110.07781. I would guess that one would really have to get their hands dirty and look at all affine subspaces of $\mathbb{F}_2^5$ to show that the stab rank of $W_5$ is 3. $\endgroup$
    – Vincent
    Commented Jun 6 at 9:22
  • $\begingroup$ The proof of theorem 4.1 in arxiv.org/pdf/2110.07781 has similar flavour arguments with the affine subspaces of $\mathbb{F}_2^4$. It's in a different situation there, but I think the flavour is kind of similar :) $\endgroup$
    – Vincent
    Commented Jun 6 at 9:27

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