What is the stabilizer rank of the W state?

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 $$$$|{\psi}\rangle = \sum_{j=1}^{r} c_j |φ_{j}\rangle.$$$$ for $$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 $$1$$.

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...

• 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 $n \geq 3$? Commented Jun 5 at 23:52
• 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. Commented Jun 6 at 9:22
• 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 :) Commented Jun 6 at 9:27