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This question is related and complementary to this one: How to get the stabilizer group for a given state?

What I want is to find the stabilizer group generators for the following state:

$$|W\rangle = \dfrac{1}{\sqrt{3}}\Big(|011\rangle + |101\rangle + |110\rangle \Big)$$

In theory, I should find $n-k = 3-0=3$ independent non trivial generators. But the only one I can find is $M_1 = Z\otimes Z\otimes Z$ because any other combination, like $-Z\otimes Z\otimes -Z$ or $iZ\otimes iZ\otimes -Z$ actually is equivalent to the first one, and there cannot be a combination with $X$ because it would alter the difference between 0s and 1s that is conserved in each sum.

Where are the other two generators?

glS
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Dani
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  • See the second answer to the question you mention. A brute-force solution is to look up a circuit which can be used for constructing such a state, and to use the corresponding unitary for conjugating single-qubit $Z$s. – mavzolej Jun 05 '20 at 17:11
  • @mavzolej I don't know if I understood that answer correctly. I'd have to start from the stabilizer $ZZZ$ and apply some unitary operator $U$ to transform it into another stabilizer? And if that's the case, how do I choose $U$?

    In addition, I have tried all the combinations of tensor products from Pauli's group and I can only find this independent generator $ZZZ$

     – Dani Jun 05 '20 at 17:31
  • BTW, see this question. Not sure if it even makes sense to use the word 'stabilizer' in this context. If not, there is no contradiction if $M_1$ you mention is the only Pauli operator whose $+1$ eignvector is the given state. – mavzolej Jun 05 '20 at 17:50
  • I mean, you may speak of the "stabilizer group" of this state, but since the state itself is not a "stabilizer state", the counting argument for the rank of the stabilizer group does not work. See the definitions here. – mavzolej Jun 05 '20 at 18:15
  • I see. So can i state that $ZZZ$ is the only one for that state, don't? By stabilizer i mean the operators (within the Pauli group) that satisfies $M_i|W\rangle = |W\rangle$ – Dani Jun 05 '20 at 19:53
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    Well, if you literally checked the action of all Paulis one by one, and $ZZZ$ is the only one — I guess you can :) I just wanted to emphasize the difference between "stabilizer group" (which can be defined for any pure state) and "stabilizer states" (which you have a finite number). – mavzolej Jun 05 '20 at 20:29
  • related: https://quantumcomputing.stackexchange.com/q/3861/55 – glS May 25 '22 at 13:59

2 Answers2

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To cite from my answer from over at physics.SE: The W state is not a stabilizer state - for a stabilizer state, the 1-site reduced density matrices must be maximally mixed or pure, which they aren't.

Or, to phrase it without reduced density matrices: For a stabilizer state, if you measure $X$, $Y$, or $Z$ for any single qubit, the probability of getting either outcome is either $0$, $1/2$, or $1$. This is clearly not the case for a $Z$ measurement on the W state above, where the probability of getting $0$ is $1/3$.

Norbert Schuch
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  • I don't quite understand what you mean by 1-site reduced density matrices. Could you send me a reference on that? is that I've only been studying this subject for a short time – Dani Jun 07 '20 at 10:29
  • @Dani Any introductory quantum information textbook will do. – Norbert Schuch Jun 07 '20 at 21:41
  • @Dani I have added an explanation which does not require density matrices. You only need to know the basics of the stabilizer formalism. – Norbert Schuch Jun 08 '20 at 08:27
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As stated in the Schuch's answer: the W state is not a stabilizer state.

We can also find some operators which can "stabilizer" $W_{3}$. You can refer to Entanglement detection in the stabilizer formalism Page 9 Eq. 59 for $W_{3}$, and Efficient estimation of multipartite quantum coherence Eq.A.8 - A.11 in Appendix A for $W_{3}$ and $W_{4}$. You can find “-ZZZ” is inside after calculating.

Curry
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