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I am reading through the paper "Direct Fidelity Estimation from Few Pauli Measurements" (arXiv:1104.4695) and it mentions 'stabilizer state'.

"The number of repetitions depends on the desired state $\rho$. In the worst case, it is $O(d)$, but in many cases of practical interest, it is much smaller. For example, for stabilizer states, the number of repetitions is constant, independent of the size of the system..."

Adam Zalcman
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Quantum Guy 123
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1 Answers1

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Let $\mathcal{G}_n$ denote the Pauli group on $n$ qubits. An $n$-qubit state $|\psi\rangle$ is called a stabilizer state if there exists a subgroup $S \subset \mathcal{G}_n$ such that $|S|=2^n$ and $A|\psi\rangle = |\psi\rangle$ for every $A\in S$.

For example, $(|00\rangle+|11\rangle)/\sqrt2$ is a stabilizer state, because it is a $+1$ eigenstate of the elements of the following four-element subgroup of $\mathcal{G}_2$: $\{II, XX, -YY, ZZ\}$.

Stabilizer states have a number of interesting properties. For example, they are exactly the states that are reachable from $|0\dots 0\rangle$ using the Clifford gates and thus, by Gottesman-Knill theorem, any quantum computation that takes place entirely in the set of stabilizer states can be simulated efficiently on a classical computer.

The significance of stabilizer states in Direct Fidelity Estimation (DFE) lies in the fact that they are a prime example of well-conditioned states. The cost of DFE on such states is relatively low.

Adam Zalcman
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    great, thanks! Very well explained. If possible, can you also provide a reference to the definition? as in, a link to a textbook or paper? – Quantum Guy 123 Jul 01 '21 at 18:05
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    The concept is part of the stabilizer formalism for which a good reference is Daniel Gottesman's PhD thesis (even though its focus is on error correction and consequently on 2- and more dimensional stabilizer subspaces rather than 1-dimensional subspaces, i.e. stabilizer states). Also, as is the case with many fundamental concepts in QC, stabilizer formalism is introduced in Nielsen & Chuang. See section 10.5.1 on p.454. Finally, here is an example of a paper focused on stabilizer states themselves. – Adam Zalcman Jul 01 '21 at 18:44
  • great thanks!!! I read most of the section in Neilson and Chuang, prior to asking the question, but there was no mention stabilizer of states anywhere in the text book. Maybe I missed it?... – Quantum Guy 123 Jul 02 '21 at 22:05
  • It seems you are familiar with DFE, maybe you can answer my other question on the paper? :) https://quantumcomputing.stackexchange.com/questions/18260/proof-for-fidelity-rewritten-in-terms-of-pauli-expectation-values – Quantum Guy 123 Jul 02 '21 at 22:06
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    I think you're right: the index in N&C does not list the term, though they do define terms such as "stabilizer code". The original book is now over two decades old (1st ed edition is from 2000, 2nd edition is from 2010), so perhaps the term "stabilizer state" has emerged following the publication. Wrote an answer to your other question. – Adam Zalcman Jul 02 '21 at 22:57
  • added an interesting remark about all pure single qubit states being stabilizer states, in case it is useful for someone – Quantum Guy 123 Jul 03 '21 at 19:50
  • It is not true that all pure single-qubit states are stabilizer states. One way to see this is to note that if $|\psi\rangle$ is a stabilizer state, then there exists a non-identity Pauli operator $P\in\mathcal{G}_1$ such that $|\psi\rangle$ is a $+1$ eigenstate of $P$. However, almost all pure single-qubit states are not eigenstates of any non-identity Pauli operator at all. – Adam Zalcman Jul 03 '21 at 20:21
  • We can easily enumerate all single-qubit pure stabilizer states. Begin by listing all sixteen operators in $\mathcal{G}_1$. Discard the nine that don't have $+1$ among their eigenvalues. Discard the identity since it adds no constraint on the states (every state is a $+1$ eigenstate of $I$). We are then left with the following six operators $\pm X, \pm Y, \pm Z$. Each of them has a single $+1$ eigenstate. Thus, we arrive at the full list of single-qubit pure stabilizer states $|+\rangle$, $|-\rangle$, $|{+i}\rangle$, $|{-i}\rangle$, $|0\rangle$ and $|1\rangle$. – Adam Zalcman Jul 03 '21 at 20:31
  • Ah yes that makes sense, my previous definition didn't allow them to always be 'stabilized' by Paul operators. – Quantum Guy 123 Jul 06 '21 at 15:02
  • is the first part of your definition just a re-wording of theorem 1 from this link: https://arxiv.org/pdf/1711.07848.pdf you sent me? I'm confused by some of the symbols that appear after "such that". It looks like it's saying there is an isomorphism between the group of stabilizers for the state and the group of vectors of dimension k composed of elements that are integers modulo 2... Is this correct? I think I'm missing something. – Quantum Guy 123 Jul 07 '21 at 20:47
  • found a reference that more closely matches the first part of your definition, (https://arxiv.org/pdf/1808.10786.pdf) second paragraph of the second page "In a nutshell, an n-qubit stabilizer state..." – Quantum Guy 123 Jul 09 '21 at 16:05
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    In your answer you describe $ { II,XX,YY,ZZ } $ as a four element subgroup, so it seems that you are using the convention $ Y=XZ $. However in your comment about single qubit stabilizer states you claim that $ \pm Y $ has a $ +1 $ eigenstate, which would imply you are using the convention $ Y=iXZ $. Maybe I'm missing something but I think there is a small inconsistency here? – Ian Gershon Teixeira Aug 19 '22 at 15:51
  • @IanGershonTeixeira Yes, I was using the convention $Y=iXZ$, but made a sign error listing group elements. Fixed now. Thanks for catching! – Adam Zalcman Aug 20 '22 at 23:43