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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 ...


6

To obtain the stabilisers of a graph state, from its adjacency matrix: Change all 1s to Zs Change all 0s to identity operators Put X operators on the diagonal Each row then represents a stabiliser of the graph state, and any nontrivial stabiliser is a product of one or more rows.


6

There are $S(n) = 2^n \prod_{i=1}^n (2^i + 1)$ $n$-qubit stabilizer states, as per Corollary 21 of D. Gross, Hudson's Theorem for finite-dimensional quantum systems, J. Math. Phys. 47, 122107 (2006). Here are some simple-to-state bounds on $S(n)$: $2^{(n^2 + 3n)/2} \leq S(n) \leq 3 \cdot 2^{(n^2 + 3n)/2}$


6

It is not necessary to define the group as commuting —$\def\ket#1{\lvert#1\rangle}$ by virtue of every element in the group stabilising the state $\ket{\psi}$, this property follows. Because we are considering subgroups of the $N$-qubit Pauli group, any two elements either commute or anti-commute. Let $P \in \mathcal P_N$ be an operator which ...


4

The state $\psi$ (this is denoting the density matrix, even though it's a pure state) can be described as a sum of all the products of the stabilizers. We are promised that $X_i$ is not in the stabilizer, so every term in the sum of $\psi$, when multiplied by $X_i$, returns a tensor product of terms that is not just identity. Hence, it has zero trace. Thus, $...


4

First of all, keep in mind that there is no efficient way of generating them, simply because the number of stabiliser states increases super-exponentially with the number of qudits $n$ (like $p^{O(n^2)}$). Second, I can only give you hints how to do it here, since a detailed treatise would take too long. Sorry. As far as I know, there is also no literature ...


3

The way that I'd do it is to write out the stabilizers in a $10\times 8$ matrix in this case (number of rows= number of stabilizers, number of columns is double the number of qubits). For each row, take a stabilizer and write out, for the first 4 columns, if there's an $X$ on a given qubit, and in the last 4 columns, if there's a $Z$ on a given qubit (...


3

I think stim is the right tool for the job here, because it gives you access to the stabilizer generators and also it defines stim.PauliString which you can use to represent the stabilizers and more easily implement the manipulations being described in the paper. Ultimately, it will be up to you to translate what the paper wants into a specific combination ...


3

You can pretty easily prove by counting that specifying a stabilizer operation or a stabilizer state requires $\Omega(n^2)$ bits. If you're not tracking some of those bits, your simulator is definitely wrong. So clearly tracking one generator or ten generators ($\Theta(n)$ bits) is not enough. The tricky trouble then is that for certain gates we don't know ...


3

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 ...


3

You need to watch out when you say 'stabilizer' or 'stabilizers' because there is a little bit of ambiguity in that terminology$^{1}$. The stabilizer $\mathcal{S}$ of a state $|\psi \rangle$ is the group of $n$-qubit Paulis of which $|\psi \rangle$ is a $+1$ eigenstate. That is, $|\psi \rangle$ is the shared $+1$ eigenspace of all these operators. We can ...


3

If you start with one graph state, which is an eigenstate of stabilizer (each of which comprises an X tensored with a bunch of Zs), then the other eigenstates of those stabilizers are the original state acted on by Z rotations (take all possible combinations). To see this note the commutation and anticommutation relations between a tensor product of Zs and ...


2

Here's a necessary condition that might help recognise potential stabilizer states. I'll state it for qubits as that's what I'm used to thinking about, but I suspect it can be generalised: all the non-zero amplitudes of a stabilizer state must have the same magnitude. To see this, let's assume that the state $|\psi\rangle$ is an $n$-qubit stabilizer ...


2

The post-selection idea can be made to work without involving an entangled state. All that matters is that the initial state, before post-selection, has some overlap with the correct final state. I'm unsure how to find or prove that a specific easy-to-prepare state has overlap with every possible stabilizer state. It seems like just ensuring each qubit is ...


2

I'm not sure I understand the question, since this seems quite straightforward. Graph states are Clifford states, so for a state on $n$ qubits, the set of stabilizers has $n$ generators looking like $$i^k P_1\otimes\ldots\otimes P_n$$ where $k\in\{0,1,2,3\}$ and the $P$'s are Pauli $X,Y,$ or $Z$ operators. (For graph states, these stabilizers look like $\...


2

Here are a couple of observations which will hopefully clarify things. Only some states have Pauli stabilisers. You have correctly identified that not all states have Pauli stabilisers. An example of a state that does not have any Pauli stabilisers (apart from the identity operator of course, which we typically ignore) is $\lvert A \rangle = \tfrac{1}{\...


2

Let's make a few observations first: Since an $N$-qubit stabilizer state can be generated starting from $| 0 \rangle^{\otimes N}$ and applying H, CNOT, and S gates, we make the following observations. By simply applying the Hadamard on all qubits, one can generate the state $ | + \rangle^{\otimes N}$ which is maximally coherent (under the free operations ...


2

A partial explanation is motivated by the proof in Theorem 1 (bottom of page 2). Assuming two locally non-commuting stabilizing operators, using the Cauchy-Schwarz inequality, for pure product states it is shown that $$\langle S_{l}^{(GHZ_N)} + S_{m}^{(GHZ_N)} \rangle \leq 1.$$ But since we also assume these are stabilizer operators (GHZ eigenstate with ...


2

As you say, non-degenerate codes have a lot of well-understood machinery that's brought in from the classical side. That helps us from a conceptual stance, and a mathematical one (making rigorous results easier to prove), although, in terms of practical implementation, doesn't necessarily mean that one is easier to implement than the other. Errors in a ...


2

You can use stim for this, although you do have to write the stabilizer projection procedure for yourself. Write some methods to project a system into the +1 eigenstate of several stabilizers: from typing import List import stim def find_compatible_tableau(stabilizers: List[stim.PauliString]) -> stim.Tableau: num_qubits = max(len(e) for e in ...


1

Suppose you have a chain of length $n$. Then the smallest amplitude in that chain is no larger than $2^{-n}$. But this implies the operations you are applying have a maximum error term $\epsilon$ that is smaller than that, since otherwise they would overwhelm that amplitude. And approximating arbitrary rotations to within $\epsilon$ requires $\Omega(\lg(1/\...


1

With the help of Craig Gidney and some others, I was able to pin down the procedure to calculate the entropy. Here are the steps. Create your circuit with a stabilizer simulator This can be done with whatever simulator you want. As Craig mentioned in his answer, Stim is a great tool for the job. The rest of the answer in this section will assume you're using ...


1

Suppose you have a logical qubit encoded using a stabilizer code with generators $g_1, g_2, \dots, g_k$. If you measure all the generators and each of the measurements yields $+1$ then you know that the state of your logical qubit belongs to the code subspace. On the other hand, if one or more of then measurements of the generators returns $-1$ then you know ...


1

These are related to simulation of a quantum device on a classical computer. If all the gates in a circuit are H, S, and CNOT, then the circuit can be simulated on a classical computer in polynomial time. This is what the stabilizer simulator does. If the circuit is composed mostly of H, S, and CNOT, but there is also a small number t of T gates, then the ...


1

If you know the quantum circuit for generating a particular state, starting from the all-zero state, it's easy enough to work out the stabilizers. You just start with stabilizers $K=III\ldots IZII\ldots I$, where you have one with a $Z$ on each qubit (i.e. the stabilizers of the all-zero state), and you just update them to $UKU^\dagger$. Particularly if you'...


1

The way that I approach the calculation is quite different. I think of $$ e^{i\frac{\theta}{2}\sum_n X_i}=\prod_ne^{i\frac{\theta}{2}X_i}. $$ So, this means that you can consider the term $$ e^{i\frac{\theta}{2}\sum_n X_i}S_Me^{-i\frac{\theta}{2}\sum_n X_i} $$ on a qubit-by-qubit basis. You'll have terms like $$ e^{i\frac{\theta}{2}X}Ze^{-i\frac{\theta}{2}X},...


1

I have written a collection of C programs together with a python interface that satisfies most of the requiremets listed above. Documentation see section https://mmgroup.readthedocs.io/en/latest/api.html#the-subgroup-2-1-24-co-1-of-the-monster-and-the-clifford-group in https://mmgroup.readthedocs.io/en/latest/ . My motivation for doing so was that high-speed ...


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