9 votes
Accepted

How can one check whether a given quantum state is a graph state?

Remember that a graph state is simply the $|+\rangle$ state on every qubit together with a bunch of controlled phases enacted between them. So, assuming you have a list of the probability amplitudes ...
DaftWullie's user avatar
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6 votes

Is there a tool that shows me all $2^n$ stabilizers for a given graph state?

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 ...
Niel de Beaudrap's user avatar
5 votes
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Query on paper on entanglement in graph states

The state of Equation * is local unitary equivalent to all of the four Bell states. That is, there exists a unitary matrix of tensor-product structure $V_a \otimes V_b$, such that $U_{ab}|+\rangle\...
Felix Huber's user avatar
5 votes

QiskitError: 'No statevector for experiment

When running on real backends you can't get a statevector since you can't return the quantum state of the device, only get measurement outcomes. The get_statevector ...
Matthew Treinish's user avatar
5 votes
Accepted

How are the two definitions of graph state mathematically equivalent?

The first definition says the nodes start in the $|+\rangle$ state and then you apply a CZ for each edge. The $|+\rangle$ state's stabilizer is $X$. When you conjugate an $X$ observable by a CZ ...
Craig Gidney's user avatar
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4 votes
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Graph States subjected to finite erasures

The idea of erasure being projection onto $|0\rangle$ is perhaps misleading in this context (my fault for mentioning it in a comment without having looked at the full details of what this specific ...
DaftWullie's user avatar
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4 votes
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Dephasing in graph states

Imgine that there is a bit string $\vec{k}\in\{0,1\}^n$. We use this to specify sites (bit value 1) where an error has occurred, and sites (bit value 0) where no error has occurred. The number of 1s ...
DaftWullie's user avatar
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4 votes

Query on paper on entanglement in graph states

Felix Huber's answer is correct, but I thought I'd fill in a few more details. You've written the two-qubit state of ($\star$) as $(|0\rangle|+\rangle+|1\rangle|-\rangle)/\sqrt{2}$. Notice that if ...
DaftWullie's user avatar
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4 votes

How to calculate the number of ebits in a graph state?

The number of Bell pairs required to construct a given graph state can easily be given an upper bound: $|V|-1$, where $V$ is the set of vertices. You do this simply by preparing the entire state at ...
DaftWullie's user avatar
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4 votes

Creating a specific cluster state

You can think of a cluster state as a graph state, where the graph's vertices are on some $d$-dimensional lattice (normally just $2$-dimensional). Each vertex represents a qubit in the $|+\rangle$ ...
JSdJ's user avatar
  • 5,469
3 votes
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Constructing an eigenbasis of graph states for a set of stabilizers

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 ...
DaftWullie's user avatar
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3 votes

Graph state and maximally entangled state

I'm not familiar with how graph states extend to qudits, so let me just answer for the specific case of qubits. Consider a graph $G$, and we create the corresponding graph state $|G\rangle$ by ...
DaftWullie's user avatar
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3 votes
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how to go from a stabilizer state to a graph

See this paper for a proof, but I will sumamrize the idea below. Suppose you have a stabilizer state $|\psi\rangle$ and a set $\mathcal{S}$ of Paulis that generate its stabilizer group, $\langle S \...
xzkxyz's user avatar
  • 451
2 votes
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The classical simulation of 2D graph state and the measurement based quantum computation

the measurement is carried out on each individual qubit and the measurement on one qubit will not change the state of another qubit This is an incorrect statement. If the state that you are ...
DaftWullie's user avatar
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2 votes
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Does local Clifford equivalence have a direct graphical representation for qudit graph states of non-prime dimension?

It is incorrect to use modulo arithmetic in this context. Instead finite field arithmetic should be applied. In $\textrm{GF}(4) = \{0, 1, x, x^2\}$ where $x^2 = x + 1$ and conjugation of $a$ is ...
SLesslyTall's user avatar
  • 1,626
2 votes

Constructing an eigenbasis of graph states for a set of stabilizers

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 ...
NNN's user avatar
  • 373
2 votes
Accepted

Cluster/Graph state teleportation

The calculation of the resulting state of the described circuit: After applying the same circuit identities described in this answer (and here) to the connected question we will obtain a "...
Davit Khachatryan's user avatar
2 votes

Quantum Circuit explaination

Because $CNOT = I\otimes H \cdot CZ \cdot I\otimes H$ as was mentioned here, and because $CZ(q_1, q_2) = CZ(q_2, q_1)$, we can rewrite the circuit in this way (by adding Hadamards as needed): The ...
Davit Khachatryan's user avatar
2 votes
Accepted

Quantum Circuit explaination

We need to go throught the gates one by one to understand what's happening. We have to keep a few things in mind. $H|0\rangle=\frac{1}{\sqrt2}(|0\rangle + |1\rangle) = |+\rangle$ and $H|1\rangle=\...
vasjain's user avatar
  • 802
2 votes
Accepted

Unitary interaction term of two-qubit graph state

Absolutely! Think about the eigenstates of $H_{ab}$: $|11\rangle$ has eigenvalue 1, while $|00\rangle, |01\rangle$ and $|10\rangle$ have eigenvalue 0. This tells us that the time evolution must be $$ ...
DaftWullie's user avatar
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2 votes

What does it mean to have 2000 qubits and 6016 couplers?

The vertices and edges are the number of qubits and connections in DWaves quantum processing unit topology. It is called Chimera Graph, see DWaves docs . The graph is not fully connected. Therefore an ...
tomtuamnuq's user avatar
2 votes

How can one check whether a given quantum state is a graph state?

If you want to find the answer experimentally, a possible (not necessarily optimal) way to do so is the following. Note that a graph state is stabilized by the generators $$g_v=X_v \prod_{w\in N(v)} ...
M. Stern's user avatar
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2 votes

Entanglement Witnesses close to GHZ states

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 ...
John Doe's user avatar
  • 881
2 votes

Possible typo in the paper "Graph states for quantum secret sharing"

Sure, I would say this is pretty clearly a typo. They very helpfully write out what they mean immediately after that statement so you can compare your understanding!
DaftWullie's user avatar
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2 votes
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Is a generalized 2D cluster state still a universal resource?

I think that a generalized 2D weighted graph state is a resource with $\textbf{specific}$ values in angles $\theta$ of the entangling gates. Not any pattern of weighted graph state is a universal ...
raf_quantum's user avatar
2 votes

Graph Limits in Quantum Computing

Because the adjacency matrix of undirected graphs are symmetric about the diagonal, these graphs are hermitian, and you are correct to suppose that quantum computing can be a natural vehicle for ...
Mark Spinelli's user avatar
1 vote

Distance of one dimensional quantum error correcting code

The distance of an $[[n, 0]]$ code is defined to be the smallest non-zero weight of any stabilizer.
esabo's user avatar
  • 196
1 vote
Accepted

Entanglement entropy for graph states defined on a tree graph

Yes. Think about the process of building a graph state: create qubits in the $|+\rangle$ state and apply controlled-phase gates along all the edges. Now take a bipartition. You can apply any unitary ...
DaftWullie's user avatar
  • 58.1k
1 vote
Accepted

Quantum Graph states hand computation

The calculation in the paper is correct. One of the easiest ways to think about this is that you start all qubits in the $|+\rangle$ state, so $|+++\rangle=(|0\rangle+|1\rangle)|++\rangle)/\sqrt{2}$. ...
DaftWullie's user avatar
  • 58.1k
1 vote
Accepted

Quantum Error Correcting Codes and Graphs

The distance of an error correcting code is the smallest number of single-qubit rotations that you have to apply to map one logical codeword into an orthogonal one. If I express it in terms of the ...
DaftWullie's user avatar
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