# Tag Info

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 ...
• 58.6k

### 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 ...
• 12.1k
Accepted

• 451
Accepted

### 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 ...
• 58.6k
Accepted

### 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 ...
• 1,626

### 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 ...
• 126

### 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 ...
• 383
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 "...
• 4,331

### 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 ...
• 4,331