# Tag Info

4

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 one site, and teleporting all the other qubits to the relevant party. I wonder if this is actually all there is to it? If we assume that the entire graph is ...

2

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 placing a qubit on every vertex in the $|+\rangle$ state, and applying a controlled-phase gate along every edge. Now, take a bipartition of $G$. On either side of ...

2

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 measuring is entangled (which it very much is for the 2D cluster state), measuring the state of one qubit absolutely changes the state of another qubit. The trivial ...

1

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 defined as $\bar{a} = a^2$. Addition, multiplication and conjugation tables are then as follows: In this picture we have $0 \equiv 0$, $1 \equiv 1$, $2 \equiv x$, ...

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