# Tag Info

5

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 $U_{ab}|+\rangle\otimes |+\rangle = V_a \otimes V_b \frac{1}{\sqrt{2}}\left(|00\rangle + |11\rangle\right)$ In terms of its entanglement properties, Equation * is completely ...

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 ...

4

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 you apply a Hadamard to the second qubit, you change it to $(|0\rangle|0\rangle+|1\rangle|1\rangle)/\sqrt{2}$, a Bell state. The point is that from the ...

2

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 in the bit string is $k$. The probability of this particular error arising is then $p^k(1-p)^{n-k}$ because there are $k$ sites with an error and $n-k$ sites ...

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 ...

2

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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