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

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

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 ...


1

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 $\...


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