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Quick question on the paper Entanglement in Graph States. On page 14. a definition of a graph state:

Given $|+\rangle=\frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$. Let $G=(V,E)$ be a graph. The graph state $|G\rangle$ that corresponds to the graph $G$ is the pure state with vector $$|G\rangle = \prod_{\{a,b\}\in E}U_{ab}|+\rangle^{V} $$ where $$U_{ab}|+\rangle^a|+\rangle^b = \frac{1}{\sqrt{2}}(|0\rangle^a|+\rangle^b+|1\rangle^a|-\rangle^b)~~~~~(*)$$ is maximally entangled (Bell state).

Why is (*) stated as being a maximally entangled Bell state, this does not seem to correspond to any of the four maximally entangled Bell states?

Lastly, as an example, on page 37, it states

This is easily seen by applying Hadamard unitaries $H^{V\setminus a}$ to all but one qubit $a$ in the GHZ state, which yields the star graph state with $a$ as the central qubit.

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If we start with an empty graph, with vertices representing states $|+\rangle$, then how would Hadamard operations which are local on each qubit site going to produce an entangled state? Why are they not using the above definition, using $U_{ab}|+\rangle$, to define the GHZ state?

Thanks for any assistance.

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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 equivalent to the Bell state. You can check this for pure bipartite states by looking at the eigenvalues of its reduced density matrices.

Page 37 again refers to local unitary equivalence, where a tensor product of Hadamards on all but one qubit maps the GHZ state to the star graph state. Again, both states have the same entanglement properties.

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  • $\begingroup$ Thanks for your response. Suppose we define the unitary $U_H$ as a tensor product of Hadamards on all but one qubit, such that $$|G\rangle = U_H|GHZ\rangle.$$ Are you stating that, the graph state $|G \rangle$, corresponds to the graph state we would obtain, using the definition for graph states above (i.e. starting with states $|+\rangle$ and then applying $U_{ab}$ for each edge) given the star graph $G = (V,E)$? $\endgroup$ – John Doe Nov 1 '19 at 0:44
  • $\begingroup$ Exactly! Also see @DaftWullie 's answer for a less terse explanation. $\endgroup$ – Felix Huber Nov 1 '19 at 11:48
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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 perspective of entanglement, you cannot change how much entanglement is there by applying unitaries on individual qubits, so ($\star$) must be just as entangled as a Bell state. We say that it's equivalent up to local unitaries.

The GHZ case is very similar. The way I find it easiest to think about this case is to start with the $|+\rangle^{\otimes 7}$ state, before we apply controlled-phases, and split up the central qubit of the star: $$ \frac{1}{\sqrt{2}}(|0\rangle|+\rangle^{\otimes 6}+|1\rangle|+\rangle^{\otimes 6}). $$ Now, we apply the controlled-phase gate 6 times, each time controlled by the central qubit. So, if that qubit is 0, nothing happens. If that qubit is $|1\rangle$, a $Z$ is applied on all the other qubits. But $Z|+\rangle=|-\rangle$, so we're left with $$ \frac{1}{\sqrt{2}}(|0\rangle|+\rangle^{\otimes 6}+|1\rangle|-\rangle^{\otimes 6}). $$ This is locally equivalent to a GHZ state because if we apply Hadamards to those 6 non-central qubits, we get $$ \frac{1}{\sqrt{2}}(|0\rangle|0\rangle^{\otimes 6}+|1\rangle|1\rangle^{\otimes 6}). $$

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  • $\begingroup$ One last question regarding the paper. First page of the section on weighted graphs (page 71). The literature refers to a standard basis '$|W\rangle_Z$' such that any weighted graph can be expressed as $$|G\rangle = 2^{-N/2}\sum_{W \subseteq V}\prod_{\{a,b\}\in E}U_{ab}|W\rangle_z$$ What is the standard basis that is being referred to here? $\endgroup$ – John Doe Nov 2 '19 at 18:53
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    $\begingroup$ @JohnDoe it’s been a while since I looked at the paper, but I assume they mean the usual 0/1 basis, I.e. W is a bit string which is a 1 on the vertices in W and 0 everywhere else. $\endgroup$ – DaftWullie Nov 2 '19 at 19:14
  • $\begingroup$ Thus it is a basis of $(\mathbb{C}^2)^V$? $\endgroup$ – John Doe Nov 3 '19 at 15:37
  • $\begingroup$ @JohnDoe yes, but that’s not entirely helpful because there are many bases. $\endgroup$ – DaftWullie Nov 3 '19 at 17:42

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