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I was reading about of entangled states and I encountered a concept which is called "fully entangled state" according to the following definition:

Consider a two-qubit state $|\Psi\rangle$. We say that $|\Psi\rangle$ is fully entangled if there exist two one-qubit unitaries $\cup, \vee \in \mathbb{C}^{2 \times 2}$ such that $\left|\phi^{+}\right\rangle=\cup \otimes \vee|\Psi\rangle,$ where $\left|\phi^{+}\right\rangle=\frac{1}{\sqrt{2}}(|00\rangle+|11\rangle)$ is the EPR-pair.

Then I considered a two-qubit general state like $a|00\rangle+b|01\rangle+c|10\rangle+d|11\rangle$ and I tried to show that under what condition this state is fully entangled. I form a $4\times 4$ matrix that represents $\cup \otimes \vee$ and then I had $4$ equations with $8$ unknowns ($4$ belongs to the $2\times 2$ matrix $\cup$ and 4 belongs to the $2\times2$ matrix $\vee$). But the proof should show that there exist $\cup$ and $\vee$ separately.

For example, there was an exercise which wants us to prove that $$\exists \text { a fully entangled state }\left|E_{2}\right\rangle \text { so that } \mathrm{CNOT}\left|E_{2}\right\rangle=\left|E_{2}\right\rangle,$$ \begin{equation} \mathrm{CNOT} = \left(\begin{array}{cc} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0 \end{array}\right) \end{equation} I found $\cup$ and $\vee$, but not in a proof manner. I considered $\left|E_{2}\right\rangle=|00\rangle-|01\rangle+|10\rangle+|11\rangle$ and I found $\cup$ and $\vee$ as follows: \begin{equation} \cup \otimes \vee = \left(\begin{array}{cc} 1 & 0\\ 0 & 1 \end{array}\right) \left(\begin{array}{cc} 1 & -1\\ 1 & 1 \end{array}\right). \end{equation} However, I do not like my procedure, and want a rigorous way to find the conditions for a general state.

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    $\begingroup$ to be clear, you are asking how to prove that a generic maximally entangled two-qubit state has the form $U\otimes V |\phi^+\rangle$ for some unitaries $U,V$, correct? I do not understand what's the connection with the exercise about the CNOT you mention later $\endgroup$
    – glS
    Oct 8, 2020 at 17:11
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    $\begingroup$ if that comment is directed to me, I have no idea what you meant to say $\endgroup$
    – glS
    Oct 8, 2020 at 17:19
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    $\begingroup$ Note that "fully entangled" is ambiguous language and doesn't actually mean anything. The concept described by your quote is known as a maximally entangled state. $\endgroup$ Oct 8, 2020 at 17:23
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    $\begingroup$ so these are two completely separate problems? I feel like you don't really know what exactly you are asking $\endgroup$
    – glS
    Oct 8, 2020 at 17:28
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    $\begingroup$ @glS My dear friend, forget maximally entangled, we want to show there exists a stat which satisfy two condithion, one is $\mathrm{CNOT}\left|E_{2}\right\rangle=\left|E_{2}\right\rangle$ and $\left|\phi^{+}\right\rangle=\cup \otimes \vee|\E_{2}\rangle$ $\endgroup$ Oct 8, 2020 at 17:39

2 Answers 2

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I'll assume the question is, given the state $|\Psi\rangle=a|00\rangle+b|01\rangle+c|10\rangle+d|11\rangle$, how can we figure out whether this state is maximally entangled, as that is the question in the title of the post.

A bipartite state $|\Psi\rangle$ is maximally entangled if and only if the corresponding reduced state $\rho_A\equiv\operatorname{Tr}_B(|\Psi\rangle\!\langle\Psi|)$ is maximally mixed, that is, $\rho_A=I/2$ where $I$ is the $2\times2$ identity matrix.

You therefore just take $|\Psi\rangle$, compute the partial trace, and verify it equals $I/2$.

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The updated question is to show that there exists a state that satisfies $CNOT\vert\psi\rangle = \vert\psi\rangle$ and $U\otimes V\vert\psi\rangle = \vert\phi^+\rangle$

The eigenstates of $CNOT$ with eigenvalue $1$ are $\vert 00\rangle$, $\vert 01\rangle$ and $\vert 1+\rangle$. Hence, our state is of the form

$$\vert\psi\rangle = a\vert 00\rangle + b\vert 01\rangle + c\vert 1+\rangle$$

You can work out the reduced states and check for what $a,b,c$ you get maximally mixed reduced states. Alternatively, you see that choosing $a = \frac{1}{2}, b = -\frac{1}{2}, c = \frac{1}{\sqrt{2}}$ gives you $$\vert\psi\rangle = \frac{1}{\sqrt{2}}\vert0-\rangle + \frac{1}{\sqrt{2}}\vert1+\rangle$$

This can be converted to $\vert\phi^+\rangle =I\otimes H\vert\psi\rangle$

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