# How can we show that an arbitrary state $|\Psi\rangle\in \mathbb{C}^{2}$ is maximally entangled?

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 $$\exists \text { a fully entangled state }\left|E_{2}\right\rangle \text { so that } \mathrm{CNOT}\left|E_{2}\right\rangle=\left|E_{2}\right\rangle,$$ $$$$\mathrm{CNOT} = \left(\begin{array}{cc} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0 \end{array}\right)$$$$ I found $$\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$$ as follows: $$$$\cup \otimes \vee = \left(\begin{array}{cc} 1 & 0\\ 0 & 1 \end{array}\right) \left(\begin{array}{cc} 1 & -1\\ 1 & 1 \end{array}\right).$$$$ However, I do not like my procedure, and want a rigorous way to find the conditions for a general state.

• 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
– glS
Commented Oct 8, 2020 at 17:11
• if that comment is directed to me, I have no idea what you meant to say
– glS
Commented Oct 8, 2020 at 17:19
• 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. Commented Oct 8, 2020 at 17:23
• so these are two completely separate problems? I feel like you don't really know what exactly you are asking
– glS
Commented Oct 8, 2020 at 17:28
• @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 $\left|\phi^{+}\right\rangle=\cup \otimes \vee|\E_{2}\rangle$ Commented Oct 8, 2020 at 17:39

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 $$I/2$$.

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 $$\vert\phi^+\rangle =I\otimes H\vert\psi\rangle$$