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From this question, gls states that given $\Pi\equiv\sum_i |\eta_i\rangle\!\langle i|$ and $\Psi\equiv\sum_i|\psi_i\rangle\!\langle i|$, if $\Pi^\dagger\Psi=I_{d\times\ell}$, then $\Psi$ is "maximally entangled", ie has rank $\ell$ and all singular values are equal to 1. By maximal entanglement, what does that mean exactly, in the context used for matrices here? Is it referring to the inability to decompose it to a product of matrices on subsystems, or something else? Because if I recall correctly, CNOT, which is non-decomposable, doesn't have 1 for all it's singular values, yet it fits the criteria of non-decomposable, but then wouldn't be maximally entangled?

Edit: I am assuming "maximally entangled" means in this context it is diagonal up to a certain dimension $\ell$

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I think the terminology makes sense if you think the matrix $ \Psi $, via the linear bijection $ vec\big( |b \rangle \langle a|\big) = |b \rangle |a \rangle$, as a pure bipartite state $ vec(\Psi) = |\psi \rangle_{AB} = \frac{1}{\sqrt{l}} \sum_{i=1}^{l} | \psi_i \rangle_A | i \rangle_B $ and observe that the reduced density matrix is $$ \rho_B = \text{Tr}_A[\rho_{AB}] = \frac{1}{l} \sum_{i,j} \langle \psi_j | \psi_i \rangle \cdot | i \rangle \langle j| = \frac{1}{l} \cdot \big(\Psi^\dagger \Psi\big)^T $$

So if $ \Psi^\dagger \Psi = I_l $, then $ \rho_{AB} = |\psi \rangle \langle \psi |_{AB} $ is "maximally entagled".

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  • $\begingroup$ I don't follow when you say "if you think $\Psi$ as a pure bipartite state". I assume $|\psi\rangle_{AB}$ is actually meant to be $\sum_{i}|\psi_{i}\rangle_{A}|i\rangle_{B}$ I also assume this isn't a "state" as such given it's not normalised. But I don't understand what you mean by "think of the matrix as a pure bipartite state". $\endgroup$ Oct 9 '20 at 12:49
  • $\begingroup$ Are you saying that if I can equate $\Psi^\dagger \Psi$ with the partial trace over a maximally entangled state, wherein the states of ket's and bra's are taken to pure states of separate subsystems, each of which exists as parts of an operator, then that operator is also maximally entangled. Is there a concrete method involved here? $\endgroup$ Oct 9 '20 at 12:50
  • $\begingroup$ Fixed some typos, added the normalization for clarity and gave a mathematical definition of "think of the matrix as a pure bipartite state". Not a concrete method that i'm aware of, just trying to make a connection with a more commonly used terminology $\endgroup$
    – tsgeorgios
    Oct 9 '20 at 14:06
  • $\begingroup$ Ah I see what you are getting at now, and given this approach, and take $\ell$ as $d$, we get the maximally mixed state $\frac{1}{\ell}I$ $\endgroup$ Oct 9 '20 at 14:07
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The other answer is already pretty on point with what I meant, but just to restate the same thing in different words:

Given a matrix $\Psi\in\mathrm{Lin}(\mathcal X,\mathcal Y)$, whose singular value decomposition reads $\Psi=\sum_i \sqrt{p_i}|u_i\rangle\!\langle v_i|$, we can define the corresponding vectorization as the vector $$\operatorname{vec}(\Psi)=\sum_i\sqrt{p_i}|u_i\rangle\otimes|v_i\rangle\in\mathcal Y\otimes\mathcal X,$$ (conventions may vary as to the order of the spaces after the vectorization). Notice that $\operatorname{vec}(\Psi)$ is a proper state (i.e. it's normalised) iff $\|\Psi\|_2^2=\sum_i p_i=1$.

The entanglement of $\operatorname{vec}(\Psi)$ as a pure bipartite state (if it is a state) is encoded in its Schmidt coefficients, which are the singular values of $\Psi$. More generally, $\Psi$ might not correspond to a state upon vectorization, in which case calling it "maximally entangled" is just abuse of notation to refer to a specific feature of its singular values. I should note that this is not by any means standard notation, it just came natural to use the term in the context because formally a state being maximally entangled is really the same thing as the singular values of a matrix being equal.

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