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With $\{ |e\rangle_j \}_{j=1}^{dim. \mathcal{H}_A}$ for $\mathcal{H}_A$ and $\{|f\rangle_j \}_{j=1}^{dim. \mathcal{H}_B}$ for $\mathcal{H}_B$, the product state reads \begin{equation} |u\rangle \otimes |v\rangle = \left(\sum\limits_{j} a_j |e\rangle_j \right) \otimes \left(\sum\limits_{k} b_k |f_k\rangle \right) = \sum\limits_{j,k} a_j b_k |e_j\rangle \otimes |f_k\rangle, \end{equation} where $a_j$, $b_k$ are complex number.

What is the expalaination of the following statement: If the vectors $|u\rangle$ and $|v\rangle$ do not belong to the respective orthonormal bases then there are at least two coefficients $a_j$ and similarly at least two coefficients $b_k$. From this, we conclude that the state is entangled.

Edit: Here is the original version from the book The Mathematical Language of Quantum Theory by Mário Ziman and Teiko Heinosaari (Chapter 6).

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  • $\begingroup$ to be clear: $|u\rangle\otimes|v\rangle$ is a product state, and thus not entangled, regardless of what $|u\rangle$ and $|v\rangle$ are. $\endgroup$
    – glS
    Oct 31 at 21:20
  • $\begingroup$ Thanks, @glS, could you explain how the authors conclude entanglement in the argument given below the equation $\eta=...$? $\endgroup$
    – Micheal
    Nov 1 at 9:43
  • $\begingroup$ I think you're misquoting the authors. The $a_j$ they refer to is not a coefficient in the expansion of $|u\rangle$ (or $|v\rangle$). Rather, those are Schmidt coefficients, which equal the singular values of the matrix $A$ with components the matrix elements of the state, so in this case I guess $A_{ij}=\langle i,j|\eta\rangle$. To be fair though, I don't quite understand the connection between first and and second part in the text you quote. $\endgroup$
    – glS
    Nov 1 at 10:25
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It essentially said that if $|u\rangle$ belongs to the orthonormal basis of $\mathcal{H}_A$ then it can be identify with a particular $|e_j\rangle \in \{ |e_j\rangle \}_{j=1}^{dim \mathcal{H}_A}$. For instance, $|u\rangle = |e_3\rangle$. Similarly, if $|v\rangle$ belongs to the orthonormal basis of $\mathcal{H}_B$, then it can be identify with a particular $|f_j\rangle \in \{ |f_j\rangle \}_{j=1}^{dim \mathcal{H}_B}$, for example $|v\rangle = |f_2\rangle $. In such case, $|u\rangle \otimes |v\rangle = |e_3\rangle\otimes|f_2\rangle$

However, if $|u\rangle$ is not one of the element of the orthonormal basis of $\mathcal{H}_A$, then you can't identity $|u\rangle$ with one of the $|e_j\rangle$ but rather express $|u\rangle$ as a linear combination of them (at least two). For instance, $|u\rangle = a_1 |e_1\rangle + a_2 |e_2 \rangle$, note $a_1, a_2 \neq 0$. Similar with $|v\rangle$, says $|v\rangle = b_1|f_1\rangle + b_2|f_2\rangle$ with $b_1,b_2 \neq 0$. So what happen now is that when you perform $u \otimes v$, you no longer have a single state $|e_j\rangle \otimes |f_k\rangle$ but rather a linear combinations.

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  • $\begingroup$ Thanks, @KAJ226, but won't $|u> \otimes |v>$ still be a separable state? $\endgroup$
    – Micheal
    Oct 31 at 11:13
  • $\begingroup$ Yeah, not sure what the author meant here. What textbook/article is this from? By def, if we can write $|\psi\rangle \in \mathcal{H}_{AB}$ as tensor product of $|u\rangle \in \mathcal{H}_A$ and $|v\rangle \in \mathcal{H}_B$ then it is separable. But without seeing a more detail description, it is hard for me to say. It probably relates to the Schmidt (SVD) decomposition here. I will delete this answer soon as it not very helpful in anyway. $\endgroup$
    – KAJ226
    Oct 31 at 17:43
  • $\begingroup$ Please see my edit. $\endgroup$
    – Micheal
    Oct 31 at 19:58

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