# What conditions on the coefficients of a bipartite pure state imply it being entangled?

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). • 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.
– glS
Oct 31 at 21:20
• Thanks, @glS, could you explain how the authors conclude entanglement in the argument given below the equation $\eta=...$? Nov 1 at 9:43
• 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.
– glS
Nov 1 at 10:25

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.
• Thanks, @KAJ226, but won't $|u> \otimes |v>$ still be a separable state? Oct 31 at 11:13
• 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. Oct 31 at 17:43