# Decomposition of the maximally entangled states

We know that the set of symmetric bipartite pure states is spanned by $$S=\{|\phi\rangle^{\otimes 2},|\phi\rangle \in \mathbb{C}^d\}$$. I want to know if the maximally entangled state $$|\psi\rangle = \frac{1}{\sqrt{d}}\sum_i |i\rangle|i\rangle$$ can be spanned by a small portion of $$S$$. I.e., is it true that $$|\psi\rangle \in \operatorname{span} \{|\phi\rangle^{\otimes 2}, \,\,|\langle0|\phi\rangle| >1-s,|\phi\rangle \in \mathbb{C}^d\}$$for some small $$s$$? Furthermore, is it true that $$|\psi\rangle$$ is in the convex cone of such set?

Edit: If the answers are no, can we get a good approximation of $$|\psi\rangle$$ from such set?

Edit 2: Does it still work if I demand $$\langle0|\phi\rangle >1-s$$?

• @glS sorry. I actually mean $|\langle0|\phi\rangle| >1-s$. I assume $|\psi \rangle = |00 \rangle + \cdots$. I want to know if $|\psi \rangle$ is in the span of $S’$ where $S’$ is a small set around $|00 \rangle$ (The diameter of $S’$ is a function of $s$). – user193108 Apr 18 '19 at 10:55
• $|\psi\rangle$ is in the span of $S’$. But I don’t know if it is in the convex cone of $S’$. – user193108 Apr 18 '19 at 11:15
• Convex cone? There is no convex structure! – Norbert Schuch Apr 18 '19 at 19:03

Answer: Yes, and it works for any $$s>0$$.

$$\newcommand{\ket}{|#1\rangle}$$ To see how, define $$\ket{\phi_\epsilon}=(\ket{0}+\epsilon\ket{1})^{\otimes 2} =\ket{00}+\epsilon(\ket{01}+\ket{10})+\epsilon^2\ket{11}\ .$$ Then, $$\ket{\phi_{\epsilon}}+\ket{\phi_{-\epsilon}} = 2\ket{00}+2\epsilon^2\ket{11}$$, and thus, $$\ket{\phi_{\epsilon}}+\ket{\phi_{-\epsilon}}-2\ket{00}=\ket{11}\ .$$

You thus have access to $$\ket{00}$$ and $$\ket{11}$$, and thus to $$\ket{00}+\ket{11}$$.

Obviously, the same can be done for the pair $$\ket{00}$$ and $$\ket{22}$$, etc., so you get all maximally entangled states.

Since this works for any $$\epsilon>0$$, it also works for any $$s>0$$.

With regard to Edit 2 to the question -- does it still work for $$\langle \phi\ket0>1-s$$ -- the answer is also yes: The procedure above already satisfies that property. (Note that it is not a very meaningful condition, since all what could be different without the absolute value is a global phase, which you can always put in the amplitudes in the superposition.)

• Yes. $|\psi\rangle = \sum_i a_i |\phi_i\rangle|\phi_i\rangle$ for $|\phi_i\rangle$ close to $|0\rangle$. Here $a_i$ can be negative. What puzzles me is that can we only take positive $a_i$? We assume $\langle \phi | 0 \rangle > 1-s$. I have edited the original question. – user193108 Apr 19 '19 at 1:33
• Please don't make edits which alter the original question. It makes it look like I was unable to read your question. I have edited your question accordingly. --- Your condition <phi|0> > 1-s does not match what you ask before. It is already satisfied. – Norbert Schuch Apr 19 '19 at 10:26
• If you want all amplitudes to be positive, it might work if you allow to use complex phases. You would have to be more precise which states exactly do you want to consider. If it is too rigid, it is probably impossible as the amplitude of |00> will always be larger than that of |11>. – Norbert Schuch Apr 19 '19 at 10:28