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 $\langle0|\phi\rangle >1-s$?
 A: Answer: Yes, and it works for any $s>0$.
$\newcommand{\ket}[1]{|#1\rangle}$
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.)
