Schmidt decomposition - example

The phase of your eigenvectors is not correct (or rather, it is not determined, so you need to make a judicious choice). If you put a minus sign in front of the 2nd term, it works out.

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To elaborate on that: If you want to find the Schmidt decomposition, you can proceed e.g. as in Preskill's lecture notes: Diagonalize the reduced state of A, which yields eigenvalues $\lambda_i$ and eigenvectors $|a_i\rangle$. Then, rewrite $$ |\psi\rangle = \sum_i |a_i\rangle\otimes |b_i\rangle\ .\tag{*} $$ ($|b_i\rangle$ can be determined e.g. as $|b_i\rangle = \langle a_i|\psi\rangle$.) Then, the $|b_i\rangle$ are orthogonal with $\langle b_i|b_i\rangle=\lambda_i$ (cf. Preskill), i.e., the form $(*)$ above is the Schmidt decomposition (upon normalizing the $|b_i\rangle$).

(The latter can be seen by computing the reduced density matrix of A from $(*)$, which yields $$ \sum |a_i\rangle \langle a_j | \; \langle b_j|b_j\rangle = \sum \lambda_i |a_i\rangle \langle a_i|\ , $$ which yields $\langle b_j|b_j\rangle = \lambda_i\delta_{ij}$ as the $|a_i\rangle\langle a_j|$ are linearly independent.)

Norbert Schuch's answer is correct. Just for fun, here's the exact decomposition: $$ |\psi\rangle\propto (3+\sqrt{5})|A\rangle\otimes |A\rangle -(3-\sqrt{5})|B\rangle\otimes |B\rangle $$ with \begin{align} |A\rangle &= 2|0\rangle+(\sqrt{5}-1)|1\rangle \\ |B\rangle &= 2|0\rangle-(\sqrt{5}+1)|1\rangle. \end{align} Using these equations, we can verify that the coefficient of $|11\rangle$ is zero and that the coefficients of $|00\rangle$, $|01\rangle$, and $|10\rangle$ are all equal to each other, and $$ \langle A|B\rangle = 0. $$ Unnormalized vectors $A,B$ are used here to simplify the coefficients in the overall expression for $|\psi\rangle$.