Consider a random quantum circuit $U$ over $n$ qubits, drawn from the Haar measure. Consider the quantum state
$$|\psi\rangle = U |0^{n}\rangle.$$
Now, partition $n$ into two and consider the Schmidt decomposition of $|\psi\rangle$ across this bipartition. Let the decomposition be
$$|\psi\rangle = \sum_{i} \alpha_i |\phi_{i} \rangle |\chi_{i} \rangle.$$
The Schmidt coefficients of $|\psi\rangle$ are well studied. For example, according to this link,
$$\sum_{i} \alpha_i^{4} = \frac{|A| + |B|}{|A| \cdot |B| + 1},$$ where $|A| = 2^{a}$, where a is the size of the first part, and $|B| = 2^{b}$, where $b$ is the size of the second part, with $a + b = n.$
I was trying to argue about the Schmidt vectors too. Do we know anything about the nature of the Schmidt vectors? My hunch is that they are very close to standard basis states. Is there any way to see this?
Note that from the same link, from Theorem $16$,
$$ \underset{|\psi\rangle}{\mathbb{E}} \left[ \bigg|\bigg|\rho_A - \frac{\mathbb{I}}{|A|}\bigg|\bigg|_{2} \right] \leq \sqrt{\frac{|A|}{|B|}}.$$
Is this sufficient to indicate that either $\{|\phi_{i} \rangle\}$ or $\{|\chi_{i} \rangle\}$ are close to standard basis states? For a particular $|\psi\rangle$, with high probability, what is the overlap between its Schmidt vectors and standard basis states?