Schmidt vectors for random quantum states

Question

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 $a + b = n.$

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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?

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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?

Answer 1

We use the Haar measure, thus for any unitary $U_0$ the distributions of $U$ and $UU_0$ are the same. Hence, the distributions of $|\psi\rangle = U|0^n\rangle$ and $|\psi\rangle = UU_0|0^n\rangle$ are also the same. Therefore, it's the same distribution $|\psi\rangle = U|\psi_0\rangle$ for any state $|\psi_0\rangle$. So, the standard basis plays absolutely no role. And we can't gain any information about what was our choice for standard basis by sampling $|\psi\rangle$. Theorem 16 doesn't depend on the choice of basis either, since maximally mixed state is the same in any basis.

Similarly, for any two unitaries $U_1,U_2$ we have that $(U_1 \otimes U_2) |\psi\rangle$ has the same distribution as $|\psi\rangle$. Thus, the distributions for $\rho_A = \text{Tr}_B(|\psi\rangle\langle\psi|)$ and $U_1\rho_AU_1^\dagger = \text{Tr}_B((U_1 \otimes U_2)|\psi\rangle\langle\psi|(U_1 \otimes U_2)^\dagger)$ also coincide.

Since the distributions for $\rho_A$ and $U_1\rho_AU_1^\dagger$ coincide for any $U_1$, it's natural to expect that distributions for $|\phi_i\rangle$ and $U_1|\phi_i\rangle$ also coincide. Though, we have a complication here – the choice of $\{|\phi_i\rangle\}$, which are eigenstates of $\rho_A$, is not unique for a given $\rho_A$. To make this procedure definite, let's assume that for any eigenvalue $\lambda$ of $\rho_A$ we pick the first eigenstate $|\phi_{\lambda,1}\rangle$ randomly from $H_\lambda$ (the corresponding eigenspace), the second $|\phi_{\lambda,2}\rangle$ we pick randomly from $H_\lambda \ominus \text{span}\langle\phi_{\lambda,1}\rangle$, and so on. Also, we sort eigenspaces in accordance to their eigenvalues.

With such a definite procedure of picking an ordered set of $\{|\phi_i\rangle\}$ for a given $\rho_A$, I believe it's then follows that distributions of $|\phi_i\rangle$ and $U_1|\phi_i\rangle$ coincide. It's a bit technical to show this precisely.