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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?

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    $\begingroup$ Surely for a random state $|\psi\rangle$, any of the states $U_1\otimes U_2|\psi\rangle$ are just as likely. So the Schmidt vectors could be absolutely anything? This essentially shows up in your "from theorem 16" bit, right, because being close to identity doesn't tell you anything - it's the non-zero departure from the identity that fixes the basis. It doesn't matter how big that departure is. (Viewed another way: identity is diagonal in any orthonormal basis, so what's special about the computational basis?) $\endgroup$
    – DaftWullie
    Nov 19 at 7:37
  • $\begingroup$ Can we say anything when that departure is small? As in, when $|B|$ is much larger than $|A|$? $\endgroup$
    – BlackHat18
    Nov 19 at 12:40
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    $\begingroup$ I don't think so. As soon as you have even a small departure from identity, that departure fixes the basis, which could be anything. $\endgroup$
    – DaftWullie
    Nov 19 at 12:54
  • $\begingroup$ You might find this paper interesting, although not directly answering your question: arxiv.org/abs/0810.4331 $\endgroup$
    – DaftWullie
    Nov 19 at 16:49
  • $\begingroup$ Might you elaborate your comments into an answer? $\endgroup$
    – BlackHat18
    Nov 19 at 21:31
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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.

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  • $\begingroup$ In other words, the Schmidt vectors could pretty much be anything? Can we say something, at least, about the Schmidt vectors with high probability --- for example what their overlap is with something like the standard basis states? $\endgroup$
    – BlackHat18
    Nov 19 at 12:34
  • $\begingroup$ Yes, I believe it can be proved that $|\phi_{i}\rangle$ is also the Haar random (under some additional assumptions). $\endgroup$
    – Danylo Y
    Nov 19 at 13:06
  • $\begingroup$ Might you add a proof of that too, in your answer? $\endgroup$
    – BlackHat18
    Nov 19 at 14:26
  • $\begingroup$ A quick sanity check of something that is still a bit unclear to me: do your results mean that when we are given a particular Haar random $|\psi\rangle$, and partition it into two, and consider the reduced density matrix for one of the parts, then, with high probability over the choice of $|\psi\rangle$, we can take any (ordered, like you did) orthonormal basis to be the eigenbasis of that reduced density matrix -- it does not matter which one we choose? $\endgroup$
    – BlackHat18
    Nov 19 at 21:39
  • $\begingroup$ For a particular $\rho_A$ we can't pick any orth. basis to be its eigenbasis. If all eigenvalues of $\rho_A$ are different then eigenbasis is unique up to the phases of eigenvectors. Most of the time eigenvalues will be different since $\rho$ is random. But if some eigenvalue of $\rho_A$ has multiple multiplicity, then there are different choices for eigenbasis, and consequently, there are different Schmidt decompositions of $|\psi\rangle$. To make statements about eigenvectors $|\phi_i\rangle$ we have to define what is it, precisely. $\endgroup$
    – Danylo Y
    Nov 19 at 22:32

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