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Hot answers tagged haar-distribution

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What is the expectation value of $|\langle \psi|U|\psi \rangle|$ over Haar random states $|\psi\rangle$?

This partial answer calculates the integral for $d=2$. In this case, every traceless unitary $U$ is equivalent to the Pauli $Z$ up to similarity and global phase, so, by rotational invariance of the ...
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Schur's lemma for quantum states

Most generally, Schur's Lemma is used as a tool in representation theory. In this answer, I'll try to explain it without talking about said theory. Preliminaries: Schur's Lemma Let us consider a ...
• 6,162
3 votes
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Two qubit Pauli expectation value of $\underset{U}{\mathbb{E}}[U^{\otimes 2} (P_1 \otimes P_2)^{\otimes 2} U^{*\otimes 2}]$

We can prove a slightly more general statement, which works for any Hermitian "sandwiched" matrix $P$ in any dimension $d$... but only for $t=2$ in the number of copies :'( Let us call $V$ ...
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• 121
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What is the expectation value of $|\langle \psi|U|\psi \rangle|$ over Haar random states $|\psi\rangle$?

As a follow-up discussion with Adam, I evaluated the integral till $d=4$ for high-dimensional unitary $Z$, with ${Z}_{kl}{=}\exp(\frac{i2\pi k}{d})\delta_{kl}$. For $d{=}4$, I haven't considered the ...
• 135
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Symmetric subspaces and Haar averaging over the Unitary group

You have a syntactic mistake in your argument. The dimensions in the equation $$(V^\dagger)^{\otimes 2} \otimes V^{\otimes 2}D(x)(V^\dagger)^{\otimes 2} \otimes V^{\otimes 2} = D(x) \,,$$ don't match. ...
• 4,987
1 vote

Show that, averaging over uniformly random unitaries, $\mathbb{E}[UXU^\dagger]=\operatorname{Tr}(X)\frac{I}{d}$

Another intuitive way of looking at it is to note that the Haar average of all pure states is a maximally mixed state, $\int |\psi\rangle\langle\psi| d\psi=\frac{\mathbb{I}}{d}$. Considering \$|\psi\...
• 135

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