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First of all, your question should be more carefully formulated, since it is not even possible to always find a non-trivial subsystem (not subspace) of $\mathcal H_A$, see also my comment here Can a s …

I know the question is already answered, but there was some question on my comment and I wanted to elaborate on that. First, let us consider one system only. The $\mathbb{R}$-span of all states $\rho$ …

I am not aware of a direct application of this projection. However, maybe the following geometrical construction is nevertheless of interest to you. Similar ray constructions appear in resource theori …

I apologise in advance. This is a rough and hand-waivy answer. You can give "information-theoretic" lower bounds by noting that the measurements can be seen as a linear map $M$ from quantum states to …

A pure state is by definition rank one, $\rho = | \psi \rangle\langle \psi |$. A state can have maximal rank and be arbitrarily close to a pure state. Just mix a pure state with the maximally mixed st …

To talk about entanglement, you have to first identify subsystems. In your $d=4$ example, you defined an isomorphism $\mathbb{C}^4\simeq \mathbb{C}^2\otimes\mathbb{C}^2$ via the identification of basi …

This is an addendum to glS's answer and concern the issue of an "axis" in higher dimensions. Let us set $D:=N^2-1$ such that we are interested in elements $R\in SO(D)$. $R$ is in general only diagonal …

The technical term is "quantum state discrimination". One has to carefully formulate the problem, because it is generally hard to identify an arbitrary state (tomography) as you noticed. However, give …

First of all: the bit-order that you use is not the conventional one in quantum computing. We usually use the convention that $$ |x_1 x_2 \dots x_n \rangle = |x_1\rangle \otimes |x_2\rangle \otimes \d …

No, such a unitary cannot exists for any state $\rho$. To see this, note that the spectrum of $\rho'$ is $(1+\lambda, 1-\lambda, 1-\lambda, 1+\lambda)/4$, in particular, $\rho'$ is full rank, except f …

The factor in your claim is wrong. It should be $\binom{d+t-1}{t}^{-1}$. The correct claim follows from the identity $$ \int |\psi\rangle\langle\psi|^{\otimes t} d\psi = \binom{d+t-1}{t}^{-1} P_{\mat …

As for any platform, one has to choose a suitable $d$-dimensional "computational" subspace. Suitability depends on your application, but generally it means that one should be able to perform operation …

The authors are certainly thinking about finite frames. In this case, your statement is correct, since the number of elements in every spanning set is at least the vector space dimension. As glS alre …

First of all, that does not imply anything for shorter (constant/logarithmic) depths. Moreover, the 2-design property does not imply that the outcome distribution is the same as for Haar-random unitar …

What the author wrote is completely correct, they did not make a mistake. The subgroup of Cliffords fixing $X_n$ and $Z_n$ is indeed isomorphic to $C_{n-1}$ as a group, this is simply because this sub …