3

I think your explanation based on the circuit is perfectly adequate. For a more rigorous "proof", why not simply take the output of the circuit? Substitute in $x_i=0$ for all $i$ and see that all the outputs are $(|0\rangle+|1\rangle)/\sqrt{2}$ for $i\neq 1$ and $(|0\rangle+(-1)^{x_1}|1\rangle)/\sqrt{2}$ for $i=1$, exactly as it would be for the ...


3

Recall that a function between two sets (say $A,B$) is simply a binary relation that associates to every element of the first set $A$ exactly one element of the second set $B$. Therefore, a function can be represented as a set $G$ of ordered pairs $(x,y), x \in A, y \in B$, satisfying the following: each element of $A$ appears exactly once in this set (to ...


2

$P$ is a projection operator in the limiting case where $P$ represents a state that is completely known, i.e. a pure state, so the entropy is zero. As a limiting case, the valid and well-defined mathematical description is $$\lim_{P \rightarrow 0^+} P \log(P)=0.$$ This is still a bit sloppy. Since $P$ is a matrix, we're actually taking the trace and $P \...


2

If it holds for hermitian matrices, it holds for all matrices due to linearity: Over $\mathbb C$, the hermitian matrices span the full matrix space.


2

This problem can be approached without regards to Kraus representations (even if the motivation is to prove the convexity of entropy) or whether A is a normal matrix or not. Rather, this is a feature of the choice of $\{ U_{j} \}$. In particular, there exists a choice such that their action is to ``coarse-grain'' all the information in a state. Here's a ...


2

We know that the Initial state $|\psi\rangle$ can be represented as $\sin\frac{\theta}{2}|\chi\rangle + \cos\frac{\theta}{2}|\xi\rangle$. We can prove the result $G^R|\psi\rangle = \sin\frac{(2R+1)\theta}{2}|\chi\rangle + \cos\frac{(2R+1)\theta}{2}|\xi\rangle$ by Induction Base Case When $R=0$, $G^R=G^0=I$ and $2R+1=2\times0+1=1$. Thus $G^0|\psi\rangle = \...


2

Remember that the partial transpose condition is generally good for detecting entanglement, i.e. a bipartite state $\rho$ is certainly entangled if the partial transpose is not non-negative. In other words, if there exists a state $|\psi\rangle$ such that $$ \langle\psi|I\otimes\text{T}(\rho)|\psi\rangle<0, $$ then the state is certainly entangled. If you ...


1

Welcome to the community Attila! I do not believe your equation 4 holds; consider the orthogonal vectors $ [\frac{i}{\sqrt{2}}, \frac{-i}{\sqrt{2}}]^T, [\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}]^T$ - their inner product should be 0 by orthogonality, but it can been seen that each of the terms you have would be nonzero, so their sum must be nonzero. The correct ...


1

Just consider the set of all functions from $\mathbb{Z}$ to a two point set $\{0,1\}$. The cardinality of the set of all functions between these two sets is $2^{|\mathbb{Z}|}$. Now ask yourself is the power set of $ \mathbb{Z}$ countable. If the answer is NO then the answer to your question is also a NO.


1

CW from self-answer, and also because this is more of an extended comment than an answer. Let $A$ be the adjacency matrix of the Cayley graph of our group $\mathcal{H}$ of order $N$. Notice that $A$ is square-hermitian. Further let $\mathbb{I}_N$ be the $N\times N$ identity matrix. It occurs to me that I am, in a sense, asking to prepare the ground state $...


1

(General result) The main thing to keep in mind is that this is a result about a type of channel, not about specific states. Suppose $\operatorname{tr}(U_i U_j^\dagger)=\delta_{ij}$ for some set of matrices $U_i$. This is equivalent to $\sum_{k\ell}(U_i)_{k\ell} (U_j^*)_{k\ell}=\delta_{ij}$. If $U_i$ form a basis (i.e. there are $n^2$ of them), then we must ...


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