7

It is true that unitary evolution cannot destroy information. This is the content of the no-cloning theorem and its time reversal - the no-deleting theorem. The no-hiding theorem says something different and more subtle. Information hiding In order to understand what it says it helps to start with the observation that classical information residing in a ...


7

No. The minimal size of the environment is just the rank of the Choi matrix of $\mathcal E$, call it $J(\mathcal E)$. Since $J(\mathcal E^{\otimes n}) = \big(J(\mathcal E)\big)^{\otimes n}$ and $\text{rank}(A \otimes B) = \text{rank}(A)\text{rank}(B)$, the minimal size of the environment is just $\text{rank}\big(J(\mathcal E)\big)^n$.


6

Short answer: Assuming you are measuring in the computational basis (Z basis), $\{|0\rangle, |1\rangle \}$, there is no randomness upon measurement in the following quantum circuit (you will always get back the state $|1\rangle$): Thus, measurement needs not be random. However, if you try this following circuit, you will have 50% to see a $|0\rangle$ and 50%...


6

Generally speaking, in order to describe elements of a set $A$ using classical information we need two ingredients: a non-empty finite alphabet $\Sigma$ and an encoding $E: A\to\Sigma^\omega$ which injectively maps objects in $A$ to the set $\Sigma^\omega$ of sequences of symbols in $\Sigma$. Infinite encodings If the set $A$ is uncountable, as is the case ...


6

Given $\rho$ and a fixed ensemble $\{ |\psi_i \rangle \}$ it might not be possible to write $\rho$ as $\sum_i p_i |\psi_i \rangle \langle \psi_i |$. For example, let $| + \rangle = \frac{1}{\sqrt{2}} (| 0 \rangle + | 1 \rangle )$. Then the state $|+\rangle \langle + |$ cannot be expressed as a convex combination in the ensemble $\{ | 0 \rangle, |1\rangle \...


5

Your suspicion is correct, even when $A=B$. Consider the Hilbert space of two qubits and let $^{T_A}$ denote the partial transpose with respect to one of them. Suppose that $$ A=B=\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}. $$ Then $$ A^{T_A}=B^{T_A}=\begin{...


5

Let's start with the second question. There is nothing special about an extension $\sigma_{AR}^{\ast}$ that allows it to be optimal for the right-hand side of (1); any extension $\sigma_{AR}$ of $\sigma_A$ could happen to be optimal for the right choice of $\rho_{AR}$. For example, if we suppose that $\sigma_{AR}$ is any given extension of $\sigma_A$, and we ...


5

Summary: Quantum computing is believed to give uncertain measurements fast. If you delve a little bit into computational theory, you'll find that there are problems that cannot be solved exactly and efficiently with traditional computational tools. However, for some of these problems, if we have a solution we can check if it is correct efficiently with a ...


5

This only holds if the two distributions are independent. In this case $$ \begin{aligned} H_{\beta}(p \times q) &= \frac{1}{1-\beta} \log\left( \sum_{i,j}(p(i) q(j))^{\beta} \right) \\ &= \frac{1}{1-\beta} \log\left( \left(\sum_{i}p(i)^{\beta}\right) \left(\sum_jq(j)^{\beta}\right) \right) \\ &= \frac{1}{1-\beta} \left(\log \left(\sum_{i}p(i)^{\...


5

It is not just the binary entropy that is denoted $H(p_i)$. The quantity that is relevant here is the Shannon entropy of the distribution $\{p_i\}$ which is defined as $$ H(p_i) = - \sum_i p_i \log p_i. $$ Note that when the distribution has only two elements we recover the binary entropy.


5

Of course you can. Take any entanglement measure that can be applied to a system whose overall description is a density matrix, and you can apply that to the density matrix describing your subsystem. (So, you wouldn't use the von Neumann entropy of one qubit because that assumes the overall state is pure.) A particularly good option for a pair of qubits is ...


4

Prohibited device Such a circuit $C$ would enable faster-than-light communication and therefore does not exist. Suppose Alice and Bob share a Bell pair $|\psi\rangle = \frac{1}{\sqrt{2}}(|01\rangle - |10\rangle)$ and Alice has a classical bit $b \in \{0, 1\}$ she wishes to communicate to Bob. She proceeds as follows. If $b=0$ Alice measures her half of $|\...


4

Indeed that is an excellent question! Since measurements always lead to a probabilistic outcome, why should we think quantum computing is useful? I definitely second the answer by Michele Amoretti and recommend the reference he pointed out, and will add a few things: Firstly we know that quantum computers can, in principle, perform all calculations that a ...


4

Entanglement + a classical channels allows you to build a quantum channel using teleportation. Thus, adding a quantum channel does not give you additional power.


4

The key feature that you've left out of the question is the requirement that the measurement is performed, but we do not learn the answer. For example, imagine starting with a $|+\rangle$ state. You measure it in the $Z$ basis, so you get answers $|0\rangle$ and $|1\rangle$ with 50:50 probability. If you knew which result you'd got, then great, you have a ...


4

The intuition you're looking for is explained using an example in the very next paragraph of the textbook. Suppose, $p_U = \frac{1}{3}$ and $p_A = \frac{5}{6}$, and Alice tells Bob that the result of her toss is heads. Then Bob immediately realizes that the coin Alice tossed was most likely the American coin. So, in this scenario, Bob gains some additional ...


4

TL;DR: The assumption of non-orthogonality is implicitly used by the linked answer. It is needed due to a "loophole" in the no-cloning theorem that allows cloning of known orthogonal states. Universal cloner is prohibited The no-cloning theorem is the statement that there is no unitary $U$ such that $$ U|\gamma\rangle|b\rangle = |\gamma\rangle|\...


4

Assuming w.l.o.g. that $p\in\mathbb{R}$, the linear map in the question may be rewritten as $$ \mathcal{E}(\rho) = p^2\rho+p^2X\rho X = 2p^2\left(\frac12\rho + \frac12 X\rho X\right) $$ where $X$ is the Pauli matrix. Thus, the action of $\mathcal{E}$ can be understood as the composition $\mathcal{E}=\mathcal{S}_{2p^2}\circ\mathcal{X}_{\frac12}$ of a scaling ...


3

We will use the upper bound on the entropy of a mixture (for proof see for example theorem 11.10 on p.518 in Nielsen & Chuang) $$ S\left(\sum_k p_k \rho_k\right) \leq H(p) + \sum_k p_k S(\rho_k)\tag1 $$ where $H(p) = -\sum_k p_k \log p_k$. Set $p_x := p(x)$. Note that if $|\psi_y^x\rangle$ is an eigenvector of $\rho_A^x$ associated to eigenvalue $q_y^x$ ...


3

I don't believe this holds even for Hermitian matrices. The main reason being is that $\mathrm{tr}[\rho X Y ] \neq \mathrm{tr}[\rho Y X]$ in general. This is a problem because we see that $$ COV_{\rho}(X,Y) = \mathrm{tr}[\rho X Y] - \mu_X \mathrm{tr}[\rho Y] - \mu_Y \mathrm{tr}[\rho X] + \mu_X \mu_Y $$ and $$ COV_{\rho}(Y,X) = \mathrm{tr}[\rho Y X] - \mu_X \...


3

They are invariant under conjugation of unitaries, i.e. under the mapping $\rho \to U \rho U^*$. To see this note that $(U \rho U^*)^{\alpha} = U \rho^\alpha U^*$. Then we have $$ \begin{aligned} S_\alpha(U \rho U^*) &= \frac{1}{1-\alpha} \log \mathrm{Tr}[(U \rho U^*)^{\alpha}] \\ &= \frac{1}{1-\alpha} \log \mathrm{Tr}[U \rho^\alpha U^*] \\ &= \...


3

Let $$ \rho(t)=U\rho U^\dagger, $$ just to simplify notation a bit. Now notice that $$ \rho(t)^2=U\rho U^\dagger U\rho U^\dagger=U\rho^2 U^\dagger $$ since $U^\dagger U=I$. Thus, similarly, $$ \rho(t)^n=U\rho^n U^\dagger. $$ So, take the trace of this, remembering that trace is invariant under permutations: $$ \text{Tr}(\rho(t)^n)=\text{Tr}(U\rho^n U^\dagger)...


3

Yes. Expression (X) is in principle more general, but it boils down to doing classical post-processing of the outcomes, and this can always be included in the POVM. So as long as you do optimize over the POVM, expression (Y) is completely general. There are four possibilities: $\lambda_0 p_0(0)>\lambda_1 p_1(0)$ and $\lambda_1 p_1(1)>\lambda_0 p_0(1)$....


3

No, such an ordering does not exist. For example, take $\rho = |\phi\rangle \langle \phi|$ with $| \phi \rangle = \cos(\theta) |00 \rangle + \sin(\theta) |11\rangle$ and $\theta \in (0,\pi/4)$. Then take $\sigma_B = I/2$, the maximally mixed qubit. A direct calculation gives $$ V(\rho_{AB}\|\rho_A \otimes \rho_B) = 8 \big(\log[\tan(\theta)] \sin(\theta) \...


3

Why not take the example of the GHZ state? $$ |GHZ\rangle=(|0\rangle^{\otimes n}+|1\rangle^{\otimes n})/\sqrt{2}, $$ such that $\rho_{A^n}=|GHZ\rangle\langle GHZ|$. The $\rho_A=I/2$ and $\rho^{\otimes n}_A=I/2^n$. Then for this specific case $$ \lambda \rho^{\otimes n}_A-\rho_{A^n}, $$ the eigenvalues are $\lambda/2^n-1$ (once) and $\lambda/2^n$ ($2^n-1$ ...


3

Assuming everything is finite dimensional. For $S_0$ we have $$S_0(\rho) = \log \mathrm{rank}(\rho).$$ It's pretty straightforward to see this is not continuous. Take $\rho_{\epsilon} = \epsilon |0\rangle \langle 0 | + (1-\epsilon) |1\rangle \langle 1 |$. Then for all $0 < \epsilon < 1$ we have $S_0(\rho) = \log 2$ but for $\epsilon \in \{0,1\}$ we ...


3

If perturbations are sufficiently small and $\rho_{AB}$ has sufficiently broad support then a desired global state $\rho_{AB}'$ exists. Define $$ \rho_{AB}' = \rho_{AB} + (\rho_A' - \rho_A) \otimes \rho_B + \rho_A \otimes (\rho_B' - \rho_B) \tag1. $$ Note that $\rho_{AB}'$ is Hermitian and trace one, but may not be positive. However, $\rho_{AB}'$ is positive ...


3

As @Rammus has mentioned in the comments one does not need a quantum channel to utilize entanglement as a resource. One can utilize quantum correlations to aptly perform tasks that are impossible to perform with classical correlations (i.e. shared randomness). For example in a nonlocal game, two players share entanglement (but don't need any kind of quantum ...


3

Let $ \rho = \sum_n \rho_n |\psi_n \rangle \langle \psi_n | $ be the eigendecomposition of $\rho$. We will calculate everything in terms of $ |\psi_n \rangle$ basis. Note that $ \frac{d \rho_\theta}{d \theta} = i [\rho_\theta, A] $ and $ \frac{d^2 \rho_\theta}{d \theta^2} = -\big[[\rho_\theta, A], A\big] $. Now we write $$ \sqrt{ \sqrt{\rho} \rho_\theta \...


3

Note that if $p_{i}q_{i} = 0\,\,\forall i$, then for all $i$ either $p_{i} = 0$, $q_{i} = 0$, or both are $0$. Divide $\{i\} = \{1,\ldots,N\}$ into those $i$ for which these three different things happen: $\{N_{p_{i}}\} = \{i|p_{i} = 0\}$, $\{N_{q_{i}}\} = \{i|q_{i} = 0\}$, $\{N_{pq_{i}}\} = \{i|p_{i} = q_{i} = 0\}$. Then \begin{equation} \begin{split} \|P-...


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