11

It appears that the statement is not true in general. Suppose $X = Y = \{0,1\}$, $\mathcal{H}$ is the Hilbert space corresponding to a single qubit, and $W$ is defined as \begin{align} W(0,0) & = | 0 \rangle \langle 0 |,\\ W(0,1) & = | 1 \rangle \langle 1 |,\\ W(1,0) & = | 1 \rangle \langle 1 |,\\ W(1,1) & = \frac{1}{2} | 0 \rangle \langle 0 |...


10

The conditional von Neumann entropy is a concave function: if $\rho$ and $\sigma$ are states of a pair of registers $(\mathsf{X},\mathsf{Y})$ and $\lambda\in[0,1]$ is a real number, then $$ \mathrm{H}(\mathsf{X}|\mathsf{Y})_{\lambda\rho + (1-\lambda)\sigma} \geq \lambda\, \mathrm{H}(\mathsf{X}|\mathsf{Y})_{\rho} + (1-\lambda)\,\mathrm{H}(\mathsf{X}|\mathsf{Y}...


9

Geometric characterization (as any other characterization) of subsets of the quantum state space in relation with their locality and entanglement properties becomes very complicated as the number of qubits rises. The geometry of the space of negative conditional entropy two qubit states, which are also locally maximally mixed (Weyl states) is known; it is ...


8

The equation at the top of the question is not correct: there is a missing factor of $1/d$ on the right-hand side. Let's eliminate this factor from the left-hand side to make it simpler, so that the equation we want is this: $$ \text{Tr}\bigl(\rho^{AB} \bigl(\sigma^A \otimes I\bigr)\bigr) = \text{Tr}\bigl(\rho^A \sigma^A\bigr). $$ To see why this is true, ...


8

No dimension-independent bound is possible. Consider states $\rho_A$ and $\sigma_A$ that are close in $p$-norm (for $p>1$) but have relatively low fidelity. Specifically, assume $$ \|\rho_A - \sigma_A\|_p = \varepsilon $$ and $$ \operatorname{F}(\rho_A,\sigma_A) = \bigl\|\sqrt{\rho_A}\sqrt{\sigma_A}\bigr\|_1 = \delta, $$ where $\varepsilon$ is small and $...


7

Here the important fact is that the maximally mixed state is in fact an identity matrix. Let me rewrite the expression on the left in index notation (the summation sign is omitted according to the Einstein convention): $$ Tr(\rho^{AB} (\sigma^A \otimes I/d)) = [\rho^{AB}]_{ijkl} [\sigma^A]_{ji} [I/d]_{lk} $$ But $[I/d]_{lk} = \frac1d \delta_{lk}$, ...


7

Operator $\rho$ is not a tensor product, it's a sum of tensor products $$ p_1|1\rangle\langle 1| \otimes \rho_1 + p_2|2\rangle\langle 2| \otimes \rho_2 + \dots + p_d|d\rangle\langle d| \otimes \rho_d. $$ This is not the same as $$ \big(\sum_ip_i|i\rangle\langle i|\big) \otimes \big(\sum_i\rho_i\big), $$ so your expansion isn't correct. Also in general $S(...


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 $\text{rank}\big(J(\mathcal E)\big)^n$.


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 ...


6

A channel $\Phi$ is said to be degradable if there exists another channel $\Xi$ such that $\Xi\Phi$ is complementary to $\Phi$. The idea here is as follows. Suppose $\Phi$ is a channel and $\Psi$ is complementary to $\Phi$. If $\Phi$ is applied to a state $\rho$, then the output of the channel is $\Phi(\rho)$ (of course), while $\Psi(\rho)$ represents ...


6

For density matrices $\rho_A$ and $\rho_B$ having eigenvalues $\lambda^{\left(A\right)}$ and $\lambda^{\left(B\right)}$, \begin{align}S\left(\rho_A\otimes\rho_B\right) &= -\rho_A\otimes\rho_B\ln\left(\rho_A\otimes\rho_B\right)\\ &= -\sum_{j, k}\lambda^{\left(A\right)}_j\lambda^{\left(B\right)}_k\ln\left(\lambda^{\left(A\right)}_j\lambda^{\left(B\...


6

There are two types of information in physics: Classical information Quantum information Physics doesn't answer the question "What is (classical or quantum) information?". This is philosophic question, and physics never answers questions of this kind. Instead, physics answers another question "How (classical or quantum) information is measured?". The ...


6

Let me quote my answer from over at physics.SE: The intuition Let us consider a channel $\mathcal E$, which we want to apply to a state $\rho$. (This could equally well be part of a larger system.) Now consider the following protocol for applying $\mathcal E$ to $\rho$: Denote the system of $\rho$ by $A$. Add a maximally entangled state $|\omega\rangle=\...


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%...


5

The conditional min-entropy $\text{H}_{\text{min}}(A | B)_{\rho}$ can be defined for an arbitrary state $\rho$ of a pair of registers $(A,B)$ as $$ - \inf_{\sigma} \,\text{D}_{\text{max}}(\rho \| \mathbb{1}\otimes \sigma), $$ where the infimum is over all states $\sigma$ of $B$ and $\text{D}_{\text{max}}$ is the quantum relative max-entropy: $$ \text{D}_{\...


5

You may be interested in the answers to this question. One example of a degenerate code beating the quantum Hamming bound is here. I also have a numerical example of a small violation in my own work, here. In Figure two, you will see a zoomed in section. Essentially, the black line is the quantum Hamming bound (that may not be entirely obvious from what is ...


5

The mutual information can be written in terms of the relative entropy, please see Nielsen and Chuang (the entropy Venn diagram figure 11.2). I am writing the equation in the question's notation: $$I(\rho^{AB}) = S(\rho^{AB}|\rho^{A} \otimes \rho^{B})$$ The relative entropy can be estimated without full tomography. The procedure is described in Bengtsson ...


5

These are not really the definitions of classical and quantum capacity, as I will explain. Before doing that, let me adjust the notation being used slightly: let $\Phi:\text{L}(\mathcal{X}) \rightarrow \text{L}(\mathcal{Y})$ be the channel whose capacities we are interested in and let $\Psi:\text{L}(\mathcal{X}) \rightarrow \text{L}(\mathcal{Z})$ be a ...


5

Right, they are quite similar. The Holevo bound is a bound on the amount of accessible information between your quantum system and your classical system. The I(X;B) object written in the HSW theorem wikipedia page is actually this bound, while the $\chi$ there is the Holevo rate, or product state capacity. What HSW showed was that if you took many copies of ...


5

The complexity that you are glossing over is that in the general case you need to store $2^n$ complex amplitudes to even represent an $n$ qubit system classically. Therefore, for a quantum computer of let's say 1000 qubits you need to store $2^{1000}$ complex amplitudes. Even if you use one atom per amplitude to do this, you still run out of atoms in the ...


5

That quantity appears to be identical to Holevo information, which turns out to be the upper bound on how much classical information you can transmit using a quantum channel [1]. More generally the Holevo information is an upper bound for a quantity called "accessible information" which is (roughly speaking) the maximum information you can learn ...


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

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

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

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

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

My favourite way of proving that the Shannon entropy is minimized for a measurement in the qubit basis is through the notion of majorizaion (see Nielsen and Chuang or the book on Matrix Analyis by Bhatia for a formal definition). Specifically $p$ and $(1-p)$ is related to $p'$ and $(1-p')$ with the following relation \begin{equation} \left(\begin{array}{c} ...


4

Posting an answer because I realised what my issue was: What I didn't realise then: When a density matrix is written in any basis, the diagonal elements correspond to the probabilities of the density matrix landing on the basis states of that basis. So, if in some basis formed by vectors $|x_1\rangle, |x_2\rangle, |x_3\rangle, |x_4 \rangle$, my density ...


4

It perhaps helps to express $P(ab|xy)$ in words: the probability that Alice gets answer A and Bob gets answer B given that choices x and y were made Now independence in classical probability holds if and only if $$ P(e_1\text{ and }e_2)=P(e_1)P(e_2) $$ where $e_1$ and $e_2$ are events, and practically, you can see what it means through Bayes' theorem $$ ...


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