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## Hot answers tagged entropy

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 $$\mathrm{H}(\mathsf{X}|\mathsf{Y})_{\lambda\rho + (1-\lambda)\sigma} \geq \lambda\, \mathrm{H}(\mathsf{X}|\mathsf{Y})_{\rho} + (1-\lambda)\,\mathrm{H}(\mathsf{X}|\mathsf{Y}... 9 Geometric characterization (as any other characterization) of subsets of the quantum state space in relation with their locality and entanglement properties becomes very complicated as the number of qubits rises. The geometry of the space of negative conditional entropy two qubit states, which are also locally maximally mixed (Weyl states) is known; it is ... 8 For two probability distributions, there is a clear notion how to say which one is more mixed: \vec p is more mixed than \vec q if it can be obtained from \vec p by a mixing process, this is, a stochastic process described by a doubly stochastic matrix (i.e. one which preserved the flat distribution). Birkhoff's theorem relates this to a concept ... 7 Operator \rho is not a tensor product, it's a sum of tensor products$$ 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(... 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 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 You are calculating the entropy of one of the marginal states and so you would not expect the answer to be independent of \theta, except in the case that |\psi\rangle = |\phi_A\rangle \otimes |\phi_B\rangle -- this will be true whenever \theta = \frac{k \pi}{2} for some k \in \mathbb{Z}. In this case the reduced state \rho_A is also pure. However, ... 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

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

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

Cirq uses numpy's pseudo random number generator to pick measurement results, e.g. here is code from XmonStepper.simulate_measurement: def simulate_measurement(self, index: int) -> bool: [...] prob_one = np.sum(self._pool.map(_one_prob_per_shard, args)) result = bool(np.random.random() <= prob_one) [...] Cirq ...

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

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. $$Note that when the distribution has only two elements we recover the binary entropy. 4 My favourite way of proving that the Shannon entropy is minimized for a measurement in the qubit basis is through the notion of majorizaion (see Nielsen and Chuang or the book on Matrix Analyis by Bhatia for a formal definition). Specifically p and (1-p) is related to p' and (1-p') with the following relation \left(\begin{array}{c} ... 4 Posting an answer because I realised what my issue was: What I didn't realise then: When a density matrix is written in any basis, the diagonal elements correspond to the probabilities of the density matrix landing on the basis states of that basis. So, if in some basis formed by vectors |x_1\rangle, |x_2\rangle, |x_3\rangle, |x_4 \rangle, my density ... 4 Suppose that \mathsf{X} is a register that can store each possible choice for x, as a classical state, while \mathsf{Y} is a register that can store each possible state \rho_x. It is then natural to associate the classical-quantum state$$ \rho = \sum_x p_x |x\rangle \langle x| \otimes \rho_x $$with the ensemble \{(p_x,\rho_x)\}. Now try taking ... 4 That seems to restrict the output probability distributions of all quantum circuits to rather high entropy distributions. The output of a typical randomly chosen quantum circuit is rather high entropy. That doesn't mean you can't construct circuits that have low entropy outputs (you can), it just means that picking random gates is a bad strategy for ... 4 Mathematically when is \sum_i p_i S(\rho_i) > 0? I am assuming that \{p_i\} form a probability distribution (and that none of the p_i = 0) and each \rho_i is a normalised state. As p_i \geq 0 and S(\rho_i) \geq 0 we have \sum_i p_i S(\rho_i) = 0 \iff S(\rho_i) = 0 for all i. Then we can ask the question under what circumstances do we ... 4 This can be seen through "twirling" with a bunch of unitaries. Call your density operator \rho. Let U_i be a unitary with \pm 1 on the diagonal, and zeros everywhere else when expressed in your basis. Consider all 2^d such unitaries where d is the dimension of your density matrix. I leave it to you to show that \rho_D = \frac{1}{2^d}\sum\... 4 Yes, you can formulate the smooth max-entropy as an SDP. The author of the book you linked notes this when they explain how to derive the SDP for the smooth min-entropy that you reference on page 91. In particular they say that the smoothing constraint \tilde{\rho}_{AB} \in B^\epsilon(\rho_{AB}) can be reformulated as the triple of constraints$$ \mathrm{...

4

Let $D_{\alpha}(\rho\|\sigma):= \frac{1}{\alpha - 1} \log \mathrm{Tr}[\rho^\alpha \sigma^{1-\alpha}]$ be the Petz-Rényi divergence for $\alpha \in (0,1)\cup(1,\infty)$. Note that for $\alpha \in (0,1)\cup(1,2]$ this quantity satisfies the data processing inequality $$D_{\alpha}(\rho\|\sigma) \geq D_{\alpha}(\mathcal{E}(\rho) \| \mathcal{E}(\sigma)),$$ ...

4

This is not true for general tripartite states. Take the trivial example where $ABE$ share a maximally mixed state and each parties subsystem is of dimension $d$. The reduced states of a two-party subsystem are maximally mixed of dimension $d^2$ and of a single party system are maximally mixed of dimension $d$. As the dimensions are different and they are ...

4

As you mention pure states will not do. So lets look at a simple example of mixed entangled states, two-qubit Werner states. Let $\rho_{AB} = q |\Psi^- \rangle \langle \Psi^-| + (1-q) I / 4$ where $| \Psi^- \rangle = \frac{1}{\sqrt{2}}(|01\rangle - |10\rangle)$ and $q \in [0,1]$. It is known that this family of states is separable iff $q \in [0, 1/3]$. So ...

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 ...

3

As @NorbertSchuch said in a comment, matlab has a function for taking the logarithm of a matrix: logm. In general, there is a standard method for calculating the function $f(\sigma)$ of a matrix $\sigma$. You first diagonalise the matrix: $$\sigma=UDU^\dagger,$$ where $U$ is a unitary and $D$ is diagonal. We then say $$f(\sigma)=Uf(D)U^\dagger,$$ where $... 3 It turns out to be a novice mistake. I was using matlab and this log is elementwise, as @Ahusain pointed out. We must take the matrix logarithm in Matlab which is denoted by$logm$. Then the calculation becomes: $$-\text{trace}(\rho \log m (\rho)) = \text{NaN}.$$ The reason is, we have to define$0 \times \log (0)$as$0$instead of$\text{NaN}$which is ... 3 It means that if you lose information from your system, that information must have been transferred to the system's surroundings. This shows up as an increase in the entropy in the surroundings. This is directly related to the 2nd law of thermodynamics which says the entropy of an isolated system is always increasing. See Wikipedia: Entropy in thermodynamics ... 3 I don't understand how they go from the first to the second line So, you're starting from$-\sum_{ij}p_i\lambda^j_i\log(p_i\lambda^j_i)$. Remember that$\log(ab)=\log(a)+\log(b)$, so this is the same as $$-\sum_{ij}p_i\lambda^j_i\log(p_i)-\sum_{ij}p_i\lambda^j_i\log(\lambda^j_i)$$ For the first term, do the sum over$j$:$\sum_j\lambda^j_i=\text{Tr}(\...

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