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9

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 |...


8

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

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}...


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\...


5

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 ...


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 ...


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} ...


3

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 ...


3

For a general $d$-dimensional system, the maximally mixed state is the one described by the normalised identity matrix: $$\rho=I/d.$$ In the specific case of a two-qubit system, this reduces to the first state you write. Why is this state more mixed than the following This is a bit hard to answer, as it depends on what your current understanding/...


3

The Von Neumann entropy of $1/2 (|00\rangle \langle 00| + |11\rangle \langle 11|)$ is one bit. For $I/4$ it's two bits of entropy instead. The entropy of states that only have entries on the diagonal of a density matrix is very easy to compute in general, because you just treat the entries as probabilities and compute the Shannon entropy. The Shannon ...


3

I'm not an expert with this sort of thing (i.e. there may be imperfections in this argument), but hopefully this will set you in the right direction... Consider $\rho_{AB}=\rho_A\otimes |0\rangle\langle 0|$ and $\sigma_{AB}=\sigma_A\otimes |0\rangle\langle 0|$. It must be that $S(\rho_A\|\sigma_A)=S(\rho_{AB}\|\sigma_{AB})$. Now, your superoperator can be ...


3

You can always do that. Subspaces $\text{Im}(V)$ and $A\otimes |0\rangle$ have the same dimension, so there must be some unitary that translates one subspace to another. That is, $\exists W \in \text{Unitary}(A\otimes B), W(\text{Im}(V)) = A\otimes |0\rangle$. Now $WV$ translates $A$ to $A\otimes |0\rangle$. Since $V$ is isometry $WV$ is also isometry, hence ...


3

You don't need any additional conditions beyond those already stated in the question. That is, for any isometry $V: A \rightarrow A\otimes B$ and any unit vector $|\psi\rangle_B$, there will always be a unitary $U$ satisfying the equation in the question (simultaneously for every choice of $\rho_A$). One way to see this is to first pick any orthonormal ...


3

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 ...


2

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 ...


2

The entanglement entropy (what you call "von Neumann entropy") is a good measure for entanglement of pure states in the asymptotic setting, i.e. when one is dealing with many copies. However, it is not a good measure for mixed states. Distillable entanglement and entanglement cost are entanglement measures which apply to both pure and mixed states. ...


2

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 ...


2

Let's write $$ p'=\frac12+m\cdot n\frac{2p-1}{2}, $$ and assume without loss of generality that $p>\frac12$, which also means that $p'>\frac12$. Note the binary entropy function $h(p')$ is symmetric about $p'=\frac12$, and is monotonically decreasing for $p'>\frac12$, meaning that we want to make $p'$ as large as possible in order to minimise $h(p')$...


2

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}_{\...


2

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 ...


2

Yes, unitarity preserves eigenvalues. This is because the definition of eigenvalues is that any Hermitian matrix $H$ can be brought into diagonal form by a unitary $V$, $$ VHV^\dagger=D, $$ and the diagonal elements of $D$ are the eigenvalues. So, now consider $UHU^\dagger$. This can be diagonalised by a unitary $VU^\dagger$ into the same matrix $D$: $$ (VU^\...


2

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 ...


1

We have, \begin{equation} \begin{aligned} S &= - \operatorname { Tr } \left( \varrho \log _ { 2 } \varrho \right) = \log _ { 2 } \left( \frac { \left| \gamma _ { B } \right| ^ { \left( 2 \left| \gamma _ { B } \right| ^ { 2 } \right) / \left( \left| \gamma _ { B } \right| ^ { 2 } - 1 \right) } } { 1 - \left| \gamma _ { B } \right| ^ { 2 } } \right) = \...


1

I think I have an answer. The following should be the CVX code for one of the formulations found in this link. cvx_begin sdp variable X(2, 2) semidefinite minimize(trace(id' * X)) % id is eye(2) subject to kron(id, X) >= rho_ab % the tensor product of two density matrices a, b X >= 0 cvx_end The optimal value found in this program is $$\text{...


1

In the sense of the entropy, the maximum entropy from a "$d$" dimensional density matrix is just $\log_2 d$ if you measure wrt two degrees of freedom. This is more than $\log_22=1 \ \text{bit}$, and hence more information. This is because we always compare the Quantum Information with the Classical Information as over a "bit". However, if we measure the ...


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