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The Cauchy-Schwarz-Bunyakovsky inequality says that $$|\langle x\vert y\rangle|^2\le \langle x\vert x\rangle\,\langle y\vert y\rangle.$$ In other words $\omega(x,y)\le 1$. Is that what you want?

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To fully answer your question, to get that $$\sum_i(p_i - q_i) \leq D(p_i, q_i) = \frac{1}{2}\sum_i|p_i - q_i|$$ (whereas naively you might be tempted to bound $\sum_i(p_i - q_i)$ by $\sum_i|p_i - q_i| = 2\times D(p_i, q_i)$), you need to use Equation 9.4, namely, $$D(p_i, q_i) = \max_{S} \sum_{i\in S}(p_i - q_i),$$ where the maximisation is over all ...

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Yes, one can break entanglement without getting entangled with something else. Consider e.g. two spins interacting via the Heisenberg interaction $\vec S_1\cdot \vec S_2$, which are initially in a state $\lvert\uparrow,\downarrow\rangle$. As time evolves, the system first gets entangled, and subsequently gets disentangled again, without any other system ...

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A general version of the holographic principle is introduced in Bousso (1999) A covariant entropy conjecture. In the simplest case of an isolated black hole, it reduces to an earlier version of the principle, which says this: The entropy (the natural log of the number of microstates compatible with that black hole macrostate, in a theory of quantum gravity) ...

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When thinking of quantum states as resources for computation, one must consider the dimension of the system. Systems with more degrees of freedom have greater capacity for computation. In the case of quantum illumination (QI), the dimensionality of the composite system (i.e. d=dS dI where dS and dI are the dimensions of the signal and idler subsystems, ...

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There is a mistake (or rather: inconsistency) in how you define the $M_i$ (which makes the condition you want to prove incorrect, so there cannot be a proof!). To be consistent, given the Kraus representation $\rho\mapsto \sum M_i\rho M_i^\dagger$, you need to define $$M_i = \langle i_B| U |0_B\rangle$$ (with $U$ the unitary in the Stinespring dilation ...

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I'm no expert in weak measurement, so I might be missing something, but I'm pretty sure you can analyze both these examples by considering what the wavefunction should be after the glass optic, applying the projective measurement, and renormalizing. In which case, 1. If I add a mode label for the three spatial modes involved in the question, (i.e. $|... 0 I think I got the answer. Writing down a problem always helps. The very definition of trace of a matrix having operators as elements implies that the trace can "commute" with matrices having numbers as elements. More precisely, for a matrix$\hat{\textbf{M}}$with operatorial elements and a real matrix$S$, since trace acts only on operators, $$\text{ij}^{... 2 If I understand well the meaning of the various symbols, the strict inequality is already false for n=1. In that case, as remarked in Norbert's answer, a generic mixed state has always the form$$\rho = \frac{1}{2}\left(t_0 \sigma_0 + \sum_{k=1}^3 t_k \sigma_k \right)\:, \quad t_\mu \in \mathbb{R}$$where \sqrt{\sum_{k=1}^3 t^2_k}\leq 1 and t_0=1. ... 1 This is nothing but a different way of stating the well-known fact that points on the surface of the Bloch sphere are pure: For a point \vec r\in\mathbb R^3 in or on the Bloch sphere, \rho=\tfrac12(I+\vec r\cdot\vec \sigma) is precisely of the form you give for n=1 and with T=(1,\vec r). 4 Two perspectives: Yes. Schrödinger's equation is just the continuous time version of unitary evolution. Thus, if you take unitary evolution from quantum information and ask to make it continuous, you arrive at the Schrödinger equation. No. There is neither continuous time in quantum information (depends on your "definition" of quantum information), nor is ... 1 The Rabi oscillation is a simplified picture of the atom-light interaction. The atom is considered to be a pure 2-level system and the energy difference of the two atomic states is given by E = \hbar \omega_0. Now, if we shine light onto the atom, the electric dipole moment of the light couples these two states. If the light is resonant with the atomic ... 0 Optical and quantum coherence are intimately related. For example, in the case of a pair of polarized photons prepared in a net-zero polarization state (a superposition of s-p and p-s pairs), both optical and quantum coherence are degraded by scattering. To scatter light, a charged particle oscillates in the incident e-field direction, thereby "observing" ... 2 No, a qudit gate must also be unitary, just as all quantum gates. What can be done, however, is to probabilistically implement a non-unitary gate (in fact, any non-unitary operation). If your gate is e.g. only implemented conditioned on the detection (or non-detection) of a photon, then this is possible. 0 Let the system be at time t_1 in a state \psi(x). Decompose this state as a superposition of position eigenstates:$$ \psi(x,t_1) = \sum_y a_y \delta(y-x)\ . $$The position expectation value at time t_1 is thus e_1=\sum y |a_y|^2. Further, any of the position eigenstates can at most evolve with the speed of light, this is, if \delta(x-x_0) ... 1 No, you will need an infinity of integrals. That is, whereas$$S = \operatorname{Tr}\rho \rho = \int \rho(x,x') \rho(x',x)dx'dx ,$$you are looking at$$S = -\operatorname{Tr}(\rho \log \rho) = \operatorname{Tr}\left ( \rho \sum_{n=1}^\infty (1\!\! 1 -\rho)^n ~/n \right ), $$which necessitates an integral per matrix multiplication. Note the sign ... 2 My last answer was wrong on the final equation. I've corrected it below. Not really. Remember the definition of the trace, it's the sum of the diagonal matrix elements, not all of them. Using a discrete basis \{|e_i \rangle\}_{i=1}^\infty, the trace of an operator \hat A \in \mathcal{L(H)} can be written as:$$\mathrm{Tr}(\hat A) = \sum_{i=1}^\infty \... 3 The no cloning theorem states that if you have two systems,$A$and$B$, and the following state $$|\psi\rangle_A|0\rangle_B$$ then there is no unitary$U$such that $$U|\psi\rangle_A|0\rangle_B=e^{i\alpha}|\psi\rangle_A|\psi\rangle_B$$ what you're asking, if I understand correctly, is whether there exists a unitary$U_\tau$such that $$U_\tau|\psi(t)\... 0 For block matrices we have the Helmert–Wolf formula$$ \left(\matrix{{\bf A}& {\bf B}\cr {\bf C}&{\bf D}}\right)^{-1} =\left(\matrix{({\bf A}^{-1}+{\bf A}^{-1}{\bf B}({\bf D}-{\bf C} {\bf A}^{-1}{\bf B}) {\bf C}{\bf A}^{-1} & -{\bf A}^{-1}{\bf B}({\bf D}-{\bf C}{\bf A}^{-1}{\bf B})^{-1} \cr -({\bf D}-{\bf C}{\bf A}^{-1}{\bf B})^{-1}{\bf C}{\bf A}^... 1 You were essentially already there with the derivation. You found that $$\sum_{q_x>p_x}(p_x-q_x)=-\sum_{p_x>q_x}(p_x-q_x),$$ which used in the last equation you wrote for$D(p_x,q_x)$gives you $$\frac12\left[\sum_{p_x>q_x} (p_x-q_x)-\sum_{q_x>p_x}(p_x-q_x)\right] =\sum_{p_x>q_x} (p_x-q_x).$$ This is what you were trying to prove, with$S_M\...

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For the first question, the relation between an information measure and correlation function is given by: $$I(A : B) \geq \frac{\mathcal{C}\left(\mathcal{O}_{A}, \mathcal{O}_{B}\right)^{2}}{2\,\left\|\mathcal{O}_{A}\right\|^{2} \left\|\mathcal{O}_{B}\right\|^{2}}$$ where $\mathcal{C}\left(\mathcal{O}_{A}, \mathcal{O}_{B}\right)$ is the correlation function ...

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