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6

No such (orthonormal) basis can exist. An orthonormal basis $\{|\psi_i\rangle\}$ requires $\langle \psi_i | \psi_j \rangle = 0$ for $i\neq j$, and so clearly \begin{align} [\rho_i, \rho_j] &= |\psi_i\rangle \langle \psi_i | \psi_j\rangle \langle \psi_j | - | \psi_j\rangle \langle \psi_j |\psi_i\rangle \langle \psi_i | \\ &= 0 \end{align} So to get a ...

6

It is indeed possible to have extremal, non-projective POVMs. Examples can be drawn from SIC-POVMs, as suggested in a comment. For example, as mentioned in the Wiki page, the only possible SIC-POVM in $d=2$ dimensions is $\{\tilde\Pi_i\}_{i=1}^4$ where $\tilde\Pi_i\equiv \frac12 \Pi_i$, $\Pi_i\equiv \lvert\psi_i\rangle\!\langle\psi_i\rvert$, and $$|\psi_1\... 5 One way of looking at the relationship between POVMs and observables arises from identifying their counterparts in the theory of probability of which quantum mechanics can be thought of as an extension. It is easier to identify the counterparts if we temporarily restrict our attention to a special type of POVMs known as projection-valued measures or PVMs. ... 4 First, recall that \mathrm{tr} A = \sum_i \langle i|A|i \rangle. Each equation is then a sum where all terms are products of P(z) and three other quantities. Further, the sum in the first equation ranges over a single index suggesting that all matrices under the trace are diagonal. In fact, since we are working with a composite system this also suggests ... 3 The optimal probability of guessing correctly is$$ \frac12 + \frac12 \Big\|\frac23 \rho_0 - \frac13 \rho_1 \Big\|_1 $$where \| X \|_1 = \mathrm{Tr}[\sqrt{X^* X}] is the Schatten 1-norm. This success probability is achieved by the POVM with operator$$ E_0 = \Pi_{[\tfrac23 \rho_0 - \tfrac13 \rho_1]_+} \qquad E_1=I-E_0 $$where [X]_+ denotes the positive ... 3 Well, since these are projective measurements on the subspace of the first m qubits, we can just list all projectors on the computational basis of this first subspace and 'pad' them with I's on the second subspace:$$ P_{j} = |j\rangle\langle j|_{m} \otimes I_{|n|},\,\,\, \forall j \in \{0,1\}^{m}, $$which gives exactly |\{0,1\}^{m}| = 2^{m} different ... 3 What is the guarantee this implementation is efficient? Is there any rule regarding when implementing such POVMs is efficient? The implementation of such a gate will only depend on the parameter k (which I assume you mean to be fixed), not n. Since efficiency is generally phrased in terms of scaling with n, and you have no dependence on that, it is ... 3 Minimalist formal proof (I'll use \mu_a\equiv \mu(a)): \textrm{(A)}\Rightarrow\textrm{(B)}: Let \Gamma\ge0. Then,$$ (\Phi_A\otimes I_B)(\Gamma_{AB}) = \sum (\sigma_a)_A\otimes\mathrm{tr}_A[((\mu_a)_A\otimes I_B)\,\Gamma_{AB}]\ , $$which is a separable decomposition, since \mathrm{tr}_A[((\mu_a)_A\otimes I_B)\,\Gamma_{AB}]\ge0 because it describes ... 3 Quantum state tomography owes its power and flexibility to the fact that it supports a wide class of measurements. Any informationally complete POVM, i.e. one whose elements span the space L_H(\mathcal{H}) of Hermitian operators on the target system's Hilbert space \mathcal{H} qualifies for use in QST. One way to highlight the generality of QST with ... 2 I'll write \rho_1 = |\psi_1\rangle \langle \psi_1| and \rho_2 = |\psi_2\rangle \langle \psi_2|. We want the discrimination to be unambiguous so we want,$$ \mathrm{tr}[\rho_1 \Pi_2] = 0 = \mathrm{tr}[\rho_2 \Pi_1]. $$That is, when we get outcome i\in \{1,2\} we know that we received \rho_i as the other state has a zero probability of obtaining that ... 2 Indeed, M_i=\sqrt{E_{i}} is wrong. The correct relationship is E_i = M_i^\dagger M_i. The possible M_i for a given E_i are M_i = U\sqrt{E_i} for any unitary U, as M_i^\dagger M_i = \sqrt{E_i}U^\dagger U \sqrt{E_i} = \sqrt{E_i}\sqrt{E_i} = E_i. 2 POVM for standard QST in the Pauli bases In standard single-qubit QST one measures in the Pauli bases, each with equal probability \frac{1}{3}. As @Rammus has pointed out, this corresponds to the POVM$$ \{E_{m}\} = \Big\{\tfrac{1}{3}|0\rangle\langle0|,\tfrac{1}{3}|1\rangle\langle1|,\tfrac{1}{3}|+\rangle\langle+|,\tfrac{1}{3}|-\rangle\langle-|,\tfrac{1}{3}...

2

That (A) implies (B) should be obvious from the physical intuition behind (A): A channel of the form (A) can be interpreted as performing a POVM measurement with elements $\mu_a$, and on obtaining outcome $a$ preparing the state $\sigma_a$. It should be obvious that this breaks any entanglement, since it (destructively) measures the input. (Note that also ...

2

That is indeed some weirdly written exposition with typos, but the result is correct. Let $\Phi(\rho) = \sum_k R_k \text{Tr}(F_k\rho)$ and $\Phi_k(\rho)=R_k \text{Tr}(F_k\rho)$. For $\Gamma = \rho_1 \otimes \rho_2$ we have $$(I \otimes \Phi_k)(\Gamma) = \rho_1 \otimes \Phi_k(\rho_2) = \rho_1 \otimes R_k\text{Tr}(F_k\rho_2) =$$ $$= \rho_1\text{Tr}(F_k\... 2 These two definitions define the same concept: the POVM measurement. The observable definition is how POVM is defined for use in the case of infinite index set and dimension (see e.g. POVM) and POVM definition in the question is how it is simplified for use in the finite case. If you are working in finite dimensions, the two constructions are equivalent. ... 1 An m-outcome positive-operator value measure or POVM consists of m positive operators A_1,\ldots,A_m on \mathbb{C}^n such that$$\sum_{i=1}^m A_i=I.$$Each A_i is positive iff A_i=B_i^*B_i for some operator B_i on \mathbb{C}^n, thus each POVM is a projection-valued measure or PVM iff each B_i is an orthogonal projection, if so then B_i=... 1 Upon some more reflection, the answer is probably as follows. Let \mathrm A be an observable according to the definition in the question, and assume \Omega is finite. Then any X\in\mathcal F is also some finite subset of \Omega. By definition of observable, we require the mapping \mathrm A_\psi to be additive and non-negative, and therefore$$\...

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