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My question is about how in general this operator be experimentally realisable in terms of some combination of measurements of $\hat{A}$ and $\hat{B}$ where $\hat{A}$ and $\hat{B}$ are hermitian operators that are experimentally realisable? There is no reason to believe that a measurement of $\hat{A}\hat{B}+\hat{B}\hat{A}$ can be realized as a combination ...

5

To understand what it means to have maximum knowledge, maybe it helps to consider a situation in which you don't have that knowledge. Suppose I was giving you a qubit and telling you that I threw a fair coin and prepared either a $|0\rangle$ if it was heads, or a $|1\rangle$ if it was tails. You would then describe the state with a density matrix $$\rho = \... 4 Measuring some observable A of a system in state |\psi\rangle does not mean looking at A|\psi\rangle. To determine probabilities of measurement outcomes, expand your state in the eigenbasis of the observable. You have already done this. So your |\psi\rangle=|0\rangle+|1\rangle tells us there is a 1/2 probability of measuring state |0\rangle and ... 4 Zurek has a more public-oriented intro/overview, if you are into the popular picture pond, W Zurek, 2003, Decoherence and the transition from quantum to classical -- REVISITED There is also a popular go-to book by Schlosshauer, M Schlosshauer, 2008, Decoherence and the Quantum-to-Classical Transition , Springer, ISBN-13: 978-3540357735 . To get ... 4 Quantum circuit setup Since the quantum circuits are all unitary operations, they are reversible too. The circuit that is used to generate Bell basis from the product basis can be used in reverse, just as shown below, to measure in the product basis, for which we already have an understanding. &... 4 First, you apply a unitary transformation (that is, an interaction) which transforms the Bell states to a product basis. Then, you measure in that product basis. For instance, a CNOT gate will transform the Bell state to the four x axis eigenstates, or a controlled-phase (generated by a \sigma_z\otimes \sigma_z interaction + local z fields) to the x... 3 The essence is the following: You can write any purification in Schmidt form (note that this is not a transformation, just rewriting the state in a different basis). Then, any two purifications of a given state will be of the form$$ |\psi\rangle = \sum \lambda_i |a_i\rangle \otimes |b_i\rangle \in \mathcal H_A\otimes \mathcal H_B$$and$$ |\phi\rangle = \...

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Measurement is usually just defined as a gate on it's own. The "well known gate" that measurement maps to is simply the measurement gate. Its one of the few places where some difficult, hard to model interactions with an environment are allowed in a quantum circuit, so we just separate that part out. For example, in this circuit You can clearly see two ...

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Not even the $d=2$ state is isotropic, for any $M$. For two spin-1/2 particles, there are four basis states, which we will label by their total spin $S$ and spin along the $z$-axis $m_s$ as $|S,m_s\rangle$: $$|1,1\rangle=|00\rangle$$ $$|1,0\rangle=\frac{1}{\sqrt{2}}(|01\rangle+|10\rangle)$$ $$|1,-1\rangle=|11\rangle$$ $$|0,0\rangle=\frac{1}{\sqrt{2}}(|01\... 2 If a unitary is chosen at random, you want every unitary to be "equally likely". So in particular if you have a subset \mathcal X\subset \mathrm{U}(1) and you multiply it with another unitary U, then U\mathcal X is just "the same set" of unitaries, just "displaced" within the unitary group. It should therefore have the same probability, \mu(U\mathcal ... 2 Whenever we have a vector space, there is a complete orthogonal basis that can be constructed for the space. And that basis can be transformed to arrive at another basis which is also complete and orthogonal. This can be done in many ways, one of which is Schmidt decomposition. Given a vector \phi, you can find a whole basis using Schmidt decomposition and ... 2 A partial isometry is mapping a sub-vector space K of a Hilbert space H onto another sub-vector space K' of the same dimension isometrically, that is$$(V\psi, V\phi) = (\psi,\phi)$$for any two vectors in the initial domain of the isometry, that is K=V^*VH. The fact that V^*VH is the initial domain of V can be proved by showing that E=V^*V ... 2 We can do this for general quantum states instead. Suppose we have two states |\phi\rangle and |\psi\rangle, which are not orthogonal. We want to construct the POVM that best distinguishes these two states, and makes no false positives. As a simpler example, we can choose |\phi\rangle = |0\rangle, and |\psi\rangle = 1/\sqrt{2}(|0\rangle+|1\rangle). ... 2 The magic transformation One way to approach this problem is to use a Klein transformation. A Klein transformation can be used to make S and B commute with each other without changing the (anti)commutation relations within S or B individually. This is done by using the S-operators to construct an operator K that commutes with everything in B ... 2 Probably the author means that a state in which you know everything you can know about that system, is represented by such a vector (or rather a ray, as Alfred Centauri remarked). When there is uncertainty in the state, which could be considered classical uncertainty (where quantum uncertainty would be the uncertainty in experimental outcomes that we still ... 2 The quantum Fisher information is always greater than or equal to the classical Fisher information, by definition. The QFI with respect to a parameter \theta is defined by$$\mathcal{H}_\theta[\hat{\rho}_\theta] = \max_{\{\hat{M}_\alpha\}} \mathcal{F}_\theta[\hat{\rho}_\theta,\{\hat{M}_\alpha\}],$$where \mathcal{F}_\theta[\hat{\rho},\{\hat{M}_\alpha\}] = ... 2 The convention usually followed is to define 0 \ln 0 = 0. That is not so bad as it looks, you may calculate \lim\limits_{x \rightarrow 0} x\ln x. Using L'Hopital's rules, \begin{array}{lll}\lim\limits_{x \rightarrow 0} x \ln(x) & = & \lim\limits_{x\rightarrow 0} \frac{\ln(x)}{x^{-1}} \\ { } & = & \lim\limits_{x \rightarrow 0} \frac{x^{-... 2 Yes, you have written the state | \psi \rangle  in the eigen basis of Z (|0 \rangle, |1 \rangle) , that is why Z is diagonal. Since Z and X do not commute with each other they cannot be simultaneously diagonalised in one basis. If the operators commute then they can be simultaneously diagonalised and will have same eigen basis. Measurement is ... 2 This cannot be true - assuming that I understand the question correctly - since for instance, for \rho=|0\rangle\langle0| and \sigma=|1\rangle\langle1|, the measurement outcomes on all n qubits in \omega are perfectly correlated. On the other hand, on any purification of the form |\psi\rangle_{AR}^{\otimes n}, the measurement outcomes on the A ... 2 This is not even true for d=2 and M=2: The state$$ |00\rangle+|11\rangle $$is not invariant under rotations - e.g., applying$$ U=\begin{pmatrix} 1 & i\\ i &1 \end{pmatrix} $$will map it to$$ (U\otimes U)(|00\rangle+|11\rangle) = |00\rangle+i|01\rangle+ i|10\rangle+|11\rangle\ . $$Indeed, the orbit of |00\rangle+|11\rangle under ... 1 Yes, it is possible to entangle more than two qubits. Nice example is GHZ state composed of three qubits:$$ |\psi_\text{GHZ}\rangle = \frac{1}{\sqrt{2}}(|000\rangle + |111\rangle) $$Another example is W state defined for n qubits followingly$$ |\psi_\text{w}\rangle = \frac{1}{\sqrt{n}}(|10\dots00\rangle + |01\dots00\rangle + \dots +|00\dots10\rangle+|...

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Instead of state vector you can also represent states of a quantum system using a density matrix. If you have a basis $\{|n\rangle\}$ and a state vector $|\Psi\rangle=\sum_nc_n|n\rangle$ you can calculate the density matrix as follows $$\rho=|\Psi\rangle\langle\Psi|=\sum_{m,n}c_m{c_n}^{\!\!*}|m\rangle\langle n|$$ Expectation values can be calculated using $$\... 1 If only 2 qubits out of these 3 are entangled how can we represent it in such a form? If none of the qubits are entangled, the state can be written as: (a_1|0\rangle + a_2|1\rangle) (a_3|0\rangle + a_4|1\rangle) (a_5|0\rangle + a_6|1\rangle) Taking the appropriate products of the a_x coefficients gives the amplitudes of each 3-bit states in your ... 1 The amount of information carried by a system is given by entropy function  H(X)  which is defined as:$$ H(X) =-\sum p(x) \log_{2}(p(x)) $$where  p(x) is the probability of random variable X, where the sum is over support of p(x) . For example: Now if you have biased coin with p({\rm heads}) = 0.8 and p({\rm tails}) = 0.2. You can calculate ... 1 The projection operator is defined by P_{\lambda}=|\lambda\rangle\langle\lambda|, here \lambda is one of the eigenstates of the system. So I believe that |00\rangle,|01\rangle,|10\rangle and |11\rangle are the eigenstates of two qubits system. So,$$\hat{M}=\lambda_{00}|00\rangle\langle00|+\lambda_{01}|01\rangle\langle01|+\lambda_{10}|10\rangle\...

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The smallest explicit example I could come up with where this can be done for sure, a bit silly maybe: let $$A = \begin{pmatrix}1&0\\0&-1\end{pmatrix},\ \ \ \ \ B = \begin{pmatrix}1&-i\\i&1\end{pmatrix}.$$ Then $A$ and $B$ don't commute, and $$AB + BA = \begin{pmatrix}2&0\\0&-2\end{pmatrix}.$$ This means that you can just measure $... 1 An isometry is a map such that $$\langle Vx,Vy\rangle=\langle x,y\rangle$$ if the image of$V$has smaller dimension than its domain, then clearly this property cannot hold, as if we have an orthonormal basis $$\langle x_i,x_j\rangle=\delta_{ij}$$ we cannot have $$\langle Vx_i,Vx_j\rangle=\delta_{ij}\tag{*}$$ because there aren't enough ... 1 As far as I can tell the link does not state that For any quantum state$\rho_A$and purifications$\vert\psi\rangle_{AB}$and$\vert\phi\rangle_{AC}$, there exists an isometry$V_{B\rightarrow C}$such that$(I_A\otimes V_{B\rightarrow C})\vert\psi\rangle_{AB} = \vert\phi\rangle_{AC}$. Which is just as well since that claim is incorrect, as your example ... 1 If$\mathcal E$is a channel, then it can be expressed in terms of Kraus operators, $$\mathcal E(\rho) = \sum K_i\rho K_i^\dagger\ .$$ Then, the map$\mathcal E\otimes I$is of the form $$(\mathcal E\otimes I)(\rho) = \sum (K_i\otimes I)\rho(K_i\otimes I)^\dagger\ .$$ It is thus also a channel, with Kraus operators$K_i\otimes I$. In particular, if$\...

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Your first two points are correct, and the fact that $\mathcal E\otimes \mathrm{id}$ is trace preserving is simply a consequence of the fact that, for any state $\rho_{AB}$, $$\mathrm{Tr}_B(\mathcal{E}_A\otimes \mathrm{id}_B(\rho_{AB}))=\mathcal{E}(\mathrm{Tr}_B(\rho_{AB}))$$ which is very easy to show either by writing down the partial trace explicitly ...

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