A pure state is said to be maximally entangled if the Von Neumann Entropy $S(\rho_{A})$, where $\rho_{A}=Tr_{B}(\rho_{AB})$ is the maximum value. ie $S(\rho_{A})=log(d)$ where $d$ is the dimension of ...

If Alice and Bob share a maximally entangled state wherein each controls one qubit: $$|\Phi^{+}\rangle=\frac{|0\rangle^{A}\otimes|0\rangle^{B}+|1\rangle^{A}\otimes|1\rangle^{B}}{\sqrt{2}}$$ and Alice ...

First, encode the bipartite ensemble $\gamma_{12}$ into a CQ state $\omega$ $$\omega_{XAB}=\sum_{x}p_{x}(x)|x\rangle\langle x|\otimes\rho_{AB}^{x}$$ now we can take the difference between $$H(A|B)\... View answer Accepted answer 2 votes$$I(A:B)=S(A)+S(B)-S(AB)J(A_{\{\Pi_{i}\}}:B)=S(A_{\{\Pi_{i}\}})+S(B)-S(A_{\{\Pi_{i}\}}B)I(A:B)-J(A_{\{\Pi_{i}\}}:B)=S(A)-S(AB)-S(A_{\{\Pi_{i}\}})+S(A_{\{\Pi_{i}\}}B)$$Since$$\rho = \sum_j ...

Given the quantum discord is always minimized for a 1 dimensional projector: $$D_{A}(A:BC)=I(A:BC) - J_{A}(A:BC)$$where $$J_{A}(A:BC)=max_{\{\Pi_{i}^{A}\}}(S(BC)-\sum_{i}p_{i}S(BC_{i}))$$ In this case,...

$U=\sum_{j}|\phi_{j}\rangle\langle\psi_{j}|$ by the preceding discussion, as $$(\langle\psi_0|U^\dagger)\ (|\psi_1\rangle U)= \langle\psi_0|UU^\dagger|\psi_1\rangle =\langle\psi_0|\psi_1\rangle$$ ...

You can do it via use of substitution, or via the expansion into vectors and comparison. However, for this and other expansions, I find the use of the identity operator, which can be diagonalised in ...

To expand on @Purva Tharke's comment, the strong subadditivity inequality states: $$H(ABC)+H(C) \le H(AC) + H(BC)$$ $$=H(ABC)+H(C) +H(C) -H(C) \le H(AC) + H(BC)$$ $$=H(AB|C) \le H(A|C) + H(B|C)$$ $$=0\... View answer 1 votes Given a quantum state |\psi\rangle, you can performa a basis change on this state using the I, the Identity operator, which of course can be expanded as \sum_{i}|i\rangle\langle i|, where |i\... View answer Accepted answer 3 votes I assume for \rho and \sigma, you meant to write \rho = \sum_i p(i) \vert i\rangle\langle i\vert \otimes \rho_i and \sigma = \sum_i q(i)\vert i\rangle\langle i\vert\otimes\sigma_i. From QIT by ... View answer 1 votes I_{acc}(\rho^{XA})=S(\rho_{A})-\sum_{x}p(x)S(\rho_{A}^x) I(X:A)=S(\rho_{x})+S(\rho_{A})-S(\rho_{XA})=S(\rho_{x})+S(\rho_{A})-(S(\rho_{X})+\sum_{x}p(x)S(\rho_{A}^x))=S(\rho_{A})-\sum_{x}p(x)S(\rho_{... View answer 1 votes When a measurement is performed in a certain basis, an orthogonal projection is performed on the state in question. So given a state |\psi\rangle, a measurement in the hadamard basis would result in ... View answer Accepted answer 2 votes$$-iZ\frac{\theta}{2} = \begin{bmatrix}-i\frac{\theta}{2} &0\\0&i\frac{\theta}{2} \end{bmatrix}$$This is alread diagonal, so now you just take the exponenital operation on each of the ... View answer 0 votes Due to \rho being a positive and adjoint, you can always spectrally decompose it into a convex combination of pure states, it's eigenvectors. However, since the measurement statistics, or the action ... View answer 1 votes The reduced density matrix is$$\begin{bmatrix}\frac{1}{2} & 0 \\ 0 & \frac{1}{2}\end{bmatrix}$$This easily seen by taking the partial trace over the subsystem of A:$$Tr_{A}\begin{bmatrix}\...

$$\langle M\rangle=\sum_{j}p_{j}\langle\psi_{j}|M|\psi_{j}\rangle=\sum_{j}p_{j}\langle\psi_{j}|M\sum_{i}|e_{j}\rangle\langle e_{j}|\psi_{j}\rangle=\sum_{j}p_{j}\sum_{i}\langle e_{j}|\psi_{j}\rangle\... View answer Accepted answer 4 votes \frac{s}{2}|0\rangle\langle1|\otimes|0\rangle\langle1|+\frac{s}{2}|1\rangle\langle0|\otimes|1\rangle\langle0| should disappear when you take the trace over them, as \langle0|1\rangle and \langle1|... View answer Accepted answer 3 votes The complex conjugation flips the sign of the imaginary part of a complex number. Transposition exchanges the row and column co-ordinates of a value in a matrix. A vector can be thought of as a matrix ... View answer 0 votes$$\sqrt{SWAP}|01\rangle\ = \frac{1+i}{2}|01\rangle + \frac{1-i}{2}|10\rangle\sqrt{SWAP}(\frac{1+i}{2}|01\rangle + \frac{1-i}{2}|10\rangle)=\frac{1+i}{2}(\frac{1+i}{2}|01\rangle + \frac{1-i}{2}|10\...

As I understand it, the key caveat to postselection, at least in regards to retrocausal effects, is that Bob already knows what state Alice will post-select. Since he already knows what she will do, ...