GaussStrife
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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, ...

For interactions between non-nearest neighbour qubits, ancilla qubits are required, together with SWAP gates. The state of one of the (in this case) two qubits is swapped with the ancilla. This ...

The gate/operator in the brackets is the non-local CNOT gate, frequently used to create bipartite entanglement. Given it itself is a 2 qubit gate, then the tensor of this with the identity is simply a ...