GaussStrife
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What is a maximal entangled multipartite state?
1 votes

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 ...

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What it means that the quantum teleportation involves projection onto a maximally entangled state?
1 votes

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 ...

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Concavity of Conditional Quantum Entropy
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1 votes

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)\...

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Why is the quantum discord of $\rho$ zero iff $\rho=\sum_j p_j \pi_j\otimes \rho_j$ for mutually orthogonal projections $\pi_j$?
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 ...

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Quantum discord of a tripartite system A:BC
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1 votes

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,...

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How is outer product an operator?
1 votes

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

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How to derive $|0\rangle=\frac{1}{\sqrt{2}}(|+\rangle+|-\rangle)$?
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4 votes

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 ...

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How to prove the positivity of the conditional quantum mutual information, $I(A;B|C)\ge0$?
1 votes

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\...

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How do we change the basis of a given qubit state?
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\...

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If $\rho,\sigma$ are classical-quantum states, can the fidelity $F(\rho,\sigma)$ be expressed in terms of $F(\rho_i,\sigma_i)$?
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 ...

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How is the additivity of accessible information, $\frac{1}{n} I_{\rm acc}(\rho^{\otimes n})=I_{\rm acc}(\rho)$, proved?
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_{...

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Measure a state in Hadamard basis
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 ...

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How do I show that $R_z(\theta)=e^{-iZ\theta/2}$?
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 ...

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Can we write the density operator as a sum of mixed states?
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 ...

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Reduced density matrix of the ensemble of two-qubit state that is possessed by Alice and Bob
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}\...

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Derivation of the identity $\sum_j p_j \langle \psi_j|M|\psi_j \rangle = \sum_j p_j \operatorname{tr}\left(|\psi_j \rangle \langle \psi_j|M\right)$
2 votes

$$\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\...

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Partial Trace of Werner State
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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|...

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Why is a conjugate transpose of $|+\rangle$ a vector $1/\sqrt{2} (\langle0| + \langle1|)$?
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 ...

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Is the square root of SWAP gate "maximally entangling"?
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\...

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Post-selection applied to quantum teleportation
1 votes

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, ...

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How do you CNOT between qubits faraway from each other?
2 votes

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 ...

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What does the unitary $[|0\rangle\langle 0|\otimes I+|1\rangle\langle1|\otimes(|1\rangle\langle 0|+|0\rangle\langle1|)]\otimes I$ represent?
1 votes

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 ...

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