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Mathematically it is a relationship between a bipartite linear operator vector space $L(X\otimes Y)$ and a superoperator vector space $C(X): L(X)\to L(Y)$ (maps of linear operators to linear operators). Bipartite density matrices are contained in the former, and quantum channels in the latter. The real "physical" meaning of the isomorphism for quantum ...


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As @NorbertSchuch said in a comment, matlab has a function for taking the logarithm of a matrix: logm. In general, there is a standard method for calculating the function $f(\sigma)$ of a matrix $\sigma$. You first diagonalise the matrix: $$ \sigma=UDU^\dagger, $$ where $U$ is a unitary and $D$ is diagonal. We then say $$ f(\sigma)=Uf(D)U^\dagger, $$ where $...


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If $|z\rangle$ are orthogonal to each other, then $$ \log(\sum_z |z\rangle\langle z| \cdot b_z) = \sum_z |z\rangle\langle z| \cdot \log(b_z) $$ So $$ \mathrm{trace}(\sum_z |z\rangle\langle z| \cdot a_z \cdot \log(\sum_{z^\prime} |z^\prime\rangle\langle z^\prime| \cdot b_{z^\prime})) $$ $$ =\mathrm{trace}(\sum_z \sum_{z^\prime} |z\rangle\langle z| \cdot |z^...


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You are running into problems because $\rho$ is not a density operator. A mixed state density operator has $\text{tr}(\rho^2) < 1$, but even a mixed state density operator must have $\text{tr}(\rho)=1$. This is necessary because $\text{tr} (\rho) = \sum \limits_i p_i \, \text{tr}\left(\vert \psi_i \rangle \langle \psi_i \vert \right) = \sum \limits_i ...


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$\def\braket#1#2{\langle#1|#2\rangle}\def\bra#1{\langle#1|}\def\ket#1{|#1\rangle}$ I was being silly :) The sign change that happens is in some sense either associated with the $\ket{a}$ or with the ancilla qubit. By taking the ancilla qubit to be (say) $\left(\frac{\ket{0}-\ket{1}}{\sqrt{2}}\right)$, upon XORing with $f(x)$ we get a sign change that we ...


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In principle, yes, you can always do it. The Bloch representation can be generalised to arbitrary dimensions, and you can always parametrise states in it by their "angle coordinates". For example, you can write an arbitrary 3-modes pure state as $$|\psi\rangle=\cos\alpha|0\rangle + e^{i\theta}\sin\alpha\cos\beta|1\rangle+e^{i\phi}\sin\alpha\sin\beta|2\...


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The state of single qubit can be described as a point on the Bloch sphere. All the allowed transformations of a single qubit can then be described as rotations on the Bloch sphere. Unfortunately, bigger quantum systems can no longer be described as fitting on a sphere like geometry. As a result, this idea of considering transformations as rotations does not ...


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