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3

Suppose $\lambda_0 = 1$ and the rest are $0$. $$ F_Q [\rho,A] = 2 \sum_{k,l} \frac{(\lambda_k-\lambda_l)^2}{\lambda_k + \lambda_l} | \langle k |A| l \rangle |^2\\ = 2 \sum_{k=0,l \neq 0} \frac{(1-0)^2}{1 + 0} | \langle 0 |A| l \rangle |^2 + 2 \sum_{k\neq 0,l = 0} \frac{(0-1)^2}{0 + 1} | \langle k |A| 0 \rangle |^2\\ = 4 \sum_{l \neq 0} | \langle l |A| 0 \...


2

Consider a bipartite state $|\psi\rangle=\sum_{ij}\psi_{ij}|i\rangle\otimes|j\rangle$. In the following, I will work directly on the matrix elements of the objects involved. Tracing out the second space amounts to the following mapping $$\psi_{ij}\rightarrow \rho_{ii'}\equiv\sum_j \psi_{ij}\bar\psi_{i'j}.\tag A$$ Now forget about the partial trace, and ...


3

Let me give you the structure for how you go about answering this question. Let $|\Psi\rangle$ be the pure state shared between Alice and Bob. Bob measures his system in an orthonormal basis $\{|\phi_i\rangle\}$. Thus, Bob gets an answer $i$ with probability $p_i$, and the overall system is left in the state $|\psi_i\rangle|\phi_i\rangle$. Now, Alice does ...


0

Qiskit does have a density matrix class. See: qiskit-terra/qiskit/quantum_info/states/densitymatrix.py Also see here for some usage examples like initializing from a pure state, adding, subtracting, etc.


2

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^...


3

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 $...


2

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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