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Fidelity is a single-number measure of how good a gate is. Since there are many ways that a gate can go wrong, there are multiple ways that the fidelity can be defined. The exact answer to your question will therefore depend on which kind of fidelity you want. Any measure of fidelity will typically involve comparing the gate that you wanted to the channel ...


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The question presupposes a misconception that the vector form of a state $|\psi\rangle$ exists independently of its density operator form $|\psi\rangle\langle\psi|$, which is often described as secondary. In reality, the density operator of a state is all that truly exists --- and even then, it only exists as statistical information. In fact, you can ...


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Density matrix of single qubit state can be estimated based on this formula \begin{equation} \rho = \frac{\text{tr}(\rho)I+\text{tr}(X\rho)X+\text{tr}(Y\rho)Y+\text{tr}(Z\rho)Z}{2}, \end{equation} where $X$, $Y$, $Z$ are Pauli matrices. Obviously $\mathrm{tr}(\rho) = 1$. Terms $\mathrm{tr}(X\rho)$, $\mathrm{tr}(Y\rho)$ and $\mathrm{tr}(Z\rho)$ can be ...


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Basically you add measurements in different bases by applying gates before the (Z-basis) measurement. See here the standard implementation: https://github.com/Qiskit/qiskit-ignis/blob/3c59f82c11e87c071bc7e84240b50e2aa995281f/qiskit/ignis/verification/tomography/basis/paulibasis.py#L31


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This can be done using the statevector_simulator provided with Qiskit Aer. It will return the statevector that describes the quantum state at the end of your circuit. It can be used in the same way as the qasm_simulator, only your circuit shouldn't have measurement gates at the end. There is more information about this simulator in this tutorial.


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I find it a little tough to understand your calculations directly. I am especially confused by the circuit diagrams in your question; why they are there and what you are using them for. If you are performing calculations on theoretical data (without noise), then I feel you can make do with an easier approach for quantum state tomography. As per my answer on ...


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Preliminary I would like to rewrite the equation that you have in a slightly different manner. Since a density matrix can be written as a matrix, we can also write it down as a linear combination of elements from a basis for the space of density matrices. We can use essentially any basis to do this, but some are preferred: most notably, the Pauli basis. For ...


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As indicated by Danylo in his anwser, eq. (32) in arXiv: 1103.2030 presents the sixteen vectors ("ignoring overall phases and normalisation") \begin{equation} \left( \begin{array}{cccc} x & 1 & 1 & 1 \\ x & 1 & -1 & -1 \\ x & -1 & 1 & -1 \\ x & -1 & -1 & 1 \\ i & x & 1 & -i \\ i & x & -...


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You can find it here Symmetric Informationally Complete Quantum Measurements or here SIC-POVMs: A new computer study, in the appendix B. Update Given a single fiducial vector $v = (a_1,a_2,a_3,a_4)^T \in \mathbb{C}^4$ it's pretty easy to write down all SIC-POVM vectors. They are just $C^kS^lv$ for $k,l \in \{0..3\}$, where $C$ and $S$ are clock and shift ...


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There isn't. A density matrix encodes all the knowledge available about a state, therefore if two states are described by the same density matrix, they are indistinguishable. Ket vectors differing by only a global phase have always the same density matrix, and represent the same physical state.


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Quick answer: You will not be able to fully recover $x$. Explanations: By design, the HHL algorithm stores $x$ in the amplitudes of a quantum state. Because of how quantum mechanics works, the vector representing the quantum state (i.e. containing all the amplitudes of the quantum state) needs to be of unit-norm (according to the Euclidean norm). Because ...


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