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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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( I copied some text from a previous answer of mine) Defining the Choi and $\chi$ matrix The Choi matrix is a direct result of the Choi-Jamiolkowski isomorphism. Some intuition on what this is can be found in this previous answer. Consider the maximally entangled state $|\Omega \rangle = \sum_{\mathrm{i}}|\mathrm{i}\rangle \otimes |\mathrm{i}\rangle$, where $... 6 The task that you describe in your question — a circuit which flips a single qubit, if and only if the two input states are different — is not possible. We can show this as follows. First, there is no way to distinguish two states which differ only by a global phase, because no quantum operations can distinguish between two state-vectors which only differ in ... 6 Let's say you have a magical machine that gives you$\langle P_{p} \rangle$(which are expectation values and therefor, well, numbers) and only the$\langle P_{p} \rangle$. It does this for all the$3$- (n)-fold tensor products of the traceless Paulis. That is:$p \in \{X,Y,Z\}^{\otimes 3 (n)}$, for a total of$3^{3 (n)}$Paulis, where$n$is the general ... 5 No, weak measurement and quantum tomography don't break BB84. I recommend that you create an explicit quantum circuit that implements the weak measurement or the quantum tomography, and check for yourself that it actually fails. The basic problem comes down to the fact that there is a trade-off between how much information you get and how likely you are to ... 5 The picture has two parts: The first goes until the dots. It is simply three$|0\rangle$states. (The ground state.) You will recognize that the same picture -- but only until the dot -- is used in panel b) and c) of the same figure. After the dot, there is a second part of the circuit -- starting with the open half-circles -- which describes the measurement/... 5 Under the assumption that the ensemble$\mathcal{U}$faithfully produces the Haar expectations at least to the second moment, the inversion can be performed as suggested in the last paragraph of the question: Define: $$\theta_b = U^{\dagger}|b\rangle\langle b| U$$ and replace the averaging over the ensemble by Haar averaging. (This step is is done only to ... 5 An empirical solution could be to use the Grover's Diffusion Operator$D$. Lets say the qubits are in an initial state$|\psi\rangle = \sum_{0}^{2^n-1}\alpha_i|i\rangle$. Since global phase/sign is irrelevant. We can assume that phase/sign of$\alpha_0$is + for the sake of convenience (If$\alpha_0=0$choose the lowest index with non-zero amplitude). We can ... 5 The mutual information can be written in terms of the relative entropy, please see Nielsen and Chuang (the entropy Venn diagram figure 11.2). I am writing the equation in the question's notation: $$I(\rho^{AB}) = S(\rho^{AB}|\rho^{A} \otimes \rho^{B})$$ The relative entropy can be estimated without full tomography. The procedure is described in Bengtsson ... 5 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 ... 5 I am sure that since you are asking this question you probably already understand this, but for future & other's reference let me give a quick recap of what we are trying to achieve. Quantum channels Any process (in an open quantum system) is some map$\Lambda$from a space of density matrices to a space of density matrices. I write a, because these ... 5 From linear algebra, if$v_1, \dots, v_n$is a basis of the vector space$V$then every vector$v\in V$can be written as a linear combination $$v = a_1 v_1 + \dots + a_n v_n\tag1$$ where the coefficients$a_k$belong to the underlying scalar field. Moreover, if$V$is an inner product space and$v_1, \dots, v_n$is an orthonormal basis then the ... 5 I apologise in advance. This is a rough and hand-waivy answer. You can give "information-theoretic" lower bounds by noting that the measurements can be seen as a linear map$M$from quantum states to outcome probabilities$y$. For instance, if you have a POVM$(E_i)_{i=1,\dots,N}$, then the probability vector is$y = \sum_i \mathrm{tr}(E_i \rho) ...

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Without really checking your arguments, there is a fundamental reason why the scaling could be fine, but it is still not strange at all. The point is that you estimate with additive precision, but your fidelity could be exponentially small. E.g. if the fidelity is $10^{-8}$, you need a much smaller error to resolve that compare to a fidelity of, say, 1/2. ...

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Denote the projections onto basis elements by $P_j^{(k)}=|u_j^{(k)}\rangle\langle u_j^{(k)}|$, where superscript indexes different bases. Tomography of a density matrix $\rho$ gives us probabilities $\text{Tr}(\rho P_j^{(k)})$. This is actually a value of the Hilbert-Schmidt inner product between $\rho$ and $P_j^{(k)}$ in the space $L(\mathcal{H})$ $-$ the ...

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I would suggest you use the code from the tutorial about quantum state tomography, adapting it to a real device of your choice. You can find the updated tutorial here Caveat: as state tomography requires 3^n circuits, you will need probably to find a method of batch processing of these circuits if they exceed the job circuit limit of your real device. See ...

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This would not be enough information to reconstruct the bi-partite state. Single-qubit case For the one-qubit case, reconstruction of the state (which we describe as $\rho$) works, because the single-qubit Pauli observables $\{\sigma_{x},\sigma_{y},\sigma_{z}\}$ together with the $\sigma_{I}$-operator creates a basis for the space of single-qubit density ...

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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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As indicated by Danylo in his anwser, eq. (32) in arXiv: 1103.2030 presents the sixteen vectors ("ignoring overall phases and normalisation") \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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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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First of all, here's a short disclaimer: I'm not an in-depth expert in this field, I'm just currently getting in contact with tomography more and more often :) So take the following with a grain of salt. It might be incomplete in the sense that better results have been shown somewhere. We consider the problem of reconstructing a $d$-dimensional quantum state ...

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Tomography generally speaking uses a collection of measurements to reproduce an underlying state. So you experimentally reproduce the same situation over and over, collect statistics and find the most likely estimate for that state. In QKD, information is sent once and doesn't repeat. So for each unit of information, you're never ever able to collect enough ...

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The technical term is "quantum state discrimination". One has to carefully formulate the problem, because it is generally hard to identify an arbitrary state (tomography) as you noticed. However, given the promise that two states are prepared at random, there's an upper bound on the probability of correctly identifying the state via a single ...

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This is not a complete answer but a couple of comments that should help clarify these details: (i) $\mathcal{M}$ is not invertible as a quantum channel but as a linear map; namely, even though the inverse exists, $\mathcal{M}^{-1}$ is not CP, even though it is a linear map. Assuming the input-output dimensions are the same, the only CP maps that are ...

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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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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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Density matrix of single qubit state can be estimated based on this formula $$\rho = \frac{\text{tr}(\rho)I+\text{tr}(X\rho)X+\text{tr}(Y\rho)Y+\text{tr}(Z\rho)Z}{2},$$ 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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Quantum state tomography owes its power and flexibility to the fact that it supports a wide class of measurements. Any informationally complete POVM, i.e. one whose elements span the space $L_H(\mathcal{H})$ of Hermitian operators on the target system's Hilbert space $\mathcal{H}$ qualifies for use in QST. One way to highlight the generality of QST with ...

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