# Tag Info

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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 ...

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

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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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 ...

4

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 ...

4

( 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 $|\Omega \rangle = \sum_{\mathrm{i}}|\mathrm{i}\rangle \otimes |\mathrm{i}\rangle$, ...

4

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 ...

4

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 ...

4

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 ...

4

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 & -...

4

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 ...

4

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/...

3

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 ...

3

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 ...

3

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 ...

3

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 ...

3

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 ...

3

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 ...

3

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 ...

3

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 ...

3

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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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 ...

2

Being restricted to real amplitudes means that you don't need to go for full on tomography. If you were looking at a single qubit, for example, to do full tomography with projective measurements, you'd need to make $X$, $Y$ and $Z$ measurements, while for the real-only version, you'd only need to make $X$ and $Z$ measurements. The question, then, is what's a ...

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Long story short: if you want to characterize a circuit or operation, you have to perform Quantum process tomography (QPT), which is a generalization of quantum state tomography, which is used to characterize states. QPT is not very easy unfortunately, there is a little math involved. Please see this answer to a previous question for a general overview. A ...

1

Similar ideas with Quantum PCA may be useful. Meaning, apply Quantum Phase Estimation on unitary $e^{-i\rho t}$ to obtain estimates of the eigenvalues of $\rho$ and finally estimate von Neumann entropy as $S(\rho) = -\sum \tilde{\lambda_i} \text{log}\tilde{\lambda_i}$. In the original paper there is the claim that you obtain estimates of the eigenvalues with ...

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About checking. I do not quite understand why you can't check how the GHZ-state was teleported in the most ordinary ways: in addition to measuring the state immediately after teleportation, as well as by YYX, YXY, XYY, XXX measurements, you can inverse your GHZ-state and make sure that all 0 are obtained, e.g. like as for this with the most usual GHZ-state ...

1

In principle, Bob here just has to guess the $2\times 2$ matrix $\sigma$. If he starts with any parametric state $\sigma(\alpha,\beta)$ with $\alpha,\beta\in\mathbb{C}$ and measures the outcome Tr$(M\sigma)$ with the post measurement state $\sigma '=M\rho M^\dagger/\text{Tr}(M\rho)$, he receives a number and he has to tune $\alpha,\beta$ to come close to ...

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