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Let's describe the one qubit pure stat in Bloch sphere notation (in order to avoid the global phase ambiguity): $$|\psi \rangle = \cos\big(\frac{\theta}{2}\big) |0\rangle + e^{i\varphi} \sin\big(\frac{\theta}{2}\big)$$ The problem can be solved with the quantum state tomography, but in this answer, I want to consider a slightly different approach for dealing ...


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 ...


3

Yes, if we have fixed backend, number of qubits, and noise model (e.g., Basic device noise model in https://qiskit.org/documentation/stubs/qiskit.providers.aer.noise.NoiseModel.html#qiskit.providers.aer.noise.NoiseModel), we would have a fixed calibration matrix. I think the advantage is that once we have this calibration matrix, we can use it to perform ...


3

You measure many times and collect statistics. E.g. you do $1000$ measurements and find $600$ times the first outcome. You can then deduce that $|\alpha|^2\simeq 0.6$ and $|\beta|^2\sim0.4$ (using appropriate statistical methods to compute the associated estimation errors). Note that this does not fully characterise the state, but only gives you the ...


3

The issue is that you are using noisy hardware with imperfect operations and measurements. In particular, the most likely problem here is that after you prepare a qubit it immediately begins decaying towards the ground state $|0\rangle$ via interactions with the environment. Each qubit will be slightly more likely to be measured as 0 instead of 1 than you'd ...


2

The reason for this viewpoint on measurement is primarily historical. Physicists often think of measurements in terms of observables (aka hermitian operators). The way this is related to the more mathematical notion of projective measurements (a type of POVM measurement) is by thinking about the spectral decomposition of a hermitian operator $$M=\sum_i \...


2

The answer is no. To this end, pick a linearly independent set $\{\sigma_k\}$ which spans the full matrix space (over $\mathbb C)$, that is, a basis. (This is always possible, as the positive operators span the hermitian ones over $\mathbb R$.) Then pick a dual basis $\sigma'_\ell$ such that $$ \mathrm{tr}[\sigma'_\ell \sigma_k]=\delta_{k\ell}\ . $$ Then, $$...


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I think your reasoning is just fine and I checked that the Bell states are indeed eigenvectors of $$M=U^\dagger(Z\otimes Z)U,$$ as $$\begin{align} M|\phi^+\rangle =& \phantom{{}-{}}|\phi^+\rangle,\\ M|\phi^-\rangle =& -|\phi^-\rangle,\\ M|\psi^+\rangle =& -|\psi^+\rangle,\\ \text{and } M|\psi^-\rangle =& \phantom{{}-{}}|\psi^-\rangle \end{...


1

For measuring 2 qubits, there are 4 possible outcomes, corresponding to projectors $$ P_{00}=|00\rangle\langle 00|,\qquad P_{01}=|01\rangle\langle 01|,\qquad P_{10}=|10\rangle\langle 10|,\qquad P_{11}=|11\rangle\langle 11|. $$ So, if you have a state $|\psi\rangle$, you get the outcome $x$ with probability $p_x=\langle\psi|P_x|\psi\rangle$. This immediately ...


1

"what are the advantages and disadvantages in the determination of the calibration matrix each time that we do an experiment and mitigate its error?" Advantage: The noise matrix will be a more accurate description of the current noise situation. My understanding is that each day, the qubtis are cooled from 300K all the way down to about 15mK, and ...


1

Let $P_{\pm}$ be the projectors onto the two orthonormal basis states of the measurement. So, $$ P_+=\frac{1}{2}(|0\rangle\langle 0|+|1\rangle\langle 1|+e^{-i\phi}|0\rangle\langle 1|+e^{i\phi}|1\rangle\langle 0|). $$ Also, let $|\psi\rangle$ be the state that you're measuring (note that this must be normalised. Yours might be depending on your constraints on ...


1

A reset gate is equivalent to a swap gate between the target qubit and a new ancilla qubit in the $|0\rangle$ state. So you can replace your question with "how does swapping a qubit Q with an fresh ancilla qubit affect the qubits Q is entangled with?" or "how does discarding Q affect the qubits Q is entangled with". And the answer is that,...


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We start from the defining form of the channel as $\Phi_\mu(X)=\sum_a \operatorname{tr}(\mu(a)X)E_{a,a}$.$\newcommand{\PP}{\mathbb{P}}\newcommand{\tr}{\operatorname{tr}}\newcommand{\calX}{\mathcal X}\newcommand{\calY}{\mathcal Y}\newcommand{\calZ}{\mathcal Z}\newcommand{\ket}[1]{\lvert #1\rangle}\newcommand{\bs}[1]{\boldsymbol{#1}}$ (Natural representations) ...


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I think unlike the relative phase in the answer you reference, it is a global phase in your case: Your XHP-circuit where P=ID, prepares the state: [0.707+0j,-0.707+0j], where P=X, prepares the state: [-0.707+0j, 0.707+0j]. These states are differ by a global phase ${e}^{i\pi}=-1$. But the global phase is undetectable $|ψ⟩:={e}^{iδ}|ψ⟩$, also see the answer.


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