# Tag Info

12

A measurement is basically a CNOT between the quantum computer and the external environment. The important distinction between this CNOT and the CNOTs entirely within the quantum computer is that the target qubit is not protected. The environment is going to spread and mix the target qubit's value all over the place. To perform the inverse of a measurement ...

9

If the simulator is saying that state 00 occurs 75% of the time then the simulator has a bug. Reordering measurements can't make certain outcomes more likely in that way. It would violate the no communication theorem.

6

Three outcomes amounts to more than one bit if the outcomes are all deterministic, and give you information about the original qubit. But suppose I have a coin (that is either heads or tails). I roll a dice, and if it comes 1 through 5, I tell you "H" or "T", depending on what the coin is. If it comes up 6, I tell you "6". There are three outcomes, but ...

5

Yes, it is possible. To obtain the state $$\vert \phi \rangle = \alpha \vert 00 \rangle + \beta \vert 11 \rangle$$ from $$\vert \psi \rangle = \alpha \vert 000 \rangle + \beta \vert 111 \rangle,$$ note that the latter can be written as \begin{aligned}\vert \psi\rangle=&\frac{1}{\sqrt{2}}\alpha \vert 00+ \rangle + \frac{1}{\sqrt{2}}\alpha \vert 00 -\... 5 Measuring the second qubit in the X basis you get one of two results, associated with the two eigenstates of X, which are: \newcommand{\ket}[1]{|#1\rangle}\sqrt2\ket\pm\equiv\ket0\pm\ket1. The post-measurement states corresponding to these two outcomes are given (up to normalisation) by the overlaps \langle\pm_2|\psi\rangle, where here \ket{\pm_2}\... 5 In general, the number of shots does not increase the accuracy of an experiment. Rather it gives a more precise answer. Attached is a figure showing the distance (in terms of Hellinger distance) for a Bell state run on the IBM Quantum Boeblingen device from the theoretical answer as a function of the number of shots taken. For each value of the shots, the ... 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 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)$$The problem can be solved with the quantum state tomography, but in this answer, I want to consider a slightly different approach for dealing ... 4 Yes, it is possible, same way as it is possible to measure a state with complex amplitudes in a basis with real amplitudes (say, a |i\rangle state in the [|0\rangle, |1\rangle] basis). Either way, the probability of measuring state |\psi\rangle and getting measurement result corresponding to basis state |a_i\rangle is defined as P_i = |\langle a_i| \... 4 Write down the two reduced density matrices of the single qubits that you have access to. Apply the Helstrom measurement (there are several descriptions of this on the site already). The problem is that, in this case, the two reduced density matrices are the same. That means that you cannot tell them apart. More explicitly,$$ |\varphi_2\rangle=(I\otimes X)|...

4

If you wish to distinguish two states $|\psi\rangle$ and $|\phi\rangle$, you can only guarantee to do this if $\langle\psi|\phi\rangle=0$. You do this by measuring in a basis defined by the two states (alternatively, you apply a unitary $U$ such that $$U|\psi\rangle=|0\rangle,\qquad U|\phi\rangle=|1\rangle,$$ and then measure in the standard $Z$ basis. ...

4

You can calculate the probability of a given answer $\pm 1$ to each measurement by evaluating $$\langle B|\frac{I+(-1)^{\eta_1}\vec{n}_1\cdot\vec{\sigma}}{2}\otimes\frac{I+(-1)^{\eta_2}\vec{n}_2\cdot\vec{\sigma}}{2}|B\rangle$$ Thus, the probability of equal measurement outcomes is $$\langle B|\left(\frac{I+\vec{n}_1\cdot\vec{\sigma}}{2}\otimes\frac{I+\vec{... 4 How it works depends on the choice of quantum system used for computation. For any choice of quantum system, the common theme is that \text{CNOT} does not collapse the wavefunction, i.e. force a choice between \vert 0 \rangle and \vert 1 \rangle, while a measurement does. A simple example (oversimplified here) uses a non-linear Kerr medium to create ... 4 In general, "operating on a state with an observable" does not have direct physical meaning (i.e. you cannot think of it as evolving the state doing something to it). What does have physical meaning, is applying a unitary operation to a state. Every unitary operator corresponds to a physical operation that you can (in principle) implement, transforming (... 4 Now If I chose the standard basis |0\rangle,|1\rangle what will be the result I will get? If I measure with respect of the projection operator \langle 0| I get \alpha and If I measure with respect of the projection operator \langle 1| I get \beta. This is wrong. First of all, arguably the most natural kind of measurement in QM consists in choosing ... 4 Look like a lot of misunderstanding. If you measure a state \alpha|0\rangle+\beta|1\rangle in computational basis, the state collapses either to |0\rangle or |1\rangle. In Quantum Information we say that we measured state |0\rangle or |1\rangle, or simply we measured 0 or 1. There is no chance to know \alpha and \beta from the measurement. ... 4 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 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 ... 4 While talking about knowing the position exactly is a nice theoretical ideal, in practice, you cannot do that. You'll really be asking: "In which 'bin' of width \delta x where x spans from x_{\min} to x_{\max} is the particle confined to?". This means that there's (x_{\max}-x_{\min})/\delta x bins, and so you basically need$$ \log_2\left((x_{\max}-...

4

Depending on your physical implementation, it may be that the two basis states are at different energies. Thus, when the state collapses, the energy changes. However, the average energy remains constant so you cannot use this for energy generation or similar. For example, assume we have a qubit with a Hamiltonian $$H=E|1\rangle\langle 1|.$$ For any state $... 4 This would not be enough information to reconstruct the bi-partite state. Single-qubit case For the one-qubit case, reconstruction of the state (which we describe as$\rho$) works, because the single-qubit Pauli observables$\{\sigma_{x},\sigma_{y},\sigma_{z}\}$together with the$\sigma_{I}$-operator creates a basis for the space of single-qubit density ... 3 Let me give you the structure for how you go about answering this question. Let$|\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$. Now, Alice does ... 3 There isn't. A density matrix encodes all the knowledge available about a state, therefore if two states are described by the same density matrix, they are indistinguishable. Ket vectors differing by only a global phase have always the same density matrix, and represent the same physical state. 3 This procedure is referred to as superdense coding protocol; you can find a lot of detailed explanations of it, starting with the Wikipedia article. If you want a hint before getting a full explanation... Effectively you need to distinguish the four Bell states. You can always do this with 100% accuracy, since all the states are orthogonal to each other. To ... 3 There are a lot of comments and objections in the question, too many in fact to go through them all. I will try to address some of the points that I think hide misconceptions, to hopefully give a clearer idea of what's going on. About "interpretations of QM in which the measurement is non-unitary" The non-unitarity of the measuring process is not a matter ... 3 When you give an observable, such as$\Omega$, that is used to define the measurement basis. It is not something that you would usually use to directly perform calculations (you can, and for$2\times 2$matrices, we often do, as I'll detail below). Normally, you want to take your observable,$\Omega$, and find the eigenvalues and eigenvectors. More ... 3 The fact is that it doesn't matter. So long as one of the four terms has the minus sign on it, you'll get a suitable test. Yes, there seems to be some inconsistency with labelling the different cases, but it's not a big deal - nothing physically changes if you swap which measurement basis you call$A$and which you call$A'$for example; you're still doing ... 3 The most direct way to do this using normal notation is to simply write the Bell projections using the same convention for subscripts:$\def\ket#1{\lvert#1\rangle}\def\bra#1{\langle#1\rvert}\def\idop{\mathbf 1}\begin{aligned} \bra{\Phi^+}_{1,5} \;&=\; \tfrac{1}{\sqrt 2}\Bigl(\,\bra{0}_1\bra{0}_5 \,+\, \bra{1}_1 \bra{1}_5\,\Bigr), \\ \bra{\Phi^-}_{1,... 3 This can actually be easily done using the Qiskit Terra qiskit.quantum.info.analysis.average.average_data function that takes the counts data returned by a backend and the desired observable defined by a dict, list, or ndarray. The doc-string for that function actually has the ZZ your looking for as an example. 3 Here is an implementation of a circuit producing state|\psi\rangle = \frac{1}{\sqrt{3}}(|00\rangle + |01\rangle + |10\rangle)$on IBM Q: Note that$\theta = 1.2310$for$\mathrm{Ry}$on$q_0$.$\theta = \frac{\pi}{4}$and$\theta = -\frac{\pi}{4}$for first and second$\mathrm{Ry}$on$q_1$. The$\mathrm{Ry}$on$q_0$prepares qubit in superposition$|...

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