# Tag Info

## Hot answers tagged decoherence

10

In simpler terms your question is: if noise/decoherence keeps entering the computation, how can a big computation possibly survive? The key concept you're missing is quantum error correction, which can pump noise/decoherence back out of the system. Of particular practical interest is the surface code.

10

What you call a black box is simply isolating the quantum system that stores (or represents) your qubits from the environment. This can be done in several ways depending on your physical realization. For example, in an ion trap based quantum computer, one uses states of a single ion to represent a qubit, and isolates that from the environment by levitating ...

8

In short: "coherence"! It's the crucial difference between quantum and classical. $\rho=|0\rangle\langle 0|+|1\rangle\langle 1|$ is just a statistical mixture, and behaves like a classical coin - any measurement that you perform on it gives a 50:50 split between the two possible outcomes. By contrast, $|+\rangle=(|0\rangle+|1\rangle)/\sqrt{2}$ is a very ...

6

The naming started in NMR and it has become the difference between the following two experiments. Experiment one: Prepare the qubit in a superposition state (apply a H gate) and vary the wait time and then measure in the superposition basis (apply another H gate). The decay time of this experiment is $T_2^*$. We commonly call this a Ramsey experiment. ...

6

To simplify things a bit, let's take a single qubit and a single qutrit for comparison. First, the amplitude damping channel (giving e.g. emission of a photon) for a qubit is $\mathcal E\left(\rho\right) = E_0\rho E_0^\dagger + E_1\rho E_1^\dagger$, where $$E_0 = \begin{pmatrix}1 && 0 \\ 0 &&\sqrt{1-\gamma}\end{pmatrix}, \quad E_1 = \begin{... 6 The statement in Wikipedia is very generic, and only cites this paper as a reference. Quoting from the abstract of the paper: We demonstrate that decoherence of many-spin systems can drastically differ from decoherence of single spin systems. The difference originates at the most basic level, being determined by parity of the central system, i.e., ... 5 There are multiple ways to mathematically express the state of a quantum system. One is to write it as a linear combination of basis states, as either a vector or a matrix, as you have here. This is useful in many cases, but it also seems to be confusing to newcomers, since things like \vert 0 \rangle + \vert 1 \rangle, \vert 0 \rangle \langle 0 \vert + \... 5 The quantum circuit model describes a quantum computer as a closed quantum system and assumes that there is a system which executes the circuit but is completely isolated from the rest of the universe. In the real world, however, there are no known mechanisms for truly isolating a quantum system from its environment. Real quantum systems are open quantum ... 5 I agree with most of what you've written in the first paragraph, though I would say that at roughly the same time (only 1 month apart!) as the Rebentrost et al. paper you mentioned, a very similar paper was posted to arXiv by Plenio and Huelga called "Dephasing assisted transport: Quantum networks in biomolecules" and it actually got published in the same ... 5 I believe the current record is held by the Jian-Wei Pan group in China, who are able to generate entanglement via a satellite. The journal article is here, while there's plenty of media coverage that is a bit more accessible, e.g. New Scientist. This claims a distance of 1203km. 4 I have worked with NVs in nanodiamonds a little bit, and you are totally right, surface characteristics have a huge influence on how far we can push them. There are definitely multiple groups working on the chemistry/material science that are working to clean up the surfaces as much as possible. I had a colleague, Carlo Bradac who worked with our chemistry ... 4 Photons travel fast, and there's often the option to transfer their entanglement to solid state. Of course, the advantage of transferring entanglement to a solid state qubit is that one is able to operate with it (one- and two-qubit gates, for example) with ease and efficiency, whereas it is very hard to effect two-qubit quantum gates on photons themselves, ... 4 Why not input one half of a maximally entangled state as the input to the black box (so that half has the same dimension as the input dimension)? Then you could test your favourite measure, such as the purity, of the full output state. If the oracle corresponds to a unitary evolution, the purity is 1. The less coherent the smaller the purity. Incidentally, ... 4 You could also look at the following webpage: https://quantumcomputingreport.com/scorecards/qubit-quality/ where they provide recent (I'm not sure how often they update this scores) values for gate fidelities and decoherece times for IBM and Rigetti chips (unfortunately they don't give any results on ion traps and photonics, since these machines are not ... 4 I guess your best shot would be to look for experimental comparisons like this one on Arxiv. But I am not aware of a tracking. I do not think we can consider having a "state of the art" in this field. The goal being to make them always better of course with better connectivity for instance (a possible factor to take into account). 4 Suppose you have a state \rho, and a random process that changes this to a state \rho_j with probability p_j. If you know what the value of j is, your knowledge of the resulting state will be described by the corresponding \rho_j. If you have no information regarding j, your knowledge will be described by$$\sum_j ~ p_j ~ \rho_j$$This is a ... 4 frequency (GHz): The frequency(energy) associated with the transition between the qubit's ground state (|0\rangle) and first excited state (|1\rangle). readout error: The probability of preparing a |0\rangle(|1\rangle) and measuring a |1\rangle(|0\rangle), ie., of having an error in your readout single qubit U2 error rate: The average error per ... 3 Josu, You might be mis-understanding something. Your equation for the Pauli channel says that when t\rightarrow \infty, all operators (X,Y,Z,I) have an equal probability (1/4) of transforming the density matrix \rho. You seem to be suggesting that t\rightarrow \infty, the probability of the I operator acting on \rho whould go to 0. Keep in mind ... 3 I cannot give you a complete answer(I am not too familiar with the IBM quantum tools) however I might be able to give you a few hints from a NMR/EPR perspective. In magnetic resonance T2 is commonly measured by generating a spin coherence, and refocusing at progressively longer times then measuring a spin echo. In quantum gate language that would be: ... 3 The length of all backend basis gates is available from backend.properties().gate_length. For example properties = backend.properties() id_gate_length_qo = properties.gate_length('id', 0) 3 T2 is so-called dephasing time. It describes how long a phase of qubit stay intact. In your words, it is time from |+\rangle= \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle) to |-\rangle= \frac{1}{\sqrt{2}}(|0\rangle - |1\rangle), or conversely. Just note that both T1 and T2 are not actually "time from state x to state y" but rather decay constants. ... 2 Let's assume that your black box processes classical inputs (i.e. a bit string) to classical outputs in a deterministic way, i.e. it defines a function f:x\mapsto y. If you can only prepare and measure separable states in that basis, all you can determine is what that function f is. Assuming that all the outputs are different, it could have been ... 2 I'm not exactly sure what you mean by quantumness of your black box. So maybe there are some more sophisticated approaches (similar to the other answer you could use an entanglement witness to show that your black box is not entanglement breaking). However, in general you could perform quantum process tomography (see e.g. arXiv:quant-ph/9611013). 2 Your question revolves implicitly around the concept of quantum decoherence and how to protect real-world implementations of qubits from it for a long time. This is an incredibly general problem, and at the same time, the details are wildly dependent on the technology used. If you have access to it, you can check chapter 5 : "Noise and decoherence" of ... 2 Decoherence is the very general term which, more or less, is anything resulting in a loss of purity during the evolution of a system. Sometimes, when people are being a bit non-specific, they might be thinking of a particular type of decoherence such as dephasing (or perhaps depolarising) when they use the term decoherence. Relaxation and dephasing are two ... 2 It's fundamentally similar to/the same as Baker–Campbell–Hausdorff (BCH). Generally, in quantum physics, this is most often used (or at least taught) with commuting Hamiltonians:$$e^{-i\left(H_1+H_2\right)t} = \sum_{n=1}^\infty\frac{\left(-it\right)^n}{n!}\left(H_1+H_2\right)^n = e^{-iH_1t}e^{-iH_2t}e^{\frac{1}{2}\left[H_1, H_2\right]t^2}\cdots$$where the ... 2 The state you've given, |\psi\rangle=a|0\rangle+b|1\rangle is a pure state. It has not been decohered. Decoherence is a process which turns pure states into mixed states. We usually write these in terms of density matrices. A pure state can be written in this form:$$ |\psi\rangle\langle\psi|=\left(\begin{array}{cc} |a|^2 & ab^\star \\ a^\star b & |...

2

You will need to know how long it takes for each gate of the circuit to be performed. Then the decoherence error rate is simply $$e^{t_{gate}/t_{decoherence}}$$

2

Is it possible for quantum computers to exhibit behavior similar to flip errors in classical computers where a state |0⟩ becomes |1⟩ due to this error? Yes. It would be a very abrupt error if you're talking about errors on physical qubits. Usually, we'd think of an error as being a little bit of an X rotation (for example). However, the effect of performing ...

2

If the measurement is an irreversible process then the probability of the resultant state is towards the initial quantum state it was in, that is, if the resultant state is $1$ with a probability of, say, $63\%$, that means that is has a $63\%$ probability that the initial state was $|\psi \rangle= |1\rangle$. This is a fundamental misinterpretation....

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