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That doesn't scale well. After a moderately long calculation you're basically left with the maximally mixed state or whatever fixed point your noise has. To scale to arbitrary long calculations you need to correct errors before they become too big.
Here's some short calculation for the intuition given above. Consider the simple white noise model (...
10
As an addition to Nat's answer, it's worth mentioning that 'noise' is a specific concept1 in quantum computing. This answer will use Preskill's lecture notes as a basis.
In essence, noise is indeed considered to be something that could be described as 'thermal noise', although it should be noted that it is an interaction with a thermal environment causing ...
9
If the error rate were low enough, you could run a computation a hundred times and take the most common answer. For instance, this would work if the error rate were low enough that the expected number of errors per computation was something very small — which means that how well this strategy works would depend on how long and complicated a computation ...
9
Now, adding to M. Stern's answer:
The primary reason as to why error correction is needed for quantum computers, is because qubits have a continuum of states (I'm considering qubit-based quantum computers only, at the moment, for sake of simplicity).
In quantum computers, unlike classical computers each bit doesn't exist in only two possible states. For ...
9
Unfortunately for analog computation it turns out that when realistic assumptions about the presence of noise in analog computers are made, their power disappears in all known instances; they cannot efficiently solve problems which are not solvable on a Turing machine.
"Noise" appears to be used in the general sense of non-idealities in a signal:
In ...
8
Throughout this answer, the norm of a matrix $A$, $\left\lVert A\right\rVert$ will be taken to be the spectral norm of $A$ (that is, the largest singular value of $A$). The solovay-Kitaev theorem states that approximating a gate to within an error $\epsilon$ requires $$\mathcal O\left(\log^c\frac 1\epsilon\right)$$ gates, for $c<4$ in any fixed number of ...
7
As far as I’m aware, the surface code is still regarded as the best. With an assumption of all elements failing with equal probability (and doing so in a certain way) it has a threshold of around 1%.
Note that the paper you linked to doesn’t have a 3D surface code. It is the decoding problem that is 3D, due to tracking changes to the 2D lattice over time. ...
7
The answer to noise (and any source of error, really) in quantum computations is quantum error correction: You choose an encoding such that discretized errors correspond not only to invalid encodings but also uniquely determine what kind of error must have occured. This is not possible for all errors but with reasonable error models (such as single qubit ...
6
How do we prevent quantum noise in a quantum computer?
Well, technically the answer is (at least for most systems): we use ridiculously low temperatures (much colder than space), we shield everything (or at least as much as possible) out, that might introduce any noise (radio waves, such as phone signals or light, magnetic fields, ...), we do everything to ...
5
A channel $\Phi$ is said to be degradable if there exists another channel $\Xi$ such that $\Xi\Phi$ is complementary to $\Phi$.
The idea here is as follows. Suppose $\Phi$ is a channel and $\Psi$ is complementary to $\Phi$. If $\Phi$ is applied to a state $\rho$, then the output of the channel is $\Phi(\rho)$ (of course), while $\Psi(\rho)$ represents ...
5
I'm going to go for an intuitive answer here, as requested. Let's s go in steps:
Your input is (often?) classical, so up to that point we're good.
Then you start doing quantum operations and achieve, for example, quantum superpositions between different states. Here you're right, you cannot look to check if you're doing OK, and that indeed is a problem, or ...
5
There are several ways that you could realise the depolarising map $
\mathcal N_p(\rho) = (1\!-\!p)\:\!\rho + p \!\!\:\cdot\!\tfrac{1}{2}\mathbf 1$ map on a quantum computer — including an idealised quantum computer, in which waiting around for the noise to do the work for you would not be an available method.$\def\ket#1{\lvert#1\rangle}$
We start ...
4
I believe that the Centre for Engineered Quantum Systems, School of Physics, The University of Sydney and the Center for Theoretical Physics, Massachusetts Institute of Technology use of a tensor network decoder of Bravyi, Suchara and Vargo (BSV), to achieve the highest error correction threshold to date.
In their whitepaper from last December, "Ultrahigh ...
4
This can be done using the 'Aer' component of Qiskit. The properties information can be turned into a noise model using
from qiskit.providers.aer import noise
properties = device.properties()
noise_model = noise.device.basic_device_noise_model(properties)
basis_gates = noise_model.basis_gates
This can then be supplied to the execute() method, as is ...
3
For amplitude damping, $\gamma$ is something like $e^{-\Delta t/T_1}$ where $\Delta t$ is how long the Kraus operator is supposed to act. But be very careful, Kraus evolution assumes your system has no initial correlations, that every qubit interacts with identical baths and that every qubit is identical. All the assumptions are most likely violated and so ...
2
I'm not sure what was the expected solution, but this also works.
First of all, note that $$ I = |\phi^+\rangle\langle\phi^+| + |\phi^-\rangle\langle\phi^-| + |\psi^+\rangle\langle\psi^+| + |\psi^-\rangle\langle\psi^-|,$$
where $|\phi^+\rangle, |\phi^-\rangle, |\psi^+\rangle, |\psi^-\rangle$ are Bell states and $I$ is 4-dimensional identity operator.
...
2
In the dim and distant past (I.e. I don’t remember the details any more), I tried to calculate an upper bound on a fault tolerant threshold. I suspect the assumptions that I made to get there wouldn’t apply to every possible scenario, but I came up with an answer of 5.3% (non-paywall version).
The idea was roughly to make use of a well-known connection ...
2
Asking the author to clarify would give you the exact answer you are looking for. However, based upon the context provided I believe this may be related to the problem quantum noise spectroscopy attempts to solve.
Noise
According to a team of Dartmouth researchers led by Professor Lorenza Viola,
These quantum properties are essential for quantum ...
2
noise should average itself out.
Noise doesn't perfectly average itself out. That's the Gambler's Fallacy. Even though noise tends to meander back and forth, it still accumulates over time.
For example, if you generate N fair coin flips and sum them up, the expected magnitude of the difference from exactly $N/2$ heads grows like $O(\sqrt N)$. That's ...
2
Why do you need error correction? My understanding is that error correction removes errors from noise, but noise should average itself out.
If you built a house or a road and noise was a variance, a difference, with respect to straightness, to direction, it's not solely / simply: "How would it look", but "How would it be?" - a superposition of both ...
1
The best I have it's this generic answer, which I put here for clarity, hoping for improvements/corrections or even to be superseded by something better:
If the limiting factor for fidelity in a given architecture+algorithm are the single-qubit gates, or the two-qubit gates, or the measurement, and if this limiting factor is not optimized in a ZEFOZ point,...
1
The whole point is that you do not want, nor need, to "look" how the computation is going.
You can ensure that the input is what it should be by a variety of means. The simplest case being that you may simply trust that your apparatus, which you previously characterized very well, will produce what you ask it to produce.
After that, you will know whether ...
1
The first paper you mention, by Tejada et al, does not actually refer to conventional nanoparticles as such, but rather to single molecule magnets. This other paper by Loss, Quantum computing in molecular magnets, is about the same systems but perhaps more clear in quantum computing terms, since it gives more details on the underlying Hamiltonian and gives a ...
1
The Barkhausen effect has to do with domain wall motion. The magnetic qubit discussed in the first reference is based on nm sized magnetic particle, which we can assume to be single domain, and therefore would not exhibit Barkhausen noise.
This paper by Kittel discusses domains in magnetic particles.
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