# Tag Info

12

Thanks for pointing this out! It turns out that this device was mis-calibrated in a way that was leading to that behavior. We just fixed the calibrations, so the problem should be gone now. I apologize for the trouble, and we will try to update our routine calibrations to detect and prevent this problem from coming up in the future :-).

6

This was a readout crosstalk error that has now been resolved.

4

That seems to restrict the output probability distributions of all quantum circuits to rather high entropy distributions. The output of a typical randomly chosen quantum circuit is rather high entropy. That doesn't mean you can't construct circuits that have low entropy outputs (you can), it just means that picking random gates is a bad strategy for ...

3

It seems like the skew is indeed high on qubit 0. I ran a single Hadamard followed by measure on this qubit, and see about 13% skew. The other qubits on this device seem fine (less than 2% skew). This is probably an error on the backend's discriminator (i.e. manifesting as high readout error). To see this, you can try applying readout error mitigation (code ...

3

So, to begin, I would point out that the 500 micosec T1 time is for a single qubit in isolation, while the GHZ results are on a 20 qubit device. This device has an avg T1 time of around ~75 microsec. The GHZ results were done by Ken Wei from IBM, and will be published shortly. In short, the circuit is a standard GHZ building circuit, with a hadamard ...

3

Your expectation here is correct. c[0] should be 0 (well, modulo some small readout errors). The difference between backends is just due to a software bug on some of them. This will get fixed, thanks for reporting. As an aside, it is important to note that on current IBM devices, there is a constraint that all measurements are done simultaneously. So both ...

2

Of course if we have unitary evolution $$|\psi_1\rangle = U|\psi_0\rangle$$ then $$|\psi_0\rangle = U^\dagger|\psi_1\rangle$$ I did not read the paper, but evidently the authors do something different, based on the following: the Schrödinger equation $$i\hbar\frac{\partial\Psi}{\partial t}=\hat{H}\Psi$$ changes its form if we substitute $t\rightarrow -t$ to ...

2

After some further consideration I think it's quite clear that the only probability mass function evaluated in the computation of $\mathcal{F}_{\text{XEB}}$ is that of the classically computed ideal distribution, denoted $P(x_i)$ in the main paper. This leads me to the conclusion that the phrasing of the following excerpt from section IV.C of the ...

2

As an initial matter, I think the Supplementary Information (linked in some other answers on this sight) has a significant amount of discussion on $\mathcal{F}_{XEB}$. However, as I understand it (misunderstandings are my own): There is indeed a concentration of outputs from a random quantum circuit, away from a state wherein the square of the coefficients ...

1

Each quantum processor has specific so-called error rate and a little bit different type of noise caused by specific conditions the processor runs in. Therefore, results produced by same circuits can be different on different quantum processors. In your case, there is apparently a bias caused by some external factors specific for ibmqx2. You can try to run ...

1

I would be careful with so-called "quantum supremacy". This term means that there are some tasks where quantum computers are faster than classical ones. However, a speed-up is different for different task, e.g. quantum algorithms allow exponential speed-up of integer factoring (Shor algorithm) or calculating values of $x^{T}Mx$, where $x$ is a solution of $... 1 The key to figuring out the probability of any measurement result is Born's rule, which says that if you have a state$\left|\psi\right\rangle$the probability of observing measurement outcome$\left|\phi\right\rangle\$ is given by \begin{align} \Pr(\phi | \psi) = \left|\left\langle \phi | \psi \right\rangle \right|^2. \end{align} In the example you ...

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