# 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

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 ...

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

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

So basically you want to distinguish the state $| + \rangle \langle + |$ from the dephased state $\frac{1}{2}(| 0 \rangle \langle 0 | + | 1 \rangle \langle 1 | ) = \frac{\mathbb{I}}{2}$. Here's a simple experiment: apply a Hadamard to both states and then measure in the $\sigma_{z}$ basis. For the true superposition'', this transforms it into the state ...

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

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

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 \$...

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