New answers tagged

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The rightmost two bits of each string correspond to a pixel (e.g. 010 and 110 both correspond to pixel 10). The frequency we measure the string with a leading 1 tells us the intensity of that pixel. Here's some quick code that decodes the counts dictionary: for pixel in range(4): # convert to pixel number to bit strings, i.e. (00, 01 etc.) bit_str = ...


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According to Equation 6 of the paper, the state one is considering is: $$\frac{\mathrm{i}\sin\left(\theta_d\right)}{2}|1\rangle\left(|\zeta\rangle|\psi\rangle-|\psi\rangle|\zeta\rangle\right)+\frac{\mathrm{e}^{-\mathrm{i}\theta_d}+\cos\left(\theta_d\right)}{2}|0\rangle|\psi\rangle|\zeta\rangle-\frac{\mathrm{i}\sin\left(\theta_d\right)}{2}|0\rangle|\zeta\...


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The correction of errors during computations is required for large-scale quantum information processing. A qubit is encoded in a subspace of many physical qubits in quantum error correction so that faults can be actively rectified without altering the stored data. We must account for the fact that errors occur not only when something like a gate or ...


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"I wonder after the first mid-circuit measurement, how can I determine if this matrix is still the same?" I had one of my students do a daily benchmarking of the various IBM chips and part of this project was to monitor the calibration matrix as a function of time. Unfortunately, the calibration matrix is not a constant with respect to time. It ...


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TL;DR No. Non-local gates and measurements between causally disconnected observers violate the no-signalling theorem and are therefore impossible. No-signalling theorem The following protocol allows Bob to send one classical bit $b$ to Alice faster than light. Alice prepares two sets of qubits $x_i$ and $y_i$ in the $|0\rangle$ state. Bob prepares one set of ...


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You can save your figure using the filepath argument, e.g. plot_histogram(counts, filepath="myfig") would save the plot as myfig.png in your current working directory. See API Reference: qiskit.visualization.plot_histogram


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$\vec{a}\cdot\vec{\sigma}$ defines an operator $$ a_xX+a_YY+a_ZZ $$ where $X,Y,Z$ are the Pauli matrices. So, for example of $\vec{a}=(0,a_Y,a_Z)$ then it has no component in the X plane, and we say it defines a measurement in the $Y-Z$ plane.


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Without loss of generality your question could also be phrased as "what would I measure in the bottom, second register if I waited to measure the second register until after performing the QFT on the upper first register?" In that case it might be seen that measuring the second register only affects the (unmeasurable) global phase of the system. ...


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The IBM histogram is correct. You seem to be under the impression that a reset is a postselection. Applying a reset to a qubit cannot change the expected measurement statistics on other qubits. That would violate the no-communication theorem. So the distribution before the reset is equal to the distribution after. The same is not true for a postselection, ...


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Yes, probabilities shown in a histogram are squares of absolute values of coefficients in a state vector (i.e. probability amplitudes). Note that the amplitudes are generally complex numbers, hence we square the absolute values and not the amplitudes themselves.


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Applying a reset to a qubit is equivalent to measuring it, and then applying a bit flip to it conditioned on the measurement result. def reset(qubit): if measure(qubit) == ON: X(qubit) For example, in this Quirk circuit, you can see that the post-reset state matches the state you'd get when conditioning on a measurement-via-ancilla+bit-flip of ...


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Bob does not know the outcome of Alice's measurement. All he knows is that Alice would have obtained $|\psi\rangle$ with probability $\frac12$ and $|\psi^\perp\rangle$ with probability $\frac12$ for some orthonormal basis $|\psi\rangle$, $|\psi^\perp\rangle$. Therefore, his state is a mixture $$ \rho_B = \frac12|\overline\psi\rangle\langle\overline\psi|+\...


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The claim does not specify what protocols for distinguishing quantum states are acceptable. In particular, it does not state whether we are allowed to err or reserve judgment. Below, we note success probability for protocols allowed to err and compute success probability for an error-free protocol. The success probability of the former is not bounded by $4\...


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I don't think Qiskit currently provides a straightforward way to exclude a certain transpiler pass. You can construct a pass manager that contains only the passes you want then use it. my_pass_manager = PassManager() my_pass_manager.append(my_list_of_passes) qc_foo_trans = transpile(qc_foo, backend=lima, pass_manager=my_pass_manager) However, this would be ...


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Yes, this is exactly what happens. Instead of saying "I want to measure this specific value," you specify a set of POVM elements asking "which of these values will I get?". When you are measuring the value of the first qubit alone to see if it is in state $|0\rangle$ or $|1\rangle$, the two POVM elements can be specified by $\Pi^{(0)}=|0\...


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2 assumptions for my answer (because I did not really understood your setup): Let's assume a pure state like |0010110> (not superposition) Assume you are doing the same experiment again and again (because in each time you collapse and project your state) The N Zs operator is actually counting if there is an odd or even number of ones in the state. Using ...


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Not in general, no. Consider the state $$ |\psi\rangle=\frac{1}{\sqrt{2}}\left(|0x\rangle+|1\bar x\rangle\right)|+\rangle $$ for $x\in\{0,1\}$. We have $\langle Z_i\rangle=0$, and $\langle Z_1Z_2Z_3\rangle=0$ (to see this most trivially, look at the third qubit). However, the first two qubits are an eigenstate of $Z_1Z_2$ of eigenvalue $(-1)^x$. From the ...


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Yes, the reasoning is correct. In fact, it can be generalized beyond pure states. By definition, every mixed quantum state $\rho$ is a positive semidefinite operator with unit trace. Since every positive semidefinite operator is Hermitian, we may interpret $\rho$ as an observable. In this case, the expectation of observable $\rho$ in state $\sigma$ $$ \...


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Yes, the measurements on the ancillary qubits are unnecessary. You can just discard those qubits instead of measuring them.


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Asymmetric readout error While its hard to make strong claims about the noise characteristics of any given quantum device, one explanation for what you're observing is readout error. For superconducting qubits readout error tends to be asymmetric: the probability $p(0|1)$ of observing a "0" after performing measurement on a computational basis ...


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I think there are two nested questions there: How can you simulate partial measurement? There are a few approaches, but one basic one is described here. In short, zero the amplitudes of the states which didn't happen (i.e. ones for which the qubit you measured has the opposite value than the one you measured) and then normalize remaining amplitudes so that $...


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This looks like expected behavior by optimizer at level 3. Whatever backend you chose, level 3 is noise-adaptive, and most probably qubits chosen as 1 and 2 are the least noisy to perform the swap on, and later reinterpret the result during measurement. The qubit identities are not wrong in itself - the optimizer/transpiler will remember its mapping of ...


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$|\text{scattered}\rangle$ refers to the state where the photon is no longer present (because it was absorbed/scattered by the detector). More generally, you can think of it as any state where the state of the photon has become entangled with things outside the system. The simplest way to model this state is to introduce an ancilla qubit to represent "...


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