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4

That looks right to me. Since, $HZH = X$ then we have that $\langle \psi | X | \psi \rangle = \langle \psi | HZH | \psi \rangle = \langle \psi H | Z | H\psi \rangle $. In your code, you generate $|\psi \rangle$ with a $U_3(\theta, \phi, \lambda) $ gate applied to $|0\rangle$. Then you applied the Hadamard gate ($H$) before measuring which is what needed ...


0

From the paper: "In Figure 5, the first three qubits are in possession of Alice, the Bank contains the fifth qubit, and the fourth qubit remains unused." Alice's bits two and three are "provided" by the bank; they are part of a GHZ state. So, the account holder gets two -- not three -- qubit of the GHZ state.


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If you use from qiskit.providers.aer.noise import NoiseModel to add noise in your U gate operation. Then it won't behave like an ideal gate, https://qiskit.org/documentation/tutorials/simulators/4_custom_gate_noise.html here you can get more information.


2

If all what you want is a list of all job IDs for the jobs in a job set: job_set = job_manager.retrieve_job_set(job_set_id = 'XXX-YYY', provider = provider) id_list = [ job.job_id() for job in job_set.jobs() ]


1

"Runtime" is not so easily quantified, it depends a lot on the compilation, the other operations in your circuit and whether you simulate or have a real backend. Generally, the different methods trade off circuit depth (more gates, but less qubits) against circuit width (more qubits, less gates). If we define the runtime by the number of gates we ...


0

Thank Steve! finally it works inicializing the qubits and save_statevector and probabilities ans.initialize([1/np.sqrt(2), -1/np.sqrt(2)], 0) ans.initialize([1/np.sqrt(2), -1/np.sqrt(2)], 1) vqc.circuit.save_state() vqc.circuit.save_probabilities() vqc.circuit.save_statevector() Regards!


1

The article of Devoret and Schoelkopf [1] and an update provided in Section 7.1 of Reagor [2] makes a comparison between Moore's law and an observed trend of exponentially improving $T_1$ and $T_2$ times for superconducting qubits. The trend they present shows a roughly exponential improvement from $10^0$ to $10^6$ nanoseconds for $T_2$ between various ...


1

This is because the last job is to extract the eigenstate. If you check the circuit, then you should see that it is just being measure in the Z basis. This circuit will give back the user the counts of the states that the Ansatz generated.


0

If you use memory=True in execute(), then you store the measurement outcomes for each individual shots. job.result().get_memory()[0] mean that you access measurement outcome of the first shot. The meaning of your output is for inp1, inp2, inp3, inp4 = 0, the measurement outcome of the first shot is 0000, for inp1, inp2, inp3 = 0, and inp4 = 1, the first shot ...


1

I don't think there is a feature to download a folder, but you can download the folder by zipping it first. To zip the folder, write in a new cell: !zip -r MyFolder.zip path/to/folder Then you can download the zip file. If you only want to download the .txt files, you can write: !zip MyFolder.zip path/to/folder/*.txt


0

QSVC is really the sklearn SVC passing the quantum kernel to the SVC. Now I see SVC has a 'random_state' argument on its constructor to control any randomness of the SVC within sklearn - see https://scikit-learn.org/stable/modules/generated/sklearn.svm.SVC.html and https://scikit-learn.org/stable/common_pitfalls.html#controlling-randomness Since QSVC ...


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Qiskit ML VQC is a lightweight sub-class of sklearn SVC that supplies the quantum kernel to the SVC. You can therefore leverage sklearns capabilities to say save/restore models see https://scikit-learn.org/stable/modules/model_persistence.html#python-specific-serialization


2

Yes, you can add at any time. The reservation system cares about the project you send the jobs from, not the individual users.


2

Firstly, you might be interested in paper Elementary gates for quantum computation explaining how complex gates can be decomposed to simpler ones. This would allow you understand how the matrix $U_j$ is decomposed. Before we proceeed further, we have to define gate $U1$ used on IBM Q computer: $$ U1(\lambda)= \begin{pmatrix} 1 & 0 \\ 0 & e^{i\lambda} ...


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No, I dont think that is possible right now with limited access and with all open API's. Could you elaborate a little what exactly you would like to achieve.


0

In simple terms, what you observe in the IBM-Q is for multi qubits, whereas in the Bloch sphere the visualization is for one qubit. Overall the concept is the same, to find the probability/position of the the qubit when measured.


1

From reading your post I assume that you calculate the CNOT error by: preparing one of the four states from {00,01,10,11} Applying a CNOT gate Measuring the final state The probalitiy of measuring {00,01,11,10}, respectively, is then what you (?) assume to be the cnot error If you follow this procedure then your 'CNOT error rate' will also be affected by ...


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