I'm using the following piece of python code:

import matplotlib.pyplot as plt
from qiskit import *
from qiskit.tools.visualization import plot_bloch_multivector, plot_histogram

from qiskit import execute, IBMQ, BasicAer
from qiskit.providers.ibmq import least_busy
from qiskit.tools.monitor import job_monitor

secret_number = "1010"
n = len(secret_number)
circuit = QuantumCircuit(n+1, n)
for i,v in enumerate(reversed(secret_number)):
    if v=="1":
        circuit.cx(i, n)

circuit.measure(range(n), range(n))
simulator = Aer.get_backend("qasm_simulator")
sim_result = execute(circuit, backend=simulator, shots=4096).result()

provider = IBMQ.get_provider("ibm-q")
# qcomp = provider.get_backend("ibmq_16_melbourne")
qcomp = provider.get_backend("ibmq_burlington")
job = execute(circuit, backend=qcomp, shots=4096)
real_result = job.result()
counts = real_result.get_counts(circuit)
import collections
c = collections.Counter(counts)

# printing
print("result simulation:")
print("results (top 5):")

The results are shown here:

enter image description here

As you can see, the expected result is 1010, which is correctly the value that the quantum simulator gives me. However, I find this result on the fifth position for this execution in the real quantum computer. Is this due to noise? How can I get around this?

As complementary information, I tried the same code with secret_number = "1011001" and the ibmq_16_melbourne server. These are the results: enter image description here

As you see. The error is amplified.

  • 1
    $\begingroup$ Does this section of the textbook help? qiskit.org/textbook/ch-quantum-hardware/… $\endgroup$
    – met927
    Apr 16, 2020 at 14:56
  • $\begingroup$ Thank you @met927 , the problem apparently was not noise. I provided an answer. I remains unclear to me the real reason the CX gate doesn't work. $\endgroup$
    – silgon
    Apr 16, 2020 at 16:06
  • $\begingroup$ You have only four qubits in your register. How are you getting seven bits out? $\endgroup$ Apr 16, 2020 at 20:21
  • $\begingroup$ the program initializes the needed gates depending on the length of the variable secret_number. $\endgroup$
    – silgon
    Apr 17, 2020 at 9:11

2 Answers 2


I am quite certain the problem here is noise. 2-qubit gates such as the CX gate are usually more prone to errors than single qubit gates like the Z gates. Also it really depends on when you run the code because with time these machines also get worse and need recalibration. As I write this answer the CNOT error rate in burlington is 2 orders of magnitude bigger than the error rate for the single qubit gate U2 which is the one probably used to implement the Z gates you useed in the second example.

enter image description here

The Oracle for the Bernstein algorithm can be implemented both using CX or Z gates. The CX gates make use of the Phase Kickback effect to create the same effect you can achieve with the Z gates (basically in this case a 180 degrees phase flip) so no black magic here. The issue is not in the gate itself but in the noise of the machines.

If you still do not believe you can try to copy the noise model from the burlington machine and apply it to the qasm simulator. You will see that you get also really messed up results in comparison to a noiseless run. Try plugging this to your code and see what happens (reference: https://qiskit.org/documentation/apidoc/aer_noise.html)

from qiskit.providers.aer.noise import NoiseModel

noise_model = NoiseModel.from_backend(qcomp)
# Get coupling map from backend
coupling_map = qcomp.configuration().coupling_map
# Get basis gates from noise model
basis_gates = noise_model.basis_gates

noisy_result = execute(circuit, Aer.get_backend('qasm_simulator'),


I got the following counts:

{'0010': 178, '0001': 2, '1000': 316, '1111': 2, '0011': 7, '0110': 1, '1001': 16, '1101': 1, '1100': 4, '1110': 49, '1010': 3331, '0000': 39, '1011': 150}

These results are still better than the results you posted but I'm quite certain the issue here is in the noise and the fact that the first implementation is using CX gates and therefore will experience more trouble than the second one.

You can also try playing with this: https://qiskit.org/textbook/ch-quantum-hardware/measurement-error-mitigation.html and see if you get any improvements there.

Hope this is helpful ;)

edit I'm not personally aware of other tricks you could play with the hardware here and I've had bad experiences with Burlington in the past as well. You probably can't apply any repetition code either (https://qiskit.org/textbook/ch-quantum-hardware/error-correction-repetition-code.html) because you'd need three times more qubits and Burlington has only 5 so I really feel there's nothing you can do to work around this using that specific device.

  • $\begingroup$ Thanks @dncolomer, it's really informative. I didn't know about the U2 error rate. As I said, I'm pretty new in this domain (I just recently studied the coursera course on quantum computing and the youtube courses of quantum mechanics). Well, in the end it's actually en error, which was kind of expected due to the difference between the simulation. $\endgroup$
    – silgon
    Apr 20, 2020 at 11:26
  • $\begingroup$ It's an amazing subject isn't it!? :) happy my answer brought some clarity. I'm not an expert either but I've seen how unpredictable noise can be in some of the smaller devices :) $\endgroup$
    – dncolomer
    Apr 21, 2020 at 23:34

This is a partial answer, in case somebody is having the same issue as I.

My code was based on this IBM tutorial video but with the implementation in the real computer, for the video they use the description of the oracle with CX gate. However in the documentation's example, they use the Z gate for reasons that are still unclear to me, since I'm still learning and I don't have the necessary background. Thus, on the question's code, it suffice to replace circuit.cx(i, n) with circuit.z(i) and then you can see the magic in the following picture:

enter image description here

The reason why CX gate doesn't work is still unclear to me. It would be nice if anyone can provide some clarification.


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