I'm currently going through the lab problems for the Qiskit course here. I'm trying to finish lab set number 3 on QPE but I can't seem to get the desired output, even from the solution notebook. I'm trying to estimate the phase of the following unitary:

$$ U_{\theta} = \left(\begin{matrix} 1&0\\ 0&e^{2\pi i \theta} \end{matrix}\right)$$

Specifically I want to implement a circuit for the case $\theta=0.5$ and I am using a 5 qubit estimation. Here is the code for the circuit:

import numpy as np
from qiskit.circuit.library import QFT
from qiskit import QuantumCircuit
pi = np.pi    

def initialize_qubits(given_circuit, measurement_qubits, target_qubit):

def unitary_operator_exponent(given_circuit, control_qubit, target_qubit, theta, exponent):
  given_circuit.cp(2*pi*theta*exponent, control_qubit, target_qubit)

def apply_iqft(given_circuit, measurement_qubits, n):
  given_circuit.append(QFT(n).inverse(), measurement_qubits)

def qpe_program(n, theta):

  qc = QuantumCircuit(n+1, n)

  # Initialize the qubits
  initialize_qubits(qc, range(n), n)


  # Apply the controlled unitary operators in sequence
  for x in range(n):
    exponent = 2**(n-x-1)
    unitary_operator_exponent(qc, x, n, theta, exponent)


  # Apply the inverse quantum Fourier transform
  apply_iqft(qc, range(n), n)

  # Measure all qubits
  qc.measure(range(n), range(n))

  return qc

n = 5; theta = 0.5
mycircuit = qpe_program(n, theta)

This is a fairly standard phase estimation circuit and I don't see why it's not working (especially since it's in the solution manual).

The problem is when I run a simulator on this and get counts, the state $|{11111}\rangle$ shows up in half the instances and I can't for the life of me find out why (it should just be $|{00001}\rangle$ plus some small error). Here is the simulation code:

from qiskit import Aer, execute
simulator = Aer.get_backend('qasm_simulator')
counts = execute(mycircuit, backend=simulator, shots=1000).result().get_counts(mycircuit)

from qiskit.visualization import plot_histogram

Here's a screenshot of the circuit and then the simulation results:

Circuit Diagram

Simulation Results

Notes: Running on Python v3.9.7 and Qiskit v0.29


1 Answer 1


In order to understand why this does not work, let us consider the circuit you've created with two measurement qubits: Two-qubits circuit Running this circuit using the following code:

backend = Aer.get_backend('aer_simulator')
job = execute(mycircuit, backend)
result = job.result()

yielded for me:

{'11': 499, '01': 525}

If we write down the state you create before applying the $IQFT$, you would get: $$\frac12|1\rangle\left(|00\rangle+|01\rangle-|10\rangle-|11\rangle\right)$$ Note that you have to consider that qiskit uses little-endian for ordering qubits. Thus, the state which would have been written in textbooks is: $$\frac12\left(|00\rangle-|01\rangle+|10\rangle-|11\rangle\right)|1\rangle$$ Now, why does this matter? If we write down the definition of the $QFT$ matrix on two qubits, according to the qiskit documentation, we get: $$QFT=\begin{pmatrix}1&1&1&1\\1&\mathrm{i}&-1&-\mathrm{i}\\1&-1&1&-1\\1&-\mathrm{i}&-1&\mathrm{i}\end{pmatrix}$$ In particular, the state we obtained using the little-endian version can be written as: $$\frac12|1\rangle\left(|00\rangle+|01\rangle-|10\rangle-|11\rangle\right)=\frac12|1\rangle\left[(1-\mathrm{i})QFT|01\rangle+(1+\mathrm{i})QFT|11\rangle\right]$$ while the texbook version can be written as: $$(QFT|10\rangle)|1\rangle$$ By linearity, you can now see that the state you get in the little-endian version is, omitting the qubit in state $|1\rangle$: $$\frac{1-\mathrm{i}}{2}|01\rangle+\frac{1+\mathrm{i}}{2}|11\rangle$$ which explains why we measured these states with equal probability. On the other hand, the textbook version yields the state $|10\rangle$.

Thus, the problem is that you've created the big-endian/textbook version of the QPE algorithm, while qiskit works with little-endian. The fix one can immediately think of is to change the qubits on which you apply the controlled phase gates. In order to do this, yuo can just replace the line:

exponent = 2**(n-x-1)


exponent = 2**x

which then yields the correct result. Another way to do it is to apply SWAP gates on the state before applying the $IQFT$. The neat property is that the $IQFT$ already has swap gates in its implementation. You can see this with:

QFT(4, inverse=True).decompose().draw("mpl")

IQFT for 4 qubits Thus, what we want to do is essentially not to apply these gates, so that the qubits ordering is reversed. You can do it using the do_swaps argument of $QFT$:

QFT(4, inverse=True, do_swaps=False).decompose().draw("mpl")

IQFT for 4 qubits without swaps Thus, if you let the line exponent = 2 ** (n - x - 1) unchanged but instead replace the line:

given_circuit.append(QFT(n).inverse(), measurement_qubits)


given_circuit.append(QFT(n, inverse=True, do_swaps=False), measurement_qubits)

Then you will get the right result, while applying less gates than with the other solution.

  • $\begingroup$ Thank you for your great answer! One more question, at risk of being off topic. When I open the notebook for solutions and don't run anything in my console, everything is already loaded and I get the right answer, even with the code as above. If I instead run it in my console I get the problems outlined above. Did Qiskit used to use big endian? $\endgroup$ Sep 10, 2021 at 15:55
  • 1
    $\begingroup$ AFAIK, no, it always has used little-endian. Maybe try to ask another question describing your setup would help? $\endgroup$
    – Tristan Nemoz
    Sep 10, 2021 at 15:58
  • $\begingroup$ My setup should be fine, I made the adjustments you suggested and everything worked out and corresponded with your explanation. It's the setup of whoever wrote the code for the solutions in the Qiskit course. Quite confusing, but unsure if anyone on this site could help there $\endgroup$ Sep 10, 2021 at 16:03
  • $\begingroup$ See the following link if you are bored enough to solve this puzzle: github.com/qiskit-community/… $\endgroup$ Sep 10, 2021 at 16:07
  • $\begingroup$ I don't have the time to look at it for now, but if I understood correctly, the notebook presents a solution which, if you run it, gives the wring answer, but which, when reading the notebook (that is, without executing it) shows the right answer? In that case it's probably an error in the solution notebook, and I would advise you to let the Qiskit team know about it $\endgroup$
    – Tristan Nemoz
    Sep 10, 2021 at 16:10

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