Implementing the quantum Fourier transform with Qiskit

The quantum Fourier transform is a key building block of many quantum algorithms, from Shor’s factoring algorithm over matrix inversion to quantum phase estimation and simulations. Time to see how this can be implemented with Qiskit.

Recall that the quantum Fourier transform (or, depending on conventions, its inverse) is given by

$|x \rangle \mapsto \frac{1}{\sqrt{2^n}} \sum_s \eta^{xs} |s \rangle$

where $\eta$ is an 2n-th root of unity with n being the number of qubits in the register, i.e.

$\eta = \exp \frac{2\pi i}{2^n}$

How can we effectively create this state with a quantum circuit? They key to this is the observation (see the references below) that the result of the quantum Fourier transform can be written as a product state, namely as

$\sum_s \eta^{xs} |s \rangle = (|0\rangle + \eta^{2^{n-1} x}|1\rangle)(|0 \rangle + \eta^{2^{n-2}x}|1 \rangle) \cdots (|0 \rangle + \eta^x|1 \rangle)$

which you can easily verify by multiplying out the product and collecting terms.

Here we use the tensor product order that is prescribed by OpenQASM, i.e. the most significant bit is q[n-1]. This bit is therefore given by

$|0 \rangle + \eta^{2^{n-1}x} |1 \rangle$

Let us analyse this expression further. For that purpose, we decompose x into its representation as a binary number, i.e. we write

$x = \sum_{i=0}^{n-1} 2^i x_i$

with xi being the binary digits of x. If we now multiply this by 2n-1, we will get $2^{n-1} x_0$ plus a multiple of 2n. As $\eta$ is a root of unity, this multiple cancels out and we obtain that

$\eta^{2^{n-1}x} = \eta^{2^{n-1}x_0} = e^{\pi i x_0} = (-1)^{x_0}$

Thus, we can write the most significant qubit of the Fourier transform as

$|0 \rangle + (-1)^{x_0} |1 \rangle$

which is nothing but

$H |x_0\rangle$

Thus we obtain the most significant qubit of the quantum Fourier transform by simply applying a Hadamard gate to the least significant qubit of the input.

This is nice and simple, but what about the next qubit, i.e. qubit n-2? From the decomposition above, we can see that this is simply

$|0 \rangle + \eta^{2^{n-2}x} |1 \rangle$

Using exactly the same arguments as for the most significant qubit, we easily find that this is

$|0 \rangle + \eta^{2^{n-2}x_0} H |x_1 \rangle$

Thus we obtain this qubit from qubit 1 by first applying a Hadamard gate and then a conditional phase gate, i.e. a conditional rotation around the z-axis, conditioned on the value of x0. In general, qubit n-j is

$|0 \rangle + \prod_{i < j - 1} \eta^{2^{n-j+i}x_i} H |x_{j-1}\rangle$

which is a Hadamard gate followed by a sequence of conditional rotations around the z-axis, conditioned on the qubits with lower significance.

So we find that each qubit of the Fourier transform is obtained by applying a Hadamard followed by a sequence of conditional rotations. However, the order of the qubits in the output is reversed, i.e. qubit n-j is obtained by letting gates act on qubit j. Therefore, at the end of the circuit, we need to revert the order of the qubits.

In OpenQASM and Qiskit, a conditional rotation around the z-axis is called CU1, and there are swap gates that we can use to implement the final reversing of the qubits. Thus, we can use the following code to build a quantum Fourier transformation circuit acting on n qubits.

def nBitQFT(q,c,n):
circuit = QuantumCircuit(q,c)
#
#
for k in range(n):
j = n - k
circuit.h(q[j-1])
#
# there is one conditional rotation for
# each qubit with lower significance
for i in reversed(range(j-1)):
circuit.cu1(2*np.pi/2**(j-i),q[i], q[j-1])
#
# Finally we need to swap qubits
#
for i in range(n//2):
circuit.swap(q[i], q[n-i-1])
return circuit


Here is the circuit that this code produces for n=4. We can clearly see the structure – on each qubit, we first act with a Hadamard gate, followed by a sequence of conditional rotations with decreasing angle, conditioned on the less significant qubits, and finally reorder the qubits.

This is already a fairly complex circuit, and we need to find a way to test it. Let us look at the options we have. First, a quantum circuit is a unitary transformation and can be described by a matrix. In our case, it is especially easy to figure out what this matrix should be. Looking at the formula for the quantum Fourier transform, we find that the matrix describing this transformation with respect to the computational basis has the elements

$U_{ij} = \frac{1}{\sqrt{2^n}} \eta^{ij}$

The Qiskit frameworks comes with a simulator called the unitary_simulator that accepts a quantum circuit as input and returns the matrix describing that circuit. Thus, one possible test approach could be to build the circuit, run the unitary simulator on it, and to compare the resulting unitary matrix with the expected result given by the formula above. In Python, the expected result is produced by the following code

def qftMatrix(n):
qft = np.zeros([2**n,2**n], dtype=complex)
for i in range(2**n):
for j in range(2**n):
qft[i,j] = np.exp(i*j*2*1j*np.pi/(2**n))
return 1/np.sqrt(2**n)*qft


and the test can be done using the following function

def testCircuit(n):
q = QuantumRegister(n,"x")
c = ClassicalRegister(n,"c")
circuit = nBitQFT(q,c,n)

backend = Aer.get_backend('unitary_simulator')
job = execute(circuit, backend)
actual = job.result().get_unitary()
expected = qftMatrix(n)
delta = actual - expected
print("Deviation: ", round(np.linalg.norm(delta),10))
return circuit


The outcome is reassuring – we find that the matrices are the same within the usual floating point rounding differences.

After passing this test, a next reasonable validation step could be to run the algorithm on a specific input. We know that the QFT will map the state $|0 \rangle$ into an equal superposition of all elements of the computational basis. Conversely, we therefore expect that if we start with such a superposition, the QFT will map the superposition onto $|0\rangle$, at least up to a phase.

Let us try this out. Our test circuit will consist of a layer of Hadamard gates to create the equal superposition, followed by the QFT circuit, followed by a measurement. The resulting circuit for n=4 is displayed below.

It we run this circuit on the QASM simulator embedded into Qiskit, the result is as expected – for 1024 shots, we get 1024 times the output ‘0000’. So our circuit works – at least theoretically. But what about real hardware?

Let us compile and run the circuit targeting the IBM Q Experience 14 qubit device. If we dump the QASM code after compilation, we see that the overall circuit will have roughly 140 gates. This is already significant, and we expect to see some noise. To see how bad it is, I have conducted several test runs and plotted the results as a histogramm (if you want to play with this yourself, you will find the notebook on Github). Here is the output of a run with n=4.

We still see a clear peak at the expected result, but also see that the noise level is close to making the result unusable – if we did not know the result upfront, we would probably not dare to postulate anything from this output. With only three qubits, the situation becomes slightly better but is still far from being satisfactory.

Of course we could now start to optimize the circuit – remove cancelling Hadamard gates, remove the final swap gates, reorder qubits to take the coupling map into account and so on – but it becomes clear that with the current noise level, we are quickly reaching a point where even comparatively simple circuit will inflict a level of noise that is at best difficult to handle. Hopefully, this is what you expected after reading my posts on quantum error correction, but I found it instructive to see noise in action in this example.

References

1. M. Nielsen, I. Chuang, Quantum Computation and Quantum Information, Cambridge University Press 2010
2. R. Cleve, A. Ekert, C. Macchiavello, M. Mosca, Quantum Algorithms revisited, arXiv:9708016