Quantum Fourier transform (QFT) is a key concept in quantum algorithm design. For example, it is a major ingredient of Shor’s algorithm for factoring integers and the quantum phase estimation algorithm (PEA) for solving eigenvalues and eigenvectors. In this post, I will explain how it works.

If you prefer to learn QFT without equations, see this blog post by Scott Aaronson: Shor, I’ll do it

## discrete Fourier transform (DFT)

QFT is closely related to discrete Fourier transform (DFT), an important tool in digital signal processing. And we will start from there.

As the name indicates, DFT is the discrete version of Fourier transform. Most commonly, the input sequence is time series data or spatial samples, and the output sequence is frequency data, i.e., the Fourier spectrum. Overall, it is

• a linear map between two sequences of complex numbers;
• a non-degenerate map with an inverse, i.e., the inverse discrete Fourier transform (IDFT);
• a map that’s cheap to compute: the fast Fourier transform (FFT) algorithm has computational complexity $$O(N\log N)$$ (instead of $$O(N^2)$$ for a general linear map)

In practice, DFT is widely used because

• In signal processing, often times the signals are band-limited. Then the Fourier spectrum provides a more succinct representation of the signals;
• In physics problem, often times the symmetry can be more easily utilized in the Fourier domain;
• It is useful for convolution calculations;
• It is efficient to calculate;

In 1D, DFT can be defined as

$X_k = \frac{1}{\sqrt{N}} \sum_{j=0}^{N-1} \omega_N ^{-jk} x_j$

where $$\omega_N\equiv\exp(2\pi i/N)$$ is the $$N$$‘th root of unity and $$i=\sqrt{-1}$$. And the IDFT is given by

$x_j = \frac{1}{\sqrt N} \sum_{j=0}^{N-1} \omega_N ^{jk} X_k$

Generalization to higher dimensions is straightforward.

Different authors use different normalization conventions for DFT and IDFT. I have the impression that physicists and engineers prefer to have $$1$$ in DFT and $$1/N$$ in IDFT, whereas mathematicians prefer this $$1/\sqrt{N}$$ normalization. For QFT, $$1/\sqrt{N}$$ is better as it normalizes the wave functions correctly.

There is also arbitrariness in the range of the summation index $$j$$: any $$N$$ consecutive integers will do. Sometimes it is more convenient to include both positive and negative values. To complicate things even more, some authors swap the definitions of DFT and IDFT. Thus one needs to be very cautious with other people’s formulas.

The DFT transformation has explicit matrix form. For example,

$DFT_2 = \frac{1}{\sqrt 2}\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix},\quad DFT_4 = \frac{1}{2}\begin{bmatrix} 1 & 1 & 1 & 1\\ 1 & -i & -1 & i\\ 1 & -1 & 1 & -1\\ 1 & i & -1 & -i \end{bmatrix}$

In general, the DFT matrix looks like

$DFT_N = \frac{1}{\sqrt N}\begin{bmatrix} 1 & 1 & 1 & \cdots & 1\\ 1 & \omega_N^{-1} & \omega_N^{-2} & \cdots & \omega_N^{-(N-1)} \\ 1 & \omega_N^{-2} & \omega_N^{-4} & \cdots & \omega_N^{-2(N-1)} \\ \vdots & \vdots & \vdots & \cdots & \vdots \\ 1 & \omega_N^{-(N-1)} & \omega_N^{-2(N-1)} & \cdots & \omega_N^{-(N-1)(N-1)} \end{bmatrix}$

You can easily check that the rows are all perpendicular to each other (remember to take complex conjugate). Thus each component of the DFT output, i.e., $$X_k$$, is a projection of the input to one of a set of orthogonal directions. If you visualize the $$\omega_N^j$$ in the complex plane, then each row can be seen as a rotating unit vector with different angular velocity. And the action of each row is to extract the component of a specific angular velocity.

There are a lot of subtleties in DFT. For example, given a continuous function (analog signal), should I do Fourier transform first, then sample the frequency domain function? Or sample the continuous function first, and then do DFT on the time domain samples? My favorite theorem along these lines is the Shannon sampling theorem, which answers the following questions:

• Does a continuous function have infinite degrees of freedom since there are infinite input values (like in classical field theory)? This is plausible but is also complicated by the continuity requirement.
• If a function is band-limited, does it still have infinite degrees of freedom?

To learn more about DFT, check out this nice textbook on Signals and Systems by Prof. Alan Oppenheim (age 81 this year).

To appreciate the significance of QFT, we need to first examine the measurement/readout problem in quantum mechanics.

Suppose we do not have any prior information about a quantum state, it is impossible to know what it is. Take 1-qubit for example, an arbitrary state takes the form

$\left|\psi\right> = \alpha\left|0\right> + \beta\left|1\right>,$

where $$\|\alpha\|^2 + \|\beta\|^2=1$$.

A measurement with respect to the computational basis (0 and 1) causes its wave-function to collapse. Thus if we have only one copy of the quantum state, no information of the probability amplitudes $$\alpha$$ or $$\beta$$ can be retrieved. The situation is slightly better if we have many copies of the same state. In that case, $$\alpha$$ and $$\beta$$ can be estimated at the cost of repeated measurements. This task is known as quantum state tomography and is not easy in general. In the 1-qubit case, there are 3 observables to be estimated: the 3 Pauli spin operators. For $$n$$ qubits, there are $$2^{2n} - 1$$ observables (i.e., the number of $$SU(2^n)$$ generators), a daunting task.

Thus to have an efficient readout of any quantum computation result, we need to have prior information of the output state. In other words, an efficient quantum algorithm has to encode its answer in special quantum states. For example, an algorithm with boolean answer could ensure the output to be one of two known states. This is indeed the strategy of many (not all though) quantum algorithms, such as Deutsch Jozsa algorithm, Simon’s algorithm, Bernstein-Vazirani algorithm, and Shor’s algorithm.

Since measurement has to be in the computational basis, these known states can be seen as intermediate results. Obviously, it’s more convenient to have these states to form an orthonormal bases. It helps even more if these states are all separable states (i.e., not entangled states). Then the transformation to the computational basis states can be done independently, potentially in parallel. From the hardware perspective, single qubit gates are usually much easier to implement and faster than multi-qubit gates too.

As a concrete example, let’s assume a boolean answer is encoded in the following 1-qubit states

\begin{align} \left|+\right> \equiv& \frac{1}{\sqrt 2}(\left|0\right> + \left|1\right>) = \frac{1}{\sqrt 2}\begin{bmatrix}1\\ 1\end{bmatrix} \\ \left|-\right> \equiv& \frac{1}{\sqrt 2}(\left|0\right> - \left|1\right>) = \frac{1}{\sqrt 2}\begin{bmatrix}1\\ -1\end{bmatrix} \end{align}

In this case, applying Hadamard gate $$H_1$$ before measurement helps to get a definite answer with one copy of the quantum state since

\begin{align} H_1 \left|+\right> =& \left|0\right>\\ H_1 \left|-\right> =& \left|1\right>\end{align}

where

$H_1 = \frac{1}{\sqrt 2}\begin{bmatrix} 1& 1 \\ 1& -1 \end{bmatrix}.$

As you may have guessed, QFT is a gate whose role is similar to $$H_1$$ in this example.

## quantum Fourier transform (QFT)

In fact, QFT is exactly DFT, with the input/output vectors being probability amplitudes of quantum states. Thus it is a transformation between quantum states, i.e., a quantum gate.

For quantum computing, the state vector is of dimension $$N=2^n$$ where $$n$$ is the number of qubits. Following the arguments in the previous section, let’s look at two sets of special states, both of which consists of only separable states:

• states corresponds to the rows of the Walsh-Hadamard matrix: $$\prod_{j=0}^{n-1}\otimes\left(\left|0\right>+(-1)^{r_j}\left|1\right>\right)$$
• states corresponds to the rows of the IQFT/IDFT matrix: $$\prod_{j=0}^{n-1}\otimes\left(\left|0\right>+\omega_{2^n}^{jr}\left|1\right>\right)$$

where $$r=0,1,..,2^n-1$$ is the row index of the $$n$$-qubit transform matrix, $$r_j$$ is the $$j$$‘th digit of $$r$$’s binary representation, and $$\otimes$$ denotes tensor product. This QFT tensor product decomposition can be easily derived from small $$n$$ cases and induction.

Note that the Walsh-Hadamard gate is particularly easy to implement since it can be decomposed as 1-qubit Hadamard gates, i.e., $$H_n = H_1^{\otimes n}$$. In the QFT case, the qubits do not fully decouple (note the coefficient before $$\left|1\right>$$ depends on $$r$$ instead of a single binary digit of $$r$$), thus a naive QFT implementation requires $$n-1$$ 1-qubit controlled-phase gate on each qubit, which is already efficient ($$O(n^2)$$ 1-qubit gates) on a quantum computer. A more careful examination will reveal that many of them can be avoided.

Recall that classical FFT has complexity $$O(N\log N)$$ where $$N=2^n$$, i.e., $$O(n2^n)$$. Thus it appears that quantum computer can calculate DFT exponentially faster. This is actually not completely true: the catch is that the DFT result (i.e., the probability amplitudes) is not accessible in general due to the measurement/readout problem.

## quantum phase estimation algorithm

In special occasions, the exponential speedup materializes. It happens when the answer corresponds to one of the QFT basis states. Then the application of QFT followed by a measurement can tell us the answer. This is the main idea of the phase estimation algorithm (PEA), a quantum algorithm for eigenvalues and eigenvectors.

A simplified version of it can be stated as follows. Suppose we are given an eigenstate of a unitary gate/operator satisfying

$U\left|\psi\right> = e^{i\phi}\left|\psi\right>$

where $$\phi$$ is guaranteed to be smaller than $$2\pi$$, estimate $$\phi$$.

For now, let’s assume that $$\phi$$ happens to be $$\omega_N^k$$ with an unknown $$k$$ (here we assume we somehow know $$N$$ and again $$N=2^n$$). Then we can use controlled-$$U$$ gates and $$N$$ ancilla qubits to do the phase kickback trick with the output state

$\prod_{j=0}^{n-1}\otimes\left(\left|0\right>+\omega_{N}^{jk}\left|1\right>\right)$

In other words, we can transform it to be one of the QFT basis states and the unknown value $$k$$ is exactly the row index. Then it only takes a QFT transformation and one measurement to reveal $$k$$.

What if $$\phi$$ is not $$\omega_N^k$$? A moment of reflection will show that the aforementioned protocol produces the best $$N$$-bit approximation of $$\phi$$ (in this case one also needs to repeat this protocol several times to be sure).