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I am new to the field and I can't help having a feeling that Hadamard and Fourier Transform are somehow related, but it is not clear to me how.

Any explanation on how these two are related would be appreciated.

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To summarize the relation in one line, the Hadamard Transform is essentially the Quantum Fourier Transform for the special case of a single qubit. Traditionally, the Quantum Fourier Transform $U_{FT}$ for $n$ qubits is defined as

$$U_{FT}|x\rangle_n = \frac{1}{2^{n/2}} \sum_{y=0}^{2^n - 1} e^{2 \pi i x y / 2^n}|y\rangle_n$$

However, this is not really the most general definition. The factor $2^n$ really comes from the fact that an $n$ qubit quantum system is essentially a $2^n$ dimensional quantum system. We can generalize this to a general $d$ level quantum system (also known as a qudit).

$$U_{FT}|x\rangle = \frac{1}{\sqrt{d}} \sum_{y=0}^{d - 1} e^{2 \pi i x y / d}|y\rangle$$

Here, $|x\rangle, |y\rangle \in \{|0\rangle, |1\rangle, \dots |d-1\rangle\}$. Now, the $H$ operator basically transfers between eigenvectors of the $Z$ and $X$ operators, that is $H|0\rangle = |+\rangle$ and $H|+\rangle = |0\rangle$ and similarly for $|1\rangle$ and $|-\rangle$. So, we may write

$$H = |+ \rangle\langle 0| + |- \rangle\langle 1|$$

So, before generalizing the $H$ operator to $d$ level systems, we will need to generalize the $Z$ and $X$ operators. The generalized version of the $Z$ operator (denoted $Z_g$, subscript $g$ for 'generalized') is written as

$$Z_g = \sum_{x=0}^{d-1} e^{2 \pi i x / d}| x\rangle\langle x|$$

It is easy to check that putting $d = 2$ reduces this to the familiar single qubit $Z = |0 \rangle\langle 0| - |1 \rangle\langle 1|$. We define the generalized version of the $X$ operator (denoted $X_g$) by

$$X_g|x\rangle = |x \oplus 1 \rangle$$

Here, $\oplus$ represents the cyclic right shift operator. It takes $0$ to $1$, $1$ to $2$, $d - 1$ to $0$ and so on. Again, it is easy to see that this mimics the familiar $X|0\rangle = |1\rangle$ and $X|1\rangle = |0\rangle$ when $d = 2$.

The eigenvectors of $Z_g$ are easily seen to be the computational basis $\{|x\rangle\}_{x=0}^{d-1}$ with corresponding eigenvalues $e^{2 \pi i x / d}$. The same is not as obvious for $X_g$, but a short and simple calculation (which I will leave to the reader) reveals the eigenvectors of $X_g$ to be

$$|\tilde{x}\rangle = \frac{1}{\sqrt{d}} \sum_{y=0}^{d - 1} e^{2 \pi i x y / d}|y\rangle$$

with corresponding eigenvalues $e^{-2 \pi i x / d}$. Since the $H$ Transform takes eigenvectors of the $Z$ operator to the eigenvectors of the $X$ operator, the generalized version of the $H$ Transform (denoted $H_g$) may be defined as

$$H_g|x\rangle = |\tilde{x}\rangle = \frac{1}{\sqrt{d}} \sum_{y=0}^{d - 1} e^{2 \pi i x y / d}|y\rangle$$

Now, this is just the Quantum Fourier Transform! It is worth pointing out that the above is a special case of the application of what is known as the Heisenberg-Weyl Operators. More details may be found in section 3.7 of Mark Wilde's textbook on Quantum Information Theory

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  • $\begingroup$ @Akonaire, Thank you for the kind explanation but I am a bit lost. here " Now, this is just the Quantum Fourier Transform!". If I look at 4x4 transformation of H vs. QFT, they look different right? Also, QFT circuit is not just made out of Hadarmard gates . Sorry I must have missed the important message here. $\endgroup$ Jun 4 at 20:28
  • $\begingroup$ @JohnParker The H transform in the last line is not your familiar 2 x 2 H transform. It is the generalized version of H for d-level systems, so it is a d x d matrix. You can think of the QFT as a family of transforms defined on d-level systems (For qubits, d = 2). Every member of this family transforms the eigenvectors of the Z operator to the eigenvectors of the X operator. So d-level QFT transforms the eigenvectors of d-level generalized Z to that of d-level generalized X in much the same way as the 2-level H transforms the eigenvectors of 2-level Z to that of 2-level X. $\endgroup$
    – Arkonaire
    Jun 5 at 6:13

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