How to generalize the relationship HXH = Z for higher dimensions

Question

Concerning the Hadamard gate and the Pauli $X$ and $Z$ gates for qubits, it is straightforward to show the following relationship via direct substitution:

$$ HXH = Z.\tag{1}$$

And I would like to demonstrate this relationship for higher dimensions (I know that this still holds for higher dimensions but I haven't found a proof for it anywhere).

I know that the $d$-dimension generalization of the X and Z Pauli gates for qudits are given by $X_d\lvert j\rangle = \lvert j\bigoplus1 \rangle$ and $Z_d\lvert j\rangle = \exp^{\frac{i2\pi j}{d}}\lvert j\rangle$ (where $j = 0,1,2,\dots,d-1 $). My approach to finding the d-dimension generalization of equation $(1)$ is to make use of these operators, but unfortunately I cannot find a $d$-dimensional analogue of the Hadamard gate like I have for the $X$ and $Z$ Pauli gates. I have considered using the relationship $H = \frac{1}{\sqrt2}(X + Z)$ but I don't know for certain if this is just true for qubits.

If anyone could provide any suggestions or hints as to how I should prove the generalization of equation $(1)$, it will be much appreciated (my biggest problem specifically is expressing H in an alternate form).

Answer 1

The appropriate $d$-dimensional analogue of $H$ turns out to be the Quantum Fourier Transform. This is obscured by the fact that even though $(1)$ is conjugation the inverse is written implicitly since $H^\dagger = H$. Thus, $d$-dimensional generalization of $(1)$ is

$$ QFT \circ X_d \circ QFT^\dagger = Z_d.\tag{1'} $$

Proof

The following calculation shows that both sides of $(1')$ agree on the computational basis and therefore are equal

$$ \begin{align} QFT \circ X_d \circ QFT^\dagger |j\rangle &= QFT \circ X_d \left(\frac{1}{\sqrt{d}}\sum_{k=0}^{d-1} \exp\left(-\frac{2\pi ijk}{d}\right)|k\rangle\right) \\ & = QFT \left(\frac{1}{\sqrt{d}}\sum_{k=0}^{d-1} \exp\left(-\frac{2\pi ijk}{d}\right)|k\oplus 1\rangle\right) \\ & = QFT \left(\frac{1}{\sqrt{d}}\sum_{k=0}^{d-1} \exp\left(-\frac{2\pi ij(k-1)}{d}\right)|k\rangle\right) \\ & = \frac{1}{d}\sum_{k=0}^{d-1} \exp\left(-\frac{2\pi ij(k-1)}{d}\right)\sum_{l=0}^{d-1}\exp\left(\frac{2\pi ikl}{d}\right)|l\rangle \\ & = \frac{1}{d}\exp\left(\frac{2\pi ij}{d}\right)\sum_{k,l=0}^{d-1} \exp\left(\frac{2\pi ik(l-j)}{d}\right)|l\rangle \\ & = \frac{1}{d}\exp\left(\frac{2\pi ij}{d}\right)\cdot d\,|j\rangle \\ & = Z_d |j\rangle. \end{align} $$

Answer 2

The correct generalization is the discrete Fourier transform $F_d$ over $d$ dimensions defined as

$$ F_d = \sum_{jk} e^{i2\pi jk / d} |j\rangle \langle k| $$

up to a sign convention. I'll show $F_d^\dagger Z_d F_d = X_d$ since its equivalent and a bit easier. Letting $\omega_d \equiv \exp(i2\pi/d)$ we have \begin{align} F_d^\dagger Z_d F_d &= \left(\sum_{j,k}\omega^{jk}|j\rangle \langle k|\right)^\dagger \left(\sum_\ell \omega^\ell |\ell \rangle \langle \ell |\right) \left(\sum_{m,n}\omega^{mn}|m\rangle \langle n|\right) \\ &= \sum_{j,k,\ell, m,n} \omega^{-jk}\omega^\ell \omega^{mn} \delta_j^\ell \delta_\ell^m |k \rangle \langle n| \\ &= \sum_{k,n} |k \rangle \langle n| \left(\sum_\ell \omega^{\ell(1 + n - k)} \right) \\ &= \sum_{k,n}|k \rangle \langle n| \delta_k^{n+1 \text{ mod d}} \\ &= \sum_{k,n}|n+1 \text{ mod d}\rangle \langle n| \end{align}

where in the third line I used an identity for the sum over roots of unity, and for simplicity I've left out a host of factors of $\frac{1}{\sqrt{d}}$ that are required for unitarity. That last operator is $X_d$, and so the same proof shows that $F_d X_d F_d^\dagger = Z_d$.