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A gate is involutory if $G^{2}=I$. This is true for all Pauli gates.

Does the definition change if it is a gate for qutrits and beyond? Is there a good article for this?

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2 Answers 2

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Let $X$ be a set. The identity function $I: X\to X$ is such that, for all $x\in X$, $I(x) = x$. An involution is a function $f: X\to X$ such that $f\circ f = I$, where $\circ$ denotes function composition. We equivalently say that $f$ is involutory if it has this property.

A qubit gate is a certain kind of function $G: \mathbb{C}^2 \to \mathbb{C}^2$. Using the above definition, $G$ being involutory is equivalent to saying that $G\circ G = G^2$ is the identity $I$. A qudit gate is a certain kind of function $G: \mathbb{C}^d \to \mathbb{C}^d$, so the definition and discussion applies equally well to qudit gates.

For articles about this, see https://en.wikipedia.org/wiki/Involution_(mathematics) and references therein.

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  • $\begingroup$ Yeah for qutrits the definition of involutory would $G^{3}=I$ and so on. $\endgroup$ Mar 8 at 0:18
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I assume you are asking if $P$ is a generalized Pauli operator for qudits, i.e. $P$ is a Heisenberg–Weyl operator, then does $P^2 \stackrel{?}{=} I$.


Generalized Pauli $X$

The action of generalized Pauli $X$ for $n$-level on the basis states is given by

$$ X(x) |j\rangle = |(x +j) \text{mod} (n) \rangle \,,$$

where

  • $\{|j\rangle\} \equiv \{|0\rangle, |1\rangle, \cdots, |n-1\rangle\} $ are the basis states of your qudit .
  • $x$ is the shift & $x\in\{0,1,\cdots n-1\} \,.$

So you can see that if we apply $X(x)$ to the state $|(x +j) \text{mod} (n) \rangle$ again,

$$\begin{align} X(x) |(x +j) \text{mod} (n)\rangle &= |(x +x+j) \text{mod} (n)\rangle\,,\\ &= |(2x+j) \text{mod} (n)\rangle\,. \end{align} $$

So, you can see, $\big( X(x) \big)^2 \neq I$

However, if you apply $X(x)$ $n$-times, you can see

$$\begin{align} \big(X(x)\big)^n |j\rangle &= |(\underbrace{x +x+\cdots x}_{n\text{ times}}+j) \text{mod} (n)\rangle\,,\\ &= |(nx+j) \text{mod} (n)\rangle\,,\\ &= |j\rangle\,. \end{align} $$

So, you can see, $\big( X(x) \big)^n = I\,.$

  • A fun exercise, see: $\big(X(x)\big)^{-1} = X(-x)\,.$

Generalized Pauli $Z$

The action of generalized Pauli $Z$ for $n$-level on the basis states is given by

$$ Z(z) |j\rangle = e^{\frac{i 2\pi zj}{n}}|j \rangle \,,$$

where

  • $\{|j\rangle\} \equiv \{|0\rangle, |1\rangle, \cdots, |n-1\rangle\} $ are the basis states of your qudit .
  • $z\in\{0,1,\cdots n-1\} \,.$

So you can see that if we apply $Z(z)$ to the state $e^{\frac{i 2\pi zj}{n}}|j \rangle$ again,

$$\begin{align} Z(z) e^{\frac{i 2\pi zj}{n}}|j \rangle &= e^{\frac{i 2\pi zj}{n}} \bigg(e^{\frac{i 2\pi zj}{n}}|j \rangle\bigg)\,,\\ &= \bigg(e^{\frac{i 2\pi zj}{n}}\bigg)^2|j \rangle\,,\\ &= e^{\frac{i 4\pi zj}{n}}|j \rangle\,. \end{align} $$

So, you can see, $\big( Z(z) \big)^2 \neq I$

However, if you apply $Z(z)$ $n$-times, you can see

$$\begin{align} \big(Z(z)\big)^n |j\rangle &= \overbrace{e^{\frac{i 2\pi zj}{n}}\cdot e^{\frac{i 2\pi zj}{n}}\cdots e^{\frac{i 2\pi zj}{n}}}^{n\text{ times}}|j \rangle\,,\\ &= \bigg(e^{\frac{i 2\pi zj}{n}}\bigg)^n |j\rangle\,,\\ &= e^{\frac{i 2n\pi zj}{n}}|j\rangle\,,\\ &= e^{i 2\pi zj}|j\rangle\,,\\ &= |j\rangle\,. \end{align} $$

So, you can see, $\big( Z(z) \big)^n = I\,.$

  • A fun exercise, see: $\big(Z(z)\big)^{-1} = Z(-z)\,.$
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