# How is involution defined for qudit gates?

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?

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.

• Yeah for qutrits the definition of involutory would $G^{3}=I$ and so on. Mar 8 at 0:18

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: $$\big(Z(z)\big)^{-1} = Z(-z)\,.$$