Matrix representation of the qutrit gate analogous to Pauli-$X$ gate

Question

What is the qutrit gate analogous to the $X$ gate for qubits, i.e. NOT gate? and what is the matrix representation of that gate?

I have searched on the Internet but couldn't find anything related.

Edit:

I would like to know the state matrix of the Controlled T-Shift gate if possible.

Answer 1

To expand a bit on what I said in the comments it seems the closest thing to a Pauli $X$ or $\text{NOT}$ operation for qutrits is the so-called $\text{TShift}$ gate.

This gate has the following action on the computational basis states...

$$ \begin{align} \text{TShift}|0 \rangle &= |1\rangle \\ \text{TShift}|1 \rangle &= |2\rangle \\ \text{TShift}|2 \rangle &= |0\rangle \\ \end{align} $$

Recall that the basis states for a qutrit system are defined as the following unit vectors

\begin{equation} |0\rangle = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\,, \quad |1\rangle = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}\,, \quad |2\rangle = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \end{equation}

A qutrit is defined as a linear combination of these basis states

\begin{equation} |\psi\rangle = \alpha |0\rangle + \beta|1\rangle = \gamma |2\rangle\,, \quad |\alpha|^2 +|\beta|^2 + |\gamma|^2 = 1 \end{equation}

You can see from the above that the $(3 \times 3)$ unitary matrix for this operation is the following

$$ \begin{equation} \text{TShift} = \begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}. \end{equation} $$

This was new to me as well actually. Hope this helps.

Answer 2

Qutrit gates equivalent to Pauli $X$ gate for qubit would be generalized Pauli $X$ gates for a $3$-level system.

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

$$ X(x) |j\rangle = |x \oplus j \rangle \tag{1}\,,$$

where

• $\{|j\rangle\} \equiv \{|0\rangle, |1\rangle, \cdots, |n-1\rangle\} $ are the basis states of your system
• $x$ is the shift & $x\in\{0,1,\cdots n-1\} $
• $\oplus$ is the cyclic addition operator, meaning that the result of the addition is $(x +j) \text{mod} (n)\,.$

---

Hence, for qutrit, we have two gates. (Technically 3, but $X(0)$ is trivial)

Gate 1 : $X(1)$

• $X(1)|0\rangle = |0\oplus1\rangle = |(0+1) \text{mod} (3)\rangle = |1\rangle$
• $X(1)|1\rangle = |1\oplus1\rangle = |(1 +1) \text{mod} (3)\rangle = |2\rangle$
• $X(1)|2\rangle = |2\oplus1\rangle = |(2+1) \text{mod} (3)\rangle = |0\rangle$

Hence, the gate is

$$X(1) = \begin{bmatrix} 0&0&1\\ 1&0&0 \\ 0&1&0 \end{bmatrix}\tag{2}\,.$$

Gate 2 : $X(2)$

• $X(2)|0\rangle = |0\oplus2\rangle = |(0+2) \text{mod} (3)\rangle = |2\rangle$
• $X(2)|1\rangle = |1\oplus2\rangle = |(1+2) \text{mod} (3)\rangle = |0\rangle$
• $X(2)|2\rangle = |2\oplus2\rangle = |(2 +2) \text{mod} (3)\rangle = |1\rangle$

Hence, the gate is

$$X(2) = \begin{bmatrix} 0&1&0 \\0&0&1\\ 1&0&0 \end{bmatrix}\tag{3}\,.$$

Answer 3

The article Elementary gates of ternary quantum logic circuit defines three qutrit NOT gates: \begin{split} X^{(01)} &= \begin{pmatrix} 0 & 1 & 0\\ 1 & 0 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix}\\ X^{(02)} &= \begin{pmatrix} 0 & 0 & 1\\ 0 & 1 & 0 \\ 1 & 0 & 0 \\ \end{pmatrix}\\ X^{(12)} &= \begin{pmatrix} 1 & 0 & 0\\ 0 & 0 & 1 \\ 0 & 1 & 0 \\ \end{pmatrix} \end{split} I found this reference while reading Qudit Dicke state preparation, which also makes use of such qutrit NOT gates.