3
$\begingroup$

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.

$\endgroup$
8
  • 1
    $\begingroup$ What would be the action of such a gate? A NOT gate for (qubit) systems makes perfect sense to me as it transforms $|0\rangle \to |1\rangle$ and $|1\rangle \to |0\rangle$. How is a NOT gate defined if you have three basis states $|0\rangle, |1\rangle, |2\rangle$? $\endgroup$
    – Callum
    Commented Dec 15, 2023 at 21:29
  • $\begingroup$ I dont know honestly.I assumed that if there is a X Gate for qubits then there must be for qutrits as well. $\endgroup$ Commented Dec 15, 2023 at 21:34
  • $\begingroup$ If you scroll for a bit this blog post discusses the TShift gate which is the closest thing I can find to a Pauli X for qutrits -> pennylane.ai/qml/demos/tutorial_qutrits_bernstein_vazirani $\endgroup$
    – Callum
    Commented Dec 15, 2023 at 21:41
  • $\begingroup$ Look at the clock and shift matrices en.wikipedia.org/wiki/…. X is a shift in higher dimensions. $\endgroup$
    – Danylo Y
    Commented Dec 15, 2023 at 23:20
  • 1
    $\begingroup$ @RootGroves Please keep one post limited to one question. You can make another post looking for a matrix representation of Controlled-NOT for qutrits. This helps other people searching for the same question in future to find the answer faster. $\endgroup$
    – FDGod
    Commented Dec 15, 2023 at 23:59

3 Answers 3

1
$\begingroup$

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

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

$\endgroup$
5
  • 1
    $\begingroup$ Cool generalisation :) thats useful to know. $\endgroup$
    – Callum
    Commented Dec 15, 2023 at 22:14
  • 1
    $\begingroup$ @Callum Thank you. You can also find generalized qudit $Z$-operators in this comment of mine $\endgroup$
    – FDGod
    Commented Dec 15, 2023 at 23:55
  • 1
    $\begingroup$ Thanks, I'll check it out. $\endgroup$
    – Callum
    Commented Dec 15, 2023 at 23:55
  • 1
    $\begingroup$ I think there's a minor typo in the last equation. It should be $X(2)$ rather than $X(1)$ right? $\endgroup$
    – Callum
    Commented Dec 15, 2023 at 23:56
  • 1
    $\begingroup$ @Callum Oh yes. My bad. Fixed it. Thanks. $\endgroup$
    – FDGod
    Commented Dec 16, 2023 at 0:05
1
$\begingroup$

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.

$\endgroup$
3
  • 1
    $\begingroup$ I believe there is a mistake in the matrix. It should be $$\text{TShift} = \begin{bmatrix} 0&0&1\\ 1&0&0 \\ 0&1&0 \end{bmatrix}\tag{2}\,.$$ Because here,$ |1\rangle \mapsto |0\rangle$ and $|2\rangle \mapsto |2\rangle$. $\endgroup$
    – FDGod
    Commented Dec 17, 2023 at 9:35
  • 1
    $\begingroup$ Ah yes, you’re correct. Thank you. $\endgroup$
    – Callum
    Commented Dec 17, 2023 at 11:52
  • 1
    $\begingroup$ Fixed it, thanks :) $\endgroup$
    – Callum
    Commented Dec 17, 2023 at 11:55
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.