For stabilizer codes we do many things like extracting error syndromes and preparing states and stuff like that by measuring the stabilizer generators. For qubit stabilizer codes measuring the stabilizer generators seems reasonable because qubit Pauli operators are all Hermitian.

But what about qutrit Pauli operators like $$ X= \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix} $$ and $$ Z= \begin{bmatrix} 1 & 0 & 0 \\ 0 & \omega & 0 \\ 0 & 0 & \omega^2 \end{bmatrix} $$ where $ \omega=e^{2 \pi i /3} $. Since the qutrit Pauli stabilizer generators are not hermitian what do we do? How do extract error syndromes? The normal way would be to measure the stabilizer generators, but they aren't hermitian.


1 Answer 1


$\def\braket#1#2{\langle#1|#2\rangle}\def\bra#1{\langle#1|}\def\ket#1{|#1\rangle}$ I think your confusion lies with measuring qudit Pauli operators. It is true that operators like $X$ and $Z$ are not hermitian for qutrits. But one can still measure them -- in essence, a von Neumann measurement just requires an orthonormal basis with some labels corresponding to measurement outcomes. One way of thinking about it is that, instead of measuring $Z$ as you defined, consider relabeling the eigenvalues of $Z$ so one is instead measuring the operator $\tilde{Z} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{pmatrix}$. For example if you measure $Z$ on the state $\ket{2}$ the outcome will be $2$ with 100 percent probability (which is why we call it $\ket{2}$). However, it is better if you agree to label the outcome by the complex number $\omega^2$ where $\omega=e^{2\pi i /3}$. Then if you measure an operator like $Z^2$ or $Z \otimes Z$ the results multiply as expected.

In principle, then, syndrome extraction works the same way as qubits. Let us denote $\ket{\bar{k}}$ as the eigenvector of $X$ with eigenvalue $\omega^k$. so $\ket{\bar{0}}=\frac{1}{\sqrt{3}} \left( \ket{0}+\ket{1}+\ket{2}\right)$. Then a simple, non-fault-tolerant way to measure a syndrome is:

  1. Initialize an ancilla in the state $\ket{\bar{0}}$.
  2. If you want to measure the stabilizer $ZZX$, act with controlled-$Z$ on the first two physical qubits, and a controlled $X$ on the third physical qubit. The ancilla is the control and the physical qubits are the targets.
  3. Measure $X$ on the ancilla. If the outcome is $\ket{\bar{0}}$, then the syndrome is $0$ (i.e. trivial, or $\omega^0$). If the outcome is $\ket{\bar{1}}$ or $\ket{\bar{2}}$ the syndrome is $1$ (i.e. $\omega^1$) or $2$ ($\omega^2$) respectively.
  • $\begingroup$ I see, so is it fair to say that any operator with an orthonormal eigenbasis is just as easy to measure as a Hermitian operator? I suppose that would include any unitary matrix since, as you point out, every unitary matrix has an orthonormal eigenbasis. Also, out of curiousity, how do we prepare the ancilla state $\ket{\bar{0}}=\frac{1}{\sqrt{3}} \left( \ket{0}+\ket{1}+\ket{2}\right)$ ? $\endgroup$ Commented May 18 at 14:34
  • $\begingroup$ Another way is to think that you can always construct a Hermitian operator with the same eigenbasis and measure that. The state can be prepared by applying a Hadamard gate to the state ket-0. $\endgroup$
    – user173611
    Commented Jul 8 at 9:42

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