I was just curious, why is the quantum gate Y-gate (Pauli-Y gate) written in terms of complex numbers? We can actually write Pauli-Y gate as

$$ Y = i * \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} $$

and still we can get result $YY = I$.

  • 3
    $\begingroup$ I suppose you could write it either way. But there’s really no good reason to factor out the “i” because it doesn’t simplify the expression at all, it actually makes it longer. $\endgroup$
    – ryanhill1
    Apr 4, 2021 at 15:30
  • $\begingroup$ Close voters: this question does not "need details or clarity", it is clear. You may not like the question, but it does not need more details or clarity, and already has 2 answers, one which got accepted by the OP! $\endgroup$ Apr 4, 2021 at 20:15

3 Answers 3


As noted by @KAJ226 in another answer, the global phase factor $i$ is unobservable and can be ignored, unless we are considering a controlled gate in which case the phase factor $i$ becomes a relative phase which may not be ignored. Consequently, we can choose to represent the single-qubit gate known as $Y$ using the real matrix

$$ Y=\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}. $$

In fact, you can occasionally see this definition in older papers (see e.g. equation $(1)$ on page 3 in this paper). This highlights an important point: quantum gates do not correspond one-to-one to matrices we use to represent them in calculations.

In certain contexts, it is preferable to use the imaginary matrix $\sigma_Y=iY$ over the real matrix $Y$, which is probably why the convention has settled on using the former. Here are some interesting properties that distinguish the two.

Pauli group

First, unlike the complex matrices

$$ \sigma_X=\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\quad \sigma_Y=\begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}\quad \sigma_Z=\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\tag1 $$

the real matrices

$$ X=\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\quad Y=\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\quad Z=\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\tag2 $$

do not generate the full one-qubit Pauli group. They do generate a useful subgroup (which contains real matrices only).


All three matrices in $(1)$ are Hermitian with eigenvalues $+1$ and $-1$. On the other hand, $Y$ is anti-Hermitian and has imaginary eigenvalues $i$ and $-i$. Consequently, matrices in $(1)$ can be used to represent both quantum gates and observables, but matrix $Y$ in $(2)$ is not an observable.

In fact, together with identity the matrices in $(1)$ form a basis of the real vector space of single-qubit Hermitian operators. Of course, similar statement is not true about $(2)$.

Stabilizer formalism

Related to the above is the fact that $Y$ can never be a member of a stabilizer group (all stabilizers must have $1$ among their eigenvalues).

  • 1
    $\begingroup$ Good answer, except you can measure both Y and i Y, so both are "observables" $\endgroup$
    – M. Stern
    Apr 5, 2021 at 7:52
  • 1
    $\begingroup$ @M.Stern By convention, we choose to represent measurement results as real numbers. In quantum mechanics this choice translates into the choice of Hermitian operators to represent observables. We could also choose to put measurement results on the imaginary axis and then we would be describing observables using anti-Hermitian operators. In fact, we could even choose to label measurement results using complex numbers and then observables would correspond to normal operators. I didn't want to go into these caveats, so I went along with what is a prevailing convention. $\endgroup$ Apr 5, 2021 at 17:16
  • 2
    $\begingroup$ I understand, of course. From time to time I just like to remind that there is nothing special about real measurement outcomes. To prevent that at some point it might be the prevailing convention just because it is presented as the only possible choice in many places, like in this answer. $\endgroup$
    – M. Stern
    Apr 6, 2021 at 8:06

$\begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} = i \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} $

There is no reason to factor out the $i$, it just make thing more cumbersome.

I think what you are trying to do or assuming is that you can ignore the phase factor $i$ since it is just an overall phase factor. That would be okay, as long as you don't do any controlled operation on it.

That is, suppose $U = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} $ and $|\psi \rangle = \alpha|0\rangle + \beta|1\rangle $, then $U|\psi \rangle $ and $Y|\psi \rangle$ are indistinguishable from one another. However, $Controlled$-$Y$ and $Controlled$-$U$ are not the same, and they will have different effect on a two qubit state $|\phi \rangle$.


The reason for holding $i$ in $\sigma_2$ is the relations between Pauli matrices. In particular, the commutation relations. The general formula for the commutation relations is \begin{equation} [\sigma_a\sigma_b]=2i\epsilon_{abc}\sigma_c \end{equation} for example \begin{equation} [\sigma_1\sigma_2]=2i\sigma_3 \end{equation} The relations are broken if we remove $i$ from $\sigma_2$


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.