Two common shorthands for eigenstates of the $Z$ operator are $\{|0\rangle,|1\rangle\}$ and $\{|1\rangle,|-1\rangle\}$, where in the first case we have $Z|z\rangle=(-1)^z|z\rangle$ and in the second case we have $Z|z\rangle=z|z\rangle$. The common shorthand for eigenstates of the $X$ operator is $\{|+\rangle,|-\rangle\}$, where $X|\pm\rangle=\pm|\pm\rangle$.

Is there any commonly used shorthand for eigenstates of the $Y$ operator?


3 Answers 3


There are several, these are the ones I have seen:

  • $|+\rangle_y,|-\rangle_y$ a bit lazy but easy to remember
  • $|+i\rangle,|-i\rangle$ the same as before but you replace the sub index with an imaginary unit $i$ inside the ket
  • $|\circlearrowleft\rangle,|\circlearrowright\rangle$ this notation is borrowed from light polarization, as you can use photons for light too, circular polarization is the equivalent of $y$-axis for photons (It is easy to mess up which is right and which is left in this notation, but it has been used by Qiskit, see section 2).
  • $|\mathrm R\rangle,|\mathrm L\rangle$ for left and right, again based on the polarization of light (example: Quantum Inspire).

In all these notations the first state is $(|0\rangle+i|1\rangle)/\sqrt{2}$ and the second one is $(|0\rangle-i|1\rangle)/\sqrt{2}$.

What I like about the second notation $|\pm i\rangle$ is that it kind of tells you the coefficient (part of it) in front of the $|1\rangle$ state (analogous to the $x$ basis $|\pm\rangle)$.

  • $\begingroup$ Just double checking. My brain equates counter-clockwise with left and clockwise with right. It's really the case that |↺⟩ is $|R\rangle$ ? Very confusing. $\endgroup$ Nov 16, 2022 at 17:56
  • 1
    $\begingroup$ @FrankYellin Is it even rotating inwards or outwards? Anyway, I had to double check at the time (more than a year ago), but take your time to verify. The rotating notation is indeed the most confusing. $\endgroup$
    – Mauricio
    Nov 16, 2022 at 22:51

As the other answer mentioned, they are often denoted as $$|+i\rangle= \dfrac{|0\rangle + i|1\rangle}{\sqrt{2}} = \dfrac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ i\end{pmatrix} \ \ \ \textrm{and} \ \ \ |-i\rangle = \dfrac{|0\rangle - i|1\rangle}{\sqrt{2}} = \dfrac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ -i\end{pmatrix} $$

but sometime you might also see them denoted as $|R\rangle$ and $|L \rangle$ correspond to the Right and Left circular polarization. This is because the matrix

$$ U = \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{pmatrix} $$

has the eigenvalues of $e^{i\theta}$ and $e^{-i\theta}$ with the corresponding eigenvectors

$$ |R\rangle = \dfrac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ i\end{pmatrix} \hspace{1.2 cm} |L\rangle = \dfrac{1}{\sqrt{2}} \begin{pmatrix} i \\ 1\end{pmatrix} = \dfrac{i}{\sqrt{2}} \begin{pmatrix} 1 \\ -i\end{pmatrix} \equiv \dfrac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ -i\end{pmatrix} \ \ \textrm{since} \ e^{i \theta}|\psi \rangle \equiv |\psi \rangle $$


They are commonly denoted as

$$ \left\{ |+i\rangle = \frac{|0\rangle + i|1\rangle}{\sqrt{2}}, |-i\rangle = \frac{|0\rangle - i|1\rangle}{\sqrt{2}}\right\} $$

So, you could use $|\pm i\rangle = \frac{|0\rangle \pm i|1\rangle}{\sqrt{2}}$, which would be very similar to the case of $X$, i.e., $Y|\pm i\rangle = \pm|\pm i\rangle$.


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